Solutions. Design and Analysis of Experiments. Montgomery

Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY
Chapter 2
Simple Comparative Experiments
Solutions
2-1 The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that
the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested. The
results are y1=145, y2=153, y3=150 and y4=147.
(a) State the hypotheses that you think should be tested in this experiment.
H0: µ = 150 H1: µ > 150
(b) Test these hypotheses using α = 0.05. What are your conclusions?
n = 4, σ = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75
148.75 150 1.25
0.8333
3 3
24
o
o
y
z
n
µ
σ
− − −
= = = = −
Since z0.05 = 1.645, do not reject.
(c) Find the P-value for the test in part (b).
From the z-table: ( )( )[ ] 20140796707995032796701 ....P =−+−≅
(d) Construct a 95 percent confidence interval on the mean breaking strength.
The 95% confidence interval is
( )( ) ( )( )2396.175.1482396.175.148
22
+≤≤−
+≤≤−
µ
σ
µ
σ
αα
n
zy
n
zy
145 81 151 69. .≤ ≤µ
2-2 The viscosity of a liquid detergent is supposed to average 800 centistokes at 25°C. A random
sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the
standard deviation of viscosity is σ = 25 centistokes.
(a) State the hypotheses that should be tested.
H0: µ = 800 H1: µ ≠ 800
2-1

812 800 12
1.92
25 25
416
o
o
y
z
n
µ
σ
− −
= = = = Since zα/2 = z0.025 = 1.96, do not reject.
(c) What is the P-value for the test? P = =2 0 0274 0 0549( . ) .
(d) Find a 95 percent confidence interval on the mean.
n
zy
n
zy
σ
µ
σ
αα
22
+≤≤−
( )( ) ( )( )
2582475799
25128122512812
425961812425961812
..
..
..
≤≤
+≤≤−
+≤≤−
µ
µ
µ
2-3 The diameters of steel shafts produced by a certain manufacturing process should have a mean
diameter of 0.255 inches. The diameter is known to have a standard deviation of σ = 0.0001 inch. A
random sample of 10 shafts has an average diameter of 0.2545 inches.
(a) Set up the appropriate hypotheses on the mean µ.
H0: µ = 0.255 H1: µ ≠ 0.255
n = 10, σ = 0.0001, y = 0.2545
0.2545 0.255
15.81
0.0001
10
o
o
y
z
n
µ
σ
− −
= = = −
Since z0.025 = 1.96, reject H0.
(c) Find the P-value for this test. P = 2.6547x10-56
(d) Construct a 95 percent confidence interval on the mean shaft diameter.
n
zy
n
zy
σ
µ
σ
αα
22
+≤≤−
( ) ( )
0.0001 0.0001
0.2545 1.96 0.2545 1.96
10 10
µ
⎛ ⎞ ⎛ ⎞
− ≤ ≤ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0 254438 0 254562. .≤ ≤µ
2-4 A normally distributed random variable has an unknown mean µ and a known variance σ2 = 9. Find
the sample size required to construct a 95 percent confidence interval on the mean, that has total length of
1.0.
2-2

Since y ∼ N(µ,9), a 95% two-sided confidence interval on µ is
y z
n
y z
n
− ≤ ≤ +α α
σ
µ
σ
2 2
y
n
y
n
− ≤ ≤ +( . ) ( . )196
3
196
3
µ
If the total interval is to have width 1.0, then the half-interval is 0.5. Since zα/2 = z0.025 = 1.96,
( )( )
( )( )
( ) 139301387611
7611503961
503961
2
≅==
==
=
..n
...n
.n.
2-5 The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested,
and the following results are obtained:
Days
108 138
124 163
124 159
106 134
115 139
(a) We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate
hypotheses for investigating this claim.
H0: µ = 120 H1: µ > 120
y = 131
S2 = 3438 / 9 = 382
382 19.54S = =
131 120
1.78
19.54 10
o
o
y
t
S n
µ− −
= = =
since t0.01,9 = 2.821; do not reject H0
Minitab Output
T-Test of the Mean
Test of mu = 120.00 vs mu > 120.00
Variable N Mean StDev SE Mean T P
Shelf Life 10 131.00 19.54 6.18 1.78 0.054
T Confidence Intervals
Variable N Mean StDev SE Mean 99.0 % CI
Shelf Life 10 131.00 19.54 6.18 ( 110.91, 151.09)
2-3

(c) Find the P-value for the test in part (b). P=0.054
(d) Construct a 99 percent confidence interval on the mean shelf life.
The 99% confidence interval is , 1 , 1
2
n
S
y t y t
n n
α µ−
− ≤ ≤ + 2
n
S
α −
with α = 0.01.
( ) ( )
1954 1954
131 3.250 131 3.250
10 10
µ
⎛ ⎞ ⎛ ⎞
− ≤ ≤ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
110 91 15109. .≤ ≤µ
2-6 Consider the shelf life data in Problem 2-5. Can shelf life be described or modeled adequately by a
normal distribution? What effect would violation of this assumption have on the test procedure you used in
solving Problem 2-5?
A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of
the normality assumption. If shelf life is not normally distributed, then the impact of this on the t-test in
problem 2-5 is not too serious unless the departure from normality is severe.
1761661561461361261161069686
99
95
90
80
70
60
50
40
30
20
10
5
1
Data
Percent
1.292AD*
Goodness of Fit
Normal Probability Plot for Shelf Life
ML Estimates
Mean
StDev
131
18.5418
ML Estimates
2-7 The time to repair an electronic instrument is a normally distributed random variable measured in
hours. The repair time for 16 such instruments chosen at random are as follows:
Hours
159 280 101 212
224 379 179 264
222 362 168 250
149 260 485 170
(a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for
investigating this issue.
2-4

H0: µ = 225 H1: µ > 225
(b) Test the hypotheses you formulated in part (a). What are your conclusions? Use α = 0.05.
y = 247.50
S2 =146202 / (16 - 1) = 9746.80
9746.8 98.73S = =
241.50 225
0.67
98.73
16
o
o
y
t
S
n
µ− −
= = =
since t0.05,15 = 1.753; do not reject H0
Minitab Output
T-Test of the Mean
Test of mu = 225.0 vs mu > 225.0
Variable N Mean StDev SE Mean T P
Hours 16 241.5 98.7 24.7 0.67 0.26
T Confidence Intervals
Variable N Mean StDev SE Mean 95.0 % CI
Hours 16 241.5 98.7 24.7 ( 188.9, 294.1)
(c) Find the P-value for this test. P=0.26
(d) Construct a 95 percent confidence interval on mean repair time.
The 95% confidence interval is , 1 , 1
2 2
n n
S S
y t y t
n n
α αµ− −
− ≤ ≤ +
( ) ( )
98.73 98.73
241.50 2.131 241.50 2.131
16 16
µ
⎛ ⎞ ⎛ ⎞
− ≤ ≤ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
12949188 .. ≤≤ µ
2-8 Reconsider the repair time data in Problem 2-7. Can repair time, in your opinion, be adequately
modeled by a normal distribution?
The normal probability plot below does not reveal any serious problem with the normality assumption.
2-5

45035025015050
99
95
90
80
70
60
50
40
30
20
10
5
1
Data
Percent
1.185AD*
Goodness of Fit
Normal Probability Plot for Hours
ML Estimates
Mean
StDev
241.5
95.5909
ML Estimates
2-9 Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling
processes can be assumed to be normal, with standard deviation of σ1 = 0.015 and σ2 = 0.018. The quality
engineering department suspects that both machines fill to the same net volume, whether or not this volume
is 16.0 ounces. An experiment is performed by taking a random sample from the output of each machine.
Machine 1 Machine 2
16.03 16.01 16.02 16.03
16.04 15.96 15.97 16.04
16.05 15.98 15.96 16.02
16.05 16.02 16.01 16.01
16.02 15.99 15.99 16.00
(a) State the hypotheses that should be tested in this experiment.
H0: µ1 = µ2 H1: µ1 ≠ µ2
(b) Test these hypotheses using α=0.05. What are your conclusions?
y
n
1
1
1
16 015
0 015
10
=
=
=
.
.σ
y
n
2
2
2
16 005
0 018
10
=
=
=
.
.σ
z
y y
n n
o =
−
+
=
−
+
=1 2
1
2
1
2
2
2
2 2
16 015 16 018
0 015
10
0 018
10
1 35
σ σ
. .
. .
.
z0.025 = 1.96; do not reject
(c) What is the P-value for the test? P = 0.1770
(d) Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines.
2-6

2
2
2
1
2
1
2121
2
2
2
1
2
1
21 22
nn
zyy
nn
zyy
σσ
µµ
σσ
αα ++−≤−≤+−−
10
018.0
10
015.0
)6.19()005.16015.16(
10
018.0
10
015.0
)6.19()005.16015.16(
22
21
22
++−≤−≤+−− µµ
0245.00045.0 21 ≤−≤− µµ
2-10 Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking
strength of this plastic is important. It is known that σ1 = σ2 = 1.0 psi. From random samples of n1 = 10
and n2 = 12 we obtain y 1 = 162.5 and y 2 = 155.0. The company will not adopt plastic 1 unless its
breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they
use plastic 1? In answering this questions, set up and test appropriate hypotheses using α = 0.01.
Construct a 99 percent confidence interval on the true mean difference in breaking strength.
H0: µ1 - µ2 =10 H1: µ1 - µ2 >10
10
1
5162
1
1
1
=
=
=
n
.y
σ
10
1
0155
2
2
2
=
=
=
n
.y
σ
z
y y
n n
o =
− −
+
=
− −
+
= −1 2
1
2
1
2
2
2
2 2
10 162 5 155 0 10
1
10
1
12
5 85
σ σ
. .
.
z0.01 = 2.225; do not reject
The 99 percent confidence interval is
2
2
2
1
2
1
2121
2
2
2
1
2
1
21 22
nn
zyy
nn
zyy
σσ
µµ
σσ
αα ++−≤−≤+−−
12
1
10
1
)575.2()0.1555.162(
12
1
10
1
)575.2()0.1555.162(
22
21
22
++−≤−≤+−− µµ
60.840.6 21 ≤−≤ µµ
2-11 The following are the burning times (in minutes) of chemical flares of two different formulations.
The design engineers are interested in both the means and variance of the burning times.
Type 1 Type 2
65 82 64 56
81 67 71 69
57 59 83 74
66 75 59 82
82 70 65 79
(a) Test the hypotheses that the two variances are equal. Use α = 0.05.
2-7

2 2
0 1 2
2
1 1 2
:
:
H
H 2
σ σ
σ σ
=
≠
S
S
1
2
9 264
9 367
=
=
.
.
F
S
S
0
1
2
2
2
8582
87 73
0 98= = =
.
.
.
F0 025 9 9 4 03. , , .= F
F
0 975 9 9
0 025 9 9
1 1
4 03
0 248. , ,
. , , .
.= = = Do not reject.
(b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use α = 0.05.
What is the P-value for this test?
S
n S n S
n n
S
t
y y
S
n n
p
p
p
2 1 1
2
2 2
2
1 2
0
1 2
1 2
1 1
2
156195
18
86 775
9 32
1 1
70 4 70 2
9 32
1
10
1
10
0 048
=
− + −
+ −
= =
=
=
−
+
=
−
+
=
( ) ( ) .
.
.
. .
.
.
t0 025 18 2101. , .= Do not reject.
From the computer output, t=0.05; do not reject. Also from the computer output P=0.96
Minitab Output
Two Sample T-Test and Confidence Interval
Two sample T for Type 1 vs Type 2
N Mean StDev SE Mean
Type 1 10 70.40 9.26 2.9
Type 2 10 70.20 9.37 3.0
95% CI for mu Type 1 - mu Type 2: ( -8.6, 9.0)
T-Test mu Type 1 = mu Type 2 (vs not =): T = 0.05 P = 0.96 DF = 18
Both use Pooled StDev = 9.32
(c) Discuss the role of the normality assumption in this problem. Check the assumption of normality for
both types of flares.
The assumption of normality is required in the theoretical development of the t-test. However, moderate
departure from normality has little impact on the performance of the t-test. The normality assumption is
more important for the test on the equality of the two variances. An indication of nonnormality would be
of concern here. The normal probability plots shown below indicate that burning time for both
formulations follow the normal distribution.
2-8

9080706050
99
95
90
80
70
60
50
40
30
20
10
5
1
Data
Percent
1.387AD*
Goodness of Fit
Normal Probability Plot for Type 1
ML Estimates
Mean
StDev
70.4
8.78863
ML Estimates
9080706050
99
95
90
80
70
60
50
40
30
20
10
5
1
Data
Percent
1.227AD*
Goodness of Fit
Normal Probability Plot for Type 2
ML Estimates
Mean
StDev
70.2
8.88594
ML Estimates
2-12 An article in Solid State Technology, "Orthogonal Design of Process Optimization and Its
Application to Plasma Etching" by G.Z. Yin and D.W. Jillie (May, 1987) describes an experiment to
determine the effect of C2F6 flow rate on the uniformity of the etch on a silicon wafer used in integrated
circuit manufacturing. Data for two flow rates are as follows:
C2F6 Uniformity Observation
(SCCM) 1 2 3 4 5 6
125 2.7 4.6 2.6 3.0 3.2 3.8
200 4.6 3.4 2.9 3.5 4.1 5.1
(a) Does the C2F6 flow rate affect average etch uniformity? Use α = 0.05.
No, C2F6 flow rate does not affect average etch uniformity.
2-9

Minitab Output
Two sample T for Uniformity
Flow Rat N Mean StDev SE Mean
125 6 3.317 0.760 0.31
200 6 3.933 0.821 0.34
95% CI for mu (125) - mu (200): ( -1.63, 0.40)
T-Test mu (125) = mu (200) (vs not =): T = -1.35 P = 0.21 DF = 10
(b) What is the P-value for the test in part (a)? From the computer printout, P=0.21
(c) Does the C2F6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use α = 0.05.
2 2
0 1 2
2 2
1 1 2
0.05,5,5
0
:
:
5.05
0.5776
0.86
0.6724
H
H
F
F
σ σ
σ σ
=
≠
=
= =
Do not reject; C2F6 flow rate does not affect wafer-to-wafer variability.
(d) Draw box plots to assist in the interpretation of the data from this experiment.
The box plots shown below indicate that there is little difference in uniformity at the two gas flow rates.
Any observed difference is not statistically significant. See the t-test in part (a).
200125
5
4
3
Flow Rate
Uniformity
2-13 A new filtering device is installed in a chemical unit. Before its installation, a random sample
yielded the following information about the percentage of impurity: y 1 = 12.5, S =101.17, and n1
2
1
= 8.
After installation, a random sample yielded y 2 = 10.2, S = 94.73, n2
2
2
= 9.
(a) Can you concluded that the two variances are equal? Use α = 0.05.
2-10

071
7394
17101
534
2
2
2
1
0
870250
2
2
2
11
2
2
2
10
.
.
.
S
S
F
.F
:H
:H
,,.
===
=
≠
=
σσ
σσ
Do Not Reject. Assume that the variances are equal.
(b) Has the filtering device reduced the percentage of impurity significantly? Use α = 0.05.
7531
4790
9
1
8
1
899
210512
11
899
7497
298
7394191710118
2
11
15050
21
21
0
21
2
22
2
112
211
210
.t
.
.
..
nn
S
yy
t
.S
.
).)(().)((
nn
S)n(S)n(
S
:H
:H
,.
p
p
p
=
=
+
−
=
+
−
=
=
=
−+
−+−
=
−+
−+−
=
≠
=
µµ
µµ
Do not reject. There is no evidence to indicate that the new filtering device has affected the mean
2-14 Photoresist is a light-sensitive material applied to semiconductor wafers so that the circuit pattern
can be imaged on to the wafer. After application, the coated wafers are baked to remove the solvent in the
photoresist mixture and to harden the resist. Here are measurements of photoresist thickness (in kÅ) for
eight wafers baked at two different temperatures. Assume that all of the runs were made in random order.
95 ºC 100 ºC
11.176 5.263
7.089 6.748
8.097 7.461
11.739 7.015
11.291 8.133
10.759 7.418
6.467 3.772
8.315 8.963
(a) Is there evidence to support the claim that the higher baking temperature results in wafers with a lower
mean photoresist thickness? Use α = 0.05.
2-11

0 1 2
1 1 2
2 2
2 1 1 2 2
1 2
1 2
0
1 2
0.05,14
:
:
( 1) ( 1) (8 1)(4.41) (8 1)(2.54)
3.48
2 8 8 2
1.86
9.37 6.89
2.65
1 1 1 1
1.86
8 8
1.761
p
p
p
H
H
n S n S
S
n n
S
y y
t
S
n n
t
µ µ
µ µ
=
≠
− + − − + −
= =
+ − + −
=
− −
= = =
+ +
=
=
Since t0.05,14 = 1.761, reject H0. There appears to be a lower mean thickness at the higher temperature. This
is also seen in the computer output.
Minitab Output
Two-Sample T-Test and CI: Thickness, Temp
Two-sample T for Thickness
Temp N Mean StDev SE Mean
95 8 9.37 2.10 0.74
100 8 6.89 1.60 0.56
Difference = mu ( 95) - mu (100)
Estimate for difference: 2.475
95% CI for difference: (0.476, 4.474)
T-Test of difference = 0 (vs not =): T-Value = 2.65 P-Value = 0.019 DF = 14
(b) What is the P-value for the test conducted in part (a)? P = 0.019
(c) Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this
interval.
From the computer output the 95% confidence interval is . This confidence interval
doesnot include 0 in it, there for there is a difference in the two temperatures on the thickness of the photo
resist.
1 20.476 4.474µ µ≤ − ≤
(d) Draw dot diagrams to assist in interpreting the results from this experiment.
2-12

Thickness
12.010.89.68.47.26.04.83.6
Temp
95
100
Dotplot of Thickness vs Temp
(e) Check the assumption of normality of the photoresist thickness.
5 10 15
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD* 1.767
Goodness of Fit
Normal Probability Plot for Thick@95
ML Estimates - 95% CI
Mean
StDev
9.36663
1.96396
ML Estimates
2-13

2 7 12
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD* 1.567
Goodness of Fit
Normal Probability Plot for Thick@100
Mean
StDev
6.89163
1.49207
ML Estimates
There are no significant deviations from the normality assumptions.
(f) Find the power of this test for detecting an actual difference in means of 2.5 kÅ.
Minitab Output
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.05 Sigma = 1.86
Sample
Difference Size Power
2.5 8 0.7056
(g) What sample size would be necessary to detect an actual difference in means of 1.5 kÅ with a power of
at least 0.9?.
Minitab Output
2-Sample t Test
Sample Target Actual
Difference Size Power Power
1.5 34 0.9000 0.9060
2-14

This result makes intuitive sense. More samples are needed to detect a smaller difference.
2-15 Front housings for cell phones are manufactured in an injection molding process. The time the part
is allowed to cool in the mold before removal is thought to influence the occurrence of a particularly
troublesome cosmetic defect, flow lines, in the finished housing. After manufacturing, the housings are
inspected visually and assigned a score between 1 and 10 based on their appearance, with 10 corresponding
to a perfect part and 1 corresponding to a completely defective part. An experiment was conducted using
two cool-down times, 10 seconds and 20 seconds, and 20 housings were evaluated at each level of cool-
down time. The data are shown below.
10 Seconds 20 Seconds
1 3 7 6
2 6 8 9
1 5 5 5
3 3 9 7
5 2 5 4
1 1 8 6
5 6 6 8
2 8 4 5
3 2 6 8
5 3 7 7
(a) Is there evidence to support the claim that the longer cool-down time results in fewer appearance
defects? Use α = 0.05.
Minitab Output
Two-Sample T-Test and CI: 10 seconds, 20 seconds
Two-sample T for 10 seconds vs 20 seconds
10 secon 20 3.35 2.01 0.45
20 secon 20 6.50 1.54 0.34
Difference = mu 10 seconds - mu 20 seconds
Estimate for difference: -3.150
95% CI for difference: (-4.295, -2.005)
T-Test of difference = 0 (vs not =): T-Value = -5.57 P-Value = 0.000 DF = 38
(b) What is the P-value for the test conducted in part (a)? From the Minitab output, P = 0.000
(c) Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this
interval.
From the computer output, . This interval does not contain 0. The two samples are
different. The 20 second cooling time gives a cosmetically better housing.
1 24.295 2.005µ µ− ≤ − ≤ −
(d) Draw dot diagrams to assist in interpreting the results from this experiment.
2-15

Ranking
8642
C4
10 sec
20 sec
Dotplot of Ranking vs C4
(e) Check the assumption of normality for the data from this experiment.
0 4 8
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD* 1.252
Goodness of Fit
Normal Probability Plot for 10 seconds
Mean
StDev
3.35
1.95640
ML Estimates
2-16

2 3 4 5 6 7 8 9 10 11
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD* 0.988
Goodness of Fit
Normal Probability Plot for 20 seconds
Mean
StDev
6.5
1.50000
ML Estimates
There are no significant departures from normality.
2-16 Twenty observations on etch uniformity on silicon wafers are taken during a qualification
experiment for a plasma etcher. The data are as follows:
5.34 6.65 4.76 5.98 7.25
6.00 7.55 5.54 5.62 6.21
5.97 7.35 5.44 4.39 4.98
5.25 6.35 4.61 6.00 5.32
(a) Construct a 95 percent confidence interval estimate of σ2.
( ) ( )
( )( ) ( )( )
2 2
2
2 2
, 1 (1 ), 1
2 2
2 2
2
2
1 1
20 1 0.88907 20 1 0.88907
32.852 8.907
0.457 1.686
n n
n S n S
α α
σ
χ χ
σ
σ
− − −
− −
≤ ≤
− −
≤ ≤
≤ ≤
(b) Test the hypothesis that σ2 = 1.0. Use α = 0.05. What are your conclusions?
H
H
0
2
1
2
1
1
:
:
σ
σ
=
≠
χ
σ
0
2
0
2
15019= =
SS
.
2-17

χ0 025 19
2
32852. , .= χ0 975 19
2
8 907. , .=
Do not reject. There is no evidence to indicate that 2
1σ ≠
(c) Discuss the normality assumption and its role in this problem.
The normality assumption is much more important when analyzing variances then when analyzing means.
A moderate departure from normality could cause problems with both statistical tests and confidence
intervals. Specifically, it will cause the reported significance levels to be incorrect.
(d) Check normality by constructing a normal probability plot. What are your conclusions?
The normal probability plot indicates that there is not any serious problem with the normality assumption.
7.86.85.84.83.8
99
95
90
80
70
60
50
40
30
20
10
5
1
Data
Percent
0.835AD*
Goodness of Fit
Normal Probability Plot for Uniformity
ML Estimates
Mean
StDev
5.828
0.866560
ML Estimates
2-17 The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of
calipers. The results were:
Inspector Caliper 1 Caliper 2 Difference Difference^2
1 0.265 0.264 .001 .000001
2 0.265 0.265 .000 0
3 0.266 0.264 .002 .000004
4 0.267 0.266 .001 .000001
5 0.267 0.267 .000 0
6 0.265 0.268 -.003 .000009
7 0.267 0.264 .003 .000009
8 0.267 0.265 .002 .000004
9 0.265 0.265 .000 0
10 0.268 0.267 .001 .000001
11 0.268 0.268 .000 0
12 0.265 0.269 -.004 .000016
=∑ 0 003. =∑ 0 000045.
2-18

(a) Is there a significant difference between the means of the population of measurements represented by
the two samples? Use α = 0.05.
H
H
0 1 2
1 1 2
:
:
µ µ
µ µ
=
≠
or equivalently
0
0
1
0
≠
=
d
d
:H
:H
µ
µ
Minitab Output
Paired T-Test and Confidence Interval
Paired T for Caliper 1 - Caliper 2
Caliper 12 0.266250 0.001215 0.000351
Caliper 12 0.266000 0.001758 0.000508
Difference 12 0.000250 0.002006 0.000579
95% CI for mean difference: (-0.001024, 0.001524)
T-Test of mean difference = 0 (vs not = 0): T-Value = 0.43 P-Value = 0.674
(b) Find the P-value for the test in part (a). P=0.674
(c) Construct a 95 percent confidence interval on the difference in the mean diameter measurements for
the two types of calipers.
( )1 2, 1 , 1
2 2
0.002 0.002
0.00025 2.201 0.00025 2.201
12 12
0.00102 0.00152
d d
Dn n
d
d
S S
d t d t
n n
α αµ µ µ
µ
µ
− −
− ≤ = − ≤ +
− ≤ ≤ +
− ≤ ≤
2-18 An article in the Journal of Strain Analysis (vol.18, no. 2, 1983) compares several procedures for
predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of
predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows:
Girder Karlsruhe Method Lehigh Method Difference Difference^2
S1/1 1.186 1.061 0.125 0.015625
S2/1 1.151 0.992 0.159 0.025281
S3/1 1.322 1.063 0.259 0.067081
S4/1 1.339 1.062 0.277 0.076729
S5/1 1.200 1.065 0.135 0.018225
S2/1 1.402 1.178 0.224 0.050176
S2/2 1.365 1.037 0.328 0.107584
S2/3 1.537 1.086 0.451 0.203401
S2/4 1.559 1.052 0.507 0.257049
Sum = 2.465 0.821151
Average = 0.274
(a) Is there any evidence to support a claim that there is a difference in mean performance between the two
methods? Use α = 0.05.
H
H
0 1 2
1 1 2
:
:
µ µ
µ µ
=
≠
or equivalently
0
0
1
0
≠
=
d
d
:H
:H
µ
µ
2-19

( )
1
1 1
2.465 0.274
9
n
i
i
d d
n =
= = =∑
1
22 1
22 2
1 1
1 1
0.821151 (2.465)
9 0.135
1 9 1
n n
i i
i i
d
d d
n
s
n
= =
⎡ ⎤⎛ ⎞ ⎡ ⎤−⎢ ⎥ −⎜ ⎟ ⎢ ⎥⎝ ⎠⎢ ⎥= = ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥
⎣ ⎦⎢ ⎥⎣ ⎦
∑ ∑
=
0
0.274
6.08
0.135
9
d
d
t
S
n
= = =
30629025012
.tt ,.n, ==−α , reject the null hypothesis.
Minitab Output
Paired T for Karlsruhe - Lehigh
Karlsruh 9 1.3401 0.1460 0.0487
Lehigh 9 1.0662 0.0494 0.0165
Difference 9 0.2739 0.1351 0.0450
95% CI for mean difference: (0.1700, 0.3777)
T-Test of mean difference = 0 (vs not = 0): T-Value = 6.08 P-Value = 0.000
(b) What is the P-value for the test in part (a)? P=0.0002
(c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load.
377770170230
9
1350
30622740
9
1350
30622740
1
2
1
2
..
.
..
.
..
n
S
td
n
S
td
d
d
d
n,d
d
n,
≤≤
+≤≤−
+≤≤− −−
µ
µ
µ αα
(d) Investigate the normality assumption for both samples.
P-Value: 0.537
A-Squared: 0.286
Anderson-Darling Normality Test
N: 9
StDev : 0.146031
Av erage: 1.34011
1.551.451.351.251.15
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Karlsruhe
Normal Probability Plot
2-20

P-Value: 0.028
A-Squared: 0.772
N: 9
StDev : 0.0493806
Av erage: 1.06622
1.151.101.051.00
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Lehigh
(e) Investigate the normality assumption for the difference in ratios for the two methods.
P-Value: 0.464
A-Squared: 0.318
N: 9
StDev : 0.135099
Av erage: 0.273889
0.520.420.320.220.12
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Difference
(f) Discuss the role of the normality assumption in the paired t-test.
As in any t-test, the assumption of normality is of only moderate importance. In the paired t-test, the
assumption of normality applies to the distribution of the differences. That is, the individual sample
measurements do not have to be normally distributed, only their difference.
2-19 The deflection temperature under load for two different formulations of ABS plastic pipe is being
studied. Two samples of 12 observations each are prepared using each formulation, and the deflection
temperatures (in °F) are reported below:
Formulation 1 Formulation 2
212 199 198 177 176 198
194 213 216 197 185 188
211 191 200 206 200 189
193 195 184 201 197 203
2-21

(a) Construct normal probability plots for both samples. Do these plots support assumptions of normality
and equal variance for both samples?
P-Value: 0.227
A-Squared: 0.450
N: 12
StDev : 10.1757
Av erage: 200.5
215205195185
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Form 1
P-Value: 0.236
A-Squared: 0.443
N: 12
StDev : 9.94949
Av erage: 193.083
205195185175
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Form 2
(b) Do the data support the claim that the mean deflection temperature under load for formulation 1
exceeds that of formulation 2? Use α = 0.05.
Minitab Output
Two sample T for Form 1 vs Form 2
Form 1 12 200.5 10.2 2.9
Form 2 12 193.08 9.95 2.9
95% CI for mu Form 1 - mu Form 2: ( -1.1, 15.9)
T-Test mu Form 1 = mu Form 2 (vs >): T = 1.81 P = 0.042 DF = 22
(c) What is the P-value for the test in part (a)? P = 0.042
2-22

2-20 Refer to the data in problem 2-19. Do the data support a claim that the mean deflection temperature
under load for formulation 1 exceeds that of formulation 2 by at least 3 °F?
Yes, formulation 1 exceeds formulation 2 by at least 3 °F.
Minitab Output
Two-Sample T-Test and CI: Form1, Form2
Two-sample T for Form1 vs Form2
Form1 12 200.5 10.2 2.9
Form2 12 193.08 9.95 2.9r
Difference = mu Form1 - mu Form2
95% lower bound for difference: 0.36
T-Test of difference = 3 (vs >): T-Value = 1.08 P-Value = 0.147 DF = 22
2-21 In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the
backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two
different etching solutionsare being evaluated. Eight randomly selected wafers have been etched in each
solution and the observed etch rates (in mils/min) are shown below:
Solution 1 Solution 2
9.9 10.6 10.2 10.6
9.4 10.3 10.0 10.2
10.0 9.3 10.7 10.4
10.3 9.8 10.5 10.3
(a) Do the data indicate that the claim that both solutions have the same mean etch rate is valid? Use α =
0.05 and assume equal variances.
See the Minitab output below.
Minitab Output
Two sample T for Solution 1 vs Solution 2
Solution 8 9.925 0.465 0.16
Solution 8 10.362 0.233 0.082
95% CI for mu Solution - mu Solution: ( -0.83, -0.043)
T-Test mu Solution = mu Solution (vs not =):T = -2.38 P = 0.032 DF = 14
(b) Find a 95% confidence interval on the difference in mean etch rate.
From the Minitab output, -0.83 to –0.043.
(c) Use normal probability plots to investigate the adequacy of the assumptions of normality and equal
variances.
2-23

P-Value: 0.743
A-Squared: 0.222
N: 8
StDev : 0.465219
Av erage: 9.925
10.510.09.5
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Solution 1
P-Value: 0.919
A-Squared: 0.158
N: 8
StDev : 0.232609
Av erage: 10.3625
10.710.610.510.410.310.210.110.0
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Solution 2
Both the normality and equality of variance assumptions are valid.
2-22 Two popular pain medications are being compared on the basis of the speed of absorption by the
body. Specifically, tablet 1 is claimed to be absorbed twice as fast as tablet 2. Assume that and
are known. Develop a test statistic for
2
1σ 2
2σ
H0: 2µ1 = µ2
H1: 2µ1 ≠ µ2
2 2
1 2
1 2 1 2
1 2
4
2 ~ 2 ,y y N
n n
σ σ
µ µ
⎛
− − +⎜
⎝ ⎠
⎞
⎟
, assuming that the data is normally distributed.
The test statistic is: z
y y
n n
o =
−
+
2
4
1 2
1
2
1
2
2
2
σ σ
, reject if z zo > α
2
2-23 Suppose we are testing
H0: µ1 = µ2
H1: µ1 ≠ µ2
2-24

where and are known. Our sampling resources are constrained such that n1 + n2 = N. How should
we allocate the N observations between the two populations to obtain the most powerful test?
2
1σ 2
2σ
The most powerful test is attained by the n1 and n2 that maximize zo for given y y1 2− .
Thus, we chose n1 and n2 to
max z
y y
n n
o =
−
+
1 2
1
2
1
2
2
2
σ σ
, subject to n1 + n2 = N.
This is equivalent to min L
n n n N n
= + = +
−
σ σ σ σ1
2
1
2
2
2
1
2
1
2
2
1
, subject to n1 + n2 = N.
Now
( )
2 2
1 2
22
1 1 1
0
dL
dn n N n
σ σ−
= + =
−
, implies that n1 / n2 = σ1 / σ2.
Thus n1 and n2 are assigned proportionally to the ratio of the standard deviations. This has
intuitive appeal, as it allocates more observations to the population with the greatest variability.
2-24 Develop Equation 2-46 for a 100(1 - α) percent confidence interval for the variance of a normal
distribution.
2
12
~ n
SS
χ
σ
−
. Thus, 2 2
1 , 1 , 1
2 2
2
1n n
SS
P α αχ χ α
σ− − −
⎧
≤ ≤ = −⎨
⎩ ⎭
⎫
⎬ . Therefore,
2 2
, 1 1 , 1
2 2
2
1
n n
SS SS
P
α α
σ α
χ χ− − −
⎧ ⎫
⎪ ⎪
≤ ≤ = −⎨ ⎬
⎪ ⎪⎩ ⎭
,
so
2 2
, 1 1 , 1
2 2
,
n n
SS SS
α αχ χ− − −
⎡ ⎤
⎢
⎢ ⎥
⎣ ⎦
⎥ is the 100(1 - α)% confidence interval on σ2.
2-25 Develop Equation 2-50 for a 100(1 - α) percent confidence interval for the ratio / , where
and are the variances of two normal distributions.
σ1
2 2
2σ 2
1σ
2
2σ
2 1
2 2
2 2
1, 12 2
1 1
~ n n
S
F
S
σ
σ
− −
2 12
2
2 2
2 2
1 , 1, 1 2 2 , 1, 12 11 1
1n n n n
S
P F F
S
α
α
σ
α
σ− − − − −
⎧ ⎫
≤ ≤ =⎨ ⎬
⎩ ⎭
− or
2 12
2
2 2 2
1 1 1
1 , 1, 12 2 2 , 1, 12 1
2 2 2
1n n n n
S S
P F F
S S
α
α
σ
α
σ− − − − −
⎧ ⎫
≤ ≤ = −⎨ ⎬
⎩ ⎭
2-26 Develop an equation for finding a 100(1 - α) percent confidence interval on the difference in the
means of two normal distributions where ≠ . Apply your equation to the portland cement
experiment data, and find a 95% confidence interval.
2
1σ 2
2σ
2-25

( ) ( )
2
1 2 1 2
,2 2
1 2
1 2
~
y y
t
S S
n n
α υ
µ µ− − −
+
( ) ( )2 2
2 2 2 2
1 2 1 2
1 2 1 2, ,
1 2 1 2
S S S S
t y y t
n n n n
α αυ υµ µ+ ≤ − − − ≤ +
( ) ( ) ( )2 2
2 2 2
1 2 1 2
1 2 1 2 1 2, ,
1 2 1 2
S S S S
y y t y y t
n n n n
α αυ υµ µ− − + ≤ − ≤ − + +
2
where
2
2 2
1 2
1 2
2 2
2 2
1 2
1 2
1 21 1
S S
n n
S S
n n
n n
υ
⎛ ⎞
+⎜ ⎟
⎝ ⎠=
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠+
− −
Using the data from Table 2-1
n
y
S
1
1
1
2
10
16 764
0100138
=
=
=
.
.
n
y
S
2
2
2
2
10
17 343
0 0614622
=
=
=
.
.
( ) ( )1 2
0.100138 0.0614622
16.764 17.343 2.110
10 10
µ µ− − + ≤ − ≤
( )
0.100138 0.0614622
16.764 17.343 2.110
10 10
− + +
where 1702417
110
10
06146220
110
10
1001380
10
06146220
10
1001380
22
2
≅=
−
⎟
⎠
⎞
⎜
⎝
⎛
+
−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+
= .
..
..
υ
( )1 21.426 0.889µ µ− ≤ − ≤ −
This agrees with the result in Table 2-2.
2-27 Construct a data set for which the paired t-test statistic is very large, but for which the usual two-
sample or pooled t-test statistic is small. In general, describe how you created the data. Does this give you
any insight regarding how the paired t-test works?
A B delta
7.1662 8.2416 1.07541
2.3590 2.4555 0.09650
19.9977 21.1018 1.10412
0.9077 2.3401 1.43239
-15.9034 -15.0013 0.90204
-6.0722 -5.5941 0.47808
2-26

9.9501 10.6910 0.74085
-1.0944 -0.1358 0.95854
-4.6907 -3.3446 1.34615
-6.6929 -5.9303 0.76256
Minitab Output
Paired T for A - B
A 10 0.59 10.06 3.18
B 10 1.48 10.11 3.20
Difference 10 -0.890 0.398 0.126
95% CI for mean difference: (-1.174, -0.605)
T-Test of mean difference = 0 (vs not = 0): T-Value = -7.07 P-Value = 0.000
Two sample T for A vs B
A 10 0.6 10.1 3.2
B 10 1.5 10.1 3.2
95% CI for mu A - mu B: ( -10.4, 8.6)
T-Test mu A = mu B (vs not =): T = -0.20 P = 0.85 DF = 18
These two sets of data were created by making the observation for A and B moderately different within
each pair (or block), but making the observations between pairs very different. The fact that the difference
between pairs is large makes the pooled estimate of the standard deviation large and the two-sample t-test
statistic small. Therefore the fairly small difference between the means of the two treatments that is present
when they are applied to the same experimental unit cannot be detected. Generally, if the blocks are very
different, then this will occur. Blocking eliminates the variabiliy associated with the nuisance variable that
they represent.
2-28 Consider the experiment described in problem 2-11. If the mean burning times of the two flames
differ by as much as 2 minutes, find the power of the test. What sample size would be required to detect an
actual difference in mean burning time of 1 minute with a power of at least 0.90?
Minitab Output
2-Sample t Test
2 458 0.9000 0.9004
2-29 Reconsider the bottle filling experiment described in Problem 2-9. Rework this problem assuming
that the two population variances are unknown but equal.
2-27

Minitab Output
Two-Sample T-Test and CI: Machine 1, Machine 2
Two-sample T for Machine 1 vs Machine 2
Machine 10 16.0150 0.0303 0.0096
Machine 10 16.0050 0.0255 0.0081
Difference = mu Machine 1 - mu Machine 2
95% CI for difference: (-0.0163, 0.0363)
T-Test of difference = 0 (vs not =): T-Value = 0.80 P-Value = 0.435 DF = 18
The hypothesis test is the same: H0: µ1 = µ2 H1: µ1 ≠ µ2
The conclusions are the same as Problem 2-9, do not reject H0. There is no difference in the machines.
The P-value for this anlysis is 0.435.
The confidence interval is (-0.0163, 0.0363). This interval contains 0. There is no difference in machines.
2-29 Consider the data from problem 2-9. If the mean fill volume of the two machines differ by as much
as 0.25 ounces, what is the power of the test used in problem 2-9? What sample size could result in a
power of at least 0.9 if the actual difference in mean fill volume is 0.25 ounces?
Minitab Output
2-Sample t Test
Alpha = 0.05 Sigma = 0.028
Sample
Difference Size Power
0.25 10 1.0000
Minitab Output
2-Sample t Test
Alpha = 0.05 Sigma = 0.028
0.25 2 0.9000 0.9805
2-28

Chapter 3
Experiments with a Single Factor: The Analysis of Variance
Solutions
3-1 The tensile strength of portland cement is being studied. Four different mixing techniques can be
used economically. The following data have been collected:
Mixing Technique Tensile Strength (lb/in2
)
1 3129 3000 2865 2890
2 3200 3300 2975 3150
3 2800 2900 2985 3050
4 2600 2700 2600 2765
(a) Test the hypothesis that mixing techniques affect the strength of the cement. Use α = 0.05.
Design Expert Output
Response: Tensile Strengthin lb/in^2
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of Mean F
Source Squares DF Square Value Prob > F
Model 4.897E+005 3 1.632E+005 12.73 0.0005 significant
A 4.897E+005 3 1.632E+005 12.73 0.0005
Residual 1.539E+005 12 12825.69
Lack of Fit 0.000 0
Pure Error 1.539E+005 12 12825.69
Cor Total 6.436E+005 15
The Model F-value of 12.73 implies the model is significant. There is only
a 0.05% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated Standard
Mean Error
1-1 2971.00 56.63
2-2 3156.25 56.63
3-3 2933.75 56.63
4-4 2666.25 56.63
Mean Standard t for H0
Treatment Difference DF Error Coeff=0 Prob > |t|
1 vs 2 -185.25 1 80.08 -2.31 0.0392
1 vs 3 37.25 1 80.08 0.47 0.6501
1 vs 4 304.75 1 80.08 3.81 0.0025
2 vs 3 222.50 1 80.08 2.78 0.0167
2 vs 4 490.00 1 80.08 6.12 < 0.0001
3 vs 4 267.50 1 80.08 3.34 0.0059
The F-value is 12.73 with a corresponding P-value of .0005. Mixing technique has an effect.
(b) Construct a graphical display as described in Section 3-5.3 to compare the mean tensile strengths for
the four mixing techniques. What are your conclusions?
62556
4
712825
.
.
n
MS
S E
y .i
===
3-1

2 7 0 0 2 8 0 0 2 9 0 0 3 0 0 0 3 1 0 0
T e n s ile S tre n g th
S c a le d t D is trib u tio n
(3 ) (2 )(1 )(4 )
Based on examination of the plot, we would conclude that 1µ and 3µ are the same; that 4µ differs from
1µ and 3µ , that 2µ differs from 1µ and 3µ , and that 2µ and 4µ are different.
(c) Use the Fisher LSD method with α=0.05 to make comparisons between pairs of means.
4951748564121792
4
7128252
2
4160250
2
...LSD
).(
tLSD
n
MS
tLSD
,.
aN,
E
==
=
=
−
−α
Treatment 2 vs. Treatment 4 = 3156.250 - 2666.250 = 490.000 > 174.495
Treatment 1 vs. Treatment 3 = 2971.000 - 2933.750 = 37.250 < 174.495
The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree
with the graphical method for this experiment.
(d) Construct a normal probability plot of the residuals. What conclusion would you draw about the
validity of the normality assumption?
There is nothing unusual about the normal probability plot of residuals.
3-2

Residual
Normal%probability
Normal plot of residuals
-181.25 -96.4375 -11.625 73.1875 158
1
5
10
20
30
50
70
80
90
95
99
(e) Plot the residuals versus the predicted tensile strength. Comment on the plot.
There is nothing unusual about this plot.
22
Predicted
Residuals
Residuals vs. Predicted
-181.25
-96.4375
-11.625
73.1875
158
2666.25 2788.75 2911.25 3033.75 3156.25
(f) Prepare a scatter plot of the results to aid the interpretation of the results of this experiment.
Design-Expert automatically generates the scatter plot. The plot below also shows the sample average for
each treatment and the 95 percent confidence interval on the treatment mean.
3-3

Technique
TensileStrength
One Factor Plot
1 2 3 4
2579.01
2759.26
2939.51
3119.75
3300
22
3-2 (a) Rework part (b) of Problem 3-1 using Tukey’s test with α = 0.05. Do you get the same
conclusions from Tukey’s test that you did from the graphical procedure and/or the Fisher LSD
method?
Minitab Output
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0117
Critical value = 4.20
Intervals for (column level mean) - (row level mean)
1 2 3
2 -423
53
3 -201 -15
275 460
4 67 252 30
543 728 505
No, the conclusions are not the same. The mean of Treatment 4 is different than the means of Treatments
1, 2, and 3. However, the mean of Treatment 2 is not different from the means of Treatments 1 and 3
according to the Tukey method, they were found to be different using the graphical method and the Fisher
LSD method.
(b) Explain the difference between the Tukey and Fisher procedures.
Both Tukey and Fisher utilize a single critical value; however, Tukey’s is based on the studentized range
statistic while Fisher’s is based on t distribution.
3-3 Reconsider the experiment in Problem 3-1. Find a 95 percent confidence interval on the mean
tensile strength of the portland cement produced by each of the four mixing techniques. Also find a 95
3-4

percent confidence interval on the difference in means for techniques 1 and 3. Does this aid in interpreting
the results of the experiment?
n
MS
ty
n
MS
ty E
aN,.ii
E
aN,.i −−
+≤≤−
22
αα µ
Treatment 1:
4
1282837
17922971 .±
3871232971 .±
38730946132847 1 .. ≤≤ µ
Treatment 2: 3156.25±123.387
63732798633032 2 .. ≤≤ µ
Treatment 3: 2933.75±123.387
13730573632810 3 .. ≤≤ µ
Treatment 4: 2666.25±123.387
63727898632542 4 .. ≤≤ µ
Treatment 1 - Treatment 3:
n
MS
tyy
n
MS
tyy E
aN,.j.iji
E
aN,.j.i
22
22
−−
+−≤−≤−− αα µµ
( )
4
7128252
1792752933002971
.
... ±−
745211245137 31 .. ≤−≤− µµ
3-4 A product developer is investigating the tensile strength of a new synthetic fiber that will be used to
make cloth for men’s shirts. Strength is usually affected by the percentage of cotton used in the blend of
materials for the fiber. The engineer conducts an experiment with five levels of cotton content and
replicated the experiment five times. The data are shown in the following table.
Cotton
Weight
Percentage
Observations
15 7 7 15 11 9
20 12 17 12 18 18
25 14 19 19 18 18
30 19 25 22 19 23
35 7 10 11 15 11
(a) Is there evidence to support the claim that cotton content affects the mean tensile strength? Use α =
0.05.
Minitab Output
One-way ANOVA: Tensile Strength versus Cotton Percentage
Analysis of Variance for Tensile
Source DF SS MS F P
Cotton P 4 475.76 118.94 14.76 0.000
Error 20 161.20 8.06
Total 24 636.96
Yes, the F-value is 14.76 with a corresponding P-value of 0.000. The percentage of cotton in the fiber
appears to have an affect on the tensile strength.
3-5

(b) Use the Fisher LSD method to make comparisons between the pairs of means. What conclusions can
you draw?
Minitab Output
Fisher's pairwise comparisons
15 20 25 30
20 -9.346
-1.854
25 -11.546 -5.946
-4.054 1.546
30 -15.546 -9.946 -7.746
-8.054 -2.454 -0.254
35 -4.746 0.854 3.054 7.054
2.746 8.346 10.546 14.546
In the Minitab output the pairs of treatments that do not contain zero in the pair of numbers indicates that
there is a difference in the pairs of the treatments. 15% cotton is different than 20%, 25% and 30%. 20%
cotton is different than 30% and 35% cotton. 25% cotton is different than 30% and 35% cotton. 30%
cotton is different than 35%.
(c) Analyze the residuals from this experiment and comment on model adequacy.
-4 -3 -2 -1 0 1 2 3 4 5 6
-2
-1
0
1
2
NormalScore
Residual
Normal Probability Plot of the Residuals
(response is Tensile)
3-6

10 15 20
-4
-3
-2
-1
0
1
2
3
4
5
6
Fitted Value
Residual
Residuals Versus the Fitted Values
(response is Tensile)
The residuals show nothing unusual.
3-5 Reconsider the experiment described in Problem 3-4. Suppose that 30 percent cotton content is a
control. Use Dunnett’s test with α = 0.05 to compare all of the other means with the control.
For this problem: a = 5, a-1 = 4, f=20, n=5 and α = 0.05
0.05
2 2(8.06)
(4,20) 2.65 4.76EMS
d
n n
= =
1. 4.
2. 4.
3. 4.
5. 4.
1 vs. 4 : 9.8 21.6 11.8*
2 vs. 4 : 15.4 21.6 6.2*
3 vs. 4 : 17.6 21.6 4.0
5 vs. 4 : 10.8 21.6 10.6*
y y
y y
y y
y y
− = − = −
− = − = −
− = − = −
− = − = −
The control treatment, treatment 4, differs from treatments 1,2 and 5.
3-6 A pharmaceutical manufacturer wants to investigate the bioactivity of a new drug. A completely
randomized single-factor experiment was conducted with three dosage levels, and the following results
were obtained.
Dosage Observations
20g 24 28 37 30
30g 37 44 31 35
3-7

40g 42 47 52 38
(a) Is there evidence to indicate that dosage level affects bioactivity? Use α = 0.05.
Minitab Output
One-way ANOVA: Activity versus Dosage
Analysis of Variance for Activity
Source DF SS MS F P
Dosage 2 450.7 225.3 7.04 0.014
Error 9 288.3 32.0
Total 11 738.9
There appears to be a different in the dosages.
(b) If it is appropriate to do so, make comparisons between the pairs of means. What conclusions can you
draw?
Because there appears to be a difference in the dosages, the comparison of means is appropriate.
Minitab Output
20g 30g
30g -18.177
4.177
40g -26.177 -19.177
-3.823 3.177
The Tukey comparison shows a difference in the means between the 20g and the 40g dosages.
(c) Analyze the residuals from this experiment and comment on the model adequacy.
3-8

50-5
1
0
-1
-2
NormalScore
Residual
(response is Activity)
30 35 40 45
-5
0
5
Fitted Value
Residual
(response is Activity)
There is nothing too unusual about the residuals.
3-7 A rental car company wants to investigate whether the type of car rented affects the length of the
rental period. An experiment is run for one week at a particular location, and 10 rental contracts are
selected at random for each car type. The results are shown in the following table.
3-9

Type of Car Observations
Sub-compact 3 5 3 7 6 5 3 2 1 6
Compact 1 3 4 7 5 6 3 2 1 7
Midsize 4 1 3 5 7 1 2 4 2 7
Full Size 3 5 7 5 10 3 4 7 2 7
(a) Is there evidence to support a claim that the type of car rented affects the length of the rental contract?
Use α = 0.05. If so, which types of cars are responsible for the difference?
Minitab Output
One-way ANOVA: Days versus Car Type
Analysis of Variance for Days
Source DF SS MS F P
Car Type 3 16.68 5.56 1.11 0.358
Error 36 180.30 5.01
Total 39 196.98
There is no difference.
(b) Analyze the residuals from this experiment and comment on the model adequacy.
-4 -3 -2 -1 0 1 2 3 4 5
-2
-1
0
1
2
NormalScore
Residual
(response is Days)
3-10

3.5 4.5 5.5
-4
-3
-2
-1
0
1
2
3
4
5
Fitted Value
Residual
(response is Days)
There is nothing unusual about the residuals.
(c) Notice that the response variable in this experiment is a count. Should the cause any potential concerns
about the validity of the analysis of variance?
Because the data is count data, a square root transformation could be applied. The analysis is shown
below. It does not change the interpretation of the data.
Minitab Output
One-way ANOVA: Sqrt Days versus Car Type
Analysis of Variance for Sqrt Day
Source DF SS MS F P
Car Type 3 1.087 0.362 1.10 0.360
Error 36 11.807 0.328
Total 39 12.893
3-8 I belong to a golf club in my neighborhood. I divide the year into three golf seasons: summer (June-
September), winter (November-March) and shoulder (October, April and May). I believe that I play my
best golf during the summer (because I have more time and the course isn’t crowded) and shoulder
(because the course isn’t crowded) seasons, and my worst golf during the winter (because all of the part-
year residents show up, and the course is crowded, play is slow, and I get frustrated). Data from the last
year are shown in the following table.
Season Observations
Summer 83 85 85 87 90 88 88 84 91 90
Shoulde
r 91 87 84 87 85 86 83
Winter 94 91 87 85 87 91 92 86
3-11

(a) Do the data indicate that my opinion is correct? Use α = 0.05.
Minitab Output
One-way ANOVA: Score versus Season
Analysis of Variance for Score
Source DF SS MS F P
Season 2 35.61 17.80 2.12 0.144
Error 22 184.63 8.39
Total 24 220.24
The data do not support the author’s opinion.
(b) Analyze the residuals from this experiment and comment on model adequacy.
-4 -3 -2 -1 0 1 2 3 4 5
-2
-1
0
1
2
NormalScore
Residual
(response is Score)
3-12

89888786
5
4
3
2
1
0
-1
-2
-3
-4
Fitted Value
Residual
(response is Score)
3-9 A regional opera company has tried three approaches to solicit donations from 24 potential sponsors.
The 24 potential sponsors were randomly divided into three groups of eight, and one approach was used
for each group. The dollar amounts of the resulting contributions are shown in the following table.
Approac
h Contributions (in $)
1 1000 1500 1200 1800 1600 1100 1000 1250
2 1500 1800 2000 1200 2000 1700 1800 1900
3 900 1000 1200 1500 1200 1550 1000 1100
(a) Do the data indicate that there is a difference in results obtained from the three different approaches?
Use α = 0.05.
Minitab Output
One-way ANOVA: Contribution versus Approach
Analysis of Variance for Contribution
Source DF SS MS F P
Approach 2 1362708 681354 9.41 0.001
Error 21 1520625 72411
Total 23 2883333
There is a difference between the approaches. The Tukey test will indicate which are different. Approach
2 is different than approach 3.
Minitab Output
3-13

1 2
2 -770
-93
3 -214 218
464 895
(b) Analyze the residuals from this experiment and comment on the model adequacy.
-500 0 500
-2
-1
0
1
2
NormalScore
Residual
(response is Contribu)
3-14

1150 1250 1350 1450 1550 1650 1750
-500
0
500
Fitted Value
Residual
(response is Contribu)
3-10 An experiment was run to determine whether four specific firing temperatures affect the density of a
certain type of brick. The experiment led to the following data:
Temperature Density
100 21.8 21.9 21.7 21.6 21.7
125 21.7 21.4 21.5 21.4
150 21.9 21.8 21.8 21.6 21.5
175 21.9 21.7 21.8 21.4
(a) Does the firing temperature affect the density of the bricks? Use α = 0.05.
No, firing temperature does not affect the density of the bricks. Refer to the Design-Expert output below.
Response: Density
Sum of Mean F
Model 0.16 3 0.052 2.02 0.1569 not significant
A 0.16 3 0.052 2.02 0.1569
Residual 0.36 14 0.026
Lack of Fit 0.000 0
Pure Error 0.36 14 0.026
Cor Total 0.52 17
The "Model F-value" of 2.02 implies the model is not significant relative to the noise. There is a
15.69 % chance that a "Model F-value" this large could occur due to noise.
3-15

Estimated Standard
Mean Error
1-100 21.74 0.072
2-125 21.50 0.080
3-150 21.72 0.072
4-175 21.70 0.080
1 vs 2 0.24 1 0.11 2.23 0.0425
1 vs 3 0.020 1 0.10 0.20 0.8465
1 vs 4 0.040 1 0.11 0.37 0.7156
2 vs 3 -0.22 1 0.11 -2.05 0.0601
2 vs 4 -0.20 1 0.11 -1.76 0.0996
3 vs 4 0.020 1 0.11 0.19 0.8552
(b) Is it appropriate to compare the means using the Fisher LSD method in this experiment?
The analysis of variance tells us that there is no difference in the treatments. There is no need to proceed
with Fisher’s LSD method to decide which mean is difference.
(c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There
is nothing unusual about the residual plots.
Residual
Normal%probability
-0.3 -0.175 -0.05 0.075 0.2
1
5
10
20
30
50
70
80
90
95
99
22
22
22
Predicted
Residuals
-0.3
-0.175
-0.05
0.075
0.2
21.50 21.56 21.62 21.68 21.74
(d) Construct a graphical display of the treatments as described in Section 3-5.3. Does this graph
adequately summarize the results of the analysis of variance in part (b). Yes.
3-16

2 1 .2 2 1 .3 2 1 .4 2 1 .5 2 1 .6 2 1 .7 2 1 .8
M e a n D e n s ity
(12 5 ) (1 7 5,15 0 ,1 0 0)
3-11 Rework Part (d) of Problem 3-10 using the Tukey method. What conclusions can you draw?
Explain carefully how you modified the procedure to account for unequal sample sizes.
When sample sizes are unequal, the appropriate formula for the Tukey method is
( , ) 1 1
2
E
i j
q a f
T MS
n n
α
α
⎛ ⎞
= +⎜ ⎟⎜ ⎟
⎝ ⎠
Treatment 1 vs. Treatment 2= 21.74 – 21.50 = 0.24 < 0.994
All pairwise comparisons do not identify differences. Notice that there are different critical values for the
comparisons depending on the sample sizes of the two groups being compared.
Because we could not reject the hypothesis of equal means using the analysis of variance, we should never
have performed the Tukey test (or any other multiple comparison procedure, for that matter). If you ignore
the analysis of variance results and run multiple comparisons, you will likely make type I errors.
3-12 A manufacturer of television sets is interested in the effect of tube conductivity of four different
types of coating for color picture tubes. The following conductivity data are obtained:
Coating Type Conductivity
1 143 141 150 146
2 152 149 137 143
3 134 136 132 127
4 129 127 132 129
(a) Is there a difference in conductivity due to coating type? Use α = 0.05.
Yes, there is a difference in means. Refer to the Design-Expert output below..
3-17

Sum of Mean F
Model 844.69 3 281.56 14.30 0.0003 significant
A 844.69 3 281.56 14.30 0.0003
Residual 236.25 12 19.69
Lack of Fit 0.000 0
Pure Error 236.25 12 19.69
Cor Total 1080.94 15
Estimated Standard
Mean Error
1-1 145.00 2.22
2-2 145.25 2.22
3-3 132.25 2.22
4-4 129.25 2.22
1 vs 2 -0.25 1 3.14 -0.080 0.9378
1 vs 3 12.75 1 3.14 4.06 0.0016
1 vs 4 15.75 1 3.14 5.02 0.0003
2 vs 3 13.00 1 3.14 4.14 0.0014
2 vs 4 16.00 1 3.14 5.10 0.0003
3 vs 4 3.00 1 3.14 0.96 0.3578
(b) Estimate the overall mean and the treatment effects.
68758937513725129
68755937513725132
31257937513725145
06257937513700145
9375137162207
44
33
22
11
...yyˆ
...yyˆ
...yyˆ
...yyˆ
./ˆ
...
...
...
...
−=−=−=
−=−=−=
=−=−=
=−=−=
==
τ
τ
τ
τ
µ
(c) Compute a 95 percent interval estimate of the mean of coating type 4. Compute a 99 percent interval
estimate of the mean difference between coating types 1 and 4.
Treatment 4:
4
6919
179225129
.
.. ±
08451344155124 4 .. ≤≤ µ
Treatment 1 - Treatment 4: ( ) ( )
4
69192
055325129145
.
.. ±−
336251646 41 .. ≤−≤ µµ
(d) Test all pairs of means using the Fisher LSD method with α=0.05.
Refer to the Design-Expert output above. The Fisher LSD procedure is automatically included in the
output.
3-18

The means of Coating Type 2 and Coating Type 1 are not different. The means of Coating Type 3 and
Coating Type 4 are not different. However, Coating Types 1 and 2 produce higher mean conductivity that
does Coating Types 3 and 4.
(e) Use the graphical method discussed in Section 3-5.3 to compare the means. Which coating produces
the highest conductivity?
2192
4
9616
.
.
n
MS
S E
y .i
=== Coating types 1 and 2 produce the highest conductivity.
1 3 0 1 3 5 1 4 0 1 4 5 1 5 0
C o n d u c tiv ity
(3 ) (2 )(1 )(4 )
(f) Assuming that coating type 4 is currently in use, what are your recommendations to the manufacturer?
We wish to minimize conductivity.
Since coatings 3 and 4 do not differ, and as they both produce the lowest mean values of conductivity, use
either coating 3 or 4. As type 4 is currently being used, there is probably no need to change.
3-13 Reconsider the experiment in Problem 3-12. Analyze the residuals and draw conclusions about
model adequacy.
There is nothing unusual in the normal probability plot. A funnel shape is seen in the plot of residuals
versus predicted conductivity indicating a possible non-constant variance.
3-19

Residual
Normal%probability
-8.25 -4.5 -0.75 3 6.75
1
5
10
20
30
50
70
80
90
95
99
22
Predicted
Residuals
-8.25
-4.5
-0.75
3
6.75
129.25 133.25 137.25 141.25 145.25
22
Coating Type
Residuals
Residuals vs. Coating Type
-8.25
-4.5
-0.75
3
6.75
1 2 3 4
3-14 An article in the ACI Materials Journal (Vol. 84, 1987. pp. 213-216) describes several experiments
investigating the rodding of concrete to remove entrapped air. A 3” x 6” cylinder was used, and the
number of times this rod was used is the design variable. The resulting compressive strength of the
concrete specimen is the response. The data are shown in the following table.
Rodding Level Compressive Strength
10 1530 1530 1440
15 1610 1650 1500
20 1560 1730 1530
25 1500 1490 1510
(a) Is there any difference in compressive strength due to the rodding level? Use α = 0.05.
There are no differences.
3-20

Sum of Mean F
A 28633.33 3 9544.44 1.87 0.2138
Residual 40933.33 8 5116.67
Lack of Fit 0.000 0
Pure Error 40933.33 8 5116.67
Cor Total 69566.67 11
Estimated Standard
Mean Error
1-10 1500.00 41.30
2-15 1586.67 41.30
3-20 1606.67 41.30
4-25 1500.00 41.30
1 vs 2 -86.67 1 58.40 -1.48 0.1761
1 vs 3 -106.67 1 58.40 -1.83 0.1052
1 vs 4 0.000 1 58.40 0.000 1.0000
2 vs 3 -20.00 1 58.40 -0.34 0.7408
2 vs 4 86.67 1 58.40 1.48 0.1761
3 vs 4 106.67 1 58.40 1.83 0.1052
(b) Find the P-value for the F statistic in part (a). From computer output, P=0.2138.
(c) Analyze the residuals from this experiment. What conclusions can you draw about the underlying
model assumptions?
There is nothing unusual about the residual plots.
Residual
Normal%probability
-86.6667 -34.1667 18.3333 70.8333 123.333
1
5
10
20
30
50
70
80
90
95
99
22
Predicted
Residuals
-86.6667
-34.1667
18.3333
70.8333
123.333
1500.00 1526.67 1553.33 1580.00 1606.67
3-21

22
Rodding Level
Residuals
Residuals vs. Rodding Level
-86.6667
-34.1667
18.3333
70.8333
123.333
1 2 3 4
(d) Construct a graphical display to compare the treatment means as describe in Section 3-5.3.
1 4 1 8 1 4 5 9 1 5 0 0 1 5 4 1 1 5 8 2 1 6 2 3 1 6 6 4
M e a n C o m p re s s ive S tre n g th
(1 0 , 2 5 ) (1 5 ) (2 0 )
3-15 An article in Environment International (Vol. 18, No. 4, 1992) describes an experiment in which the
amount of radon released in showers was investigated. Radon enriched water was used in the experiment
and six different orifice diameters were tested in shower heads. The data from the experiment are shown in
the following table.
Orifice Dia. Radon Released (%)
0.37 80 83 83 85
0.51 75 75 79 79
0.71 74 73 76 77
1.02 67 72 74 74
1.40 62 62 67 69
1.99 60 61 64 66
(a) Does the size of the orifice affect the mean percentage of radon released? Use α = 0.05.
3-22

Yes. There is at least one treatment mean that is different.
Response: Radon Released in %
Sum of Mean F
Model 1133.38 5 226.68 30.85 < 0.0001 significant
A 1133.38 5 226.68 30.85 < 0.0001
Residual 132.25 18 7.35
Lack of Fit 0.000 0
Pure Error 132.25 18 7.35
EstimatedStandard
Mean Error
1-0.37 82.75 1.36
2-0.51 77.00 1.36
3-0.71 75.00 1.36
4-1.02 71.75 1.36
5-1.40 65.00 1.36
6-1.99 62.75 1.36
1 vs 2 5.75 1 1.92 3.00 0.0077
1 vs 3 7.75 1 1.92 4.04 0.0008
1 vs 4 11.00 1 1.92 5.74 < 0.0001
1 vs 5 17.75 1 1.92 9.26 < 0.0001
1 vs 6 20.00 1 1.92 10.43 < 0.0001
2 vs 3 2.00 1 1.92 1.04 0.3105
2 vs 4 5.25 1 1.92 2.74 0.0135
2 vs 5 12.00 1 1.92 6.26 < 0.0001
2 vs 6 14.25 1 1.92 7.43 < 0.0001
3 vs 4 3.25 1 1.92 1.70 0.1072
3 vs 5 10.00 1 1.92 5.22 < 0.0001
3 vs 6 12.25 1 1.92 6.39 < 0.0001
4 vs 5 6.75 1 1.92 3.52 0.0024
4 vs 6 9.00 1 1.92 4.70 0.0002
5 vs 6 2.25 1 1.92 1.17 0.2557
(b) Find the P-value for the F statistic in part (a). P=3.161 x 10-8
(c) Analyze the residuals from this experiment.
3-23

Residual
Normal%probability
-4.75 -2.5625 -0.375 1.8125 4
1
5
10
20
30
50
70
80
90
95
99
22
22
22
22
22
Predicted
Residuals
-4.75
-2.5625
-0.375
1.8125
4
62.75 67.75 72.75 77.75 82.75
22
22
22
22
22
Orifice Diameter
Residuals
Residuals vs. Orifice Diameter
-4.75
-2.5625
-0.375
1.8125
4
1 2 3 4 5 6
(d) Find a 95 percent confidence interval on the mean percent radon released when the orifice diameter is
1.40.
Treatment 5 (Orifice =1.40):
4
357
10126
.
.±
8486715262 .. ≤≤ µ
(e) Construct a graphical display to compare the treatment means as describe in Section 3-5.3. What
conclusions can you draw?
3-24

6 0 6 5 7 0 7 5 8 0
C o n d u c tivity
(6 ) (5 ) (3 ) (2 ) (1 )(4 )
Treatments 5 and 6 as a group differ from the other means; 2, 3, and 4 as a group differ from the other
means, 1 differs from the others.
3-16 The response time in milliseconds was determined for three different types of circuits that could be
used in an automatic valve shutoff mechanism. The results are shown in the following table.
Circuit Type Response Time
1 9 12 10 8 15
2 20 21 23 17 30
3 6 5 8 16 7
(a) Test the hypothesis that the three circuit types have the same response time. Use α = 0.01.
From the computer printout, F=16.08, so there is at least one circuit type that is different.
Response: Response Time in ms
Sum of Mean F
A 543.60 2 271.80 16.08 0.0004
Residual 202.80 12 16.90
Lack of Fit 0.000 0
Pure Error 202.80 12 16.90
Cor Total 746.40 14
Estimated Standard
Mean Error
1-1 10.80 1.84
2-2 22.20 1.84
3-3 8.40 1.84
3-25

1 vs 2 -11.40 1 2.60 -4.38 0.0009
1 vs 3 2.40 1 2.60 0.92 0.3742
2 vs 3 13.80 1 2.60 5.31 0.0002
(b) Use Tukey’s test to compare pairs of treatment means. Use α = 0.01.
83851
5
1690
.
n
MS
S E
y .i
===
( ) 045123010 .q ,,. =
( ) 2669045838510 ...t ==
1
e computer output) gives the same results.
(c) eans. What conclusions can
you draw? How do they compare with the conclusions from part (a).
case, the large difference between the mean of treatment 2
vs. 2: ⏐10.8-22.2⏐=11.4 > 9.266
1 vs. 3: ⏐10.8-8.4⏐=2.4 < 9.266
2 vs. 3: ⏐22.2-8.4⏐=13.8 > 9.266
1 and 2 are different. 2 and 3 are different.
Notice that the results indicate that the mean of treatment 2 differs from the means of both treatments 1 and
3, and that the means for treatments 1 and 3 are the same. Notice also that the Fisher LSD procedure (see
th
Use the graphical procedure in Section 3-5.3 to compare the treatment m
The scaled-t plot agrees with part (b). In this
and the other two treatments is very obvious.
5 1 0 1 5 2 0 2 5
T e n s ile S tre n g th
(3 ) (2 )(1 )
Construct a set of orthogonal contrasts, assuming that at the outset(d) of the experiment you suspected the
response time of circuit type 2 to be different from the other two.
( )
0 1 2 3
1 1 2 3
1 1. 2. 3.
1
2 0
2 0
2
54 2 111 42 126
H
H
C y y y
C
µ µ µ
µ µ µ
= − + =
= − + ≠
= − +
= − + = −
3-26

( )
( )
2
1
1
126
529.2
5 6
529.2
31.31
16.9
C
C
SS
F
−
= =
= =
Type 2 differs from the average of type 1 and type 3.
(e) If you were a design engineer and you wished to minimize the response time, which circuit type would
you select?
Either type 1 or type 3 as they are not different from each other and have the lowest response time.
(f) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
The normal probability plot has some points that do not lie along the line in the upper region. This may
indicate potential outliers in the data.
Residual
Normal%probability
-5.2 -1.95 1.3 4.55 7.8
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-5.2
-1.95
1.3
4.55
7.8
8.40 11.85 15.30 18.75 22.20
Circuit Type
Residuals
Residuals vs. Circuit Type
-5.2
-1.95
1.3
4.55
7.8
1 2 3
3-27

3-17 The effective life of insulating fluids at an accelerated load of 35 kV is being studied. Test data
have been obtained for four types of fluids. The results were as follows:
Fluid Type Life (in h) at 35 kV Load
1 17.6 18.9 16.3 17.4 20.1 21.6
2 16.9 15.3 18.6 17.1 19.5 20.3
3 21.4 23.6 19.4 18.5 20.5 22.3
4 19.3 21.1 16.9 17.5 18.3 19.8
(a) Is there any indication that the fluids differ? Use α = 0.05.
At α = 0.05 there are no difference, but at since the P-value is just slightly above 0.05, there is probably a
difference in means.
Response: Life in in h
Sum of Mean F
A 30.16 3 10.05 3.05 0.0525
Residual 65.99 20 3.30
Lack of Fit 0.000 0
Pure Error 65.99 20 3.30
Cor Total 96.16 23
The Model F-value of 3.05 implies there is a 5.25% chance that a "Model F-Value"
this large could occur due to noise.
Estimated Standard
Mean Error
1-1 18.65 0.74
2-2 17.95 0.74
3-3 20.95 0.74
4-4 18.82 0.74
1 vs 2 0.70 1 1.05 0.67 0.5121
1 vs 3 -2.30 1 1.05 -2.19 0.0403
1 vs 4 -0.17 1 1.05 -0.16 0.8753
2 vs 3 -3.00 1 1.05 -2.86 0.0097
2 vs 4 -0.87 1 1.05 -0.83 0.4183
3 vs 4 2.13 1 1.05 2.03 0.0554
(b) Which fluid would you select, given that the objective is long life?
Treatment 3. The Fisher LSD procedure in the computer output indicates that the fluid 3 is different from
the others, and it’s average life also exceeds the average lives of the other three fluids.
(c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
There is nothing unusual in the residual plots.
3-28

Residual
Normal%probability
-2.65 -1.25 0.15 1.55 2.95
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-2.65
-1.25
0.15
1.55
2.95
17.95 18.70 19.45 20.20 20.95
Fluid Type
Residuals
Residuals vs. Fluid Type
-2.65
-1.25
0.15
1.55
2.95
1 2 3 4
3-18 Four different designs for a digital computer circuit are being studied in order to compare the
amount of noise present. The following data have been obtained:
Circuit Design Noise Observed
1 19 20 19 30 8
2 80 61 73 56 80
3 47 26 25 35 50
4 95 46 83 78 97
(a) Is the amount of noise present the same for all four designs? Use α = 0.05.
No, at least one treatment mean is different.
Response: Noise
3-29

Sum of Mean F
A 12042.00 3 4014.00 21.78 < 0.0001
Residual 2948.80 16 184.30
Lack of Fit 0.000 0
Pure Error 2948.80 16 184.30
Cor Total 14990.80 19
Estimated Standard
Mean Error
1-1 19.20 6.07
2-2 70.00 6.07
3-3 36.60 6.07
4-4 79.80 6.07
1 vs 2 -50.80 1 8.59 -5.92 < 0.0001
1 vs 3 -17.40 1 8.59 -2.03 0.0597
1 vs 4 -60.60 1 8.59 -7.06 < 0.0001
2 vs 3 33.40 1 8.59 3.89 0.0013
2 vs 4 -9.80 1 8.59 -1.14 0.2705
3 vs 4 -43.20 1 8.59 -5.03 0.0001
(b) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
There is nothing too unusual about the residual plots, although there is a mild outlier present.
Residual
Normal%probability
-33.8 -21.05 -8.3 4.45 17.2
1
5
10
20
30
50
70
80
90
95
99
22
22
Predicted
Residuals
-33.8
-21.05
-8.3
4.45
17.2
19.20 34.35 49.50 64.65 79.80
3-30

22
22
Circuit Design
Residuals
Residuals vs. Circuit Design
-33.8
-21.05
-8.3
4.45
17.2
1 2 3 4
(c) Which circuit design would you select for use? Low noise is best.
From the Design Expert Output, the Fisher LSD procedure comparing the difference in means identifies
Type 1 as having lower noise than Types 2 and 4. Although the LSD procedure comparing Types 1 and 3
has a P-value greater than 0.05, it is less than 0.10. Unless there are other reasons for choosing Type 3,
Type 1 would be selected.
3-19 Four chemists are asked to determine the percentage of methyl alcohol in a certain chemical
compound. Each chemist makes three determinations, and the results are the following:
Chemist Percentage of Methyl Alcohol
1 84.99 84.04 84.38
2 85.15 85.13 84.88
3 84.72 84.48 85.16
4 84.20 84.10 84.55
(a) Do chemists differ significantly? Use α = 0.05.
There is no significant difference at the 5% level, but chemists differ significantly at the 10% level.
Response: Methyl Alcohol in %
Sum of Mean F
A 1.04 3 0.35 3.25 0.0813
Residual 0.86 8 0.11
Lack of Fit 0.000 0
Pure Error 0.86 8 0.11
Cor Total 1.90 11
Estimated Standard
3-31

Mean Error
1-1 84.47 0.19
2-2 85.05 0.19
3-3 84.79 0.19
4-4 84.28 0.19
1 vs 2 -0.58 1 0.27 -2.18 0.0607
1 vs 3 -0.32 1 0.27 -1.18 0.2703
1 vs 4 0.19 1 0.27 0.70 0.5049
2 vs 3 0.27 1 0.27 1.00 0.3479
2 vs 4 0.77 1 0.27 2.88 0.0205
3 vs 4 0.50 1 0.27 1.88 0.0966
(b) Analyze the residuals from this experiment.
Residual
Normal%probability
-0.43 -0.1925 0.045 0.2825 0.52
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.43
-0.1925
0.045
0.2825
0.52
84.28 84.48 84.67 84.86 85.05
Chemist
Residuals
Residuals vs. Chemist
-0.43
-0.1925
0.045
0.2825
0.52
1 2 3 4
3-32

(c) If chemist 2 is a new employee, construct a meaningful set of orthogonal contrasts that might have
been useful at the start of the experiment.
Chemists Total C1 C2 C3
1 253.41 1 -2 0
2 255.16 -3 0 0
3 254.36 1 1 -1
4 252.85 1 1 1
Contrast Totals: -4.86 0.39 -1.51
( )
( )
( )
( )
( )
( )
543
107270
3800
3800
23
511
0750
107270
0080
0080
63
390
1156
107270
6560
6560
123
864
3
2
3
2
2
2
1
2
1
.
.
.
F.
.
SS
.
.
.
F.
.
SS
*.
.
.
F.
.
SS
CC
CC
CC
===
−
=
====
===
−
=
Only contrast 1 is significant at 5%.
3-20 Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the three
brands are different. Five batteries of each brand are tested with the following results:
Weeks of Life
Brand 1 Brand 2 Brand 3
100 76 108
96 80 100
92 75 96
96 84 98
92 82 100
(a) Are the lives of these brands of batteries different?
Yes, at least one of the brands is different.
Response: Life in Weeks
Sum of Mean F
A 1196.13 2 598.07 38.34 < 0.0001
Residual 187.20 12 15.60
Lack of Fit 0.000 0
Pure Error 187.20 12 15.60
Estimated Standard
Mean Error
1-1 95.20 1.77
2-2 79.40 1.77
3-3 100.40 1.77
3-33

1 vs 2 15.80 1 2.50 6.33 < 0.0001
1 vs 3 -5.20 1 2.50 -2.08 0.0594
2 vs 3 -21.00 1 2.50 -8.41 < 0.0001
Residual
Normal%probability
-4.4 -1.4 1.6 4.6 7.6
1
5
10
20
30
50
70
80
90
95
99
2
2
2
2
22
Predicted
Residuals
-4.4
-1.4
1.6
4.6
7.6
79.40 84.65 89.90 95.15 100.40
2
2
2
2
22
Brand
Residuals
Residuals vs. Brand
-4.4
-1.4
1.6
4.6
7.6
1 2 3
(c) Construct a 95 percent interval estimate on the mean life of battery brand 2. Construct a 99 percent
interval estimate on the mean difference between the lives of battery brands 2 and 3.
n
MS
ty E
aN,.i −
±
2
α
3-34

Brand 2:
5
6015
1792479
.
.. ±
84934079 .. ±
2498355175 2 .. ≤≤ µ
Brand 2 - Brand 3:
n
MS
tyy E
aN,.j.i
2
2
−
±− α
( )
5
60152
05534100479
.
... ±−
3691363128 32 .. −≤−≤− µµ
(d) Which brand would you select for use? If the manufacturer will replace without charge any battery
that fails in less than 85 weeks, what percentage would the company expect to replace?
Chose brand 3 for longest life. Mean life of this brand in 100.4 weeks, and the variance of life is estimated
by 15.60 (MSE). Assuming normality, then the probability of failure before 85 weeks is:
( ) 000050903
6015
410085
..
.
.
=−=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
ΦΦ
That is, about 5 out of 100,000 batteries will fail before 85 week.
3-21 Four catalysts that may affect the concentration of one component in a three component liquid
mixture are being investigated. The following concentrations are obtained:
Catalyst
1 2 3 4
58.2 56.3 50.1 52.9
57.2 54.5 54.2 49.9
58.4 57.0 55.4 50.0
55.8 55.3 51.7
54.9
(a) Do the four catalysts have the same effect on concentration?
No, their means are different.
Response: Concentration
Sum of Mean F
A 85.68 3 28.56 9.92 0.0014
Residual 34.56 12 2.88
Lack of Fit 0.000 0
Pure Error 34.56 12 2.88
Cor Total 120.24 15
Estimated Standard
3-35

Mean Error
1-1 56.90 0.76
2-2 55.77 0.85
3-3 53.23 0.98
4-4 51.13 0.85
1 vs 2 1.13 1 1.14 0.99 0.3426
1 vs 3 3.67 1 1.24 2.96 0.0120
1 vs 4 5.77 1 1.14 5.07 0.0003
2 vs 3 2.54 1 1.30 1.96 0.0735
2 vs 4 4.65 1 1.20 3.87 0.0022
3 vs 4 2.11 1 1.30 1.63 0.1298
Residual
Normal%probability
-3.13333 -1.80833 -0.483333 0.841667 2.16667
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-3.13333
-1.80833
-0.483333
0.841667
2.16667
51.13 52.57 54.01 55.46 56.90
Catalyst
Residuals
Residuals vs. Catalyst
-3.13333
-1.80833
-0.483333
0.841667
2.16667
1 2 3 4
(c) Construct a 99 percent confidence interval estimate of the mean response for catalyst 1.
3-36

n
MS
ty E
aN,.i −
±
2
α
Catalyst 1:
5
882
0553956
.
.. ±
31862956 .. ±
218659581454 1 .. ≤≤ µ
3-22 An experiment was performed to investigate the effectiveness of five insulating materials. Four
samples of each material were tested at an elevated voltage level to accelerate the time to failure. The
failure times (in minutes) is shown below.
Material Failure Time (minutes)
1 110 157 194 178
2 1 2 4 18
3 880 1256 5276 4355
4 495 7040 5307 10050
5 7 5 29 2
(a) Do all five materials have the same effect on mean failure time?
No, at least one material is different.
Response: Failure Timein Minutes
Sum of Mean F
A 1.032E+008 4 2.580E+007 6.19 0.0038
Residual 6.251E+00715 4.167E+006
Lack of Fit 0.000 0
Pure Error 6.251E+00715 4.167E+006
Cor Total 1.657E+00819
Estimated Standard
Mean Error
1-1 159.75 1020.67
2-2 6.25 1020.67
3-3 2941.75 1020.67
4-4 5723.00 1020.67
5-5 10.75 1020.67
1 vs 2 153.50 1 1443.44 0.11 0.9167
1 vs 3 -2782.00 1 1443.44 -1.93 0.0731
1 vs 4 -5563.25 1 1443.44 -3.85 0.0016
1 vs 5 149.00 1 1443.44 0.10 0.9192
2 vs 3 -2935.50 1 1443.44 -2.03 0.0601
2 vs 4 -5716.75 1 1443.44 -3.96 0.0013
2 vs 5 -4.50 1 1443.44 -3.118E-003 0.9976
3 vs 4 -2781.25 1 1443.44 -1.93 0.0732
3 vs 5 2931.00 1 1443.44 2.03 0.0604
3-37

4 vs 5 5712.25 1 1443.44 3.96 0.0013
(b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals.
What information do these plots convey?
Predicted
Residuals
-5228
-2839.25
-450.5
1938.25
4327
6.25 1435.44 2864.62 4293.81 5723.00
Residual
Normal%probability
-5228 -2839.25 -450.5 1938.25 4327
1
5
10
20
30
50
70
80
90
95
99
The plot of residuals versus predicted has a strong outward-opening funnel shape, which indicates the
variance of the original observations is not constant. The normal probability plot also indicates that the
normality assumption is not valid. A data transformation is recommended.
(c) Based on your answer to part (b) conduct another analysis of the failure time data and draw
appropriate conclusions.
A natural log transformation was applied to the failure time data. The analysis in the log scale identifies
that there exists at least one difference in treatment means.
Response: Failure Timein Minutes Transform: Natural log Constant: 0.000
Sum of Mean F
A 165.06 4 41.26 37.66 < 0.0001
Residual 16.44 15 1.10
Lack of Fit 0.000 0
Pure Error 16.44 15 1.10
Cor Total 181.49 19
Estimated Standard
Mean Error
1-1 5.05 0.52
2-2 1.24 0.52
3-3 7.72 0.52
4-4 8.21 0.52
5-5 1.90 0.52
3-38

1 vs 2 3.81 1 0.74 5.15 0.0001
1 vs 3 -2.66 1 0.74 -3.60 0.0026
1 vs 4 -3.16 1 0.74 -4.27 0.0007
1 vs 5 3.15 1 0.74 4.25 0.0007
2 vs 3 -6.47 1 0.74 -8.75 < 0.0001
2 vs 4 -6.97 1 0.74 -9.42 < 0.0001
2 vs 5 -0.66 1 0.74 -0.89 0.3856
3 vs 4 -0.50 1 0.74 -0.67 0.5116
3 vs 5 5.81 1 0.74 7.85 < 0.0001
4 vs 5 6.31 1 0.74 8.52 < 0.0001
There is nothing unusual about the residual plots when the natural log transformation is applied.
Residual
Normal%probability
-2.00945 -1.09511 -0.180766 0.733576 1.64792
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-2.00945
-1.09511
-0.180766
0.733576
1.64792
1.24 2.99 4.73 6.47 8.21
Material
Residuals
Residuals vs. Material
-2.00945
-1.09511
-0.180766
0.733576
1.64792
1 2 3 4 5
3-23 A semiconductor manufacturer has developed three different methods for reducing particle counts
on wafers. All three methods are tested on five wafers and the after-treatment particle counts obtained.
The data are shown below.
3-39

Method Count
1 31 10 21 4 1
2 62 40 24 30 35
3 58 27 120 97 68
(a) Do all methods have the same effect on mean particle count?
No, at least one method has a different effect on mean particle count.
Response: Count
Sum of Mean F
A 8963.73 2 4481.87 7.91 0.0064
Residual 6796.00 12 566.33
Lack of Fit 0.000 0
Pure Error 6796.00 12 566.33
Cor Total 15759.73 14
Estimated Standard
Mean Error
1-1 13.40 10.64
2-2 38.20 10.64
3-3 73.00 10.64
1 vs 2 -24.80 1 15.05 -1.65 0.1253
1 vs 3 -59.60 1 15.05 -3.96 0.0019
2 vs 3 -34.80 1 15.05 -2.31 0.0393
(b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals.
Are there potential concerns about the validity of the assumptions?
The plot of residuals versus predicted appears to be funnel shaped. This indicates the variance of the
original observations is not constant. The residuals plotted in the normal probability plot do not fall along
a straight line, which suggests that the normality assumption is not valid. A data transformation is
recommended.
3-40

Predicted
Residuals
-46
-22.75
0.5
23.75
47
13.40 28.30 43.20 58.10 73.00
Residual
Normal%probability
-46 -22.75 0.5 23.75 47
1
5
10
20
30
50
70
80
90
95
99
(c) Based on your answer to part (b) conduct another analysis of the particle count data and draw
appropriate conclusions.
For count data, a square root transformation is often very effective in resolving problems with inequality of
variance. The analysis of variance for the transformed response is shown below. The difference between
methods is much more apparent after applying the square root transformation.
Response: Count Transform: Square root Constant: 0.000
Sum of Mean F
A 63.90 2 31.95 9.84 0.0030
Residual 38.96 12 3.25
Lack of Fit 0.000 0
Pure Error 38.96 12 3.25
Cor Total 102.86 14
Estimated Standard
Mean Error
1-1 3.26 0.81
2-2 6.10 0.81
3-3 8.31 0.81
1 vs 2 -2.84 1 1.14 -2.49 0.0285
1 vs 3 -5.04 1 1.14 -4.42 0.0008
2 vs 3 -2.21 1 1.14 -1.94 0.0767
3-24 Consider testing the equality of the means of two normal populations, where the variances are
unknown but are assumed to be equal. The appropriate test procedure is the pooled t test. Show that the
pooled t test is equivalent to the single factor analysis of variance.
3-41

22
21
0
2
−
−
= n
p
..
t~
n
S
yy
t assuming n1 = n2 = n
( ) ( ) ( )
E
i
n
j
.ij
n
j
.j
n
j
.j
p MS
n
yy
n
yyyy
S ≡
−
−
=
−
−+−
=
∑∑∑∑ = ===
2222
2
1 1
2
1
1
2
22
1
2
11
for a=2
Furthermore, ( )
n
y
n
yn
yy ..
i
.i
..
22
22
1
2
2
21 −=⎟
⎠
⎞
⎜
⎝
⎛
− ∑=
, which is exactly the same as SSTreatments in a one-way
classification with a=2. Thus we have shown that
E
Treatments
MS
SS
t =2
0 . In general, we know that so
that . Thus the square of the test statistic from the pooled t-test is the same test statistic that
results from a single-factor analysis of variance with a=2.
u,u Ft 1
2
=
221
2
0 −n,F~t
3-25 Show that the variance of the linear combination ∑ is .
=
a
i
.ii yc
1
∑=
a
i
ii cn
1
22
σ
( ) ( )∑∑∑ ∑∑∑ === ===
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
==
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ ii n
j
.ij
a
i
i
a
i
n
j
iji
a
i
.ii
a
i
.ii yVcyVcycVycV
11
2
1 1
2
11
, ( ) 2
σ=ijyV
∑=
=
a
i
ii nc
1
22
σ
3-26 In a fixed effects experiment, suppose that there are n observations for each of four treatments. Let
be single-degree-of-freedom components for the orthogonal contrasts. Prove that
.
2
3
2
2
2
1 Q,Q,Q
2
3
2
2
2
1 QQQSSTreatments ++=
2
33433
2
224322
2
1143211
2
3
QSSyyC
QSSyyyC
QSSyyyyC
C..
C...
C....
=−=
=−−=
=−−−=
n
)yy(
Q
n
)yyy(
Q
n
)yyyy(
Q
..
...
....
2
6
2
12
3
2
432
3
2
4322
2
2
43212
1
−
=
−−
=
−−−
=
n
yyy
QQQ
i ji
.j.i.i
12
69
4
1
2
2
3
2
2
2
1
∑ ∑∑= <
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=++ and since
3-42

⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−= ∑∑∑ =<
4
1
22
2
1
i
.i..
ji
.j.i yyyy , we have Treatments
..
i
.ii
...i
SS
n
y
n
y
n
yy
QQQ =−=
−
=++ ∑
∑
=
=
412
312 24
1
2
4
1
22
2
3
2
2
2
1
for a=4.
3-27 Use Bartlett's test to determine if the assumption of equal variances is satisfied in Problem 3-14.
Use α = 0.05. Did you reach the same conclusion regarding the equality of variance by examining the
residual plots?
c
q
.302622
0 =χ , where
( ) ( )
( )
( ) ( )
( )
aN
Sn
S
aNn
a
c
SlognSlogaNq
a
i
ii
p
a
i
i
a
i
iip
−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−−
−
+=
−−−=
∑
∑
∑
=
=
−−
=
1
2
2
1
11
1
2
10
2
10
1
1
13
1
1
1
820
814
211
2
3
2
2
2
1
.S
.S
.S
=
=
= ( ) ( ) ( )
( ) ( ) ( ) 615
315
820158141521115
315
820158141521115
2
2
.
...
S
...
S
p
p
=
−
−+−+−
=
−
−+−+−
=
( )
( ) ( )
( )
13891
12
1
4
3
133
1
1
31515
133
1
1
1
11
.c
c
a
i
=⎟
⎠
⎞
⎜
⎝
⎛
+
−
+=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−−
−
+= ∑=
−−
( ) ( )
( ) (
1675015014317514
8208142114615315
1
10101010
1
2
10
2
10
...q
.log.log.log.logq
SlognSlogaNq
a
i
iip
=−=
++−−=
−−−= ∑=
)
33860
13891
16750
30262302622
0 .
.
.
.
c
q
. ===χ 4992
4050 .,. =χ
Cannot reject null hypothesis; conclude that the variance are equal. This agrees with the residual plots in
Problem 3-16.
3-28 Use the modified Levene test to determine if the assumption of equal variances is satisfied on
Problem 3-20. Use α = 0.05. Did you reach the same conclusion regarding the equality of variances by
examining the residual plots?
The absolute value of Battery Life – brand median is:
ij iy y−
3-43

Brand 1 Brand 2 Brand 3
4 4 8
0 0 0
4 5 4
0 4 2
4 2 0
The analysis of variance indicates that there is not a difference between the different brands and therefore
the assumption of equal variances is satisfired.
Response: Mod Levine
Sum of Mean F
Model 0.93 2 0.47 0.070 0.9328
A 0.93 2 0.47 0.070 0.9328
Pure Error 80.00 12 6.67
Cor Total 80.93 14
3-29 Refer to Problem 3-16. If we wish to detect a maximum difference in mean response times of 10
milliseconds with a probability of at least 0.90, what sample size should be used? How would you obtain a
preliminary estimate of ?2
σ
2
2
2
2 σ
Φ
a
nD
= , use MSE from Problem 3-10 to estimate .2
σ
( )
( )( )
n.
.
n
9860
91632
10 2
2
==Φ
Letting 050.=α , P(accept) = 0.1 , 211 =−= aυ
Trial and Error yields:
n 2υ Φ P(accept)
5 12 2.22 0.17
6 15 2.43 0.09
7 18 2.62 0.04
Choose n ≥ 6, therefore N ≥ 18
Notice that we have used an estimate of the variance obtained from the present experiment. This indicates
that we probably didn’t use a large enough sample (n was 5 in problem 3-10) to satisfy the criteria
specified in this problem. However, the sample size was adequate to detect differences in one of the circuit
types.
When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of
possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert
will be able to bound the variability in the response, by statements such as “the standard deviation is going
to be at least…” or “the standard deviation shouldn’t be larger than…”.
3-30 Refer to Problem 3-20.
3-44

(a) If we wish to detect a maximum difference in mean battery life of 0.5 percent with a probability of at
least 0.90, what sample size should be used? Discuss how you would obtain a preliminary estimate of
σ2
for answering this question.
Use the MSE from Problem 3-14.
Φ2
2
2
2
=
nD
aσ
( )
( )( )
n.
.
..n
0022440
601532
6667910050 2
2
=
×
=Φ
Letting α = 0 05. , P(accept) = 0.1 , 211 =−= aυ
n 2υ Φ P(accept)
40 117 1.895 0.18
45 132 2.132 0.10
50 147 2.369 0.05
See the discussion from the previous problem about the estimate of variance.
(b) If the difference between brands is great enough so that the standard deviation of an observation is
increased by 25 percent, what sample size should be used if we wish to detect this with a probability of
at least 0.90?
211 =−= aυ 123152 =−=−= aNυ 050.=α 10.)accept(P ≤
( )[ ] ( )( )[ ] n..nP.n 56250112501011101011 22
+=−++=−++=λ
n 2υ λ P(accept)
40 117 4.84 0.13
45 132 5.13 0.11
50 147 5.40 0.10
3-31 Consider the experiment in Problem 3-20. If we wish to construct a 95 percent confidence interval
on the difference in two mean battery lives that has an accuracy of ±2 weeks, how many batteries of each
brand must be tested?
050.=α 615.MSE =
n
MS
twidth E
aN,.
2
0250 −=
n υ2 t width
3-45

5 12 2.179 5.44
10 27 2.05 3.62
31 90 1.99 1.996
32 93 1.99 1.96
3-32 Suppose that four normal populations have means of µ1=50, µ2=60, µ3=50, and µ4=60. How many
observations should be taken from each population so that the probability or rejecting the null hypothesis
of equal population means is at least 0.90? Assume that α=0.05 and that a reasonable estimate of the error
variance is =25.2
σ
100
5555
55
4
220
4
4321
4
1
2
4321
4
1
=
=−==−=
===
=+=
∑
∑
=
=
i
i
i
i
ii
,,,
,,,i,
τ
ττττ
µ
µ
τµµ
( )
n
n
n
a
n i
=
===
∑
Φ
σ
τ
Φ
254
100
2
2
2
( ) 050143 21 .,n, =−== αυυ , From the O.C. curves we can construct the following:
n Φ υ2 β 1-β
4 2.00 12 0.18 0.82
5 2.24 16 0.08 0.92
Therefore, select n=5
(a) How would your answer change if a reasonable estimate of the experimental error variance were =
36?
2
σ
( )
n.
n.
n
a
n i
69440
69440
364
100
2
2
2
=
===
∑
Φ
σ
τ
Φ
n Φ υ2 β 1-β
5 1.863 16 0.24 0.76
6 2.041 20 0.15 0.85
7 2.205 24 0.09 0.91
3-46

(b) How would your answer change if a reasonable estimate of the experimental error variance were =
49?
2
σ
( )
n.
n.
n
a
n i
51020
51020
494
100
2
2
2
=
===
∑
Φ
σ
τ
Φ
n Φ υ2 β 1-β
7 1.890 24 0.16 0.84
8 2.020 28 0.11 0.89
9 2.142 32 0.09 0.91
(c) Can you draw any conclusions about the sensitivity of your answer in the particular situation about
how your estimate of σ affects the decision about sample size?
As our estimate of variability increases the sample size must increase to ensure the same power of the test.
(d) Can you make any recommendations about how we should use this general approach to choosing n in
practice?
When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of
possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert
will be able to bound the variability in the response, by statements such as “the standard deviation is going
to be at least…” or “the standard deviation shouldn’t be larger than…”.
3-34 Refer to the aluminum smelting experiment described in Section 3-8. Verify that ratio control
methods do not affect average cell voltage. Construct a normal probability plot of residuals. Plot the
residuals versus the predicted values. Is there an indication that any underlying assumptions are violated?
Response: Cell Average
Sum of Mean F
Model 2.746E-003 3 9.153E-004 0.20 0.8922 not significant
A 2.746E-003 3 9.153E-004 0.20 0.8922
Residual 0.090 20 4.481E-003
Lack of Fit 0.000 0
Pure Error 0.090 20 4.481E-003
Cor Total 0.092 23
Estimated Standard
Mean Error
1-1 4.86 0.027
2-2 4.83 0.027
3-3 4.85 0.027
4-4 4.84 0.027
3-47

1 vs 2 0.027 1 0.039 0.69 0.4981
1 vs 3 0.013 1 0.039 0.35 0.7337
1 vs 4 0.025 1 0.039 0.65 0.5251
2 vs 3 -0.013 1 0.039 -0.35 0.7337
2 vs 4 -1.667E-003 1 0.039 -0.043 0.9660
3 vs 4 0.012 1 0.039 0.30 0.7659
The following residual plots are satisfactory.
Residual
Normal%probability
-0.11 -0.05625 -0.0025 0.05125 0.105
1
5
10
20
30
50
70
80
90
95
99
333
22
Predicted
Residuals
-0.11
-0.05625
-0.0025
0.05125
0.105
4.833 4.840 4.847 4.853 4.860
333
22
Algorithm
Residuals
Residuals vs. Algorithm
-0.11
-0.05625
-0.0025
0.05125
0.105
1 2 3 4
3-35 Refer to the aluminum smelting experiment in Section 3-8. Verify the ANOVA for pot noise
summarized in Table 3-13. Examine the usual residual plots and comment on the experimental validity.
Response: Cell StDev Transform: Natural log Constant: 0.000
3-48

Sum of Mean F
A 6.17 3 2.06 21.96 < 0.0001
Residual 1.87 20 0.094
Lack of Fit 0.000 0
Pure Error 1.87 20 0.094
Cor Total 8.04 23
Estimated Standard
Mean Error
1-1 -3.09 0.12
2-2 -3.51 0.12
3-3 -2.20 0.12
4-4 -3.36 0.12
1 vs 2 0.42 1 0.18 2.38 0.0272
1 vs 3 -0.89 1 0.18 -5.03 < 0.0001
1 vs 4 0.27 1 0.18 1.52 0.1445
2 vs 3 -1.31 1 0.18 -7.41 < 0.0001
2 vs 4 -0.15 1 0.18 -0.86 0.3975
3 vs 4 1.16 1 0.18 6.55 < 0.0001
The following residual plots identify the residuals to be normally distributed, randomly distributed through
the range of prediction, and uniformly distributed across the different algorithms. This validates the
assumptions for the experiment.
Residual
Normal%probability
-0.55611 -0.288858 -0.0216069 0.245645 0.512896
1
5
10
20
30
50
70
80
90
95
99
333
2
22
2
2
22
2
Predicted
Residuals
-0.55611
-0.288858
-0.0216069
0.245645
0.512896
-3.51 -3.18 -2.85 -2.53 -2.20
3-49

333
2
22
2
2
22
2
Algorithm
Residuals
-0.55611
-0.288858
-0.0216069
0.245645
0.512896
1 2 3 4
3-36 Four different feed rates were investigated in an experiment on a CNC machine producing a
component part used in an aircraft auxiliary power unit. The manufacturing engineer in charge of the
experiment knows that a critical part dimension of interest may be affected by the feed rate. However,
prior experience has indicated that only dispersion effects are likely to be present. That is, changing the
feed rate does not affect the average dimension, but it could affect dimensional variability. The engineer
makes five production runs at each feed rate and obtains the standard deviation of the critical dimension (in
10-3
mm). The data are shown below. Assume that all runs were made in random order.
Feed Rate Production Run
(in/min) 1 2 3 4 5
10 0.09 0.10 0.13 0.08 0.07
12 0.06 0.09 0.12 0.07 0.12
14 0.11 0.08 0.08 0.05 0.06
16 0.19 0.13 0.15 0.20 0.11
(a) Does feed rate have any effect on the standard deviation of this critical dimension?
Because the residual plots were not acceptable for the non-transformed data, a square root transformation
was applied to the standard deviations of the critical dimension. Based on the computer output below, the
feed rate has an effect on the standard deviation of the critical dimension.
Response: Run StDev Transform: Square root Constant: 0.000
Sum of Mean F
A 0.040 3 0.013 7.05 0.0031
Residual 0.030 16 1.903E-003
Lack of Fit 0.000 0
Pure Error 0.030 16 1.903E-003
Cor Total 0.071 19
3-50

Estimated Standard
Mean Error
1-10 0.30 0.020
2-12 0.30 0.020
3-14 0.27 0.020
4-16 0.39 0.020
1 vs 2 4.371E-003 1 0.028 0.16 0.8761
1 vs 3 0.032 1 0.028 1.15 0.2680
1 vs 4 -0.088 1 0.028 -3.18 0.0058
2 vs 3 0.027 1 0.028 0.99 0.3373
2 vs 4 -0.092 1 0.028 -3.34 0.0042
3 vs 4 -0.12 1 0.028 -4.33 0.0005
(b) Use the residuals from this experiment of investigate model adequacy. Are there any problems with
experimental validity?
The residual plots are satisfactory.
Residual
Normal%probability
-0.0608614 -0.0310256 -0.00118983 0.028646 0.0584817
1
5
10
20
30
50
70
80
90
95
99 22
22
Predicted
Residuals
-0.0608614
-0.0310256
-0.00118983
0.028646
0.0584817
0.27 0.30 0.33 0.36 0.39
22
22
Feed Rate
Residuals
Residuals vs. Feed Rate
-0.0608614
-0.0310256
-0.00118983
0.028646
0.0584817
1 2 3 4
3-51

3-37 Consider the data shown in Problem 3-16.
(a) Write out the least squares normal equations for this problem, and solve them for and , using the
usual constraint . Estimate
µ τi
⎟
⎠
⎞
⎜
⎝
⎛ =∑=
3
1
0
i iˆτ 21 ττ − .
µˆ15 15τˆ+ 25τˆ+ 35τˆ+ =207
µˆ5 15τˆ+ =54
µˆ5 25τˆ+ =111
µˆ15 35τˆ+ =42
Imposing , therefore0
3
1
=∑=i
iˆτ 8013.ˆ =µ , 0031 .ˆ −=τ , 4082 .ˆ =τ , 4053 .ˆ −=τ
401140800321 ...ˆˆ −=−−=−ττ
(b) Solve the equations in (a) using the constraint 03 =τˆ . Are the estimators iˆτ and µˆ the same as you
found in (a)? Why? Now estimate 21 ττ − and compare your answer with that for (a). What statement
can you make about estimating contrasts in the iτ ?
Imposing the constraint, 03 =τˆ we get the following solution to the normal equations: 408.ˆ =µ ,
4021 .ˆ =τ , 8132 .ˆ =τ , and 03 =τˆ . These estimators are not the same as in part (a). However,
4011801340221 ...ˆˆ −=−=−ττ , is the same as in part (a). The contrasts are estimable.
(c) Estimate 1τµ + , 3212 τττ −− and 21 ττµ ++ using the two solutions to the normal equations.
Compare the results obtained in each case.
Contrast Estimated from Part (a) Estimated from Part (b)
1 1τµ + 10.80 10.80
2 3212 τττ −− -9.00 -9.00
3 21 ττµ ++ 19.20 24.60
Contrasts 1 and 2 are estimable, 3 is not estimable.
3-38 Apply the general regression significance test to the experiment in Example 3-1. Show that the
procedure yields the same results as the usual analysis of variance.
From Table 3-3:
.. 12355y =
from Example 3-1, we have:
1 2
3 4
ˆ ˆ ˆ617.75 66.55 30.35
ˆ ˆ7.65 89.25
µ τ τ
τ τ
= = − = −
= =
3-52

4 5
2
1 1
7,704,511ij
i j
y
= =
=∑∑ , with 20 degrees of freedom.
( )
( )( ) ( )( ) ( )( ) ( )( ) ( )(
5
.. .
1
ˆ ˆ,
617.75 12355 66.55 2756 30.35 2937 7.65 3127 89.25 3535
7,632,301.25 66,870.55 7,699,172.80
i
i
R y yµ τ µ τ
=
= +
= + − + − + +
= + =
∑
)
with 4 degrees of freedom.
( )
4 5
2
1 1
, 7,704,511 7,699,172.80 5339.2E ij
i j
SS y R µ τ
= =
= − = − =∑∑
with 20-4 degrees of freedom.
This is identical to the SSE found in Example 3-1.
The reduced model:
( ) ( )( )..
ˆ 617.75 12355 7,632,301.25R yµ µ= = = , with 1 degree of freedom.
( ) ( ) ( ), 7,699,172.80 7,632,301.25 66,870.55R R Rτ µ µ τ µ= − = − = , with 4-1=3 degrees of
freedom.
Note: ( ) TreatmentSSR =µτ from Example 3-1.
Finally,
( )
0
66,870.55
22290.83 3 66.8
5339.2 333.7
1616
E
R
F
SS
τ µ
= = = =
which is the same as computed in Example 3-1.
3-39 Use the Kruskal-Wallis test for the experiment in Problem 3-17. Are the results comparable to those
found by the usual analysis of variance?
From Design Expert Output of Problem 3-17
Response: Life in in h
Sum of Mean F
A 30.16 3 10.05 3.05 0.0525
Residual 65.99 20 3.30
Lack of Fit 0.000 0
Pure Error 65.99 20 3.30
Cor Total 96.16 23
( )
( )
( )
[ ] ( ) 815124354040
12424
12
13
1
12
1
2
..N
n
R
NN
H
a
i i
.i
=+−
+
=+−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
= ∑=
3-53

8172
3050 .,. =χ
Accept the null hypothesis; the treatments are not different. This agrees with the analysis of variance.
3-40 Use the Kruskal-Wallis test for the experiment in Problem 3-18. Compare conclusions obtained
with those from the usual analysis of variance?
From Design Expert Output of Problem 3-12
Response: Noise
Sum of Mean F
A 12042.00 3 4014.00 21.78 < 0.0001
Residual 2948.80 16 184.30
Lack of Fit 0.000 0
Pure Error 2948.80 16 184.30
Cor Total 14990.80 19
( )
( )
( )
[ ] ( ) 9013120362691
12020
12
13
1
12
1
2
..N
n
R
NN
H
a
i i
.i
=+−
+
=+−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
= ∑=
84122
4050 .,. =χ
Reject the null hypothesis because the treatments are different. This agrees with the analysis of variance.
3-41 Consider the experiment in Example 3-1. Suppose that the largest observation on etch rate is
incorrectly recorded as 250A/min. What effect does this have on the usual analysis of variance? What
effect does it have on the Kruskal-Wallis test?
The incorrect observation reduces the analysis of variance F0 from 66.8 to 0.50. It does change the value
of the Kruskal-Wallis test.
Minitab Output
One-way ANOVA: Etch Rate 2 versus Power
Analysis of Variance for Etch Rat
Source DF SS MS F P
Power 3 15927 5309 0.50 0.685
Error 16 168739 10546
Total 19 184666
3-54

Chapter 4
Randomized Blocks, Latin Squares, and Related Designs
Solutions
4-1 A chemist wishes to test the effect of four chemical agents on the strength of a particular type of
cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized
block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four
chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze the data from this
experiment (use α = 0.05) and draw appropriate conclusions.
Bolt
Chemical 1 2 3 4 5
1 73 68 74 71 67
2 73 67 75 72 70
3 75 68 78 73 68
4 73 71 75 75 69
Response: Strength
Sum of Mean F
Block 157.00 4 39.25
A 12.95 3 4.32 2.38 0.1211
Residual 21.80 12 1.82
Cor Total 191.75 19
Std. Dev. 1.35 R-Squared 0.3727
Mean 71.75 Adj R-Squared 0.2158
C.V. 1.88 Pred R-Squared -0.7426
PRESS 60.56 Adeq Precision 10.558
Estimated Standard
Mean Error
1-1 70.60 0.60
2-2 71.40 0.60
3-3 72.40 0.60
4-4 72.60 0.60
1 vs 2 -0.80 1 0.85 -0.94 0.3665
1 vs 3 -1.80 1 0.85 -2.11 0.0564
1 vs 4 -2.00 1 0.85 -2.35 0.0370
2 vs 3 -1.00 1 0.85 -1.17 0.2635
2 vs 4 -1.20 1 0.85 -1.41 0.1846
3 vs 4 -0.20 1 0.85 -0.23 0.8185
There is no difference among the chemical types at α = 0.05 level.
4-2 Three different washing solutions are being compared to study their effectiveness in retarding
bacteria growth in five-gallon milk containers. The analysis is done in a laboratory, and only three trials
4-1

can be run on any day. Because days could represent a potential source of variability, the experimenter
decides to use a randomized block design. Observations are taken for four days, and the data are shown
here. Analyze the data from this experiment (use α = 0.05) and draw conclusions.
Days
Solution 1 2 3 4
1 13 22 18 39
2 16 24 17 44
3 5 4 1 22
Response: Growth
Sum of Mean F
Block 1106.92 3 368.97
A 703.50 2 351.75 40.72 0.0003
Residual 51.83 6 8.64
C.V. 15.68 Pred R-Squared 0.7255
Estimated Standard
Mean Error
1-1 23.00 1.47
2-2 25.25 1.47
3-3 8.00 1.47
1 vs 2 -2.25 1 2.08 -1.08 0.3206
1 vs 3 15.00 1 2.08 7.22 0.0004
2 vs 3 17.25 1 2.08 8.30 0.0002
There is a difference between the means of the three solutions. The Fisher LSD procedure indicates that
solution 3 is significantly different than the other two.
4-3 Plot the mean tensile strengths observed for each chemical type in Problem 4-1 and compare them to
a scaled t distribution. What conclusions would you draw from the display?
4-2

7 0 .0 7 1 .0 7 2 .0 7 3 .0
M e a n S tre n g th
(1 ) (2) (3 ,4 )
6030
5
821
.
.
b
MS
S E
y .i
===
There is no obvious difference between the means. This is the same conclusion given by the analysis of
variance.
4-4 Plot the average bacteria counts for each solution in Problem 4-2 and compare them to an
appropriately scaled t distribution. What conclusions can you draw?
5 1 0 1 5 2 0 2 5
B a c te ria G ro w th
(1 ) (2 )(3 )
471
4
648
.
.
b
MS
S E
y .i
===
There is no difference in mean bacteria growth between solutions 1 and 2. However, solution 3 produces
significantly lower mean bacteria growth. This is the same conclusion reached from the Fisher LSD
procedure in Problem 4-4.
4-3

4-5 Consider the hardness testing experiment described in Section 4-1. Suppose that the experiment
was conducted as described and the following Rockwell C-scale data (coded by subtracting 40 units)
obtained:
Coupon
Tip 1 2 3 4
1 9.3 9.4 9.6 10.0
2 9.4 9.3 9.8 9.9
3 9.2 9.4 9.5 9.7
4 9.7 9.6 10.0 10.2
(a) Analyize the data from this experiment.
There is a difference between the means of the four tips.
Response: Hardness
Analysis of variance table [Terms added sequentially (first to last)]
Sum of Mean F
Bock 0.82 3 0.27
A 0.38 3 0.13 14.44 0.0009
Residual 0.080 9 8.889E-003
Cor Total 1.29 15
Estimated Standard
Mean Error
1-1 9.57 0.047
2-2 9.60 0.047
3-3 9.45 0.047
4-4 9.88 0.047
1 vs 2 -0.025 1 0.067 -0.38 0.7163
1 vs 3 0.13 1 0.067 1.87 0.0935
1 vs 4 -0.30 1 0.067 -4.50 0.0015
2 vs 3 0.15 1 0.067 2.25 0.0510
2 vs 4 -0.27 1 0.067 -4.12 0.0026
3 vs 4 -0.43 1 0.067 -6.37 0.0001
(b) Use the Fisher LSD method to make comparisons among the four tips to determine specifically which
tips differ in mean hardness readings.
Based on the LSD bars in the Design Expert plot below, the mean of tip 4 differs from the means of tips 1,
2, and 3. The LSD metod identifies a marginal difference between the means of tips 2 and 3.
4-4

A: Tip
Hardness
One Factor Plot
1 2 3 4
9.2
9.45
9.7
9.95
10.2
The residual plots below do not identify any violations to the assumptions.
Residual
Normal%Probability
Normal Plot of Residuals
-0.1 -0.0375 0.025 0.0875 0.15
1
5
10
20
30
50
70
80
90
95
99
2
2
2
2
Predicted
Residuals
-0.1
-0.0375
0.025
0.0875
0.15
9.22 9.47 9.71 9.96 10.20
4-5

Tip
Residuals
Residuals vs. Tip
-0.1
-0.0375
0.025
0.0875
0.15
1 2 3 4
4-6 A soncumer products company relies on direct mail marketing pieces as a major component of its
advertising campaigns. The company yhas three different designs for a new brochure and want to evaluate
their effectiveness, as there are substantial differences in costs between the three designs. The company
decides to test the three designs by mailing 5,000 samples of each to potential customers in four different
regions of the country. Since there are known regional differences in the customer base, regions are
considered as blocks. The number of responses to each mailing is shown below.
Region
Design NE NW SE SW
1 250 350 219 375
2 400 525 390 580
3 275 340 200 310
The residuals of the analsysis below identify concerns with the normality and equality of variance
assumptions. As a result, a squreroot transformation was applied as shown in the second analsysis table.
The residuals of both analysis are presented for comparison in part (c) of this problem. The analysis
concludes that there is a difference between the mean number of responses for the three designs.
Response: Number of responses
Sum of Mean F
Block 49035.67 3 16345.22
A 90755.17 2 45377.58 50.15 0.0002
Residual 5428.83 6 904.81
Cor Total 1.452E+005 11
4-6

Estimated Standard
Mean Error
1-1 298.50 15.04
2-2 473.75 15.04
3-3 281.25 15.04
1 vs 2 -175.25 1 21.27 -8.24 0.0002
1 vs 3 17.25 1 21.27 0.81 0.4483
2 vs 3 192.50 1 21.27 9.05 0.0001
Design Expert Output for Transformed Data
Response: Number of responses Transform: Square root Constant: 0
Sum of Mean F
Block 35.89 3 11.96
A 60.73 2 30.37 60.47 0.0001
Cor Total 99.64 11
Estimated Standard
Mean Error
1-1 17.17 0.35
2-2 21.69 0.35
3-3 16.69 0.35
1 vs 2 -4.52 1 0.50 -9.01 0.0001
1 vs 3 0.48 1 0.50 0.95 0.3769
2 vs 3 4.99 1 0.50 9.96 < 0.0001
(b) Use the Fisher LSD method to make comparisons among the three designs to determine specifically
which designs differ in mean response rate.
Based on the LSD bars in the Design Expert plot below, designs 1 and 3 do not differ; however, design 2 is
different than designs 1 and 3.
4-7

A: Design
Sqrt(Numberofresponses)
One Factor Plot
1 2 3
14.142
16.627
19.113
21.598
24.083
The first set of residual plots presented below represent the untransformed data. Concerns with the
normality as well as inequality of variance are presented. The second set of residual plots represent
transformed data and do not identify significant violations to the assumptions. The residuals vs. design
identify a slight inequality; however, not a strong violation and an improvement to the non-transformed
data.
Residual
Normal%Probability
-41.75 -22.1667 -2.58333 17 36.5833
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-41.75
-22.1667
-2.58333
17
36.5833
199.75 285.88 372.00 458.13 544.25
4-8

Design
Residuals
Residuals vs. Design
-41.75
-22.1667
-2.58333
17
36.5833
1 2 3
The following are the square root transformed data residual plots.
Residual
Normal%Probability
-0.921041 -0.455263 0.0105142 0.476292 0.942069
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.921041
-0.455263
0.0105142
0.476292
0.942069
14.41 16.68 18.96 21.24 23.52
4-9

Design
Residuals
Residuals vs. Design
-0.921041
-0.455263
0.0105142
0.476292
0.942069
1 2 3
4-7 The effect of three different lubricating oils on fuel economy is diesel truck engines is being
studied. Fuel economy is measured using brake-specific fuel consumption after the engine has been
running for 15 minutes. Five different truck engines are available for the study, and the experimenters
conduct the following randomized complete block design.
Truck
Oil 1 2 3 4 5
1 0.500 0.634 0.487 0.329 0.512
2 0.535 0.675 0.520 0.435 0.540
3 0.513 0.595 0.488 0.400 0.510
The residuals of the analsysis below identify concerns with the normality and equality of variance
assumptions. As a result, a squreroot transformation was applied as shown in the second analsysis table.
The residuals of both analysis are presented for comparison in part (c) of this problem. The analysis
concludes that there is a difference between the mean number of responses for the three designs.
Response: Fuel consumption
Sum of Mean F
Block 0.092 4 0.023
Model 6.706E-003 2 3.353E-003 6.35 0.0223 significant
A 6.706E-003 2 3.353E-003 6.35 0.0223
Residual 4.222E-003 8 5.278E-004
Cor Total 0.10 14
4-10

Estimated Standard
Mean Error
1-1 0.49 0.010
2-2 0.54 0.010
3-3 0.50 0.010
1 vs 2 -0.049 1 0.015 -3.34 0.0102
1 vs 3 -8.800E-003 1 0.015 -0.61 0.5615
2 vs 3 0.040 1 0.015 2.74 0.0255
(b) Use the Fisher LSD method to make comparisons among the three lubricating oils to determine
specifically which oils differ in break-specific fuel consumption.
Based on the LSD bars in the Design Expert plot below, the means for break-specific fuel consumption for
oils 1 and 3 do not differ; however, oil 2 is different than oils 1 and 3.
A: Oil
Fuelconsumption
One Factor Plot
1 2 3
0.329
0.4155
0.502
0.5885
0.675
The residual plots below do not identify any violations to the assumptions.
4-11

Residual
Normal%Probability
-0.0398667 -0.0243167 -0.00876667 0.00678333 0.0223333
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.0398667
-0.0243167
-0.00876667
0.00678333
0.0223333
0.37 0.44 0.52 0.59 0.66
Oil
Residuals
Residuals vs. Oil
-0.0398667
-0.0243167
-0.00876667
0.00678333
0.0223333
1 2 3
4-8 An article in the Fire Safety Journal (“The Effect of Nozzle Design on the Stability and Performance
of Turbulent Water Jets,” Vol. 4, August 1981) describes an experiment in which a shape factor was
determined for several different nozzle designs at six levels of efflux velocity. Interest focused on potential
differences between nozzle designs, with velocity considered as a nuisance variable. The data are shown
below:
Jet Efflux Velocity (m/s)
Nozzle
Design 11.73 14.37 16.59 20.43 23.46 28.74
1 0.78 0.80 0.81 0.75 0.77 0.78
2 0.85 0.85 0.92 0.86 0.81 0.83
3 0.93 0.92 0.95 0.89 0.89 0.83
4 1.14 0.97 0.98 0.88 0.86 0.83
5 0.97 0.86 0.78 0.76 0.76 0.75
4-12

(a) Does nozzle design affect the shape factor? Compare nozzles with a scatter plot and with an analysis
of variance, using α = 0.05.
Response: Shape
Sum of Mean F
Block 0.063 5 0.013
A 0.10 4 0.026 8.92 0.0003
Residual 0.057 20 2.865E-003
Cor Total 0.22 29
Estimated Standard
Mean Error
1-1 0.78 0.022
2-2 0.85 0.022
3-3 0.90 0.022
4-4 0.94 0.022
5-5 0.81 0.022
1 vs 2 -0.072 1 0.031 -2.32 0.0311
1 vs 3 -0.12 1 0.031 -3.88 0.0009
1 vs 4 -0.16 1 0.031 -5.23 < 0.0001
1 vs 5 -0.032 1 0.031 -1.02 0.3177
2 vs 3 -0.048 1 0.031 -1.56 0.1335
2 vs 4 -0.090 1 0.031 -2.91 0.0086
2 vs 5 0.040 1 0.031 1.29 0.2103
3 vs 4 -0.042 1 0.031 -1.35 0.1926
3 vs 5 0.088 1 0.031 2.86 0.0097
4 vs 5 0.13 1 0.031 4.21 0.0004
Nozzle design has a significant effect on shape factor.
4-13

Nozzle Design
Shape
One Factor Plot
1 2 3 4 5
0.749435
0.847076
0.944718
1.04236
1.14
22
22
22
22
(b) Analyze the residual from this experiment.
The plots shown below do not give any indication of serious problems. Thre is some indication of a mild
outlier on the normal probability plot and on the plot of residuals versus the predicted velocity.
Residual
Normal%probability
-0.0786667 -0.0286667 0.0213333 0.0713333 0.121333
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.0786667
-0.0286667
0.0213333
0.0713333
0.121333
0.73 0.80 0.87 0.95 1.02
4-14

22
22
Nozzle Design
Residuals
Residuals vs. Nozzle Design
-0.0786667
-0.0286667
0.0213333
0.0713333
0.121333
1 2 3 4 5
(c) Which nozzle designs are different with respect to shape factor? Draw a graph of average shape factor
for each nozzle type and compare this to a scaled t distribution. Compare the conclusions that you
draw from this plot to those from Duncan’s multiple range test.
0218520
6
0028650
.
.
b
MS
S E
y .i
===
R2= r0.05(2,20) Syi.
= (2.95)(0.021852)= 0.06446
R3= r0.05(3,20) Syi.
= (3.10)(0.021852)= 0.06774
R4= r0.05(4,20) Syi.
= (3.18)(0.021852)= 0.06949
R5= r0.05(5,20) Syi.
= (3.25)(0.021852)= 0.07102
Mean Difference R
1 vs 4 0.16167 > 0.07102 different
1 vs 3 0.12000 > 0.06949 different
1 vs 2 0.07167 > 0.06774 different
1 vs 5 0.03167 < 0.06446
5 vs 4 0.13000 > 0.06949 different
5 vs 3 0.08833 > 0.06774 different
5 vs 2 0.04000 < 0.06446
2 vs 4 0.09000 > 0.06774 different
2 vs 3 0.04833 < 0.06446
3 vs 4 0.04167 < 0.06446
4-15

0 .7 5 0 .8 0 0 .8 5 0 .9 0 0 .9 5
S h a p e F a c to r
(2 ) (3 ) (4 )(5 )(1 )
4-9 Consider the ratio control algorithm experiment described in Chapter 3, Section 3-8. The
experiment was actually conducted as a randomized block design, where six time periods were selected as
the blocks, and all four ratio control algorithms were tested in each time period. The average cell voltage
and the standard deviation of voltage (shown in parentheses) for each cell as follows:
Ratio Control Time Period
Algorithms 1 2 3 4 5 6
1 4.93 (0.05) 4.86 (0.04) 4.75 (0.05) 4.95 (0.06) 4.79 (0.03) 4.88 (0.05)
2 4.85 (0.04) 4.91 (0.02) 4.79 (0.03) 4.85 (0.05) 4.75 (0.03) 4.85 (0.02)
3 4.83 (0.09) 4.88 (0.13) 4.90 (0.11) 4.75 (0.15) 4.82 (0.08) 4.90 (0.12)
4 4.89 (0.03) 4.77 (0.04) 4.94 (0.05) 4.86 (0.05) 4.79 (0.03) 4.76 (0.02)
(a) Analyze the average cell voltage data. (Use α = 0.05.) Does the choice of ratio control algorithm
affect the cell voltage?
Response: Average
Sum of Mean F
Block 0.017 5 3.487E-003
Model 2.746E-003 3 9.153E-004 0.19 0.9014 not significant
A 2.746E-003 3 9.153E-004 0.19 0.9014
Residual 0.072 15 4.812E-003
Cor Total 0.092 23
Mean 4.84 Adj R-Squared -0.1560
Estimated Standard
Mean Error
1-1 4.86 0.028
4-16

2-2 4.83 0.028
3-3 4.85 0.028
4-4 4.84 0.028
1 vs 2 0.027 1 0.040 0.67 0.5156
1 vs 3 0.013 1 0.040 0.33 0.7438
1 vs 4 0.025 1 0.040 0.62 0.5419
2 vs 3 -0.013 1 0.040 -0.33 0.7438
2 vs 4 -1.667E-003 1 0.040 -0.042 0.9674
3 vs 4 0.012 1 0.040 0.29 0.7748
The ratio control algorithm does not affect the mean cell voltage.
(b) Perform an appropriate analysis of the standard deviation of voltage. (Recall that this is called “pot
noise.”) Does the choice of ratio control algorithm affect the pot noise?
Response: StDev Transform: Natural log Constant: 0.000
Sum of Mean F
Block 0.94 5 0.19
A 6.17 3 2.06 33.26 < 0.0001
Residual 0.93 15 0.062
Cor Total 8.04 23
Mean -3.04 Adj R-Squared 0.8432
C.V. -8.18 Pred R-Squared 0.6654
Estimated Standard
Mean Error
1-1 -3.09 0.10
2-2 -3.51 0.10
3-3 -2.20 0.10
4-4 -3.36 0.10
1 vs 2 0.42 1 0.14 2.93 0.0103
1 vs 3 -0.89 1 0.14 -6.19 < 0.0001
1 vs 4 0.27 1 0.14 1.87 0.0813
2 vs 3 -1.31 1 0.14 -9.12 < 0.0001
2 vs 4 -0.15 1 0.14 -1.06 0.3042
3 vs 4 1.16 1 0.14 8.06 < 0.0001
A natural log transformation was applied to the pot noise data. The ratio control algorithm does affect the
pot noise.
(c) Conduct any residual analyses that seem appropriate.
4-17

Residual
Normal%probability
-0.359093 -0.19708 -0.0350673 0.126945 0.288958
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.359093
-0.19708
-0.0350673
0.126945
0.288958
-3.73 -3.26 -2.78 -2.31 -1.84
Algorithm
Residuals
-0.359093
-0.19708
-0.0350673
0.126945
0.288958
1 2 3 4
The normal probability plot shows slight deviations from normality; however, still acceptable.
(d) Which ratio control algorithm would you select if your objective is to reduce both the average cell
voltage and the pot noise?
Since the ratio control algorithm has little effect on average cell voltage, select the algorithm that
minimizes pot noise, that is algorithm #2.
4-10 An aluminum master alloy manufacturer produces grain refiners in ingot form. This company
produces the product in four furnaces. Each furnace is known to have its own unique operating
characteristics, so any experiment run in the foundry that involves more than one furnace will consider
furnace a nuisance variable. The process engineers suspect that stirring rate impacts the grain size of the
product. Each furnace can be run at four different stirring rates. A randomized block design is run for a
particular refiner and the resulting grain size data is shown below.
Furnace
Stirring Rate 1 2 3 4
4-18

5 8 4 5 6
10 14 5 6 9
15 14 6 9 2
20 17 9 3 6
(a) Is there any evidence that stirring rate impacts grain size?
Response: Grain Size
Sum of Mean F
Block 165.19 3 55.06
A 22.19 3 7.40 0.85 0.4995
Residual 78.06 9 8.67
Cor Total 265.44 15
Mean 7.69 Adj R-Squared -0.0382
Estimated Standard
Mean Error
1-5 5.75 1.47
2-10 8.50 1.47
3-15 7.75 1.47
4-20 8.75 1.47
1 vs 2 -2.75 1 2.08 -1.32 0.2193
1 vs 3 -2.00 1 2.08 -0.96 0.3620
1 vs 4 -3.00 1 2.08 -1.44 0.1836
2 vs 3 0.75 1 2.08 0.36 0.7270
2 vs 4 -0.25 1 2.08 -0.12 0.9071
3 vs 4 -1.00 1 2.08 -0.48 0.6425
The analysis of variance shown above indicates that there is no difference in mean grain size due to the
different stirring rates.
(b) Graph the residuals from this experiment on a normal probability plot. Interpret this plot.
4-19

Residual
Normal%probability
-3.8125 -2.0625 -0.3125 1.4375 3.1875
1
5
10
20
30
50
70
80
90
95
99
The plot indicates that normality assumption is valid.
(c) Plot the residuals versus furnace and stirring rate. Does this plot convey any useful information?
Stirring Rate
Residuals
Residuals vs. Stirring Rate
-3.8125
-2.0625
-0.3125
1.4375
3.1875
1 2 3 4
The variance is consistent at different stirring rates. Not only does this validate the assumption of uniform
variance, it also identifies that the different stirring rates do not affect variance.
(d) What should the process engineers recommend concerning the choice of stirring rate and furnace for
this particular grain refiner if small grain size is desirable?
There really is no effect due to the stirring rate.
4-11 Analyze the data in Problem 4-2 using the general regression significance test.
µ : µˆ12 14τˆ+ 24τˆ+ 34τˆ+ 13βˆ+ 23βˆ+ 33βˆ+ 43βˆ+ =225
1τ : µˆ4 14τˆ+ 1βˆ+ 2βˆ+ 3βˆ+ 4βˆ+ =92
2τ : µˆ4 24τˆ+ 1βˆ+ 2βˆ+ 3βˆ+ 4βˆ+ =101
4-20

3τ : µˆ4 34τˆ+ 1βˆ+ 2βˆ+ 3βˆ+ 4βˆ+ =32
1β : µˆ3 1τˆ+ 2τˆ+ 3τˆ+ 13βˆ+ =34
2β : µˆ3 1τˆ+ 2τˆ+ 3τˆ+ 23βˆ+ =50
3β : µˆ3 1τˆ+ 2τˆ+ 3τˆ+ 33βˆ+ =36
4β : µˆ3 1τˆ+ 2τˆ+ 3τˆ+ 43βˆ+ =105
Applying the constraints ∑ ∑ == 0ji
ˆˆ βτ , we obtain:
12
225
=µˆ ,
12
51
1 =τˆ ,
12
78
2 =τˆ ,
12
129
3
−
=τˆ ,
12
89
1
−
=βˆ ,
12
25
2
−
=βˆ ,
12
81
3
−
=βˆ ,
12
195
4 =βˆ
( ) ( ) ( ) ( ) ( ) ( ) ( )+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
= 50
12
25
34
12
89
32
12
129
101
12
78
92
12
51
225
12
225
βτµ ,,R
( ) ( ) 176029105
12
195
36
12
81
.=⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛ −
∑∑ = 60812
ijy , ( )∑∑ =−=−= 835117602960812
..,,RySS ijE βτµ
Model Restricted to τi = 0 :
µ : µˆ12 13βˆ+ 23βˆ+ 33βˆ+ 43βˆ+ =225
1β : µˆ3 13βˆ+ =34
2β : µˆ3 23βˆ+ =50
3β : µˆ3 33βˆ+ =36
4β : µˆ3 43βˆ+ =105
Applying the constraint ∑ , we obtain:= 0j
ˆβ
12
225
=µˆ , 12891 /ˆ −=β ,
12
25
2
−
=βˆ ,
12
81
3
−
=βˆ ,
12
195
4 =βˆ . Now:
( ) ( ) ( ) ( ) ( ) ( ) 675325105
12
195
36
12
81
50
12
25
34
12
89
225
12
225
.,R =⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛
=βµ
( ) ( ) ( ) TreatmentsSS...,R,,R,R ==−=−= 50703675325176029βµβτµβµτ
Model Restricted to :0=jβ
µ : µˆ12 14τˆ+ 24τˆ+ 34τˆ+ =225
1τ : µˆ4 14τˆ+ =92
2τ : µˆ4 24τˆ+ =101
3τ : µˆ4 34τˆ+ =32
Applying the constraint ∑ , we obtain:= 0iˆτ
12
225
=µˆ ,
12
51
1 =τˆ ,
12
78
2 =τˆ ,
12
129
3
−
=τˆ
4-21

( ) ( ) ( ) ( ) ( ) 25492232
12
129
101
12
78
92
12
51
225
12
225
.,R =⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=τµ
( ) ( ) ( ) BlocksSS...,R,,R,R ==−=−= 921106254922176029τµβτµτµβ
4-12 Assuming that chemical types and bolts are fixed, estimate the model parameters τi and βj in
Problem 4-1.
Using Equations 4-14, Applying the constraints, we obtain:
20
35
=µˆ ,
20
23
1
−
=τˆ ,
20
7
2
−
=τˆ ,
20
13
3 =τˆ ,
20
17
4 =τˆ ,
20
35
1 =βˆ ,
20
65
2
−
=βˆ ,
20
75
3 =βˆ ,
20
20
4 =βˆ ,
20
65
5
−
=βˆ
4-13 Draw an operating characteristic curve for the design in Problem 4-2. Does this test seem to be
sensitive to small differences in treatment effects?
Assuming that solution type is a fixed factor, we use the OC curve in appendix V. Calculate
( )6983
4 2
2
2
2
.a
b ii ∑∑ ==
τ
σ
τ
Φ
using MSE to estimate σ2
. We have:
211 =−= aυ ( )( ) ( )( ) 632112 ==−−= baυ .
If ∑ , then:== EMSˆ
i
22
στ
( )
151
13
4
.==Φ and 700.≅β
If ∑ , then:== Ei MSˆ 22 2
στ
( )
631
23
4
.==Φ and 550.≅β , etc.
This test is not very sensitive to small differences.
4-14 Suppose that the observation for chemical type 2 and bolt 3 is missing in Problem 4-1. Analyze the
problem by estimating the missing value. Perform the exact analysis and compare the results.
y23 is missing.
( )( )
( ) ( )
( )( )
2575
34
136022752824
11
32
23 .
ba
ybyay
yˆ
'
..
'
.
'
.
=
−+
=
−−
−+
=
Thus, y2.=357.25, y.3=3022.25, and y..=1435.25
Source SS DF MS F0
Chemicals 12.7844 3 4.2615 2.154
4-22

Bolts 158.8875 4
Error 21.7625 11 1.9784
Total 193.4344 18
F0.10,3,11=2.66, Chemicals are not significant.
4-12 Two missing values in a randomized block. Suppose that in Problem 4-1 the observations for
chemical type 2 and bolt 3 and chemical type 4 and bolt 4 are missing.
(a) Analyze the design by iteratively estimating the missing values as described in Section 4-1.3.
12
54 32
23
'
..
'
.
'
. yyy
yˆ
−+
= and
12
54 44
44
'
..
'
.
'
. yyy
yˆ
−+
=
Data is coded y-70. As an initial guess, set equal to the average of the observations available for
chemical 2. Thus,
0
23y
50
4
20
23 .y == . Then ,
( ) ( ) 043
12
52565840
44 .
.
yˆ =
−+
=
( ) ( ) 415
12
0428175241
23 .
.
yˆ =
−+
=
( ) ( ) 632
12
413065841
44 .
.
yˆ =
−+
=
( ) ( ) 445
12
6327175242
44 .
.
yˆ =
−+
=
( ) ( ) 632
12
443065842
44 .
.
yˆ =
−+
=
44523 .yˆ =∴ 63244 .yˆ =
Sum of Mean F
Block 156.83 4 39.21
A 9.59 3 3.20 2.08 0.1560
Residual 18.41 12 1.53
Cor Total 184.83 19
(b) Differentiate SSE with respect to the two missing values, equate the results to zero, and solve for
estimates of the missing values. Analyze the design using these two estimates of the missing values.
∑∑∑∑ ∑ +−−= 2
20
12
4
12
5
12
..j..iijE yyyySS
Ryy.y.y.y.y.SSE ++−−+= 44234423
2
44
2
23 1073866060
From 0
4423
==
y
SS
y
SS EE
∂
∂
∂
∂
, we obtain:
4-23

732110
861021
4423
4423
.yˆ.yˆ.
.yˆ.yˆ.
=+
=+
45523 .yˆ =⇒ , 63244 .yˆ =
These quantities are almost identical to those found in part (a). The analysis of variance using these new
data does not differ substantially from part (a).
(c) Derive general formulas for estimating two missing values when the observations are in different
blocks.
( ) ( ) ( ) ( ) ( )
ab
yyy
a
yyyy
b
yyyy
yySS kviu..kvv.iuu.kv.kiu.i
kviuE
22222
22 ++′
+
+′++′
−
+′+′+′
−+=
From 0
4423
==
y
SS
y
SS EE
∂
∂
∂
∂
, we obtain:
( )( )
ab
yˆ
ab
'y'by'ay
ab
ba
yˆ kv..j..i
iu −
−+
=⎥
⎦
⎤
⎢
⎣
⎡ −− 11
ab
yˆ
ab
'y'by'ay
ab
)b)(a(
yˆ iu..v..k
kv −
−+
=⎥
⎦
⎤
⎢
⎣
⎡ −− 11
whose simultaneous solution is:
( ) ( ) ( ) ( ) ( ) ( )
( )( ) ( ) ( )
2 2 2 2 2 2
. . ..
2 2
' 1 1 1 ' 1 1 1 ' 1 1 1
ˆ
1 1 1 1 1
i u
iu
y a a b ab y b a b ab y ab a b
y
a b a b
⎡ ⎤ ⎡ ⎤ ⎡− − − − + − − − − − − − −
⎣ ⎦ ⎣ ⎦ ⎣= +
⎡ ⎤− − − − −
⎣ ⎦
⎤
⎦
[ ]
( ) ( )
. . ..
2 2
' ' '
1 1 1
k vab ay by y
a b
+ −
⎡ ⎤− − −
⎣ ⎦
( )( )[ ]
( ) ( )
. . .. . . .
2 2
' ' ' 1 1 ' ' '
ˆ
1 1 1
i u k v
kv
ay by y b a ay by y
y
a b
+ − − − − + −
=
⎡ ⎤− − −
⎣ ⎦
.
(d) Derive general formulas for estimating two missing values when the observations are in the same
block. Suppose that two observations yij and ykj are missing, i≠k (same block j).
( ) ( ) ( ) ( )
ab
yyy
a
yyy
b
yyyy
yySS
kjij..kjijj.kj.kij.i
kjijE
2222
22
++′
+
++′
−
+′++′
−+=
From 0
4423
==
y
SS
y
SS EE
∂
∂
∂
∂
, we obtain
( )( )
( )( )2
11
11
−−+
−−
′−′+′
= bayˆ
ba
yybya
yˆ kj
..j..i
ij
( )( )
( )( )2
11
11
−−+
−−
′−′+′
= bayˆ
ba
yybya
yˆ ij
..j..k
kj
whose simultaneous solution is:
4-24

( )( )
( ) ( )( ) ( )
( ) ( )
2
. . .. . . ... . ..
22
1 1 1
ˆ
1 1 1 1 1
k j i ji j
ij
b ay by y a b ay by yay by y
y
a b a b
⎡ ⎤′ ′ ′ ′ ′ ′− + − + − − + −′ ′ ′+ − ⎣ ⎦= +
− − ⎡ ⎤− − −
⎣ ⎦
( ) ( )
( )( ) ( ) ( )
2
. . .. . . ..
2 4
1 1
ˆ
1 1 1 1 1
k j i j
kj
ay by y b a ay by y
y
a b a b
′ ′ ′ ′ ′ ′⎡ ⎤+ − − − − + −⎣ ⎦=
⎡ ⎤− − − − −
⎣ ⎦
4-17 An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect
of the distance of the object from the eye on the focus time. Four different distances are of interest. He has
five subjects available for the experiment. Because there may be differences among individuals, he decides
to conduct the experiment in a randomized block design. The data obtained follow. Analyze the data from
this experiment (use α = 0.05) and draw appropriate conclusions.
Subject
Distance (ft) 1 2 3 4 5
4 10 6 6 6 6
6 7 6 6 1 6
8 5 3 3 2 5
10 6 4 4 2 3
Response: Focus Time
Sum of Mean F
Block 36.30 4 9.07
A 32.95 3 10.98 8.61 0.0025
Residual 15.30 12 1.27
Cor Total 84.55 19
Estimated Standard
Mean Error
1-4 6.80 0.50
2-6 5.20 0.50
3-8 3.60 0.50
4-10 3.80 0.50
1 vs 2 1.60 1 0.71 2.24 0.0448
1 vs 3 3.20 1 0.71 4.48 0.0008
1 vs 4 3.00 1 0.71 4.20 0.0012
2 vs 3 1.60 1 0.71 2.24 0.0448
2 vs 4 1.40 1 0.71 1.96 0.0736
3 vs 4 -0.20 1 0.71 -0.28 0.7842
Distance has a statistically significant effect on mean focus time.
4-25

4-18 The effect of five different ingredients (A, B, C, D, E) on reaction time of a chemical process is
being studied. Each batch of new material is only large enough to permit five runs to be made.
Furthermore, each runs requires approximately 1 1/2 hours, so only five runs can be made in one day. The
experimenter decides to run the experiment as a Latin square so that day and batch effects can be
systematically controlled. She obtains the data that follow. Analyze the data from this experiment (use α =
0.05) and draw conclusions.
Day
Batch 1 2 3 4 5
1 A=8 B=7 D=1 C=7 E=3
2 C=11 E=2 A=7 D=3 B=8
3 B=4 A=9 C=10 E=1 D=5
4 D=6 C=8 E=6 B=6 A=10
5 E=4 D=2 B=3 A=8 C=8
Minitab Output
General Linear Model
Factor Type Levels Values
Batch random 5 1 2 3 4 5
Day random 5 1 2 3 4 5
Catalyst fixed 5 A B C D E
Analysis of Variance for Time, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Catalyst 4 141.440 141.440 35.360 11.31 0.000
Batch 4 15.440 15.440 3.860 1.23 0.348
Day 4 12.240 12.240 3.060 0.98 0.455
Error 12 37.520 37.520 3.127
Total 24 206.640
4-19 An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the
assembly time for a color television component. Four operators are selected for the study. Furthermore,
the engineer knows that each assembly method produces such fatigue that the time required for the last
assembly may be greater than the time required for the first, regardless of the method. That is, a trend
develops in the required assembly time. To account for this source of variability, the engineer uses the
Latin square design shown below. Analyze the data from this experiment (α = 0.05) draw appropriate
conclusions.
Order of Operator
Assembly 1 2 3 4
1 C=10 D=14 A=7 B=8
2 B=7 C=18 D=11 A=8
3 A=5 B=10 C=11 D=9
4 D=10 A=10 B=12 C=14
Minitab Output
Order random 4 1 2 3 4
Operator random 4 1 2 3 4
Method fixed 4 A B C D
4-26

Method 3 72.500 72.500 24.167 13.81 0.004
Order 3 18.500 18.500 6.167 3.52 0.089
Operator 3 51.500 51.500 17.167 9.81 0.010
Error 6 10.500 10.500 1.750
Total 15 153.000
4-20 Suppose that in Problem 4-18 the observation from batch 3 on day 4 is missing. Estimate the
missing value from Equation 4-24, and perform the analysis using this value.
y354 is missing.
[ ]
( )( )
[ ] ( )
( )( )
583
43
14622415285
12
2
354 .
pp
yyyyp
yˆ
...k...j...i
=
−++
=
−−
′−′+′+′
=
Minitab Output
Day random 5 1 2 3 4 5
Catalyst 4 128.676 128.676 32.169 11.25 0.000
Batch 4 16.092 16.092 4.023 1.41 0.290
Day 4 8.764 8.764 2.191 0.77 0.567
Error 12 34.317 34.317 2.860
Total 24 187.849
4-21 Consider a p x p Latin square with rows (αi), columns (βk), and treatments (τj) fixed. Obtain least
squares estimates of the model parameters αi, βk, τj.
...
p
k
k
p
j
j
p
i
i yˆpˆpˆpˆp: =+++ ∑∑∑ === 111
2
βταµµ
..i
p
k
k
p
j
jii yˆpˆpˆpˆp: =+++ ∑∑ == 11
βταµα , p,...,,i 21=
.j.
p
k
kj
p
i
ij yˆpˆpˆpˆp: =+++ ∑∑ == 11
βταµτ , p,...,,j 21=
k..k
p
j
j
p
i
ik yˆpˆpˆpˆp: =+++ ∑∑ ==
βταµβ
11
, p,...,,k 21=
There are 3p+1 equations in 3p+1 unknowns. The rank of the system is 3p-2. Three side conditions are
necessary. The usual conditions imposed are: . The solution is then:0
111
=== ∑∑∑ ===
p
k
k
p
j
j
p
i
i
ˆˆˆ βτα
...
...2
.. ...
ˆ
ˆ , 1,2,...,i i
y
y
p
y y i
µ
α
= =
= − = p
4-27

. . ...
.. ...
ˆ , 1,2,...,
ˆ , 1,2,...,
j j
k i
y y j
y y k
p
p
τ
β
= − =
= − =
4-22 Derive the missing value formula (Equation 4-24) for the Latin square design.
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+−−−= ∑∑∑∑∑ ∑ 2
2222
2
2
p
y
p
y
p
y
p
y
ySS ...k...j...i
ijkE
Let yijk be missing. Then
( ) ( ) ( ) ( ) R
p
yy
p
yy
p
yy
p
yy
ySS
ijk...ijkk..ijk.j.ijk..i
ijkE +
+′
+
+′
−
+′
−
+′
−= 2
222
2
2
where R is all terms without yijk.. From 0=
ijk
E
y
SS
∂
∂
, we obtain:
( )( ) ( )
22
221
p
'y'y'y'yp
p
pp
y
...k...j...i
ijk
−++
=
−−
, or
( )
( )( )21
2
−−
−++
=
pp
'y'y'y'yp
y
...k...j...i
ijk
4-23 Designs involving several Latin squares. [See Cochran and Cox (1957), John (1971).] The p x p
Latin square contains only p observations for each treatment. To obtain more replications the experimenter
may use several squares, say n. It is immaterial whether the squares used are the same are different. The
appropriate model is
ijkhjh)h(kj)h(ihijkh )(y ετρβταρµ ++++++=
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
=
=
=
n,...,,h
p,...,,k
p,...,,j
p,...,,i
21
21
21
21
where yijkh is the observation on treatment j in row i and column k of the hth square. Note that αi h( ) and
βk h( ) are row and column effects in the hth square, and ρh is the effect of the hth square, and ( )τρ jh is the
interaction between treatments and squares.
(a) Set up the normal equations for this model, and solve for estimates of the model parameters. Assume
that appropriate side conditions on the parameters are 0=∑h h
ˆρ , ( ) 0=∑i hiˆα , and ( ) 0=∑k hk
ˆβ
for each h, , for each h, and0=∑j jˆτ ( ) 0=∑j jh
ˆρτ ( ) 0=∑h jh
ˆρτ for each j.
4-28

....h.....j.h.j.
jh
^
h...kh..)h(k
h...h..i)h(i
......j.j
....h...h
....
yyyy
yyˆ
yyˆ
yyˆ
yyˆ
yˆ
+−−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
−=
−=
−=
=
τρ
β
α
τ
ρ
µ
(b) Write down the analysis of variance table for this design.
Source SS DF
Treatments ∑ − 2
22
np
y
np
y ......j.
p-1
Squares ∑ − 2
2
2
2
np
y
p
y ....h...
n-1
Treatment x Squares SquaresTreatments
....h.j.
SSSS
np
y
p
y
−−−∑ 2
22
(p-1)(n-1)
Rows ∑ − 2
22
np
y
p
y h...h..i
n(p-1)
Columns ∑ − 2
22
np
y
p
y h...kh..
n(p-1)
Error subtraction n(p-1)(p-2)
Total 2
2
2
np
y
y ....
ijkh −∑∑∑∑ np2
-1
4-24 Discuss how the operating characteristics curves in the Appendix may be used with the Latin square
design.
For the fixed effects model use:
∑∑ == 2
2
2
2
2
σ
τ
σ
τ
Φ jj
p
p
, 11 −= pυ ( )( )122 −−= ppυ
For the random effects model use:
λ
σ
σ
τ
= +1
2
2
p
, υ1 1= −p ( )( )122 −−= ppυ
4-25 Suppose that in Problem 4-14 the data taken on day 5 were incorrectly analyzed and had to be
discarded. Develop an appropriate analysis for the remaining data.
Two methods of analysis exist: (1) Use the general regression significance test, or (2) recognize that the
design is a Youden square. The data can be analyzed as a balanced incomplete block design with a = b =
5, r = k = 4 and λ = 3. Using either approach will yield the same analysis of variance.
4-29

Minitab Output
Day random 4 1 2 3 4
Catalyst 4 119.800 120.167 30.042 7.48 0.008
Batch 4 11.667 11.667 2.917 0.73 0.598
Day 3 6.950 6.950 2.317 0.58 0.646
Error 8 32.133 32.133 4.017
Total 19 170.550
4-26 The yield of a chemical process was measured using five batches of raw material, five acid
concentrations, five standing times, (A, B, C, D, E) and five catalyst concentrations (α, β, γ, δ, ε). The
Graeco-Latin square that follows was used. Analyze the data from this experiment (use α = 0.05) and draw
conclusions.
Acid Concentration
Batch 1 2 3 4 5
1 Aα=26 Bβ=16 Cγ=19 Dδ=16 Eε=13
2 Bγ=18 Cδ=21 Dε=18 Eα=11 Aβ=21
3 Cε=20 Dα=12 Eβ=16 Aγ=25 Bδ=13
4 Dβ=15 Eγ=15 Aδ=22 Bε=14 Cα=17
5 Eδ=10 Aε=24 Bα=17 Cβ=17 Dγ=14
Time fixed 5 A B C D E
Catalyst random 5 a b c d e
Acid random 5 1 2 3 4 5
Analysis of Variance for Yield, using Adjusted SS for Tests
Time 4 342.800 342.800 85.700 14.65 0.001
Catalyst 4 12.000 12.000 3.000 0.51 0.729
Batch 4 10.000 10.000 2.500 0.43 0.785
Acid 4 24.400 24.400 6.100 1.04 0.443
Error 8 46.800 46.800 5.850
Total 24 436.000
4-27 Suppose that in Problem 4-19 the engineer suspects that the workplaces used by the four operators
may represent an additional source of variation. A fourth factor, workplace (α, β, γ, δ) may be introduced
and another experiment conducted, yielding the Graeco-Latin square that follows. Analyze the data from
this experiment (use α = 0.05) and draw conclusions.
Order of Operator
Assembly 1 2 3 4
1 Cβ=11 Bγ=10 Dδ=14 Aα=8
2 Bα=8 Cδ=12 Aγ=10 Dβ=12
3 Aδ=9 Dα=11 Bβ=7 Cγ=15
4-30

4 Dγ=9 Aβ=8 Cα=18 Bδ=6
Minitab Output
Method fixed 4 A B C D
Order random 4 1 2 3 4
Workplac random 4 a b c d
Method 3 95.500 95.500 31.833 3.47 0.167
Order 3 0.500 0.500 0.167 0.02 0.996
Operator 3 19.000 19.000 6.333 0.69 0.616
Workplac 3 7.500 7.500 2.500 0.27 0.843
Error 3 27.500 27.500 9.167
Total 15 150.000
However, there are only three degrees of freedom for error, so the test is not very sensitive.
4-28 Construct a 5 x 5 hypersquare for studying the effects of five factors. Exhibit the analysis of
variance table for this design.
Three 5 x 5 orthogonal Latin Squares are:
ABCDE
BCDEA
CDEAB
DEABC
EABCD
αβγδε
γδεαβ
εαβγδ
βγδεα
δεαβγ
12345
45123
23451
51234
34512
Let rows = factor 1, columns = factor 2, Latin letters = factor 3, Greek letters = factor 4 and numbers =
factor 5. The analysis of variance table is:
Source DF
Rows 4
Columns 4
Latin Letters 4
Greek Letters 4
Numbers 4
Error 4
Total 24
4-29 Consider the data in Problems 4-19 and 4-27. Suppressing the Greek letters in 4-27, analyze the data
using the method developed in Problem 4-23.
Square 1 - Operator
Batch 1 2 3 4 Row Total
1 C=10 D=14 A=7 B=8 (39)
2 B=7 C=18 D=11 A=8 (44)
3 A=5 B=10 C=11 D=9 (35)
4 D=10 A=10 B=12 C=14 (46)
(32) (52) (41) (36) 164=y…1
4-31

Square 2 - Operator
Batch 1 2 3 4 Row Total
1 C=11 B=10 D=14 A=8 (43)
2 B=8 C=12 A=10 D=12 (42)
3 A=9 D=11 B=7 C=15 (42)
4 D=9 A=8 C=18 B=6 (41)
(37) (41) (49) (41) 168=y…2
Assembly Methods Totals
A y.1..=65
B y.2..=68
C y.3..=109
D y.4..=90
Source SS DF MS F0
Assembly Methods 159.25 3 53.08 14.00*
Squares 0.50 1 0.50
A x S 8.75 3 2.92 0.77
Assembly Order (Rows) 19.00 6 3.17
Operators (columns) 70.50 6 11.75
Error 45.50 12 3.79
Total 303.50 31
Significant at 1%.
4-30 Consider the randomized block design with one missing value in Problem 4-15. Analyze this data
by using the exact analysis of the missing value problem discussed in Section 4-1.4. Compare your results
to the approximate analysis of these data given in Table 4-15.
µ : 15µ +4 1τ +3 2τ +4 3τ +4 4τ + 4 1β + 4 2β + 3 3β + 4 4β =17
τ1 : 4µ +4 1τ + β1 + β2 + β3 + β4
=3
τ2 : 3µ +3 2τ + β1 + β2 + β4
=1
τ3 : 4µ +4 3τ + β1 + β2 + β3 + β4
=-2
τ4 : 4µ +4 4τ + β1 + β2 + β3 + β4
=15
β1 : 4µ + τ1 + τ2 + τ3 +τ4 + 4 1β =-4
β2 : 4µ + τ1 + τ2 + τ3 +τ4 + 3 2β =-3
β3 : 3µ + τ1 + τ3 +τ4 + 4 3β =6
β4 : 4µ + τ1 + τ2 + τ3 +τ4 + 4 4β =19
Applying the constraints ∑ ∑ == 0ji
ˆˆ βτ , we obtain:
36
41
=µˆ ,
36
14
1
−
=τˆ ,
36
24
2
−
=τˆ ,
36
59
3
−
=τˆ ,
36
94
4 =τˆ ,
36
77
1
−
=βˆ ,
36
68
2
−
=βˆ ,
36
24
3 =βˆ ,
36
121
4 =βˆ
4-32

( ) ∑ ∑= =
=++=
4
1
4
1
78138
i j
j.j.ii.. .yˆyˆyˆ,,R βτµβτµ
With 7 degrees of freedom.
∑∑ = 001452
.yij , ( )∑∑ =−=−= 22678138001452
...,,RySS ijE βτµ
which is identical to SSE obtained in the approximate analysis. In general, the SSE in the exact and
approximate analyses will be the same.
To test Ho: 0=iτ the reduced model is . The normal equations used are:ijjijy εβµ ++=
µ : ˆ15µ 1
ˆ4β+ 2
ˆ4β+ 3
ˆ3β+ 4
ˆ4β+ =17
β1 : 4µ 1
ˆ4β+ =-4
β2 : 4µ 2
ˆ4β+ =-3
β3 : 3µ 3
ˆ3β+ =6
β4 : 4µ 4
ˆ4β+ =18
Applying the constraint ∑ = 0j
ˆβ , we obtain:
16
19
=µˆ ,
16
35
1
−
=βˆ ,
16
31
2
−
=βˆ ,
16
13
3 =βˆ ,
16
53
4 =βˆ . Now ( ) ∑=
=+=
4
1
2599
j
j.j.. .yˆyˆ,R βµβµ
with 7-4=3 degrees of freedom. ( )βµτ ,R is used to test Ho:τi = 0 .
The sum of squares for blocks is found from the reduced model . The normal equations
used are:
ijiijy ετµ ++=
µ : 15µ +4 1τ +3 2τ +4 3τ +4 4τ =17
τ1 : ˆ4µ +4 1τ =3
τ2 : ˆ3µ +3 2τ =1
τ3 : ˆ4µ +4 3τ =-2
τ4 : ˆ4µ +4 4τ =15
Applying the constraint , we obtain:ˆ 0iτ =∑
13
ˆ
12
µ = , 1
4
ˆ
12
τ
−
= , 2
9
ˆ
12
τ
−
= , 3
19
ˆ
12
τ
−
= , 4
32
ˆ
12
τ =
4-33

( ) ∑=
=+=
4
1
8359
i
.ii.. .yˆyˆ,R τµτµ
( ) ( ) ( ) BlocksSS...,R,,R,R ==−=−= 9578835978138τµβτµτµβ
with 7-4=3 degrees of freedom.
Source DF SS(exact) SS(approximate)
Tips 3 39.53 39.98
Blocks 3 78.95 79.53
Error 8 6.22 6.22
Total 14 125.74 125.73
Note that for the exact analysis, .EBlocksTipsT SSSSSSSS ++≠
4-31 An engineer is studying the mileage performance characteristics of five types of gasoline additives.
In the road test he wishes to use cars as blocks; however, because of a time constraint, he must use an
incomplete block design. He runs the balanced design with the five blocks that follow. Analyze the data
from this experiment (use α = 0.05) and draw conclusions.
Car
Additive 1 2 3 4 5
1 17 14 13 12
2 14 14 13 10
3 14 13 14 9
4 13 11 11 12
5 11 12 10 8
There are several computer software packages that can analyze the incomplete block designs discussed in
this chapter. The Minitab General Linear Model procedure is a widely available package with this
capability. The output from this routine for Problem 4-27 follows. The adjusted sums of squares are the
appropriate sums of squares to use for testing the difference between the means of the gasoline additives.
Minitab Output
Additive fixed 5 1 2 3 4 5
Car random 5 1 2 3 4 5
Analysis of Variance for Mileage, using Adjusted SS for Tests
Additive 4 31.7000 35.7333 8.9333 9.81 0.001
Car 4 35.2333 35.2333 8.8083 9.67 0.001
Error 11 10.0167 10.0167 0.9106
Total 19 76.9500
4-32 Construct a set of orthogonal contrasts for the data in Problem 4-31. Compute the sum of squares
for each contrast.
4-34

One possible set of orthogonal contrasts is:
21540 µµµµ +=+:H (1)
210 µµ =:H (2)
540 µµ =:H (3)
215430 4 µµµµµ +++=:H (4)
The sums of squares and F-tests are:
Brand -> 1 2 3 4 5
Qi 33/4 11/4 -3/4 -14/4 -27/4 c Qi i∑ SS F0
(1) -1 -1 0 1 1 -85/4 30.10 39.09
(2) 1 -1 0 0 0 -22/4 4.03 5.23
(3) 0 0 0 -1 1 -13/4 1.41 1.83
(4) -1 -1 4 -1 -1 -15/4 0.19 0.25
Contrasts (1) and (2) are significant at the 1% and 5% levels, respectively.
4-33 Seven different hardwood concentrations are being studied to determine their effect on the strength
of the paper produced. However the pilot plant can only produce three runs each day. As days may differ,
the analyst uses the balanced incomplete block design that follows. Analyze this experiment (use α = 0.05)
and draw conclusions.
Hardwood Days
Concentration (%) 1 2 3 4 5 6 7
2 114 120 117
4 126 120 119
6 137 114 134
8 141 129 149
10 145 150 143
12 120 118 123
14 136 130 127
There are several computer software packages that can analyze the incomplete block designs discussed in
this chapter. The Minitab General Linear Model procedure is a widely available package with this
capability. The output from this routine for Problem 4-33 follows. The adjusted sums of squares are the
appropriate sums of squares to use for testing the difference between the means of the hardwood
concentrations.
Minitab Output
Concentr fixed 7 2 4 6 8 10 12 14
Days random 7 1 2 3 4 5 6 7
Analysis of Variance for Strength, using Adjusted SS for Tests
Concentr 6 2037.62 1317.43 219.57 10.42 0.002
Days 6 394.10 394.10 65.68 3.12 0.070
Error 8 168.57 168.57 21.07
Total 20 2600.29
4-35

4-34 Analyze the data in Example 4-6 using the general regression significance test.
µ : 12µ + 3 1τ +3 2τ + 3 3τ + 3 4τ + 3 1β + 3 2β + 3 3β + 3 4β =870
τ1 : 3µ + 3 1τ + β1 + β3 + β4
=218
τ2 : 3µ +3 2τ + β2 + β3 + β4
=214
τ3 : 3µ + 3 3τ + β1 + β2 + β3
=216
τ4 : 3µ + 3 4τ + β1 + β2 + β4
=222
β1 : 3µ + τ1 + τ3 +τ4 + 3 1β =221
β2 : 3µ + τ2 + τ3 +τ4 + 3 2β =207
β3 : 3µ + τ1 + τ2 + τ3 + 3 3β =224
β4 : 3µ + τ1 + τ2 +τ4 + 3 4β =218
Applying the constraints , we obtain:τ βi j= =∑∑ 0
/µ = 870 12 , , , , ,/τ1 9 8= − /τ2 7 8= − /τ3 4 8= − /τ4 20 8=
/β1 7 8= , , ,/β2 31 8= − /β3 24 8= /β4 0 8=
( )
4 4
.. . .
1 1
ˆˆ ˆ, , 63,152.75i i j j
i j
R y y yµ τ β µ τ β
= =
= + + =∑ ∑
yij
2
6315600=∑∑ , .
SS y RE ij= − = − =∑∑ 2
63156 00 63152 75 325( , , ) . . .µ τ β .
To test Ho:τi = 0 the reduced model is yij j ij= + +µ β ε . The normal equations used are:
µ : 12µ + 3 1β + 3 2β + 3 3β + 3 4β =870
β1 : 3µ + 3 1β =221
β2 : 3µ + 3 2β =207
β3 : 3µ + 3 3β =224
β4 : 3µ + 3 4β =218
Applying the constraint ∑ = 0j
ˆβ , we obtain:
12
870
=µˆ ,
6
7
1 =βˆ ,
6
21
2
−
=βˆ ,
6
13
3 =βˆ ,
6
1
4 =βˆ
( ) ∑=
=+=
4
1
0013063
j
j.j.. .,yˆyˆ,R βµβµ
4-36

with 7 – 4 = 3 degrees of freedom. ( )βµτ ,R is used to test Ho: 0=iτ .
The sum of squares for blocks is found from the reduced model . The normal equations
used are:
ijiijy ετµ ++=
µ : 12µ + 3 1τ +3 2τ + 3 3τ + 3 4τ =870
τ1 : 3µ + 3 1τ =218
τ2 : 3µ +3 2τ =214
τ3 : 3µ + 3 3τ =216
τ4 : 3µ + 3 4τ =222
The sum of squares for blocks is found as in Example 4-6. We may use the method shown above to find an
adjusted sum of squares for blocks from the reduced model, .ijiijy ετµ ++=
4-35 Prove that
( )a
Qk
a
i i
λ
∑=1
2
is the adjusted sum of squares for treatments in a BIBD.
We may use the general regression significance test to derive the computational formula for the adjusted
treatment sum of squares. We will need the following:
( )a
kQ
ˆ i
i
λ
τ = , ∑=
−=
b
i
j.ij.ii ynkykQ
1
( ) ∑ ∑= =
++=
a
i
b
j
j.j.ii.. yˆyˆyˆ,,R
1 1
βτµβτµ
and the sum of squares we need is:
( ) ∑∑ ∑ == =
−++=
b
j
j.
a
i
b
j
j.j.ii..
k
y
yˆyˆyˆ,R
1
2
1 1
βτµβµτ
The normal equation for β is, from equation (4-35),
∑=
=++
a
i
j.jiij yˆkˆnˆk:
1
βτµβ
and from this we have:
∑=
−−=
a
i
iijj.j.j.jj. ˆnyˆkyyˆky
1
2
τµβ
4-37

therefore,
( ) ∑ ∑
∑
= =
=
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−−++=
a
i
b
j
j.
a
i
iijj.
j.j.
.ii..
k
y
k
ˆny
k
yˆk
k
y
yˆyˆ,R
1 1
2
1
2
τ
µ
τµβµτ
)adjusted(Treatments
a
i
a
i
i
a
i
a
i
j.ij.ii SS
a
Q
k
a
kQ
Qyn
k
yˆ),(R i
≡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−= ∑∑∑ ∑ === = 1
2
11 1
1
λλ
τβµτ
4-36 An experimenter wishes to compare four treatments in blocks of two runs. Find a BIBD for this
experiment with six blocks.
Treatment Block 1 Block 2 Block 3 Block 4 Block 5 Block 6
1 X X X
2 X X X
3 X X X
4 X X X
Note that the design is formed by taking all combinations of the 4 treatments 2 at a time. The parameters of
the design are λ = 1, a = 4, b = 6, k = 3, and r = 2
4-37 An experimenter wishes to compare eight treatments in blocks of four runs. Find a BIBD with 14
blocks and λ = 3.
The design has parameters a = 8, b = 14, λ = 3, r = 2 and k = 4. It may be generated from a 23
factorial
design confounded in two blocks of four observations each, with each main effect and interaction
successively confounded (7 replications) forming the 14 blocks. The design is discussed by John (1971,
pg. 222) and Cochran and Cox (1957, pg. 473). The design follows:
Blocks 1=(I) 2=a 3=b 4=ab 5=c 6=ac 7=bc 8=abc
1 X X X X
2 X X X X
3 X X X X
4 X X X X
5 X X X X
6 X X X X
7 X X X X
8 X X X X
9 X X X X
10 X X X X
11 X X X X
12 X X X X
13 X X X X
14 X X X X
4-38

4-38 Perform the interblock analysis for the design in Problem 4-31.
The interblock analysis for Problem 4-31 uses and . A summary of the interblock,
intrablock and combined estimates is:
7702
.ˆ =σ 1422
.ˆ =βσ
Parameter Intrablock Interblock Combined
τ1 2.20 -1.80 2.18
τ2 0.73 0.20 0.73
τ3 -0.20 -5.80 -0.23
τ4 -0.93 9.20 -0.88
τ5 -1.80 -1.80 -1.80
4-39 Perform the interblock analysis for the design in Problem 4-33. The interblock analysis for problem
4-33 uses and07212
.ˆ =σ
( )
( )
[ ]( )
( )
( )2
1 65.68 21.07 6
19.12
1 7 2
Blocks adj EMS MS b
a r
βσ
⎡ ⎤− − −⎣ ⎦= =
−
= . A
summary of the interblock, intrablock, and combined estimates is give below
Parameter Intrablock Interblock Combined
τ1 -12.43 -11.79 -12.38
τ2 -8.57 -4.29 -7.92
τ3 2.57 -8.79 1.76
τ4 10.71 9.21 10.61
τ5 13.71 21.21 14.67
τ6 -5.14 -22.29 -6.36
τ7 -0.86 10.71 -0.03
4-40 Verify that a BIBD with the parameters a = 8, r = 8, k = 4, and b = 16 does not exist. These
conditions imply that λ =
−
−
= =
r k
a
( ) ( )1
1
8 3
7
24
7
, which is not an integer, so a balanced design with these
parameters cannot exist.
4-41 Show that the variance of the intra block estimators {τi } is
( )
( )2
2
1
a
)a(k
λ
σ−
.
Note that
( )a
kQ
ˆ i
i
λ
τ = , and ∑=
−=
b
j
j.ij.ii yn
k
yQ
1
1
, and ( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−−=−= ∑∑ ==
.i
b
j
j.ij.i
b
j
j.ij.ii yynykynkykQ
11
1
yi. contains r observations, and the quantity in the parenthesis is the sum of r(k-1) observations, not
including treatment i. Therefore,
( ) ( ) ( ) ( ) 2222
11 σσ −+−== krkrQVkkQV ii
4-39

or
( ) ( ) ( ){ }[ ] ( )
k
kr
kkr
k
QV i
2
2
2
1
111
1 σ
σ
−
=+−−=
To find ( )iˆV τ , note that:
( ) ( ) ( ) ( )
( )
2
2
2
22
11
σ
λ
σ
λλ
τ
a
kkr
k
kr
a
k
QV
a
k
ˆV ii
−
=
−
⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
However, since ( ) ( 11 −=− kra )λ , we have:
( ) ( ) 2
2
1
σ
λ
τ
a
ak
ˆV i
−
=
Furthermore, the { }iˆτ are not independent, this is required to show that ( ) 22
σ
λ
ττ
a
k
ˆˆV ji =−
4-42 Extended incomplete block designs. Occasionally the block size obeys the relationship a < k < 2a.
An extended incomplete block design consists of a single replicate or each treatment in each block along
with an incomplete block design with k* = k-a. In the balanced case, the incomplete block design will have
parameters k* = k-a, r* = r-b, and λ*. Write out the statistical analysis. (Hint: In the extended incomplete
block design, we have λ = 2r-b+λ*.)
As an example of an extended incomplete block design, suppose we have a=5 treatments, b=5 blocks and
k=9. A design could be found by running all five treatments in each block, plus a block from the balanced
incomplete block design with k* = k-a=9-5=4 and λ*=3. The design is:
Block Complete Treatment Incomplete Treatment
1 1,2,3,4,5 2,3,4,5
2 1,2,3,4,5 1,2,4,5
3 1,2,3,4,5 1,3,4,5
4 1,2,3,4,5 1,2,3,4
5 1,2,3,4,5 1,2,3,5
Note that r=9, since the augmenting incomplete block design has r*=4, and r= r* + b = 4+5=9, and λ = 2r-
b+λ*=18-5+3=16. Since some treatments are repeated in each block it is possible to compute an error sum
of squares between repeat observations. The difference between this and the residual sum of squares is due
to interaction. The analysis of variance table is shown below:
Source SS DF
Treatments
(adjusted) ∑ λa
Q
k i
2
a-1
Blocks
N
y
k
y ..j.
22
−∑ b-1
Interaction Subtraction (a-1)(b-1)
Error [SS between repeat observations] b(k-a)
4-40

Total ∑∑ −
N
y
y ..
ij
2
2
N-1
4-41

Chapter 5
Introduction to Factorial Designs
Solutions
5-1 The yield of a chemical process is being studied. The two most important variables are thought to
be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment
with two replicates is performed. The yield data follow:
Pressure
Temperature 200 215 230
150 90.4 90.7 90.2
90.2 90.6 90.4
160 90.1 90.5 89.9
90.3 90.6 90.1
170 90.5 90.8 90.4
90.7 90.9 90.1
(a) Analyze the data and draw conclusions. Use α = 0.05.
Both pressure (A) and temperature (B) are significant, the interaction is not.
Response:Surface Finish
Sum of Mean F
A 0.77 2 0.38 21.59 0.0004
B 0.30 2 0.15 8.47 0.0085
AB 0.069 4 0.017 0.97 0.4700
Residual 0.16 9 0.018
Lack of Fit 0.000 0
Pure Error 0.16 9 0.018
Cor Total 1.30 17
The Model F-value of 8.00 implies the model is significant. There is only a 0.26% chance that a
"Model F-Value" this large could occur due to noise.
Values of "Prob > F" less than 0.0500 indicate model terms are significant.
In this case A, B are significant model terms.
Values greater than 0.1000 indicate the model terms are not significant.
If there are many insignificant model terms (not counting those required to support hierarchy),
model reduction may improve your model.
(b) Prepare appropriate residual plots and comment on the model’s adequacy.
The residual plots show no serious deviations from the assumptions.
5-1

Predicted
Residuals
-0.15
-0.075
4.26326E-014
0.075
0.15
90.00 90.21 90.43 90.64 90.85
Residual
Normal%probability
-0.15 -0.075 -4.26326E-014 0.075 0.15
1
5
10
20
30
50
70
80
90
95
99
22
Temperature
Residuals
Residuals vs. Temperature
-0.15
-0.075
4.26326E-014
0.075
0.15
1 2 3
3
2
22
2
3
22
2
2
3
Pressure
Residuals
Residuals vs. Pressure
-0.15
-0.075
4.26326E-014
0.075
0.15
1 2 3
(c) Under what conditions would you operate this process?
5-2

DESIGN-EXPERT Plot
Yield
X = A: Pressure
Y = B: Temperature
Design Points
B1 150
B2 160
B3 170
Te m p e ra tu re
Interaction G raph
P re s s u re
Yield
2 0 0 2 1 5 2 3 0
8 9 .8 4 9 2
9 0 .1 3 7 1
9 0 .4 2 5
9 0 .7 1 2 9
9 1 .0 0 0 8
2
2
2
2
2
2
Set pressure at 215 and Temperature at the high level, 170 degrees C, as this gives the highest yield.
The standard analysis of variance treats all design factors as if they were qualitative. In this case, both
factors are quantitative, so some further analysis can be performed. In Section 5-5, we show how response
curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor.
Since both factors in this problem are quantitative and have three levels, we can fit linear and quadratic
effects of both temperature and pressure, exactly as in Example 5-5 in the text. The Design-Expert output,
including the response surface plots, now follows.
Response:Surface Finish
Sum of Mean F
A 0.10 1 0.10 7.22 0.0198
B 0.067 1 0.067 4.83 0.0483
A2 0.67 1 0.67 47.74 < 0.0001
B2 0.23 1 0.23 16.72 0.0015
AB 0.061 1 0.061 4.38 0.0582
Residual 0.17 12 0.014
Lack of Fit 7.639E-003 3 2.546E-003 0.14 0.9314 not significant
Pure Error 0.16 9 0.018
Cor Total1.30 17
In this case A, B, A2, B2 are significant model terms.
Coefficient Standard 95% CI 95% CI
5-3

Factor Estimate DF Error Low High VIF
Intercept 90.52 1 0.062 90.39 90.66
A-Pressure -0.092 1 0.034 -0.17 -0.017 1.00
B-Temperature 0.075 1 0.034 6.594E-004 0.15 1.00
A2 -0.41 1 0.059 -0.54 -0.28 1.00
B2 0.24 1 0.059 0.11 0.37 1.00
AB -0.087 1 0.042 -0.18 3.548E-003 1.00
Final Equation in Terms of Coded Factors:
Yield =
+90.52
-0.092 * A
+0.075 * B
-0.41 * A2
+0.24 * B2
-0.087 * A * B
Final Equation in Terms of Actual Factors:
Yield =
+48.54630
+0.86759 * Pressure
-0.64042 * Temperature
-1.81481E-003 * Pressure2
+2.41667E-003 * Temperature2
-5.83333E-004 * Pressure * Temperature
Yield
A: Pres sure
B:Temperature
200.00 207.50 215.00 222.50 230.00
150.00
155.00
160.00
165.00
170.00
90.1
90.2
90.3
90.3
90.4
90.4
90.5
90.5
90.6
90.6
90.7
90.8
2 2 2
2 2 2
2 2 2
90
90.2
90.4
90.6
90.8
91
Yield
200.00
207.50
215.00
222.50
230.00 150.00
155.00
160.00
165.00
170.00
A: Pressure
B: Tem perature
5-2 An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the
depth of cut. She selects three feed rates and four depths of cut. She then conducts a factorial experiment
and obtains the following data:
Depth of Cut (in)
Feed Rate (in/min) 0.15 0.18 0.20 0.25
74 79 82 99
0.20 64 68 88 104
60 73 92 96
92 98 99 104
5-4

0.25 86 104 108 110
88 88 95 99
99 104 108 114
0.30 98 99 110 111
102 95 99 107
The depth (A) and feed rate (B) are significant, as is the interaction (AB).
Response: Surface Finish
Sum of Mean F
A 2125.11 3 708.37 24.66 < 0.0001
B 3160.50 2 1580.25 55.02 < 0.0001
AB 557.06 6 92.84 3.23 0.0180
Residual 689.33 24 28.72
Lack of Fit 0.000 0
Pure Error 689.33 24 28.72
In this case A, B, AB are significant model terms.
The residual plots shown indicate nothing unusual.
Predicted
Residuals
-8.66667
-4.5
-0.333333
3.83333
8
66.00 77.17 88.33 99.50 110.67
Residual
Normal%probability
-8.66667 -4.5 -0.333333 3.83333 8
1
5
10
20
30
50
70
80
90
95
99
5-5

22
22
Feed Rate
Residuals
Residuals vs. Feed Rate
-8.66667
-4.5
-0.333333
3.83333
8
1 2 3
Depth of Cut
Residuals
Residuals vs. Depth of Cut
-8.66667
-4.5
-0.333333
3.83333
8
1 2 3 4
(c) Obtain point estimates of the mean surface finish at each feed rate.
Feed Rate Average
0.20 81.58
0.25 97.58
0.30 103.83
DESIGN-EXPERT Plot
Surface Finish
X = B: Feed Rate
Actual Factor
A: Depth of Cut = Average
Fe e d R a te
SurfaceFinish
O ne F actor P lot
0 .2 0 0 .2 5 0 .3 0
6 0
7 3 .5
8 7
1 0 0 .5
1 1 4
W a rn in g ! Fa cto r in vo lve d in a n in te ra ctio n .
(d) Find P-values for the tests in part (a).
The P-values are given in the computer output in part (a).
5-3 For the data in Problem 5-2, compute a 95 percent interval estimate of the mean difference in
response for feed rates of 0.20 and 0.25 in/min.
5-6

We wish to find a confidence interval on µ µ1 2− , where µ1 is the mean surface finish for 0.20 in/min and
µ2 is the mean surface finish for 0.25 in/min.
( ) ( )
n
MS
tyy
n
MS
tyy E
nab
E
nab
22
)1,2..2..1211,2..2..1 −− +−≤−≤−− αα µµ
032.916
3
)7222.28(2
)064.2()5833.975833.81( ±−=±−
Therefore, the 95% confidence interval for µ µ1 2− is -16.000 ± 9.032.
5-4 An article in Industrial Quality Control (1956, pp. 5-8) describes an experiment to investigate the
effect of the type of glass and the type of phosphor on the brightness of a television tube. The response
variable is the current necessary (in microamps) to obtain a specified brightness level. The data are as
follows:
Glass Phosphor Type
Type 1 2 3
280 300 290
1 290 310 285
285 295 290
230 260 220
2 235 240 225
240 235 230
(a) Is there any indication that either factor influences brightness? Use α = 0.05.
Both factors, phosphor type (A) and Glass type (B) influence brightness.
Response: Current in microamps
Sum of Mean F
A 933.33 2 466.67 8.84 0.0044
B 14450.00 1 14450.00 273.79 < 0.0001
AB 133.33 2 66.67 1.26 0.3178
Residual 633.33 12 52.78
Lack of Fit 0.000 0
Pure Error 633.33 12 52.78
Cor Total 16150.00 17
(b) Do the two factors interact? Use α = 0.05.
There is no interaction effect.
5-7

The residual plot of residuals versus phosphor content indicates a very slight inequality of variance. It is
not serious enough to be of concern, however.
22
Predicted
Residuals
-10
-3.75
2.5
8.75
15
225.00 244.17 263.33 282.50 301.67
Residual
Normal%probability
-10 -3.75 2.5 8.75 15
1
5
10
20
30
50
70
80
90
95
99
22
2
2
2
2
Glass Type
Residuals
Residuals vs. Glass Type
-10
-3.75
2.5
8.75
15
1 2
2
22
2
Phosphor Type
Residuals
Residuals vs. Phosphor Type
-10
-3.75
2.5
8.75
15
1 2 3
5-5 Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences,
Wiley 1977) describe an experiment to investigate the warping of copper plates. The two factors studies
were the temperature and the copper content of the plates. The response variable was a measure of the
amount of warping. The data were as follows:
Copper Content (%)
Temperature (°C) 40 60 80 100
50 17,20 16,21 24,22 28,27
75 12,9 18,13 17,12 27,31
100 16,12 18,21 25,23 30,23
125 21,17 23,21 23,22 29,31
5-8

(a) Is there any indication that either factor affects the amount of warping? Is there any interaction
between the factors? Use α = 0.05.
Both factors, copper content (A) and temperature (B) affect warping, the interaction does not.
Response: Warping
Sum of Mean F
A 698.34 3 232.78 34.33 < 0.0001
B 156.09 3 52.03 7.67 0.0021
AB 113.78 9 12.64 1.86 0.1327
Residual 108.50 16 6.78
Lack of Fit 0.000 0
Pure Error 108.50 16 6.78
Residual
Normal%probability
-3.5 -1.75 -1.06581E-014 1.75 3.5
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-3.5
-1.75
1.06581E-014
1.75
3.5
10.50 15.38 20.25 25.13 30.00
5-9

2
2
2
2
2
2
2
2
Copper Content
Residuals
Residuals vs. Copper Content
-3.5
-1.75
1.06581E-014
1.75
3.5
1 2 3 4
Temperature
Residuals
-3.5
-1.75
1.06581E-014
1.75
3.5
1 2 3 4
(c) Plot the average warping at each level of copper content and compare them to an appropriately scaled t
distribution. Describe the differences in the effects of the different levels of copper content on
warping. If low warping is desirable, what level of copper content would you specify?
Factor Name Level Low Level High Level
A Copper Content 40 40 100
B Temperature Average 50 125
Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high
Warping15.50 1.84 11.60 19.40 3.19 8.74 22.26
Warping18.88 1.84 14.97 22.78 3.19 12.11 25.64
Warping21.00 1.84 17.10 24.90 3.19 14.24 27.76
Warping28.25 1.84 24.35 32.15 3.19 21.49 35.01
Use a copper content of 40 for the lowest warping.
6.78125
0.92
8
EMS
S
b
= = =
5-10

1 5 .0 1 8 .0 2 1 .0 2 4 .0 2 7 .0
W a rp in g
C u = 4 0 C u = 6 0 C u = 8 0 C u = 1 0 0
(d) Suppose that temperature cannot be easily controlled in the environment in which the copper plates are
to be used. Does this change your answer for part (c)?
Use a copper of content of 40. This is the same as for part (c).
DESIGN-EXPERT Plot
Warping
X = A: Copper Content
Y = B: Temperature
Design Points
B1 50
B2 75
B3 100
B4 125
Te m p e ra tu re
Interaction G raph
C o p p e r C o n te n t
Warping
4 0 6 0 8 0 1 0 0
7 .7 3 9 7 9
1 3 .9 9 4 9
2 0 .2 5
2 6 .5 0 5 1
3 2 .7 6 0 2
2
3
2
2
2
2
2
2
2
2
3
2
2
3
2
2
2
5-6 The factors that influence the breaking strength of a synthetic fiber are being studied. Four
production machines and three operators are chosen and a factorial experiment is run using fiber from the
same production batch. The results are as follows:
Machine
Operator 1 2 3 4
1 109 110 108 110
110 115 109 108
2 110 110 111 114
112 111 109 112
3 116 112 114 120
5-11

114 115 119 117
Only the Operator (A) effect is significant.
Response:Stength
Sum of Mean F
A 160.33 2 80.17 21.14 0.0001
B 12.46 3 4.15 1.10 0.3888
AB 44.67 6 7.44 1.96 0.1507
Residual 45.50 12 3.79
Lack of Fit 0.000 0
Pure Error 45.50 12 3.79
Cor Total 262.96 23
The Model F-value of 5.21 implies the model is significant.
There is only a 0.41% chance that a "Model F-Value" this large could occur due to noise.
Values of "Prob > F" less than 0.0500 indicate model terms aresignificant.
In this case A are significant model terms.
The residual plot of residuals versus predicted shows that variance increases very slightly with strength.
There is no indication of a severe problem.
Predicted
Residuals
-2.5
-1.25
7.81597E-014
1.25
2.5
108.50 111.00 113.50 116.00 118.50
Residual
Normal%probability
-2.5 -1.25 -7.81597E-014 1.25 2.5
1
5
10
20
30
50
70
80
90
95
99
5-12

3
3
2
2
3
3
3
3
2
2
Operator
Residuals
Residuals vs. Operator
-2.5
-1.25
7.81597E-014
1.25
2.5
1 2 3
5-7 A mechanical engineer is studying the thrust force developed by a drill press. He suspects that the
drilling speed and the feed rate of the material are the most important factors. He selects four feed rates
and uses a high and low drill speed chosen to represent the extreme operating conditions. He obtains the
following results. Analyze the data and draw conclusions. Use α = 0.05.
(A) Feed Rate
(B)
Drill Speed 0.015 0.030 0.045 0.060
125 2.70 2.45 2.60 2.75
2.78 2.49 2.72 2.86
200 2.83 2.85 2.86 2.94
2.86 2.80 2.87 2.88
Response: Force
Sum of Mean F
A 0.15 1 0.15 57.01 < 0.0001
B 0.092 3 0.031 11.86 0.0026
AB 0.042 3 0.014 5.37 0.0256
Residual 0.021 8 2.600E-003
Lack of Fit 0.000 0
Pure Error 0.021 8 2.600E-003
Cor Total 0.30 15
The factors speed and feed rate, as well as the interaction is important.
5-13

DESIGN-EXPERT Plot
Force
X = B: Feed Rate
Y = A: Drill Speed
Design Points
A1 125
D rill S p e e d
Interaction G raph
Fe e d R a te
Force
0 .0 1 5 0 .0 3 0 0 .0 4 5 0 .0 6 0
2 .4 1 1 2 1
2 .5 5 0 6
2 .6 9
2 .8 2 9 4
2 .9 6 8 7 9
A2 200
The standard analysis of variance treats all design factors as if they were qualitative. In this case, both
Since both factors in this problem are quantitative, we can fit polynomial effects of both speed and feed
rate, exactly as in Example 5-5 in the text. The Design-Expert output with only the significant terms
retained, including the response surface plots, now follows.
Response: Force
Sum of Mean F
A 0.15 1 0.15 22.70 0.0005
B 0.019 1 0.019 2.94 0.1119
B2 0.058 1 0.058 8.82 0.0117
Residual 0.078 12 6.530E-003
Lack of Fit 0.058 4 0.014 5.53 0.0196 significant
Pure Error 0.021 8 2.600E-003
Cor Total 0.30 15
In this case A, B2 are significant model terms.
Intercept 2.69 1 0.032 2.62 2.76
A-Drill Speed 0.096 1 0.020 0.052 0.14 1.00
B-Feed Rate 0.047 1 0.027 -0.013 0.11 1.00
B2 0.13 1 0.045 0.036 0.23 1.00
5-14

Force =
+2.69
+0.096 * A
+0.047 * B
+0.13 * B2
Force =
+2.48917
+3.06667E-003 * Drill Speed
-15.76667 * Feed Rate
+266.66667 * Feed Rate2
Force
A: Drill Speed
B:FeedRate
125.00 143.75 162.50 181.25 200.00
0.02
0.03
0.04
0.05
0.06
2.6 2.65
2.7
2.75
2.8
2.8
2.85
2.85
2.9
22 22
22 22
22 22
22 22
2.5
2.6
2.7
2.8
2.9
3
Force
125.00
143.75
162.50
181.25
200.00
0.02
0.03
0.04
0.05
0.06
A: Drill Speed
B: Feed Rate
5-8 An experiment is conducted to study the influence of operating temperature and three types of face-
plate glass in the light output of an oscilloscope tube. The following data are collected:
Temperature
Glass Type 100 125 150
580 1090 1392
1 568 1087 1380
570 1085 1386
550 1070 1328
2 530 1035 1312
579 1000 1299
546 1045 867
3 575 1053 904
599 1066 889
5-15

Use α = 0.05 in the analysis. Is there a significant interaction effect? Does glass type or temperature affect
the response? What conclusions can you draw? Use the method discussed in the text to partition the
temperature effect into its linear and quadratic components. Break the interaction down into appropriate
components.
Response: Light Output
Sum of Mean F
Model 2.412E+006 8 3.015E+005 824.77 < 0.0001 significant
A 1.509E+005 2 75432.26 206.37 < 0.0001
B 1.970E+006 2 9.852E+005 2695.26 < 0.0001
AB 2.906E+005 4 72637.93 198.73 < 0.0001
Residual 6579.33 18 365.52
Lack of Fit 0.000 0
Pure Error 6579.33 18 365.52
Cor Total 2.418E+006 26
Both factors, Glass Type (A) and Temperature (B) are significant, as well as the interaction (AB). For glass
types 1 and 2 the response is fairly linear, for glass type 3, there is a quadratic effect.
DESIGN-EXPERT Plot
Light Output
X = B: Temperature
Y = A: Glass Type
Design Points
A1 1
A2 2
A3 3
G la s s Typ e
Interaction G raph
Te m p e ra tu re
LightOutput
1 0 0 1 2 5 1 5 0
5 3 0
7 4 8 .0 9 9
9 6 6 .1 9 9
1 1 8 4 .3
1 4 0 2 .4
Response: Light Output
Sum of Mean F
A 1.509E+005 2 75432.26 206.37 < 0.0001
B 1.780E+006 1 1.780E+006 4869.13 < 0.0001
B2 1.906E+005 1 1.906E+005 521.39 < 0.0001
AB 2.262E+005 2 1.131E+005 309.39 < 0.0001
AB2 64373.93 2 32186.96 88.06 < 0.0001
5-16

Pure Error 9.33 18657 365.52
Cor Total 2.418E+006 26
In this case A, B, B2, AB, AB2 are significant model terms.
Coefficien Standa 95% CI 95% Ct rd I
Intercept 059.00 1 6.37 1045.61 1072.391
A[1] 28.33 1 9.01 9.40 47.27
A[2] -24.00 1 9.01 -42.93 -5.07
B-Temperature 314.44 1 4.51 304.98 323.91 1.00
B2 78.22 1 7.81 94.62 -161.82-1 -1 1.00
A[1]B 92.22 1 6.37 78.83 105.61
A[2]B 65.56 1 6.37 52.17 78.94
A[1]B2 70.22 1 11.04 47.03 93.41
A[2]B2 76.22 1 11.04 53.03 99.41
Light Output =
+1059.00
+28.33 * A[1]
-24.00 * A[2]
+314.44 * B
-178.22 * B2
+92.22 * A[1]B
+65.56 * A[2]B
+70.22 * A[1]B2
+76.22 * A[2]B2
Glass Type 1
Light Output =
-3646.00000
+59.46667 * Temperature
-0.17280 * Temperature2
Glass Type 2
Light Output =
-3415.00000
Glass Type 3
Light Output =
-7845.33333
5-9 Consider the data in Problem 5-1. Use the method described in the text to compute the linear and
quadratic effects of pressure.
See the alternative analysis shown in Problem 5-1 part (c).
5-17

5-10 Use Duncan’s multiple range test to determine which levels of the pressure factor are significantly
different for the data in Problem 5-1.
18903 .y .. = 37901 .y .. = 68902 .y .. =
( )( )
05430
23
017770
.
.
an
MS
S E
y .j.
===
( ) 60492010 .,r . = ( ) 86493010 .,r . =
( )( ) 24980054306042 ...R == ( )( ) 26400054308643 ...R ==
2 vs. 3 = 0.50 > 0.2640 (R3)
2 vs. 1 = 0.31 > 0.2498 (R2)
1 vs. 3 = 0.19 < 0.2498 (R2)
Therefore, 2 differs from 1 and 3.
5-11 An experiment was conducted to determine if either firing temperature or furnace position affects
the baked density of a carbon anode. The data are shown below.
Temperature (°C)
Position 800 825 850
570 1063 565
1 565 1080 510
583 1043 590
528 988 526
2 547 1026 538
521 1004 532
Suppose we assume that no interaction exists. Write down the statistical model. Conduct the analysis of
variance and test hypotheses on the main effects. What conclusions can be drawn? Comment on the
model’s adequacy.
The model for the two-factor, no interaction model is . Both factors, furnace
position (A) and temperature (B) are significant. The residual plots show nothing unusual.
ijkjiijky εβτµ +++=
Response: Density
Sum of Mean F
A 7160.06 1 7160.06 16.20 0.0013
B 9.453E+005 2 4.727E+005 1069.26 < 0.0001
Residual 6188.78 14 442.06
Lack of Fit 818.11 2 409.06 0.91 0.4271 not significant
Pure Error 5370.67 12 447.56
Cor Total 9.587E+005 17
5-18

Predicted
Residuals
-53.4444
-33.4444
-13.4444
6.55556
26.5556
523.56 656.15 788.75 921.35 1053.94
Position
Residuals
Residuals vs. Position
-53.4444
-33.4444
-13.4444
6.55556
26.5556
1 2
Temperature
Residuals
-53.4444
-33.4444
-13.4444
6.55556
26.5556
1 2 3
5-12 Derive the expected mean squares for a two-factor analysis of variance with one observation per
cell, assuming that both factors are fixed.
Degrees of Freedom
( )
( )∑=
−
+=
a
i
i
A
a
bMSE
1
2
2
1
τ
σ a-1
( )
( )∑=
−
+=
b
j
j
B
b
aMSE
1
2
2
1
β
σ b-1
( )
( )
( )( )∑∑= =
−−
+=
a
i
b
j
ij
AB
ba
MSE
1 1
2
2
11
τβ
σ
( )( )
1
11
−
−−
ab
ba
5-19

5-13 Consider the following data from a two-factor factorial experiment. Analyze the data and draw
conclusions. Perform a test for nonadditivity. Use α = 0.05.
Column Factor
Row Factor 1 2 3 4
1 36 39 36 32
2 18 20 22 20
3 30 37 33 34
Response: data
Sum of Mean F
A 580.50 2 290.25 60.40 0.0001
B 28.92 3 9.64 2.01 0.2147
Residual 28.83 6 4.81
Cor Total 638.25 11
The row factor (A) is significant.
The test for nonadditivity is as follows:
( )
( )( )
( )( )( )( )
2927925540513833328
540513
91667285058034
34
357
9166728505803574010014
2
2
2
1 1
2
...SSSSSS
.SS
..
..
SS
SSabSS
ab
y
SSSSyyyy
SS
NsidualReError
N
N
BA
a
i
b
j
..
BA..j..iij
N
=−=−=
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++−
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++−
=
∑∑= =
Source of Sum of Degrees of Mean
Variation Squares Freedom Square F0
Row 580.50 2 290.25 57.3780
Column 28.91667 3 9.63889 1.9054
Nonadditivity 3.54051 1 3.54051 0.6999
Error 25.29279 5 5.058558
Total 638.25 11
5-14 The shear strength of an adhesive is thought to be affected by the application pressure and
temperature. A factorial experiment is performed in which both factors are assumed to be fixed. Analyze
the data and draw conclusions. Perform a test for nonadditivity.
5-20

Temperature (°F)
Pressure (lb/in2) 250 260 270
120 9.60 11.28 9.00
130 9.69 10.10 9.57
140 8.43 11.01 9.03
150 9.98 10.44 9.80
Response: Strength
Sum of Mean F
A 0.58 3 0.19 0.54 0.6727
B 4.66 2 2.33 6.49 0.0316
Cor Total 7.39 11
The "Model F-value" of 2.92 implies the model is not significant relative to the noise.
There is a 11.24 % chance that a "Model F-value" this large could occur due to noise.
In this case B are significant model terms.
Temperature (B) is a significant factor.
( )
( )( )
( )( )( )( )
66440148948015388332
489480
6576545806917034
34
93117
6576545806917093117777415113
2
2
2
1 1
2
...SSSSSS
.SS
..
.
....
SS
SSabSS
ab
y
SSSSyyyy
SS
NsidualReError
N
N
BA
a
i
b
j
..
BA..j..iij
N
=−=−=
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++−
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++−
=
∑∑= =
Source of Sum of Degrees of Mean
Variation Squares Freedom Square F0
Row 0.5806917 3 0.1935639 0.5815
Column 4.65765 2 2.328825 6.9960
Nonadditivity 0.48948 1 0.48948 1.4704
Error 1.6644 5 0.33288
Total 7.392225 11
5-15 Consider the three-factor model
( ) ( ) ijkjkijkjiijky εβγτβγβτµ ++++++=
⎪
⎩
⎪
⎨
⎧
=
=
=
c,...,,k
b,...,,j
a,...,,i
21
21
21
5-21

Notice that there is only one replicate. Assuming the factors are fixed, write down the analysis of variance
table, including the expected mean squares. What would you use as the “experimental error” in order to
test hypotheses?
Source Degrees of Freedom Expected Mean Square
A a-1
( )∑=
−
+
a
i
i
a
bc
1
2
2
1
τ
σ
B b-1
( )∑=
−
+
b
j
j
b
ac
1
2
2
1
β
σ
C c-1
( )∑=
−
+
c
k
k
c
ab
1
2
2
1
γ
σ
AB (a-1)(b-1)
( )
( )( )∑∑= =
−−
+
a
i
b
j
ij
ba
c
1 1
2
2
11
βτ
σ
BC (b-1)(c-1)
( )
( )( )∑∑= =
−−
+
b
j
c
k
jk
cb
a
1 1
2
2
11
βγ
σ
Error (AC + ABC) b(a-1)(c-1) σ 2
Total abc-1
5-16 The percentage of hardwood concentration in raw pulp, the vat pressure, and the cooking time of the
pulp are being investigated for their effects on the strength of paper. Three levels of hardwood
concentration, three levels of pressure, and two cooking times are selected. A factorial experiment with
two replicates is conducted, and the following data are obtained:
Percentage Cooking Time 3.0 Hours Cooking Time 4.0 Hours
of Hardwood Pressure Pressure
Concentration 400 500 650 400 500 650
2 196.6 197.7 199.8 198.4 199.6 200.6
196.0 196.0 199.4 198.6 200.4 200.9
4 198.5 196.0 198.4 197.5 198.7 199.6
197.2 196.9 197.6 198.1 198.0 199.0
8 197.5 195.6 197.4 197.6 197.0 198.5
196.6 196.2 198.1 198.4 197.8 199.8
Response: strength
Sum of Mean F
A 7.76 2 3.88 10.62 0.0009
B 20.25 1 20.25 55.40 < 0.0001
C 19.37 2 9.69 26.50 < 0.0001
AB 2.08 2 1.04 2.85 0.0843
AC 6.09 4 1.52 4.17 0.0146
5-22

BC 2.19 2 1.10 3.00 0.0750
ABC 1.97 4 0.49 1.35 0.2903
Residual 6.58 18 0.37
Lack of Fit 0.000 0
Pure Error 6.58 18 0.37
Cor Total 66.31 35
In this case A, B, C, AC are significant model terms.
All three main effects, concentration (A), pressure (C) and time (B), as well as the concentration x pressure
interaction (AC) are significant at the 5% level. The concentration x time (AB) and pressure x time
interactions (BC) are significant at the 10% level.
Pressure
Residuals
-0.85
-0.425
0
0.425
0.85
1 2 3
2
2
2
2
Cooking Time
Residuals
Residuals vs. Cooking Time
-0.85
-0.425
0
0.425
0.85
1 2
Predicted
Residuals
-0.85
-0.425
0
0.425
0.85
195.90 197.11 198.33 199.54 200.75
2
2
2
2
Hardwood
Residuals
Residuals vs. Hardwood
-0.85
-0.425
0
0.425
0.85
1 2 3
5-23

(c) Under what set of conditions would you run the process? Why?
DESIGN-EXPERT Plot
strength
X = B: Cooking Time
Y = C: Pressure
C1 400
C2 500
C3 650
Actual Factor
A: Hardwood = Average
P re s s u re
Interaction G raph
C o o kin g Tim e
strength
3 4
1 9 5 .6
1 9 6 .9 2 5
1 9 8 .2 5
1 9 9 .5 7 5
2 0 0 .9
DESIGN-EXPERT Plot
strength
X = B: Cooking Time
Y = A: Hardwood
A1 2
A2 4
A3 8
Actual Factor
C: Pressure = Average
H a rd w o o d
Interaction G raph
C o o kin g Tim e
strength
3 4
1 9 5 .6
1 9 6 .9 2 5
1 9 8 .2 5
1 9 9 .5 7 5
2 0 0 .9
DESIGN-EXPERT Plot
strength
X = C: Pressure
Y = A: Hardwood
A1 2
A2 4
A3 8
Actual Factor
B: Cooking Time = Average
H a rd w o o d
Interaction G raph
P re s s u re
strength
4 0 0 5 0 0 6 5 0
1 9 5 .6
1 9 6 .9 2 5
1 9 8 .2 5
1 9 9 .5 7 5
2 0 0 .9
For the highest strength, run the process with the percentage of hardwood at 2, the pressure at 650, and the
time at 4 hours.
The standard analysis of variance treats all design factors as if they were qualitative. In this case, all three
Since the factors in this problem are quantitative and two of them have three levels, we can fit a linear term
for the two-level factor and linear and quadratic components for the three-level factors. The Minitab
output, with the ABC interaction removed due to insignificance, now follows. Also included is the Design
Expert output; however, if the student choses to use Design Expert, sequential sum of squares must be
selected to assure that the sum of squares for the model equals the total of the sum of squares for each
factor included in the model.
Minitab Output
General Linear Model: Strength versus
5-24

Hardwood 1 6.9067 4.9992 4.9992 13.23 0.001
Time 1 20.2500 1.3198 1.3198 3.49 0.074
Pressure 1 15.5605 1.5014 1.5014 3.97 0.058
Hardwood*Hardwood 1 0.8571 2.7951 2.7951 7.40 0.012
Pressure*Pressure 1 3.8134 1.8232 1.8232 4.83 0.038
Hardwood*Time 1 0.7779 1.5779 1.5779 4.18 0.053
Hardwood*Pressure 1 2.1179 3.4564 3.4564 9.15 0.006
Time*Pressure 1 0.0190 2.1932 2.1932 5.81 0.024
Hardwood*Hardwood*Time 1 1.3038 1.3038 1.3038 3.45 0.076
Hardwood*Hardwood*
Pressure 1 2.1885 2.1885 2.1885 5.79 0.025
Hardwood*Pressure*
Pressure 1 1.6489 1.6489 1.6489 4.36 0.048
Time*Pressure*Pressure 1 2.1760 2.1760 2.1760 5.76 0.025
Error 23 8.6891 8.6891 0.3778
Total 35 66.3089
Term Coef SE Coef T P
Constant 236.92 29.38 8.06 0.000
Hardwood 10.728 2.949 3.64 0.001
Time -14.961 8.004 -1.87 0.074
Pressure -0.2257 0.1132 -1.99 0.058
Hardwood*Hardwood -0.6529 0.2400 -2.72 0.012
Pressure*Pressure 0.000234 0.000107 2.20 0.038
Hardwood*Time -1.1750 0.5749 -2.04 0.053
Hardwood*Pressure -0.020533 0.006788 -3.02 0.006
Time*Pressure 0.07450 0.03092 2.41 0.024
Hardwood*Hardwood*Time 0.10278 0.05532 1.86 0.076
Hardwood*Hardwood*Pressure 0.000648 0.000269 2.41 0.025
Hardwood*Pressure*Pressure 0.000012 0.000006 2.09 0.048
Time*Pressure*Pressure -0.000070 0.000029 -2.40 0.025
Unusual Observations for Strength
Obs Strength Fit SE Fit Residual St Resid
6 198.500 197.461 0.364 1.039 2.10R
R denotes an observation with a large standardized residual.
Response: Strength
Sum of Mean F
A 6.91 1 6.91 18.28 0.0003
B 20.25 1 20.25 53.60 < 0.0001
C 15.56 1 15.56 41.19 < 0.0001
A2 0.86 1 0.86 2.27 0.1456
C2 3.81 1 3.81 10.09 0.0042
AB 0.78 1 0.78 2.06 0.1648
AC 2.12 1 2.12 5.61 0.0267
BC 0.019 1 0.019 0.050 0.8245
A2B 1.30 1 1.30 3.45 0.0761
A2C 2.19 1 2.19 5.79 0.0245
AC2 1.65 1 1.65 4.36 0.0479
BC2 2.18 1 2.18 5.76 0.0249
Residual 8.69 23 0.38
Pure Error 6.58 18 0.37
Cor Total 66.31 35
5-25

In this case A, B, C2, AC, A2C, AC2, BC2 are significant model terms.
Intercept 197.21 1 0.26 196.67 197.74
A-Hardwood -0.98 1 0.23 -1.45 -0.52 3.36
B-Cooking Time 0.78 1 0.26 0.24 1.31 6.35
C-Pressure 0.19 1 0.25 -0.33 0.71 4.04
A2 0.42 1 0.25 -0.093 0.94 1.04
C2 0.79 1 0.23 0.31 1.26 1.03
AB -0.22 1 0.13 -0.48 0.039 1.06
AC -0.46 1 0.15 -0.78 -0.14 1.08
BC 0.062 1 0.13 -0.20 0.32 1.02
A2B 0.46 1 0.25 -0.053 0.98 3.96
A2C 0.73 1 0.30 0.10 1.36 3.97
AC2 0.57 1 0.27 5.625E-003 1.14 3.32
BC2 -0.55 1 0.23 -1.02 -0.075 3.30
Strength =
+197.21
-0.98 * A
+0.78 * B
+0.19 * C
+0.42 * A2
+0.79 * C2
-0.22 * A * B
-0.46 * A * C
+0.062 * B * C
+0.46 * A2 * B
+0.73 * A2 * C
+0.57 * A * C2
-0.55 * B * C2
Strength =
+236.91762
+10.72773 * Hardwood
-14.96111 * Cooking Time
-0.22569 * Pressure
-0.65287 * Hardwood2
+2.34333E-004 * Pressure2
-1.17500 * Hardwood * Cooking Time
-0.020533 * Hardwood * Pressure
+0.074500 * Cooking Time * Pressure
+0.10278 * Hardwood2 * Cooking Time
+6.48026E-004 * Hardwood2 * Pressure
+1.22143E-005 * Hardwood * Pressure2
-7.00000E-005 * Cooking Time * Pressure2
5-26

Strength
A: Hardwood
C:Pressure
2.00 3.50 5.00 6.50 8.00
400.00
450.00
500.00
550.00
600.00
650.00
198
198.5
198.5
199
199.5
200
200.5
22 22 22
22 22 22
22 22 22
197
197.5
198
198.5
199
199.5
200
200.5
201
201.5
Strength
2.00
3.50
5.00
6.50
8.00 400.00
450.00
500.00
550.00
600.00
650.00
A: Hardwood
C: Pressure
Cooking Time: B = 4.00
5-17 The quality control department of a fabric finishing plant is studying the effect of several factors on
the dyeing of cotton-synthetic cloth used to manufacture men’s shirts. Three operators, three cycle times,
and two temperatures were selected, and three small specimens of cloth were dyed under each set of
conditions. The finished cloth was compared to a standard, and a numerical score was assigned. The
results follow. Analyze the data and draw conclusions. Comment on the model’s adequacy.
Temperature
300° 350°
Operator Operator
Cycle Time 1 2 3 1 2 3
23 27 31 24 38 34
40 24 28 32 23 36 36
25 26 29 28 35 39
36 34 33 37 34 34
50 35 38 34 39 38 36
36 39 35 35 36 31
28 35 26 26 36 28
60 24 35 27 29 37 26
27 34 25 25 34 24
All three main effects, and the AB, AC, and ABC interactions are significant. There is nothing unusual
about the residual plots.
Response: Score
Sum of Mean F
A 436.00 2 218.00 66.51 < 0.0001
B 261.33 2 130.67 39.86 < 0.0001
C 50.07 1 50.07 15.28 0.0004
AB 355.67 4 88.92 27.13 < 0.0001
5-27

AC 78.81 2 39.41 12.02 0.0001
BC 11.26 2 5.63 1.72 0.1939
ABC 46.19 4 11.55 3.52 0.0159
Residual 118.00 36 3.28
Lack of Fit 0.000 0
Pure Error 118.00 36 3.28
In this case A, B, C, AB, AC, ABC are significant model terms.
DESIGN-EXPERT Plot
Score
X = A: Cycle Time
Y = C: Temperature
C1 300
C2 350
Actual Factor
B: Operator = Average
Te m p e ra tu re
Interaction G raph
C ycle T im e
Score
4 0 5 0 6 0
2 3
2 7
3 1
3 5
3 9
22
DESIGN-EXPERT Plot
Score
X = A: Cycle Time
Y = B: Operator
B1 1
B2 2
B3 3
Actual Factor
C: Temperature = Average
O p e ra to r
Interaction G raph
C ycle Tim e
Score
4 0 5 0 6 0
2 3
2 7
3 1
3 5
3 9
2
22
2
22
22
2
3
2
3
2
2
2 2 3
Operator
Residuals
Residuals vs. Operator
-3
-1.5
4.26326E-014
1.5
3
1 2 3
2
33
2
2
22
3 2
2
3
2
3
2
3
2
Cycle Time
Residuals
Residuals vs. Cycle Time
-3
-1.5
4.26326E-014
1.5
3
1 2 3
5-28

2222
2
22
2
Predicted
Residuals
-3
-1.5
4.26326E-014
1.5
3
24.00 27.25 30.50 33.75 37.00
4
22
4
22
22
2
4
2
4
2
2
3
2
2
2
3
2
2
3
2
Temperature
Residuals
-3
-1.5
4.26326E-014
1.5
3
1 2
5-18 In Problem 5-1, suppose that we wish to reject the null hypothesis with a high probability if the
difference in the true mean yield at any two pressures is as great as 0.5. If a reasonable prior estimate of
the standard deviation of yield is 0.1, how many replicates should be run?
( )( )
( )( )
n.
.
.n
b
naD
512
1032
503
2 2
2
2
2
2
===
σ
Φ
n 2
Φ Φ ( )11 −= bυ ( )12 −= nabυ β
2 25 5 2 (3)(3)(1) 0.014
2 replications will be enough to detect the given difference.
5-19 The yield of a chemical process is being studied. The two factors of interest are temperature and
pressure. Three levels of each factor are selected; however, only 9 runs can be made in one day. The
experimenter runs a complete replicate of the design on each day. The data are shown in the following
table. Analyze the data assuming that the days are blocks.
Day 1 Day 2
Pressure Pressure
Temperature 250 260 270 250 260 270
Low 86.3 84.0 85.8 86.1 85.2 87.3
Medium 88.5 87.3 89.0 89.4 89.9 90.3
High 89.1 90.2 91.3 91.7 93.2 93.7
Response: Yield
Sum of Mean F
Block 13.01 1 13.01
A 5.51 2 2.75 5.18 0.0360
5-29

B 99.85 2 49.93 93.98 < 0.0001
AB 4.45 4 1.11 2.10 0.1733
Cor Total 127.07 17
Both main effects, temperature and pressure, are significant.
5-20 Consider the data in Problem 5-5. Analyze the data, assuming that replicates are blocks.
Response: Warping
Sum of Mean F
Block 11.28 1 11.28
A 698.34 3 232.78 35.92 < 0.0001
B 156.09 3 52.03 8.03 0.0020
AB 113.78 9 12.64 1.95 0.1214
Residual 97.22 15 6.48
Both temperature and copper content are significant. This agrees with the analysis in Problem 5-5.
5-21 Consider the data in Problem 5-6. Analyze the data, assuming that replicates are blocks.
Design-Expert Output
Response: Stength
Sum of Mean F
Block 1.04 1 1.04
A 160.33 2 80.17 19.84 0.0002
B 12.46 3 4.15 1.03 0.4179
AB 44.67 6 7.44 1.84 0.1799
Residual 44.46 11 4.04
Cor Total 262.96 23
In this case A are significant model terms.
Only the operator factor (A) is significant. This agrees with the analysis in Problem 5-6.
5-30

5-22 An article in the Journal of Testing and Evaluation (Vol. 16, no.2, pp. 508-515) investigated the
effects of cyclic loading and environmental conditions on fatigue crack growth at a constant 22 MPa stress
for a particular material. The data from this experiment are shown below (the response is crack growth
rate).
Environment
Frequency Air H2O Salt H2O
2.29 2.06 1.90
10 2.47 2.05 1.93
2.48 2.23 1.75
2.12 2.03 2.06
2.65 3.20 3.10
1 2.68 3.18 3.24
2.06 3.96 3.98
2.38 3.64 3.24
2.24 11.00 9.96
0.1 2.71 11.00 10.01
2.81 9.06 9.36
2.08 11.30 10.40
(a) Analyze the data from this experiment (use α = 0.05).
Response: Crack Growth
Sum of Mean F
A 209.89 2 104.95 522.40 < 0.0001
B 64.25 2 32.13 159.92 < 0.0001
AB 101.97 4 25.49 126.89 < 0.0001
Residual 5.42 27 0.20
Lack of Fit 0.000 0
Pure Error 5.42 27 0.20
Cor Total 381.53 35
Both frequency and environment, as well as their interaction are significant.
(b) Analyze the residuals.
The residual plots indicate that there may be some problem with inequality of variance. This is particularly
noticable on the plot of residuals versus predicted response and the plot of residuals versus frequency.
5-31

22
22
Predicted
Residuals
-1.53
-0.97
-0.41
0.15
0.71
1.91 4.08 6.25 8.42 10.59
Residual
Normal%probability
-1.53 -0.97 -0.41 0.15 0.71
1
5
10
20
30
50
70
80
90
95
99
22
22
Environment
Residuals
Residuals vs. Environment
-1.53
-0.97
-0.41
0.15
0.71
1 2 3
22
22
Frequency
Residuals
Residuals vs. Frequency
-1.53
-0.97
-0.41
0.15
0.71
1 2 3
(c) Repeat the analyses from parts (a) and (b) using ln(y) as the response. Comment on the results.
Response: Crack Growth Transform: Natural log Constant: 0.000
Sum of Mean F
A 7.57 2 3.79 404.09 < 0.0001
B 2.36 2 1.18 125.85 < 0.0001
AB 3.53 4 0.88 94.17 < 0.0001
Residual 0.25 27 9.367E-003
Lack of Fit 0.000 0
Pure Error 0.25 27 9.367E-003
Cor Total 13.71 35
5-32

Both frequency and environment, as well as their interaction are significant. The residual plots of the
based on the transformed data look better.
22
22
Predicted
Residuals
-0.16484
-0.0822988
0.000242214
0.0827832
0.165324
0.65 1.07 1.50 1.93 2.36
Residual
Normal%probability
-0.16484 -0.0822988 0.000242214 0.0827832 0.165324
1
5
10
20
30
50
70
80
90
95
99
22
22
Environment
Residuals
Residuals vs. Environment
-0.16484
-0.0822988
0.000242214
0.0827832
0.165324
1 2 3
22
22
Frequency
Residuals
Residuals vs. Frequency
-0.16484
-0.0822988
0.000242214
0.0827832
0.165324
1 2 3
5-23 An article in the IEEE Transactions on Electron Devices (Nov. 1986, pp. 1754) describes a study on
polysilicon doping. The experiment shown below is a variation of their study. The response variable is
base current.
Polysilicon Anneal Temperature (°C)
Doping (ions) 900 950 1000
4.60 10.15 11.011 x 10 20
4.40 10.20 10.58
3.20 9.38 10.812 x 10 20
3.50 10.02 10.60
5-33

(a) Is there evidence (with α = 0.05) indicating that either polysilicon doping level or anneal temperature
affect base current?
Response: Base Current
Sum of Mean F
A 0.98 1 0.98 15.26 0.0079
B 111.19 2 55.59 865.16 < 0.0001
AB 0.58 2 0.29 4.48 0.0645
Residual 0.39 6 0.064
Lack of Fit 0.000 0
Pure Error 0.39 6 0.064
Cor Total 113.13 11
Both factors, doping and anneal are significant. Their interaction is significant at the 10% level.
(b) Prepare graphical displays to assist in interpretation of this experiment.
Doping
Interaction Graph
Anneal
BaseCurrent
900 950 1000
3.03986
5.05618
7.0725
9.08882
11.1051
A-
A+
(c) Analyze the residuals and comment on model adequacy.
5-34

Predicted
Residuals
-0.32
-0.16
8.88178E-016
0.16
0.32
3.35 5.21 7.07 8.93 10.80
Residual
Normal%probability
-0.32 -0.16 -8.88178E-016 0.16 0.32
1
5
10
20
30
50
70
80
90
95
99
Doping
Residuals
Residuals vs. Doping
-0.32
-0.16
8.88178E-016
0.16
0.32
1 2
Anneal
Residuals
Residuals vs. Anneal
-0.32
-0.16
8.88178E-016
0.16
0.32
1 2 3
There is a funnel shape in the plot of residuals versus predicted, indicating some inequality of variance.
(d) Is the model supported by this experiment (xεβββββ +++++= 2112
2
22222110 xxxxxy 1 = doping
level, x2 = temperature)? Estimate the parameters in this model and plot the response surface.
Response: Base Current
ANOVA for Response Surface Reduced Quadratic Model
Sum of Mean F
A 0.98 1 0.98 17.18 0.0043
B 93.16 1 93.16 1632.09 < 0.0001
B2 18.03 1 18.03 315.81 < 0.0001
AB 0.56 1 0.56 9.84 0.0164
Residual 0.40 7 0.057
Pure Error 0.39 6 0.064
5-35

Cor Total 113.13 11
In this case A, B, B2, AB are significant model terms.
Intercept 9.94 1 0.12 9.66 10.22
A-Doping -0.29 1 0.069 -0.45 -0.12 1.00
B-Anneal 3.41 1 0.084 3.21 3.61 1.00
B2 -2.60 1 0.15 -2.95 -2.25 1.00
AB 0.27 1 0.084 0.065 0.46 1.00
All of the coefficients in the assumed model are significant. The quadratic effect is easily observable in the
response surface plot.
Base Current
Doping
Anneal
1.00E+20 1.25E+20 1.50E+20 1.75E+20 2.00E+20
900.00
925.00
950.00
975.00
1000.00
4
5
6
7
8
9
10
11
2 2
2 2
2 2
3
4
5
6
7
8
9
10
11
12
BaseCurrent
1.00E+20
1.25E+20
1.50E+20
1.75E+20
2.00E+20 900.00
925.00
950.00
975.00
1000.00
Doping
Anneal
5-24 An experiment was conducted to study the life (in hours) of two different brands of batteries in three
different devices (radio, camera, and portable DVD player). A completely randomized two-factor
experiment was conducted, and the following data resulted.
Brand Device
of Battery Radio Camera DVD Player
8.6 7.9 5.4A
8.2 8.4 5.7
9.4 8.5 5.8B
8.8 8.9 5.9
(a) Analyze the data and draw conclusions, using α = 0.05.
Both brand of battery (A) and type of device (B) are significant, the interaction is not.
Response: Life
5-36

Sum of Mean F
A 0.80 1 0.80 9.33 0.0224
B 22.45 2 11.22 130.75 < 0.0001
AB 0.082 2 0.041 0.48 0.6430
Pure Error 0.52 6 0.086
Cor Total 23.84 11
(b) Investigate model adequacy by plotting the residuals.
The residual plots show no serious deviations from the assumptions.
Predicted
Residuals
-0.3
-0.15
0
0.15
0.3
5.55 6.44 7.33 8.21 9.10
Residual
Normal%probability Normal plot of residuals
-0.3 -0.15 0 0.15 0.3
1
5
10
20
30
50
70
80
90
95
99
5-37

Battery
Residuals
Residuals vs. Battery
-0.3
-0.15
0
0.15
0.3
1 2
Device
Residuals
Residuals vs. Device
-0.3
-0.15
0
0.15
0.3
1 2 3
(c) Which brand of batteries would you recommend?
Battery brand B is recommended.
DESIGN-EXPERT Plot
Lif e
X = A: Battery
Y = B: Dev ice
Design Points
B1 Radio
B2 Camera
B3 DVD
B: Device
Interaction Graph
A: Battery
Life
A B
5
6
7
8
9
10
22
5-25 The author has recently purchased new golf clubs, which he believes witll significantly improve e
his game. Below are the scores of three rounds of golf played at three different golf courses with the old
and the new clubs.
CourseClubs
Ahwatukee Karsten Foothills
90 91 88
87 93 86
Old
86 90 90
88 90 86New
87 91 85
5-38

85 88 88
(a) Conduct an anlysis of variance. Using α = 0.05, what conclusions can you draw?
Although there is a significant difference between the golf courses, there is not a significant difference
between the old and new clubs.
Response: Score
Sum of Mean F
A 9.39 1 9.39 3.19 0.0994
B 44.44 2 22.22 7.55 0.0075
AB 0.44 2 0.22 0.075 0.9277
Pure Error 35.33 12 2.94
Cor Total 89.61 17
The residual plots show no deviations from the assumptions.
Predicted
Residuals
-3
-1.5
0
1.5
3
86.33 87.58 88.83 90.08 91.33
Residual
Normal%Probability
-2 -0.916667 0.166667 1.25 2.33333
1
5
10
20
30
50
70
80
90
95
99
5-39

2
2
2
2
2
2
Clubs
Residuals
Residuals vs. Clubs
-3
-1.5
0
1.5
3
1 2
22
B
Residuals
Residuals vs. B
-3
-1.5
0
1.5
3
1 2 3
5-26 A manufacturer of laundry products is investigating the performance of a newly formulated stain
remover. The new formulation is compared to the original formulation with respect to its ability to remove
a standard tomato-like stain in a test article of cotton cloth using a factorial experiment. The other factors
in the experiment are the number of times the test article is washed (1 or 2), and whether or not a detergent
booster is used. The response variable is the stain shade after washing (12 is the darkest, 0 is the lightest).
The data are shown in the table below.
Number of Washings Number of Washings
1 2 1 2
Booster Booster
Formulation
Yes No Yes No
6 6 3 4New
5 5 2 1
10 11 10 9Original
9 11 9 10
(a) Conduct an anlysis of variance. Using α = 0.05, what conclusions can you draw?
The formulation, number of washings, and the interaction between these to factors appear to be significant.
Continued analysis is required as a result of the residual plots in part (b). Conclusions are presented in part
(b).
Response: Stain Shade
Sum of Mean F
A 138.06 1 138.06 147.27 < 0.0001
B 14.06 1 14.06 15.00 0.0047
C 0.56 1 0.56 0.60 0.4609
AB 5.06 1 5.06 5.40 0.0486
AC 0.56 1 0.56 0.60 0.4609
BC 0.56 1 0.56 0.60 0.4609
5-40

ABC 0.56 1 0.56 0.60 0.4609
Cor Total 166.94 15
The residual plots shown below identify a violation from our assumptions; nonconstant variance. A power
transformation was chosen to correct the violation. λ can be found through trial and error; or the use of a
Box-Cox plot that is described in a later chapter. A Box-Cox plot is shown below that identifies a power
transformation λ of 1.66.
Residual
Normal%probability
-1.5 -0.75 0 0.75 1.5
1
5
10
20
30
50
70
80
90
95
99
2
2
3
3
3
3
2
2
22
3
3
Predicted
Residuals
-1.5
-0.75
0
0.75
1.5
2.50 4.62 6.75 8.88 11.00
3
3
3
3
3
3
3
3
3
3
22
3
3
Formulation
Residuals
Residuals vs. Formulation
-1.5
-0.75
0
0.75
1.5
1 2
3
3
3
3
3
3
3
3
3
3
22
3
3
Washings
Residuals
Residuals vs. Washings
-1.5
-0.75
0
0.75
1.5
1 2
5-41

4
4
4
4
4
4
4
4
2
2
22
2
2
Booster
Residuals
Residuals vs. Booster
-1.5
-0.75
0
0.75
1.5
1 2
DESIGN-EXPERT Plot
Stain Shade
Lambda
Current = 1
Best = 1.66
Low C.I. = 0.88
High C.I. = 2.66
Recommend transform:
None
(Lambda = 1)
Lambda
Ln(ResidualSS)
Box-Cox Plot for Power Transforms
1.68
4.07
6.45
8.84
11.22
-3 -2 -1 0 1 2 3
The analysis of variance was performed with the transformed data and is shown below. This time, only the
formulation and number of washings appear to be significant; the interaction between these two factors is
no longer significant after the data transformation. The residual plots show no deviations from the
assumptions. The plot of the effects below identfies the new formulation along with two washings
produces the best results. The booster is not significant.
Response: Stain Shade
Sum of Mean F
A 4587.21 1 4587.21 241.74 < 0.0001
B 312.80 1 312.80 16.48 0.0036
C 37.94 1 37.94 2.00 0.1951
AB 38.24 1 38.24 2.01 0.1935
AC 28.55 1 28.55 1.50 0.2548
BC 28.55 1 28.55 1.50 0.2548
ABC 37.94 1 37.94 2.00 0.1951
Pure Error 151.81 8 18.98
In this case A, B, are significant model terms.
5-42

Residual
Normal%Probability
-4.49332 -2.24666 0 2.24666 4.49332
1
5
10
20
30
50
70
80
90
95
99
2
2
3
3
3
3
2
2
22
3
3
Predicted
Residuals
-4.49332
-2.24666
0
2.24666
4.49332
4.68 16.89 29.11 41.33 53.54
2
2
3
3
3
3
2
2
22
3
3
Formulation
Residuals
Residuals vs. Formulation
-4.49332
-2.24666
0
2.24666
4.49332
1 2
2
2
2
2
2
2
22
2
2
Washings
Residuals
Residuals vs. Washings
-4.49332
-2.24666
0
2.24666
4.49332
1 2
2
2
2
2
22
Booster
Residuals
Residuals vs. Booster
-4.49332
-2.24666
0
2.24666
4.49332
1 2
DESIGN-EXPERT Plot
(Stain Shade)^1.66
X = A: Formulation
Y = B: Washings
Design Points
B1 1
B2 2
Actual Factor
C: Booster = Yes
B: Washings
Interaction Graph
A: Formulation
(StainShade)^1.66
New Original
-1
9
19
29
39
49
22
2
2
2
2
5-43

Chapter 9
Three-Level and Mixed-Level
Factorial and Fractional Factorial Design
Solutions
9-1 The effects of developer concentration (A) and developer time (B) on the density of photographic
plate film are being studied. Three strengths and three times are used, and four replicates of a 32 factorial
experiment are run. The data from this experiment follow. Analyze the data using the standard methods
for factorial experiments.
Development Time (minutes)
Developer Concentration 10 14 18
10% 0 2 1 3 2 5
5 4 4 2 4 6
12% 4 6 6 8 9 10
7 5 7 7 8 5
14% 7 10 10 10 12 10
8 7 8 7 9 8
Response: Data
Sum of Mean F
A 198.22 2 99.11 37.69 < 0.0001
B 22.72 2 11.36 4.32 0.0236
AB 3.28 4 0.82 0.31 0.8677
Residual 71.00 27 2.63
Lack of Fit 0.000 0
Pure Error 71.00 27 2.63
Cor Total 295.22 35
Concentration and time are significant. The interaction is not significant. By letting both A and B be
treated as numerical factors, the analysis can be performed as follows:
Response: Data
Sum of Mean F
A 192.67 1 192.67 77.88 < 0.0001
B 22.04 1 22.04 8.91 0.0056
A2 5.56 1 5.56 2.25 0.1444
B2 0.68 1 0.68 0.28 0.6038
AB 0.062 1 0.062 0.025 0.8748
Residual 74.22 30 2.47
Pure Error 71.00 27 2.63
Cor Total 295.22 35
9-1

9-2 Compute the I and J components of the two-factor interaction in Problem 9-1.
B
11 10 17
A 22 28 32
32 35 39
AB Totals = 77, 78, 71; ( )ABISSAB ==−
++
= 39.2
36
226
12
717877 2222
AB2
Totals = 78, 74, 74; ( )ABJSSAB
==−
++
= 89.0
36
226
12
747478 2222
2
( ) ( ) 28.3=+= ABJABISSAB
9-3 An experiment was performed to study the effect of three different types of 32-ounce bottles (A) and
three different shelf types (B) -- smooth permanent shelves, end-aisle displays with grilled shelves, and
beverage coolers -- on the time it takes to stock ten 12-bottle cases on the shelves. Three workers (factor
C) were employed in this experiment, and two replicates of a 33 factorial design were run. The observed
time data are shown in the following table. Analyze the data and draw conclusions.
Replicate I Replicate 2
Worker Bottle Type Permanent EndAisle Cooler Permanent EndAisle Cooler
1 Plastic 3.45 4.14 5.80 3.36 4.19 5.23
28-mm glass 4.07 4.38 5.48 3.52 4.26 4.85
38-mm glass 4.20 4.26 5.67 3.68 4.37 5.58
2 Plastic 4.80 5.22 6.21 4.40 4.70 5.88
28-mm glass 4.52 5.15 6.25 4.44 4.65 6.20
38-mm glass 4.96 5.17 6.03 4.39 4.75 6.38
3 Plastic 4.08 3.94 5.14 3.65 4.08 4.49
28-mm glass 4.30 4.53 4.99 4.04 4.08 4.59
38-mm glass 4.17 4.86 4.85 3.88 4.48 4.90
Response: Time
Sum of Mean F
A 0.33 2 0.16 1.95 0.1618
B 17.91 2 8.95 107.10 < 0.0001
C 7.91 2 3.96 47.33 < 0.0001
AB 0.11 4 0.027 0.33 0.8583
AC 0.11 4 0.027 0.32 0.8638
BC 1.59 4 0.40 4.76 0.0049
ABC 0.43 8 0.053 0.64 0.7380
Residual 2.26 27 0.084
Lack of Fit 0.000 0
Pure Error 2.26 27 0.084
Cor Total 30.64 53
9-2

In this case B, C, BC are significant model terms.
Factors B and C, shelf type and worker, and the BC interaction are significant. For the shortest time
regardless of worker chose the permanent shelves. This can easily be seen in the interaction plot below.
DESIGN-EXPERT Plot
T im e
X = C: Worker
Y = B: Shelf T ype
Design Points
B1 Perm anent
B2 End Aisle
B3 Cooler
Actual Factor
A: Bottle T ype = 28m m glass
Shelf Type
Interaction Graph
Worker
Time
1 2 3
3.36
4.1504
4.94081
5.73121
6.52162
9-4 A medical researcher is studying the effect of lidocaine on the enzyme level in the heart muscle of
beagle dogs. Three different commercial brands of lidocaine (A), three dosage levels (B), and three dogs
(C) are used in the experiment, and two replicates of a 33 factorial design are run. The observed enzyme
levels follow. Analyze the data from this experiment.
Replicate I Replicate 2
Lidocaine Dosage Dog Dog
Brand Strength 1 2 3 1 2 3
1 1 86 84 85 84 85 86
2 94 99 98 95 97 90
3 101 106 98 105 104 103
2 1 85 84 86 80 82 84
2 95 98 97 93 99 95
3 108 114 109 110 102 100
3 1 84 83 81 83 80 79
2 95 97 93 92 96 93
3 105 100 106 102 111 108
Response: Enzyme Level
Sum of Mean F
A 31.00 2 15.50 1.52 0.2359
B 4260.78 2 2130.39 209.55 < 0.0001
C 28.00 2 14.00 1.38 0.2695
AB 69.56 4 17.39 1.71 0.1768
AC 3.33 4 0.83 0.082 0.9872
BC 36.89 4 9.22 0.91 0.4738
9-3

ABC 60.78 8 7.60 0.75 0.6502
Residual 274.50 27 10.17
Lack of Fit 0.000 0
Pure Error 274.50 27 10.17
The dosage is significant.
9-5 Compute the I and J components of the two-factor interactions for Example 9-1.
A
134 188 44
B -155 -348 -289
176 127 288
I totals = 74,75,16 J totals = -128,321,-28
I(AB) = 126.78 J(AB) = 6174.12
SSAB = 6300.90
A
-190 -58 -211
C 399 230 394
6 -205 -140
I totals = -100,342,-77 J totals = 25,141,-1
I(AC) = 6878.78 J(AC) = 635.12
SSAC = 7513.90
B
-93 -350 -16
C -155 -133 533
-104 -309 74
I totals = -152,79,238 J totals =-253,287,131
I(BC) = 4273.00 J(BC) = 8581.34
SSBC = 12854.34
9-6 An experiment is run in a chemical process using a 32 factorial design. The design factors are
temperature and pressure, and the response variable is yield. The data that result from this experiment are
shown below.
Pressure, psig
Temperature, °C 100 120 140
80 47.58, 48.77 64.97, 69.22 80.92, 72.60
90 51.86, 82.43 88.47, 84.23 93.95, 88.54
100 71.18, 92.77 96.57, 88.72 76.58, 83.04
(a) Analyze the data from this experiment by conducting an analysis of variance. What conclusions can
you draw?
9-4

Response: Yield
Sum of Mean F
A 1096.93 2 548.47 6.02 0.0219
B 1503.56 2 751.78 8.25 0.0092
AB 586.64 4 146.66 1.61 0.2536
Pure Error 819.98 9 91.11
Temperature and pressure are significant. Their interaction is not. An alternate analysis is performed
below with A and B treated as numeric factors:
Response: Yield
Sum of Mean F
A 850.76 1 850.76 10.93 0.0063
B 1297.92 1 1297.92 16.68 0.0015
A2 246.18 1 246.18 3.16 0.1006
B2 205.64 1 205.64 2.64 0.1300
AB 472.78 1 472.78 6.08 0.0298
Residual 933.83 12 77.82
Pure Error 819.98 9 91.11
(b) Graphically analyze the residuals. Are there any concerns about underlying assumptions or model
adequacy?
9-5

Res idual
Normal%probability
-15.285 -7.6425 0 7.6425 15.285
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-15.285
-7.6425
0
7.6425
15.285
48.18 59.29 70.41 81.53 92.65
Pres s ure
Residuals
-15.285
-7.6425
0
7.6425
15.285
1 2 3
Tem perature
Residuals
-15.285
-7.6425
0
7.6425
15.285
1 2 3
The plot of residuals versus pressure shows a decreasing funnel shape indicating a non-constant variance.
(c) Verify that if we let the low, medium and high levels of both factors in this experiment take on the
levels -1, 0, and +1, then a least squares fit to a second order model for yield is
. . . . . .y x x x x= + + − − −86 81 10 4 8 42 7 17 7 86 7 691 2 1
2
2
2
1 2x x
The coefficients can be found in the following table of computer output.
Yield =
+86.81
+8.42 * A
+10.40 * B
-7.84 * A2
-7.17 * B2
-7.69 * A * B
9-6

(d) Confirm that the model in part (c) can be written in terms of the natural variables temperature (T) and
pressure (P) as
. . . . . .y T P T P= − + + − − −1335 63 18 56 8 59 0 072 0 0196 0 03842 2
TP
The coefficients can be found in the following table of computer output.
Yield =
-1335.62500
+8.58737 * Pressure
-0.019612 * Pressure2
-0.038437 * Pressure * Temperature
(e) Construct a contour plot for yield as a function of pressure and temperature. Based on the examination
of this plot, where would you recommend running the process.
Yield
A: Pres sure
B:Temperature
100.00 110.00 120.00 130.00 140.00
80.00
85.00
90.00
95.00
100.00
50
55
60
65 70
75
80
85
85
90
2 2 2
2 2 2
2 2 2
Run the process in the oval region indicated by the yield of 90.
9-7
(a) Confound a 33 design in three blocks using the ABC2 component of the three-factor interaction.
Compare your results with the design in Figure 9-7.
L = X1 + X2 + 2X3
Block 1 Block 2 Block 3
000 100 200
112 212 012
210 010 110
120 220 020
9-7

022 122 222
202 002 102
221 021 121
101 201 001
011 111 211
The new design is a 180° rotation around the Factor B axis.
(b) Confound a 33 design in three blocks using the AB2C component of the three-factor interaction.
Compare your results with the design in Figure 9-7.
L = X1 + 2X2 + X3
000 210 112
022 202 120
011 221 101
212 100 010
220 122 002
201 111 021
110 012 200
102 020 222
121 001 211
The new design is a 180° rotation around the Factor C axis.
(c) Confound a 33 design three blocks using the ABC component of the three-factor interaction. Compare
your results with the design in Figure 9-7.
L = X1 + X2 + X3
000 112 221
210 022 101
120 202 011
021 100 212
201 010 122
111 220 002
012 121 200
222 001 110
102 211 020
The new design is a 90° rotation around the Factor C axis along with switching layer 0 and layer 1 in the C
axis.
(d) After looking at the designs in parts (a), (b), and (c) and Figure 9-7, what conclusions can you draw?
All four designs are relatively the same. The only differences are rotations and swapping of layers.
9-8 Confound a 34 design in three blocks using the AB2CD component of the four-factor interaction.
L = X1 + 2X2 + X3 + X4
9-8

Block 1
0000 1100 0110 0101 2200 0220 0202 1210 1201
0211 1222 2212 2221 0122 2111 1121 1112 2010
2102 0021 2001 2120 1011 2022 0012 1002 1020
Block 2
1021 1110 1202 0001 0120 0212 1012 1101 1220
0200 0022 0111 2002 2121 2210 0010 0102 0221
1000 1122 1211 2112 2201 2020 2011 2100 2222
Block 3
2012 2101 2220 1022 1111 1200 2000 2121 2211
1221 1010 1102 0020 0112 0201 1001 1120 1212
2021 2110 2202 0100 0222 0011 0002 0121 0210
9-9 Consider the data from the first replicate of Problem 9-3. Assuming that all 27 observations could
not be run on the same day, set up a design for conducting the experiment over three days with AB2C
confounded with blocks. Analyze the data.
000 = 3.45 100 = 4.07 200 = 4.20
110 = 4.38 210 = 4.26 010 = 4.14
011 = 5.22 111 = 4.14 211 = 5.17
102 = 4.30 202 = 4.17 002 = 4.08
201 = 4.96 001 = 4.80 101 = 4.52
212 = 4.86 012 = 3.94 112 = 4.53
121 = 6.25 221 = 4.99 021 = 6.21
022 = 5.14 122 = 6.03 222 = 4.85
220 = 5.67 020 = 5.80 120 = 5.48
Totals = 44.23 43.21 43.18
Response: Time
Sum of Mean F
Block 0.23 2 0.11
A 0.048 2 0.024 0.14 0.8723
B 8.92 2 4.46 26.02 0.0011
C 1.57 2 0.78 4.57 0.0622
AB 1.31 4 0.33 1.91 0.2284
AC 0.87 4 0.22 1.27 0.3774
BC 0.45 4 0.11 0.66 0.6410
Cor Total 14.43 26
9-10 Outline the analysis of variance table for the 34 design in nine blocks. Is this a practical design?
9-9

Source DF
A 2
B 2
C 2
D 2
AB 4
AC 4
AD 4
BC 4
BD 4
CD 4
ABC (AB2
C,ABC2
,AB2
C2
) 6
ABD (ABD,AB2
D,ABD2
) 6
ACD (ACD,ACD2
,AC2
D2
) 6
BCD (BCD,BC2
D,BCD2
) 6
ABCD 16
Blocks (ABC,AB2
C2
,AC2
D,BC2
D2
) 8
Total 80
Any experiment with 81 runs is large. Instead of having three full levels of each factor, if two levels of
each factor could be used, then the overall design would have 16 runs plus some center points. This two-
level design could now probably be run in 2 or 4 blocks, with center points in each block. Additional
curvature effects could be determined by augmenting the experiment with the axial points of a central
composite design and additional enter points. The overall design would be less than 81 runs.
9-11 Consider the data in Problem 9-3. If ABC is confounded in replicate I and ABC2 is confounded in
replicate II, perform the analysis of variance.
L1 = X1 + X2 + X3 L2 = X1 + X2 + 2X2
Block 1 Block 2 Block 3 Block 1 Block 2 Block 3
000 = 3.45 001 = 4.80 002 = 4.08 000 = 3.36 100 = 3.52 200 = 3.68
111 = 5.15 112 = 4.53 110 = 4.38 101 = 4.44 201 = 4.39 001 = 4.40
222 = 4.85 220 = 5.67 221 = 6.03 011 = 4.70 111 = 4.65 211 = 4.75
120 = 5.48 121 = 6.25 122 = 4.99 221 = 6.38 021 = 5.88 121 = 6.20
102 = 4.30 100 = 4.07 101 = 4.52 202 = 3.88 002 = 3.65 102 = 4.04
210 = 4.26 211 = 5.17 212 = 4.86 022 = 4.49 122 = 4.59 222 = 4.90
201 = 4.96 202 = 4.17 200 = 4.20 120 = 4.85 220 = 5.58 020 = 5.23
012 = 3.94 010 = 4.14 011 = 5.22 210 = 4.37 010 = 4.19 110 = 4.26
021 = 6.21 022 = 5.14 020 = 5.80 112 = 4.08 212 = 4.48 012 = 4.08
The sums of squares for A, B, C, AB, AC, and BC are calculated as usual. The only sums of squares
presenting difficulties with calculations are the four components of the ABC interaction (ABC, ABC2
,
AB2
C, and AB2
C2
). ABC is computed using replicate I and ABC2
is computed using replicate II. AB2
C and
AB2
C2
are computed using data from both replicates.
We will show how to calculate AB2
C and AB2
C2
from both replicates. Form a two-way table of A x B at
each level of C. Find the I(AB) and J(AB) totals for each third of the A x B table.
A
C B 0 1 2 I J
0 6.81 7.59 7.88 26.70 27.55
0 1 8.33 8.64 8.63 27.25 27.17
2 11.03 10.33 11.25 26.54 25.77
0 9.20 8.96 9.35 31.41 31.25
1 1 9.92 9.80 9.92 30.97 31.29
9-10

2 12.09 12.45 12.41 31.72 31.57
0 7.73 8.34 8.05 26.09 26.29
2 1 8.02 8.61 9.34 27.31 26.11
2 9.63 9.58 9.75 25.65 26.65
The I and J components for each third of the above table are used to form a new table of diagonal totals.
C I(AB) J(AB)
0 2.670 27.25 26.54 27.55 27.17 25.77
1 31.41 30.97 31.72 31.25 31.29 31.57
2 26.09 27.31 25.65 26.29 26.11 26.65
I Totals: I Totals:
85.06,85.26,83.32 85.99,85.03,83.12
J Totals: J Totals:
85.73,83.60,84.31 83.35,85.06,85.23
Now, AB2
C2
= I[C x I(AB)] =
(85. ) (85. ) (83. ) ( . )
.
06 26 32
18
25364
54
01265
2 2 2 2
+ +
− =
and, AB2
C = J[C x I(AB)]=
(85. ) (83. ) (84. ) ( . )
.
73 60 31
18
25364
54
01307
2 2 2 2
+ +
− =
If it were necessary, we could find ABC2
as ABC2
= I[C x J(AB)] and ABC as J[C x J(AB)]. However, these
components must be computed using the data from the appropriate replicate.
The analysis of variance table:
Source SS DF MS F0
Replicates 1.06696 1
Blocks within Replicates 0.2038 4
A 0.4104 2 0.2052 5.02
B 17.7514 2 8.8757 217.0
C 7.6631 2 3.8316 93.68
AB 0.1161 4 0.0290 <1
AC 0.1093 4 0.0273 <1
BC 1.6790 4 0.4198 10.26
ABC (rep I) 0.0452 2 0.0226 <1
ABC2
(rep II) 0.1020 2 0.0510 1.25
AB2
C 0.1307 2 0.0754 1.60
AB2
C2
0.1265 2 0.0633 1.55
Error 0.8998 22 0.0409
Total 30.3069 53
9-12 Consider the data from replicate I in Problem 9-3. Suppose that only a one-third fraction of this
design with I=ABC is run. Construct the design, determine the alias structure, and analyze the design.
The design is 000, 012, 021, 102, 201, 111, 120, 210, 222.
The alias structure is: A = BC = AB2
C2
B = AC = AB2
C
C = AB = ABC2
9-11

AB2
= AC2
= BC2
C
A B 0 1 2
0 3.45
0 1 5.48
2 4.26
0 6.21
1 1 5.15
2 4.96
0 3.94
2 1 4.30
2 4.85
Source SS DF
A 2.25 2
B 0.30 2
C 2.81 2
AB2
0.30 2
Total 5.66 8
9-13 From examining Figure 9-9, what type of design would remain if after completing the first 9 runs,
one of the three factors could be dropped?
A full 32
factorial design results.
9-14 Construct a 3 design with I=ABCD. Write out the alias structure for this design.4 1
IV
−
The 27 runs for this design are as follows:
0000 1002 2001
0012 1011 2010
0021 1020 2022
0102 1101 2100
0111 1110 2112
0120 1122 2121
0201 1200 2202
0210 1212 2211
0222 1221 2220
A = AB2
C2
D2
= BCD B = AB2
CD = ACD C = ABC2
D = ABD D = ABCD2
= ABC
AB = ABC2
D2
= CD AB2
= AC2
D2
= BC2
D2
AC = AB2
CD2
= BD AC2
= AB2
D2
= BC2
D
BC = AB2
C2
D = AD BC2
= AB2
D = AC2
D BD2
= AB2
C = ACD2
CD2
= ABC2
= ABCD2
AD2
= AB2
C2
= BCD2
9-15 Verify that the design in Problem 9-14 is a resolution IV design.
The design in Problem 9-14 is a Resolution IV design because no main effect is aliased with a component
of a two-factor interaction, but some two-factor interaction components are aliased with each other.
9-12

9-16 Construct a 35-2 design with I=ABC and I=CDE. Write out the alias structure for this design. What
is the resolution of this design?
The complete defining relation for this design is : I = ABC = CDE = ABC2
DE = ABD2
E2
This is a resolution III design. The defining contrasts are L1 = X1 + X2 + X3 and L2 = X3 + X4 + X5.
00000 11120 20111
00012 22111 22222
00022 21021 01210
01200 02111 12000
02100 01222 20120
10202 12012 11111
20101 02120 22201
11102 10210 21012
21200 12021 10222
To find the alias of any effect, multiply the effect by I and I2
. For example, the alias of A is:
A = AB2
C2
= ACDE = AB2
CDE = AB2
DE = BC = AC2
D2
E2
= BC2
DE = BD2
E2
9-17 Construct a 39-6 design, and verify that is a resolution III design.
Use the generators I = AC2
D2
, I = AB2
C2
E, I = BC2
F2
, I = AB2
CG, I = ABCH2
, and I = ABJ2
000000000 021201102 102211001
022110012 212012020 001212210
011220021 100120211 211100110
221111221 122200220 020022222
210221200 010011111 222020101
202001212 201122002 200210122
112222112 002121120 121021010
101002121 111010202 110101022
120112100 220202011 012102201
To find the alias of any effect, multiply the effect by I and I2
. For example, the alias of C is:
C = C(BC2
F2
) = BF2
, At least one main effect is aliased with a component of a two-factor interaction.
9-18 Construct a 4 x 23 design confounded in two blocks of 16 observations each. Outline the analysis of
variance for this design.
Design is a 4 x 23
, with ABC at two levels, and Z at 4 levels. Represent Z with two pseudo-factors D and E
as follows:
Factor Pseudo- Factors
Z D E
Z1 0 0 = (1)
Z2 1 0 = d
Z3 0 1 = e
Z4 1 1 = de
9-13

The 4 x 23
is now a 25
in the factors A, B, C, D and E. Confound ABCDE with blocks. We have given
both the letter notation and the digital notation for the treatment combinations.
Block 1 Block 2
(1) = 000 a = 1000
ab = 1100 b = 0100
ac = 1010 c = 0010
bc = 0110 abc = 1110
abcd = 1111 bcd = 0111
abce = 1112 bce = 0112
cd = 0011 acd = 1011
ce = 0012 ace = 1012
de = 0003 ade = 1003
abde = 1103 bde = 0103
bcde = 0113 abcde = 1113
be = 0102 abd = 1101
ad = 1001 abe = 1102
ae = 1002 d = 0001
acde = 1013 e = 0002
bd = 0101 cde = 0013
Source DF
A 1
B 1
C 1
Z (D+E+DE) 3
AB 1
AC 1
AZ (AD+AE+ADE) 3
BC 1
BZ (BD+BE+BDE) 3
CZ (CD+CE+CDE) 3
ABC 1
ABZ (ABD+ABE+ABDE) 3
ACZ (ACD+ACE+ACDE) 3
BCZ (BCD+BCE+BCDE) 3
ABCZ (ABCD+ABCE) 2
Blocks (or ABCDE) 1
Total 31
9-19 Outline the analysis of variance table for a 2232 factorial design. Discuss how this design may be
confounded in blocks.
Suppose we have n replicates of a 22
32
factorial design. A and B are at 2 levels, and C and D are at 3
levels.
Source DF Components for Confounding
A 1 A
B 1 B
C 2 C
D 2 D
AB 1 AB
AC 2 AC
AD 2 AD
9-14

BC 2 BD
BD 2 CD,CD2
CD 4 ABC
ABC 2 ABD
ABD 2 ACD,ACD2
ACD 4 BCD,BCD2
BCD 4 ABCD,ABCD2
ABCD 4
Error 36(n-1)
Total 36n-1
Confounding in this series of designs is discussed extensively by Margolin (1967). The possibilities for a
single replicate of the 22
32
design are:
2 blocks of 18 observations 6 blocks of 6 observations
3 blocks of 12 observations 9 blocks of 4 observations
4 blocks of 9 observations
For example, one component of the four-factor interaction, say ABCD2
, could be selected to confound the
design in 3 blocks for 12 observations each, while to confound the design in 2 blocks of 18 observations 3
each we would select the AB interaction. Cochran and Cox (1957) and Anderson and McLean (1974)
discuss confounding in these designs.
9-20 Starting with a 16-run 24 design, show how two three-level factors can be incorporated in this
experiment. How many two-level factors can be included if we want some information on two-factor
interactions?
Use column A and B for one three-level factor and columns C and D for the other. Use the AC and BD
columns for the two, two-level factors. The design will be of resolution V.
9-21 Starting with a 16-run 24 design, show how one three-level factor and three two-level factors can be
accommodated and still allow the estimation of two-factor interactions.
Use columns A and B for the three-level factor, and columns C and D and ABCD for the three two-level
factors. This design will be of resolution V.
9-22 In Problem 9-26, you met Harry and Judy Peterson-Nedry, two friends of the author who have a
winery and vineyard in Newberg, Oregon. That problem described the application of two-level fractional
factorial designs to their 1985 Pinor Noir product. In 1987, they wanted to conduct another Pinot Noir
experiment. The variables for this experiment were
Variable Levels
Clone of Pinot Noir Wadenswil, Pommard
Berry Size Small, Large
Fermentation temperature 80F, 85F, 90/80F, 90F
Whole Berry None, 10%
Maceration Time 10 days, 21 days
Yeast Type Assmanhau, Champagne
Oak Type Troncais, Allier
Harry and Judy decided to use a 16-run two-level fractional factorial design, treating the four levels of
fermentation temperature as two two-level variables. As in Problem 8-26, they used the rankings from a
taste-test panel as the response variable. The design and the resulting average ranks are shown below:
Berry Ferm. Whole Macer. Yeast Oak Average
9-15

Run Clone Size Temp. Berry Time Type Type Rank
1 - - - - - - - - 4
2 + - - - - + + + 10
3 - + - - + - + + 6
4 + + - - + + - - 9
5 - - + - + + + - 11
6 + - + - + - - + 1
7 - + + - - + - + 15
8 + + + - - - + - 5
9 - - - + + + - + 12
10 + - - + + - + - 2
11 - + - + - + + - 16
12 + + - + - - - + 3
13 - - + + - - + + 8
14 + - + + - + - - 14
15 - + + + + - - - 7
16 + + + + + + + + 13
(a) Describe the aliasing in this design.
The design is a resolution IV design such that the main effects are aliased with three factor interactions.
Term Aliases
Intercept ABCG ABDH ABEF ACDF ACEH ADEG AFGH BCDE BCFH BDFG BEGH CDGH CEFG DEFH
A BCG BDH BEF CDF CEH DEG FGH ABCDE
B ACG ADH AEF CDE CFH DFG EGH
C ABG ADF AEH BDE BFH DGH EFG
D ABH ACF AEG BCE BFG CGH EFH
E ABF ACH ADG BCD BGH CFG DFH
F ABE ACD AGH BCH BDG CEG DEH
G ABC ADE AFH BDF BEH CDH CEF
H ABD ACE AFG BCF BEG CDG DEF
AB CG DH EF ACDE ACFH ADFG AEGH BCDF BCEH BDEG BFGH
AC BG DF EH ABDE ABFH ADGH AEFG BCDH BCEF CDEG CFGH
AD BH CF EG ABCE ABFG ACGH AEFH BCDG BDEF CDEH DFGH
AE BF CH DG ABCD ABGH ACFG ADFH BCEG BDEH CDEF EFGH
AF BE CD GH ABCH ABDG ACEG ADEH BCFG BDFH CEFH DEFG
AG BC DE FH ABDF ABEH ACDH ACEF BDGH BEFG CDFG CEGH
AH BD CE FG ABCF ABEG ACDG ADEF BCGH BEFH CDFH DEGH
(b) Analyze the data and draw conclusions.
All of the main effects except Yeast and Oak are significant. The Macer Time is the most significant.
None of the interactions were significant.
9-16

DESIGN-EXPERT Plot
Rank
A: Clone
B: Berry Size
C: Ferm T em p 1
D: Ferm T em p 2
E: Whole Berry
F: M acer T im e
G: Yeast
H: Oak
Normal plot
Normal%probability
Effect
-2.75 -0.06 2.63 5.31 8.00
1
5
10
20
30
50
70
80
90
95
99
A
B
C
D
E
F
Response: Rank
Sum of Mean F
A 30.25 1 30.25 24.20 0.0008
B 9.00 1 9.00 7.20 0.0251
C 9.00 1 9.00 7.20 0.0251
D 12.25 1 12.25 9.80 0.0121
E 12.25 1 12.25 9.80 0.0121
F 256.00 1 256.00 204.80 < 0.0001
Residual 11.25 9 1.25
Cor Total 340.00 15
Intercept 8.50 1 0.28 7.87 9.13
A-Clone -1.38 1 0.28 -2.01 -0.74 1.00
B-Berry Size 0.75 1 0.28 0.12 1.38 1.00
C-Ferm Temp 1 0.75 1 0.28 0.12 1.38 1.00
D-Ferm Temp 2 0.88 1 0.28 0.24 1.51 1.00
E-Whole Berry -0.87 1 0.28 -1.51 -0.24 1.00
F-Macer Time 4.00 1 0.28 3.37 4.63 1.00
Rank =
+8.50
-1.38 * A
+0.75 * B
+0.75 * C
+0.88 * D
-0.87 * E
+4.00 * F
9-17

(c) What comparisons can you make between this experiment and the 1985 Pinot Noir experiment from
Problem 8-26?
The experiment from Problem 8-26 indicates that yeast, barrel, whole cluster and the clone x yeast
interactions were significant. This experiment indicates that maceration time, whole berry, clone and
fermentation temperature are significant.
9-23 An article by W.D. Baten in the 1956 volume of Industrial Quality Control described an experiment
to study the effect of three factors on the lengths of steel bars. Each bar was subjected to one of two heat
treatment processes, and was cut on one of four machines at one of three times during the day (8 am, 11
am, or 3 pm). The coded length data are shown below
(a) Analyze the data from this experiment assuming that the four observations in each cell are replicates.
The Machine effect (C) is significant, the Heat Treat Process (B) is also significant, while the Time of Day
(A) is not significant. None of the interactions are significant.
Response: Length
Sum of Mean F
A 26.27 2 13.14 2.11 0.1283
B 42.67 1 42.67 6.86 0.0107
C 393.42 3 131.14 21.10 < 0.0001
AB 23.77 2 11.89 1.91 0.1552
AC 42.15 6 7.02 1.13 0.3537
BC 13.08 3 4.36 0.70 0.5541
ABC 48.98 6 8.16 1.31 0.2623
Pure Error 447.50 72 6.22
Time of
Day
Heat Treat
Process
6 9 7 9 1 2 6 6
1 3 5 5 0 4 7 3
4 6 6 5 -1 0 4 5
0 1 3 4 0 1 5 4
6 3 8 7 3 2 7 9
1 -1 4 8 1 0 11 6
3 1 6 4 2 0 9 4
1 -2 1 3 -1 1 6 3
5 4 10 11 -1 2 10 5
9 6 6 4 6 1 4 8
6 0 8 7 0 -2 4 3
3 7 10 0 4 -4 7 0
3 4
Machine
2
1
2
1
2
1
1 2
8am
11 am
3 pm
9-18

Term Estimate DF Error Low High VIF
Intercept 3.96 1 0.25 3.45 4.47
A[1] 0.010 1 0.36 -0.71 0.73
A[2] -0.65 1 0.36 -1.36 0.071
B-Process -0.67 1 0.25 -1.17 -0.16 1.00
C[1] -0.54 1 0.44 -1.42 0.34
C[2] 1.92 1 0.44 1.04 2.80
C[3] -3.08 1 0.44 -3.96 -2.20
A[1]B 0.010 1 0.36 -0.71 0.73
A[2]B 0.60 1 0.36 -0.11 1.32
A[1]C[1] 0.32 1 0.62 -0.92 1.57
A[2]C[1] -1.27 1 0.62 -2.51 -0.028
A[1]C[2] -0.39 1 0.62 -1.63 0.86
A[2]C[2] -0.10 1 0.62 -1.35 1.14
A[1]C[3] 0.24 1 0.62 -1.00 1.48
A[2]C[3] 0.77 1 0.62 -0.47 2.01
BC[1] -0.25 1 0.44 -1.13 0.63
BC[2] -0.46 1 0.44 -1.34 0.42
BC[3] 0.46 1 0.44 -0.42 1.34
A[1]BC[1] -0.094 1 0.62 -1.34 1.15
A[2]BC[1] -0.44 1 0.62 -1.68 0.80
A[1]BC[2] 0.11 1 0.62 -1.13 1.36
A[2]BC[2] -1.10 1 0.62 -2.35 0.14
A[1]BC[3] -0.43 1 0.62 -1.67 0.82
A[2]BC[3] 0.60 1 0.62 -0.64 1.85
Length =
+3.96
+0.010 * A[1]
-0.65 * A[2]
-0.67 * B
-0.54 * C[1]
+1.92 * C[2]
-3.08 * C[3]
+0.010 * A[1]B
+0.60 * A[2]B
+0.32 * A[1]C[1]
-1.27 * A[2]C[1]
-0.39 * A[1]C[2]
-0.10 * A[2]C[2]
+0.24 * A[1]C[3]
+0.77 * A[2]C[3]
-0.25 * BC[1]
-0.46 * BC[2]
+0.46 * BC[3]
-0.094 * A[1]BC[1]
-0.44 * A[2]BC[1]
+0.11 * A[1]BC[2]
-1.10 * A[2]BC[2]
-0.43 * A[1]BC[3]
+0.60 * A[2]BC[3]
(b) Analyze the residuals from this experiment. Is there any indication that there is an outlier in one cell?
If you find an outlier, remove it and repeat the analysis from part (a). What are your conclusions?
Standard Order 84, Time of Day at 3:00pm, Heat Treat #2, Machine #2, and length of 0, appears to be an
outlier.
9-19

Res idual
Normal%probability
-6.25 -3.5625 -0.875 1.8125 4.5
1
5
10
20
30
50
70
80
90
95
99
3
3
3
2
2
2
2
2
3 3
2
2
32
2
33
2
2
3
32
2
3
2
3
Predicted
Residuals
-6.25
-3.5625
-0.875
1.8125
4.5
-0.50 1.69 3.88 6.06 8.25
The following analysis was performed with the outlier described above removed. As with the original
analysis, Machine is significant and Heat Treat Process is also significant, but now Time of Day, factor A,
is also significant with an F-value of 3.05 (the P-value is just above 0.05).
Response: Length
Sum of Mean F
A 34.03 2 17.02 3.06 0.0533
B 33.06 1 33.06 5.94 0.0173
C 411.89 3 137.30 24.65 < 0.0001
AB 16.41 2 8.20 1.47 0.2361
AC 50.19 6 8.37 1.50 0.1900
BC 8.38 3 2.79 0.50 0.6824
ABC 67.00 6 11.17 2.01 0.0762
Pure Error 395.42 71 5.57
Intercept 4.05 1 0.24 3.z56 4.53
A[1] -0.076 1 0.34 -0.76 0.61
A[2] -0.73 1 0.34 -1.41 -0.051
B-Process -0.58 1 0.24 -1.06 -0.096 1.00
C[1] -0.63 1 0.42 -1.46 0.21
C[2] 2.18 1 0.43 1.33 3.03
C[3] -3.17 1 0.42 -4.00 -2.34
A[1]B -0.076 1 0.34 -0.76 0.61
A[2]B 0.52 1 0.34 -0.16 1.20
A[1]C[1] 0.41 1 0.59 -0.77 1.59
A[2]C[1] -1.18 1 0.59 -2.36 -6.278E-003
A[1]C[2] -0.65 1 0.60 -1.83 0.54
9-20

A[2]C[2] -0.36 1 0.60 -1.55 0.82
A[1]C[3] 0.33 1 0.59 -0.85 1.50
A[2]C[3] 0.86 1 0.59 -0.32 2.04
BC[1] -0.34 1 0.42 -1.17 0.50
BC[2] -0.20 1 0.43 -1.05 0.65
BC[3] 0.37 1 0.42 -0.46 1.21
A[1]BC[1] -6.944E-003 1 0.59 -1.18 1.17
A[2]BC[1] -0.35 1 0.59 -1.53 0.83
A[1]BC[2] -0.15 1 0.60 -1.33 1.04
A[2]BC[2] -1.36 1 0.60 -2.55 -0.18
A[1]BC[3] -0.34 1 0.59 -1.52 0.84
A[2]BC[3] 0.69 1 0.59 -0.49 1.87
Length =
+4.05
-0.076 * A[1]
-0.73 * A[2]
-0.58 * B
-0.63 * C[1]
+2.18 * C[2]
-3.17 * C[3]
-0.076 * A[1]B
+0.52 * A[2]B
+0.41 * A[1]C[1]
-1.18 * A[2]C[1]
-0.65 * A[1]C[2]
-0.36 * A[2]C[2]
+0.33 * A[1]C[3]
+0.86 * A[2]C[3]
-0.34 * BC[1]
-0.20 * BC[2]
+0.37 * BC[3]
-6.944E-003 * A[1]BC[1]
-0.35 * A[2]BC[1]
-0.15 * A[1]BC[2]
-1.36 * A[2]BC[2]
-0.34 * A[1]BC[3]
+0.69 * A[2]BC[3]
The following residual plots are acceptable. Both the normality and constant variance assumptions are
satisfied
Res idual
Normal%probability
-4 -1.875 0.25 2.375 4.5
1
5
10
20
30
50
70
80
90
95
99
3
3
3
2
2
2
2
2
3 3
2
2
32
2
33
2
2
3
32
2
3
2
3
Predicted
Residuals
-4
-1.875
0.25
2.375
4.5
-0.50 1.71 3.92 6.12 8.33
9-21

(c) Suppose that the observations in the cells are the lengths (coded) of bars processed together in heat
treating and then cut sequentially (that is, in order) on the three machines. Analyze the data to
determine the effects of the three factors on mean length.
The analysis with all effects and interactions included:
Response: Length
Sum of Mean F
Model 147.58 23 6.42
A 6.57 2 3.28
B 10.67 1 10.67
C 98.35 3 32.78
AB 5.94 2 2.97
AC 10.54 6 1.76
BC 3.27 3 1.09
ABC 12.24 6 2.04
Pure Error 0.000 0
Cor Total 147.58 23
The by removing the three factor interaction from the model and applying it to the error, the analysis
identifies factor C as being significant and B as being mildly significant.
Response: Length
Sum of Mean F
A 6.57 2 3.28 1.61 0.2757
B 10.67 1 10.67 5.23 0.0623
C 98.35 3 32.78 16.06 0.0028
AB 5.94 2 2.97 1.46 0.3052
AC 10.54 6 1.76 0.86 0.5700
BC 3.27 3 1.09 0.53 0.6756
Residual 12.24 6 2.04
Cor Total 147.58 23
When removing the remaining insignificant factors from the model, C, Machine, is the most significant
factor while B, Heat Treat Process, is moderately significant. Factor A, Time of Day, is not significant.
Response: Avg
Sum of Mean F
B 10.67 1 10.67 5.26 0.0335
C 98.35 3 32.78 16.15 < 0.0001
Residual 38.56 19 2.03
Cor Total 147.58 23
9-22

Intercept 3.96 1 0.29 3.35 4.57
B-Process -0.67 1 0.29 -1.28 -0.058 1.00
C[1] -0.54 1 0.50 -1.60 0.51
C[2] 1.92 1 0.50 0.86 2.97
C[3] -3.08 1 0.50 -4.14 -2.03
Avg =
+3.96
-0.67 * B
-0.54 * C[1]
+1.92 * C[2]
-3.08 * C[3]
The following residual plots are acceptable. Both the normality and uniformity of variance assumptions
are verified.
Res idual
Normal%probability
-2 -1.02083 -0.0416667 0.9375 1.91667
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-2
-1.02083
-0.0416667
0.9375
1.91667
0.21 1.79 3.38 4.96 6.54
(d) Calculate the log variance of the observations in each cell. Analyze the average length and the log
variance of the length for each of the 12 bars cut at each machine/heat treatment process combination.
What conclusions can you draw?
Factor B, Heat Treat Process, has an affect on the log variance of the observations while Factor A, Time of
Day, and Factor C, Machine, are not significant at the 5 percent level. However, A is significant at the 10
percent level, so Tome of Day has some effect on the variance.
Response: Log(Var)
Sum of Mean F
A 0.58 2 0.29 2.86 0.0966
B 0.50 1 0.50 4.89 0.0471
C 0.59 3 0.20 1.95 0.1757
AB 0.49 2 0.24 2.40 0.1324
BC 0.64 3 0.21 2.10 0.1538
9-23

Residual 1.22 12 0.10
Cor Total 4.01 23
Intercept 0.65 1 0.065 0.51 0.79
A[1] -0.054 1 0.092 -0.25 0.15
A[2] -0.16 1 0.092 -0.36 0.043
B-Process 0.14 1 0.065 2.181E-003 0.29 1.00
C[1] 0.22 1 0.11 -0.025 0.47
C[2] 0.066 1 0.11 -0.18 0.31
C[3] -0.19 1 0.11 -0.44 0.052
A[1]B -0.20 1 0.092 -0.40 3.237E-003
A[2]B 0.14 1 0.092 -0.065 0.34
BC[1] -0.18 1 0.11 -0.42 0.068
BC[2] -0.15 1 0.11 -0.39 0.098
BC[3] 0.14 1 0.11 -0.10 0.39
Log(Var) =
+0.65
-0.054 * A[1]
-0.16 * A[2]
+0.14 * B
+0.22 * C[1]
+0.066 * C[2]
-0.19 * C[3]
-0.20 * A[1]B
+0.14 * A[2]B
-0.18 * BC[1]
-0.15 * BC[2]
+0.14 * BC[3]
are verified.
9-24

Res idual
Normal%probability
-0.414972 -0.226114 -0.0372556 0.151602 0.340461
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.414972
-0.226114
-0.0372556
0.151602
0.340461
-0.12 0.20 0.52 0.84 1.16
(e) Suppose the time at which a bar is cut really cannot be controlled during routine production. Analyze
the average length and the log variance of the length for each of the 12 bars cut at each machine/heat
treatment process combination. What conclusions can you draw?
The analysis of the average length is as follows:
Response: Avg
Sum of Mean F
Model 37.43 7 5.35
A 3.56 1 3.56
B 32.78 3 10.93
AB 1.09 3 0.36
Pure Error 0.000 0
Cor Total 37.43 7
Because the Means Square of the AB interaction is much less than the main effects, it is removed from the
model and placed in the error. The average length is strongly affected by Factor B, Machine, and
moderately affected by Factor A, Heat Treat Process. The interaction effect was small and removed from
the model.
Response: Avg
Sum of Mean F
A 3.56 1 3.56 9.78 0.0522
B 32.78 3 10.93 30.07 0.0097
Cor Total 37.43 7
9-25

Intercept 3.96 1 0.21 3.28 4.64
A-Process -0.67 1 0.21 -1.34 0.012 1.00
B[1] -0.54 1 0.37 -1.72 0.63
B[2] 1.92 1 0.37 0.74 3.09
B[3] -3.08 1 0.37 -4.26 -1.91
Avg =
+3.96
-0.67 * A
-0.54 * B[1]
+1.92 * B[2]
-3.08 * B[3]
are verified.
Res idual
Normal%probability
-0.458333 -0.229167 0 0.229167 0.458333
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals Residuals vs. Predicted
-0.458333
-0.229167
0
0.229167
0.458333
0.21 1.79 3.38 4.96 6.54
The Log(Var) is analyzed below:
Response: Log(Var)
Sum of Mean F
Model 0.32 7 0.046
A 0.091 1 0.091
B 0.13 3 0.044
AB 0.098 3 0.033
Pure Error 0.000 0
Cor Total 0.32 7
Because the AB interaction has the smallest Mean Square, it was removed from the model and placed in the
error. From the following analysis of variance, neither Heat Treat Process, Machine, nor the interaction
affect the log variance of the length.
9-26

Response: Log(Var)
Sum of Mean F
A 0.091 1 0.091 2.80 0.1926
B 0.13 3 0.044 1.34 0.4071
Residual 0.098 3 0.033
Cor Total 0.32 7
Intercept 0.79 1 0.064 0.59 0.99
A-Process 0.11 1 0.064 -0.096 0.31 1.00
B[1] 0.15 1 0.11 -0.20 0.51
B[2] 0.030 1 0.11 -0.32 0.38
B[3] -0.20 1 0.11 -0.55 0.15
Log(Var) =
+0.79
+0.11 * A
+0.15 * B[1]
+0.030 * B[2]
-0.20 * B[3]
are verified.
Res idual
Normal%probability
-0.160958 -0.0804791 0 0.0804791 0.160958
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.160958
-0.0804791
0
0.0804791
0.160958
0.48 0.62 0.76 0.91 1.05
9-27

Chapter 10
Fitting Regression Models
Solutions
10-1 The tensile strength of a paper product is related to the amount of hardwood in the pulp. Ten
samples are produced in the pilot plant, and the data obtained are shown in the following table.
Strength Percent Hardwood Strength Percent Hardwood
160 10 181 20
171 15 188 25
175 15 193 25
182 20 195 28
184 20 200 30
(a) Fit a linear regression model relating strength to percent hardwood.
Minitab Output
Regression Analysis: Strength versus Hardwood
The regression equation is
Strength = 144 + 1.88 Hardwood
Predictor Coef SE Coef T P
Constant 143.824 2.522 57.04 0.000
Hardwood 1.8786 0.1165 16.12 0.000
S = 2.203 R-Sq = 97.0% R-Sq(adj) = 96.6%
PRESS = 66.2665 R-Sq(pred) = 94.91%
302010
200
190
180
170
160
Hardwood
Strength
S = 2.20320 R-Sq = 97.0 % R-Sq(adj) = 96.6 %
Strength = 143.824 + 1.87864 Hardwood
Regression Plot
(b) Test the model in part (a) for significance of regression.
Minitab Output
10-1

Analysis of Variance
Source DF SS MS F P
Regression 1 1262.1 1262.1 260.00 0.000
Residual Error 8 38.8 4.9
Lack of Fit 4 13.7 3.4 0.54 0.716
Pure Error 4 25.2 6.3
Total 9 1300.9
3 rows with no replicates
No evidence of lack of fit (P > 0.1)
(c) Find a 95 percent confidence interval on the parameter β1.
The 95 percent confidence interval is:
( ) ( )1,2111,21
ˆˆˆˆ βββββ αα setset pnpn −− +≤≤−
( ) ( )0.11653060.21.87860.11653060.21.8786 1 +≤≤− β
1473.26900.1 1 ≤≤ β
10-2 A plant distills liquid air to produce oxygen , nitrogen, and argon. The percentage of impurity in the
oxygen is thought to be linearly related to the amount of impurities in the air as measured by the “pollution
count” in part per million (ppm). A sample of plant operating data is shown below.
Purity(%) 93.3 92.0 92.4 91.7 94.0 94.6 93.6 93.1 93.2 92.9 92.2 91.3 90.1 91.6 91.9
Pollution count (ppm) 1.10 1.45 1.36 1.59 1.08 0.75 1.20 0.99 0.83 1.22 1.47 1.81 2.03 1.75 1.68
(a) Fit a linear regression model to the data.
Minitab Output
Regression Analysis: Purity versus Pollution
Purity = 96.5 - 2.90 Pollution
Predictor Coef SE Coef T P
Constant 96.4546 0.4282 225.24 0.000
Pollutio -2.9010 0.3056 -9.49 0.000
S = 0.4277 R-Sq = 87.4% R-Sq(adj) = 86.4%
PRESS = 3.43946 R-Sq(pred) = 81.77%
10-2

2.01.51.0
95
94
93
92
91
90
Pollution
Purity
S = 0.427745 R-Sq = 87.4 % R-Sq(adj) = 86.4 %
Purity= 96.4546 - 2.90096 Pollution
Regression Plot
(b) Test for significance of regression.
Minitab Output
Source DF SS MS F P
Regression 1 16.491 16.491 90.13 0.000
Total 14 18.869
No replicates. Cannot do pure error test.
(c) Find a 95 percent confidence interval on β1.
The 95 percent confidence interval is:
( ) ( )1,2111,21
ˆˆˆˆ βββββ αα setset pnpn −− +≤≤−
( ) ( )0.30561604.29010.2-0.30561604.29010.2- 1 +≤≤− β
2408.25612.3 1 −≤≤− β
10-3 Plot the residuals from Problem 10-1 and comment on model adequacy.
10-3

3210-1-2-3
1
0
-1
NormalScore
Residual
(response is Strength)
200190180170160
3
2
1
0
-1
-2
-3
Fitted Value
Residual
10-4

10987654321
3
2
1
0
-1
-2
-3
Observation Order
Residual
Residuals Versus the Order of the Data
There is nothing unusual about the residual plots. The underlying assumptions have been met.
10-4 Plot the residuals from Problem 10-2 and comment on model adequacy.
0.50.0-0.5-1.0
2
1
0
-1
-2
NormalScore
Residual
(response is Purity)
10-5

94.593.592.591.590.5
0.5
0.0
-0.5
-1.0
Fitted Value
Residual
1412108642
0.5
0.0
-0.5
-1.0
Observation Order
Residual
There is nothing unusual about the residual plots. The underlying assumptions have been met.
10-5 Using the results of Problem 10-1, test the regression model for lack of fit.
Minitab Output
Source DF SS MS F P
Regression 1 1262.1 1262.1 260.00 0.000
10-6

Lack of Fit 4 13.7 3.4 0.54 0.716
Pure Error 4 25.2 6.3
Total 9 1300.9
10-6 A study was performed on wear of a bearing y and its relationship to x1 = oil viscosity and x2 = load.
The following data were obtained.
y x1 x2
193 1.6 851
230 15.5 816
172 22.0 1058
91 43.0 1201
113 33.0 1357
125 40.0 1115
(a) Fit a multiple linear regression model to the data.
Minitab Output
Regression Analysis: Wear versus Viscosity, Load
Wear = 351 - 1.27 Viscosity - 0.154 Load
Predictor Coef SE Coef T P VIF
Constant 350.99 74.75 4.70 0.018
Viscosit -1.272 1.169 -1.09 0.356 2.6
Load -0.15390 0.08953 -1.72 0.184 2.6
S = 25.50 R-Sq = 86.2% R-Sq(adj) = 77.0%
PRESS = 12696.7 R-Sq(pred) = 10.03%
(b) Test for significance of regression.
Minitab Output
Source DF SS MS F P
Regression 2 12161.6 6080.8 9.35 0.051
Total 5 14112.0
Source DF Seq SS
Viscosit 1 10240.4
Load 1 1921.2
* Not enough data for lack of fit test
(c) Compute t statistics for each model parameter. What conclusions can you draw?
Minitab Output
Regression Analysis: Wear versus Viscosity, Load
Wear = 351 - 1.27 Viscosity - 0.154 Load
Constant 350.99 74.75 4.70 0.018
Viscosit -1.272 1.169 -1.09 0.356 2.6
10-7

Load -0.15390 0.08953 -1.72 0.184 2.6
S = 25.50 R-Sq = 86.2% R-Sq(adj) = 77.0%
PRESS = 12696.7 R-Sq(pred) = 10.03%
The t-tests are shown in part (a). Notice that overall regression is significant (part(b)), but neither variable
has a large t-statistic. This could be an indicator that the regressors are nearly linearly dependent.
10-7 The brake horsepower developed by an automobile engine on a dynomometer is thought to be a
function of the engine speed in revolutions per minute (rpm), the road octane number of the fuel, and the
engine compression. An experiment is run in the laboratory and the data that follow are collected.
Brake Horsepower rpm Road Octane Number Compression
225 2000 90 100
212 1800 94 95
229 2400 88 110
222 1900 91 96
219 1600 86 100
278 2500 96 110
246 3000 94 98
237 3200 90 100
233 2800 88 105
224 3400 86 97
223 1800 90 100
230 2500 89 104
(a) Fit a multiple linear regression model to the data.
Minitab Output
Regression Analysis: Horsepower versus rpm, Octane, Compression
Horsepower = - 266 + 0.0107 rpm + 3.13 Octane + 1.87 Compression
Constant -266.03 92.67 -2.87 0.021
rpm 0.010713 0.004483 2.39 0.044 1.0
Octane 3.1348 0.8444 3.71 0.006 1.0
Compress 1.8674 0.5345 3.49 0.008 1.0
S = 8.812 R-Sq = 80.7% R-Sq(adj) = 73.4%
PRESS = 2494.05 R-Sq(pred) = 22.33%
(b) Test for significance of regression. What conclusions can you draw?
Minitab Output
Source DF SS MS F P
Regression 3 2589.73 863.24 11.12 0.003
Total 11 3211.00
r No replicates. Cannot do pure error test.
Source DF Seq SS
rpm 1 509.35
Octane 1 1132.56
Compress 1 947.83
Lack of fit test
Possible interactions with variable Octane (P-Value = 0.028)
Possible lack of fit at outer X-values (P-Value = 0.000)
Overall lack of fit test is significant at P = 0.000
(c) Based on t tests, do you need all three regressor variables in the model?
10-8

Yes, all of the regressor variables are important.
10-8 Analyze the residuals from the regression model in Problem 10-7. Comment on model adequacy.
100-10
2
1
0
-1
-2
NormalScore
Residual
(response is Horsepow)
270260250240230220210
10
0
-10
Fitted Value
Residual
10-9

12108642
10
0
-10
Observation Order
Residual
The normal probability plot is satisfactory, as is the plot of residuals versus run order (assuming that
observation order is run order). The plot of residuals versus predicted response exhibits a slight “bow”
shape. This could be an indication of lack of fit. It might be useful to consider adding some ineraction
terms to the model.
10-9 The yield of a chemical process is related to the concentration of the reactant and the operating
temperature. An experiment has been conducted with the following results.
Yield Concentration Temperature
81 1.00 150
89 1.00 180
83 2.00 150
91 2.00 180
79 1.00 150
87 1.00 180
84 2.00 150
90 2.00 180
(a) Suppose we wish to fit a main effects model to this data. Set up the X’X matrix using the data exactly
as it appears in the table.
⎥
⎥
⎥
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⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
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⎣
⎡
⎥
⎥
⎥
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⎢
⎢
⎣
⎡
21960019801320
19802012
1320128
18000.21
15000.21
18000.11
15000.11
18000.21
15000.21
18000.11
15000.11
180150180150180150180150
00.200.200.100.100.200.200.100.1
11111111
(b) Is the matrix you obtained in part (a) diagonal? Discuss your response.
10-10

The X’X is not diagonal, even though an orthogonal design has been used. The reason is that we have
worked with the natural factor levels, not the orthogonally coded variables.
(c) Suppose we write our model in terms of the “usual” coded variables
5.0
5.1
1
−
=
Conc
x ,
15
165
2
−
=
Temp
x
Set up the X’X matrix for the model in terms of these coded variables. Is this matrix diagonal? Discuss
your response.
1 1 1
1 1 1
1 1 1
1 1 1 1 1 1 1 1 8 0 0
1 1 1
1 1 1 1 1 1 1 1 0 8 0
1 1 1
1 1 1 1 1 1 1 1 0 0 8
1 1 1
1 1 1
1 1 1
− −⎡ ⎤
⎢ ⎥−⎢ ⎥
⎢ ⎥−
⎡ ⎤ ⎢ ⎥
⎢ ⎥ ⎢ ⎥− − − − =⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ −⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥⎣ ⎦
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥⎣ ⎦
The X’X matrix is diagonal because we have used the orthogonally coded variables.
(d) Define a new set of coded variables
0.1
0.1
1
−
=
Conc
x ,
30
150
2
−
=
Temp
x
Set up the X’X matrix for the model in terms of this set of coded variables. Is this matrix diagonal?
Discuss your response.
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
424
244
448
111
011
101
001
111
011
101
001
10101010
11001100
11111111
The X’X is not diagonal, even though an orthogonal design has been used. The reason is that we have not
used orthogonally coded variables.
(e) Summarize what you have learned from this problem about coding the variables.
If the design is orthogonal, use the orthogonal coding. This not only makes the analysis somewhat easier,
but it also results in model coefficients that are easier to interpret because they are both dimensionless and
uncorrelated.
10-11

10-10 Consider the 24
factorial experiment in Example 6-2. Suppose that the last observation in missing.
Reanalyze the data and draw conclusions. How do these conclusions compare with those from the original
example?
The regression analysis with the one data point missing indicates that the same effects are important.
Minitab Output
Regression Analysis: Rate versus A, B, C, D, AB, AC, AD, BC, BD, CD
Rate = 69.8 + 10.5 A + 1.25 B + 4.63 C + 7.00 D - 0.25 AB - 9.38 AC + 8.00 AD
+ 0.87 BC - 0.50 BD - 0.87 CD
Constant 69.750 1.500 46.50 0.000
A 10.500 1.500 7.00 0.002 1.1
B 1.250 1.500 0.83 0.452 1.1
C 4.625 1.500 3.08 0.037 1.1
D 7.000 1.500 4.67 0.010 1.1
AB -0.250 1.500 -0.17 0.876 1.1
AC -9.375 1.500 -6.25 0.003 1.1
AD 8.000 1.500 5.33 0.006 1.1
BC 0.875 1.500 0.58 0.591 1.1
BD -0.500 1.500 -0.33 0.756 1.1
CD -0.875 1.500 -0.58 0.591 1.1
S = 5.477 R-Sq = 97.6% R-Sq(adj) = 91.6%
PRESS = 1750.00 R-Sq(pred) = 65.09%
Source DF SS MS F P
Regression 10 4893.33 489.33 16.31 0.008
Total 14 5013.33
Source DF Seq SS
A 1 1414.40
B 1 4.01
C 1 262.86
D 1 758.88
AB 1 0.06
AC 1 1500.63
AD 1 924.50
BC 1 16.07
BD 1 1.72
CD 1 10.21
10-12

50-5
2
1
0
-1
-2
NormalScore
Residual
(response is Rate)
100908070605040
5
0
-5
Fitted Value
Residual
(response is Rate)
10-13

1412108642
5
0
-5
Observation Order
Residual
(response is Rate)
The residual plots are acceptable; therefore, the underlying assumptions are valid.
10-11 Consider the 24
factorial experiment in Example 6-2. Suppose that the last two observations are
missing. Reanalyze the data and draw conclusions. How do these conclusions compare with those from the
original example?
The regression analysis with the one data point missing indicates that the same effects are important.
Minitab Output
Regression Analysis: Rate versus A, B, C, D, AB, AC, AD, BC, BD, CD
Rate = 71.4 + 10.1 A + 2.87 B + 6.25 C + 8.62 D - 0.66 AB - 9.78 AC + 7.59 AD
+ 2.50 BC + 1.12 BD + 0.75 CD
Constant 71.375 1.673 42.66 0.000
A 10.094 1.323 7.63 0.005 1.1
B 2.875 1.673 1.72 0.184 1.7
C 6.250 1.673 3.74 0.033 1.7
D 8.625 1.673 5.15 0.014 1.7
AB -0.656 1.323 -0.50 0.654 1.1
AC -9.781 1.323 -7.39 0.005 1.1
AD 7.594 1.323 5.74 0.010 1.1
BC 2.500 1.673 1.49 0.232 1.7
BD 1.125 1.673 0.67 0.549 1.7
CD 0.750 1.673 0.45 0.684 1.7
S = 4.732 R-Sq = 98.7% R-Sq(adj) = 94.2%
PRESS = 1493.06 R-Sq(pred) = 70.20%
Source DF SS MS F P
Regression 10 4943.17 494.32 22.07 0.014
Total 13 5010.36
Source DF Seq SS
A 1 1543.50
B 1 1.52
C 1 177.63
10-14

D 1 726.01
AB 1 1.17
AC 1 1702.53
AD 1 738.11
BC 1 42.19
BD 1 6.00
CD 1 4.50
3210-1-2-3
2
1
0
-1
-2
NormalScore
Residual
(response is Rate)
100908070605040
3
2
1
0
-1
-2
-3
Fitted Value
Residual
(response is Rate)
10-15

1412108642
3
2
1
0
-1
-2
-3
Observation Order
Residual
(response is Rate)
The residual plots are acceptable; therefore, the underlying assumptions are valid.
10-12 Given the following data, fit the second-order polynomial regression model
εββββββ ++++++= 2112
2
222
2
11122110 xxxxxxy
y x1 x2
26 1.0 1.0
24 1.0 1.0
175 1.5 4.0
160 1.5 4.0
163 1.5 4.0
55 0.5 2.0
62 1.5 2.0
100 0.5 3.0
26 1.0 1.5
30 0.5 1.5
70 1.0 2.5
71 0.5 2.5
After you have fit the model, test for significance of regression.
Minitab Output
Regression Analysis: y versus x1, x2, x1^2, x2^2, x1x2
y = 24.4 - 38.0 x1 + 0.7 x2 + 35.0 x1^2 + 11.1 x2^2 - 9.99 x1x2
Constant 24.41 26.59 0.92 0.394
x1 -38.03 40.45 -0.94 0.383 89.6
x2 0.72 11.69 0.06 0.953 52.1
x1^2 34.98 21.56 1.62 0.156 103.9
x2^2 11.066 3.158 3.50 0.013 104.7
x1x2 -9.986 8.742 -1.14 0.297 105.1
S = 6.042 R-Sq = 99.4% R-Sq(adj) = 98.9%
PRESS = 1327.71 R-Sq(pred) = 96.24%
r Analysis of Variance
Source DF SS MS F P
10-16

Regression 5 35092.6 7018.5 192.23 0.000
Lack of Fit 3 91.1 30.4 0.71 0.607
Pure Error 3 128.0 42.7
Total 11 35311.7
Source DF Seq SS
x1 1 11552.0
x2 1 22950.3
x1^2 1 21.9
x2^2 1 520.8
x1x2 1 47.6
1050-5
2
1
0
-1
-2
NormalScore
Residual
(response is y)
1701207020
10
5
0
-5
Fitted Value
Residual
(response is y)
10-17

12108642
10
5
0
-5
Observation Order
Residual
(response is y)
10-13
(a) Consider the quadratic regression model from Problem 10-12. Compute t statistics for each model
parameter and comment on the conclusions that follow from the quantities.
Minitab Output
Constant 24.41 26.59 0.92 0.394
x1 -38.03 40.45 -0.94 0.383 89.6
x2 0.72 11.69 0.06 0.953 52.1
x1^2 34.98 21.56 1.62 0.156 103.9
x2^2 11.066 3.158 3.50 0.013 104.7
x1x2 -9.986 8.742 -1.14 0.297 105.1
2
2x is the only model parameter that is statistically significant with a t-value of 3.50. A logical model
might also include x2 to preserve model hierarchy.
(b) Use the extra sum of squares method to evaluate the value of the quadratic terms, and to
the model.
2
2
2
1 , xx 21xx
The extra sum of squares due to isβ2
( ) ( ) ( ) ( ) ( )01021102101,02 ,,,, βββββββββββββ RRRRR SSSSSSSSSS −=−=
( )021, βββRSS sum of squares of regression for the model in Problem 10-12 = 35092.6
( )01 ββRSS =34502.3
( ) 3.5903.345026.350921,02 =−=βββRSS
( ) 3892.5
511.36
33.59031,02
0 ===
E
R
MS
SS
F
βββ
10-18

Since , then the addition of the quadratic terms to the model is significant. The P-values
indicate that it’s probably the term
76.46,3,05.0 =F
2
2x that is responsible for this.
10-14 Relationship between analysis of variance and regression. Any analysis of variance model can be
expressed in terms of the general linear model y = Xβ + ε , where the X matrix consists of zeros and ones.
Show that the single-factor model ijiijy ετµ ++= , i=1,2,3, j=1,2,3,4 can be written in general linear
model form. Then
(a) Write the normal equations ˆ( )′X X X′yβ = and compare them with the normal equations found for
the model in Chapter 3.
The normal equations are ˆ( )′ ′X X X yβ =
⎥
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=
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⎣
⎡
.3
.2
.1
..
3
2
1
ˆ
ˆ
ˆ
ˆ
4004
0404
0044
44412
y
y
y
y
τ
τ
τ
µ
which are in agreement with the results of Chapter 3.
(b) Find the rank of . Can′X X 1
( )−
′X X be obtained?
′X X is a 4 x 4 matrix of rank 3, because the last three columns add to the first column. Thus (X’X)-1
does
not exist.
(c) Suppose the first normal equation is deleted and the restriction is added. Can the
resulting system of equations be solved? If so, find the solution. Find the regression sum of squares
∑=
=
3
1
0ˆ
i inτ
ˆ′ ′X yβ , and compare it to the treatment sum of squares in the single-factor model.
Imposing yields the normal equations∑=
=
3
1
0ˆ
i inτ
⎥
⎥
⎥
⎥
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⎤
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⎢
⎢
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⎡
=
⎥
⎥
⎥
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⎡
.3
.2
.1
..
3
2
1
ˆ
ˆ
ˆ
ˆ
4004
0404
0044
4440
y
y
y
y
τ
τ
τ
µ
The solution to this set of equations is
..
..
12
ˆ y
y
==µ
...ˆ yyii −=τ
This solution was found be solving the last three equations for , yieldingiτˆ µτ ˆˆ . −= ii y , and then
substituting in the first equation to find ..ˆ y=µ
10-19

The regression sum of squares is
( ) ββ ˆ′=RSS X’y = ( ) ∑∑∑
===
=−+=−+
a
i
i
a
i
i
a
i
i
n
y
an
y
n
y
an
y
yyyy
1
2
.
1
2
..
2
.
2
..
1
2
.......
with a degrees of freedom. This is the same result found in Chapter 3. For more discussion of the
relationship between analysis of variance and regression, see Montgomery and Peck (1992).
10-15 Suppose that we are fitting a straight line and we desire to make the variance of as small as possible.
Restricting ourselves to an even number of experimental points, where should we place these points so as
to minimize ( )1
ˆβV ? (Note: Use the design called for in this exercise with great caution because, even
though it minimized ( )1
ˆβV , it has some undesirable properties; for example, see Myers and Montgomery
(1995). Only if you are very sure the true functional relationship is linear should you consider using this
design.
Since ( )
xxS
V
2
1
ˆ σ
β = , we may minimize ( )1
ˆβV by making Sxx as large as possible. Sxx is maximized by
spreading out the xj’s as much as possible. The experimenter usually has a “region of interest” for x. If n is
even, n/2 of the observations should be run at each end of the “region of interest”. If n is odd, then run one
of the observations in the center of the region and the remaining (n-1)/2 at either end.
10-16 Weighted least squares. Suppose that we are fitting the straight line , but the
variance of the y’s now depends on the level of x; that is,
εββ ++= xy 10
( ) ni
w
xyV
i
i ,...,2,1,
2
2
===
σ
σ
where the wi are known constants, often called weights. Show that if we choose estimates of the regression
coefficients to minimize the weighted sum of squared errors given by , the
resulting least squares normal equations are
( )∑
=
+−
n
i
iii xyw
1
2
10 ββ
∑ ∑∑
= ==
=+
n
i
n
i
iii
n
i
ii ywxww
1 11
10
ˆˆ ββ
∑ ∑∑
= ==
=+
n
i
n
i
iiii
n
i
iii yxwxwxw
1 1
2
1
10
ˆˆ ββ
The least squares normal equations are found:
10-20

( )
( )
( ) 0ˆˆ2
0ˆˆ2
1
1110
1
1
110
0
1
2
110
=−−−=
=−−−=
−−=
∑
∑
∑
=
=
=
n
i
ii
n
i
ii
n
i
ii
wxxy
L
wxy
L
wxyL
ββ
∂β
∂
ββ
∂β
∂
ββ
which simplify to
i
n
i
i
n
i
n
i
ii
i
n
i
i
n
i
n
i
ii
yxwwxwx
ywwxw
∑∑ ∑
∑∑ ∑
== =
== =
=+
=+
1
1
1 1
2
1110
11 1
110
ˆˆ
ˆˆ
ββ
ββ
10-17 Consider the design discussed in Example 10-5.14
2 −
IV
(a) Suppose you elect to augment the design with the single run selected in that example. Find the
variances and covariances of the regression coefficients in the model (ignoring blocks):
εβββββββ +++++++= 43342112443322110 xxxxxxxxy
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−
−−−
−−−−
−−−
−−−
−−−
−−−−
=
⎥
⎥
⎥
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⎥
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⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−−−
−−−−
−−−−
−−
−−
−−−−
−−−−
−−−−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
−−−−−
−−−−
−−−−
−−−−−
−−−−−
−−−−−
9711111
7911111
1191111
1119111
1111911
1111191
1111119
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
111111111
111111111
111111111
111111111
111111111
111111111
111111111
XX'
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−−−
−−
−
−
−
−
−
=−
4375.0375.00625.00625.00625.00625.00625.0
375.04375.00625.00625.00625.00625.00625.0
0625.00625.0125.00000
0625.00625.00125.0000
0625.00625.000125.000
0625.00625.0000125.00
0625.00625.00000125.0
)( 1
XX'
(b) Are there any other runs in the alternate fraction that would de-alias AB from CD?
Any other run from the alternate fraction will de-alias AB from CD.
(c) Suppose you augment the design with four runs suggested in Example 10-5. Find the variance and the
covariances of the regression coefficients (ignoring blocks) for the model in part (a).
10-21

Choose 4 runs that are one of the quarter fractions not used in the principal half fraction.
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−−−
−−
−−−−
−−−−
−−−−
−−
−−
−−−−
−−−−
−−−−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−−−−−
−−−−−−
−−−−−−
−−−−−−
−−−−−−
−−−−−−
=
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
1111111
111111111111
111111111111
111111111111
111111111111
111111111111
111111111111
111111111111
XX'
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
12000000
01240040
04120040
00012400
00041200
04400120
00000012
XX'
( )
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−−
−−
=−
1785.01429.00357.0000357.00
1429.02142.00536.0000536.00
0357.00536.01071.0000179.00
0000938.00313.000
0000313.00938.000
0357.00536.00179.0001071.00
0000000833.0
1
XX'
(d) Considering parts (a) and (c), which augmentation strategy would you prefer and why?
If you only have the resources to run one more run, then choose the one-run augmentation. But if
resources are not scarce, then augment the design in multiples of two runs, to keep the design orthogonal.
Using four runs results in smaller variances of the regression coefficients and a simpler covariance
structure.
10-18 Consider the . Suppose after running the experiment, the largest observed effects are A + BD,
B + AD, and D + AB. You wish to augment the original design with a group of four runs to de-alias these
effects.
47
2 −
III
(a) Which four runs would you make?
Take the first four runs of the original experiment and change the sign on A.
10-22

Factor 1 Factor 2 Factor 3 Factor 4 Factor 5 Factor 6 Factor 7
Std Run Block A:x1 B:x2 C:x3 D:x4 E:x5 F:x6 G:x7
1 1 Block 1 -1.00 -1.00 -1.00 1.00 1.00 1.00 -1.00
2 2 Block 1 1.00 -1.00 -1.00 -1.00 -1.00 1.00 1.00
3 3 Block 1 -1.00 1.00 -1.00 -1.00 1.00 -1.00 1.00
4 4 Block 1 1.00 1.00 -1.00 1.00 -1.00 -1.00 -1.00
5 5 Block 1 -1.00 -1.00 1.00 1.00 -1.00 -1.00 1.00
6 6 Block 1 1.00 -1.00 1.00 -1.00 1.00 -1.00 -1.00
7 7 Block 1 -1.00 1.00 1.00 -1.00 -1.00 1.00 -1.00
8 8 Block 1 1.00 1.00 1.00 1.00 1.00 1.00 1.00
9 9 Block 2 1.00 1.00 1.00 -1.00 -1.00 -1.00 -1.00
10 10 Block 2 1.00 -1.00 -1.00 1.00 -1.00 -1.00 -1.00
11 11 Block 2 -1.00 -1.00 1.00 1.00 -1.00 -1.00 -1.00
12 12 Block 2 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00
Main effects and interactions of interest are:
x1 x2 x4 x1x2 x1x4 x2x4
-1 -1 1 1 -1 -1
1 -1 -1 -1 -1 1
-1 1 -1 -1 1 -1
1 1 1 1 1 1
-1 -1 1 1 -1 -1
1 -1 -1 -1 -1 1
-1 1 -1 -1 1 -1
1 1 1 1 1 1
1 -1 1 -1 1 -1
-1 -1 -1 1 1 1
1 1 -1 1 -1 -1
-1 1 1 -1 -1 1
(b) Find the variances and covariances of the regression coefficients in the model
εβββββββ +++++++= 4224411421124422110 xxxxxxxxxy
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
12000040
01200400
00124000
00412000
04001200
40000120
00000012
XX'
( )
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−
−−
−−
−
=−
2143.01607.0000536.00714.00
1607.02143.0000714.00536.00
000938.00313.0000
000313.00938.0000
0536.00714.0001071.00179.00
0714.00536.0000178.01071.00
0000000833.0
1
XX'
10-23

(c) Is it possible to de-alias these effects with fewer than four additional runs?
is possible to de-alias these effects in only two runs. By utilizing Design Expert’s design augmentation
esign Expert Output
Factor 1 Factor 2 Factor 3 Factor 4 Factor 5 Factor 6 Factor 7
It
function, the runs 9 and 10 (Block 2) were generated as follows:
D
Std A:x1 B:x2 C:x3 D:x4 E:x5 F:x6 G:x7Run Block
1 1 Block 1 -1.00 -1.00 -1.00 1.00 1.00 1.00 -1.00
2 2 Block 1 1.00 -1.00 -1.00 -1.00 -1.00 1.00 1.00
3 3 Block 1 -1.00 1.00 -1.00 -1.00 1.00 -1.00 1.00
4 4 Block 1 1.00 1.00 -1.00 1.00 -1.00 -1.00 -1.00
5 5 Block 1 -1.00 -1.00 1.00 1.00 -1.00 -1.00 1.00
6 6 Block 1 1.00 -1.00 1.00 -1.00 1.00 -1.00 -1.00
7 7 Block 1 -1.00 1.00 1.00 -1.00 -1.00 1.00 -1.00
8 8 Block 1 1.00 1.00 1.00 1.00 1.00 1.00 1.00
9 9 Block 2 -1.00 1.00 -1.00 1.00 -1.00 -1.00 -1.00
10 0 Block 2 1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.001
10-24

Chapter 11
Response Surface Methods and Designs
Solutions
11-1 A chemical plant produces oxygen by liquefying air and separating it into its component gases by
fractional distillation. The purity of the oxygen is a function of the main condenser temperature and the
pressure ratio between the upper and lower columns. Current operating conditions are temperature =)( 1ξ
-220°C and pressure ratio =)( 2ξ 1.2. Using the following data find the path of steepest ascent.
Temperature (x1) Pressure Ratio (x2) Purity
-225 1.1 82.8
-225 1.3 83.5
-215 1.1 84.7
-215 1.3 85.0
-220 1.2 84.1
-220 1.2 84.5
-220 1.2 83.9
-220 1.2 84.3
Response: Purity
Sum of Mean F
A 2.89 1 2.89 48.17 0.0023
B 0.25 1 0.25 4.17 0.1108
Curvature 0.080 1 0.080 1.33 0.3125 not significant
Residual 0.24 4 0.060
Pure Error 0.20 3 0.067
Cor Total 3.46 7
Intercept 84.00 1 0.12 83.66 84.34
A-Temperature 0.85 1 0.12 0.51 1.19 1.00
B-Pressure Ratio 0.25 1 0.12 -0.090 0.59 1.00
Center Point 0.20 1 0.17 -0.28 0.68 1.00
Purity =
+84.00
+0.85 * A
+0.25 * B
Purity =
+118.40000
11-1

+2.50000 * Pressure Ratio
From the computer output use the model 21 25085084 x.x.yˆ ++= as the equation for steepest ascent.
Suppose we use a one degree change in temperature as the basic step size. Thus, the path of steepest ascent
passes through the point (x1=0, x2=0) and has a slope 0.25/0.85. In the coded variables, one degree of
temperature is equivalent to a step of =1x∆ 1/5=0.2. Thus, =2x∆ (0.25/0.85)0.2=0.059. The path of
steepest ascent is:
Coded Variables Natural Variables
x1 x2 1ξ 2ξ
Origin 0 0 -220 1.2
∆ 0.2 0.059 1 0.0059
Origin + ∆ 0.2 0.059 -219 1.2059
Origin +5 ∆ 1.0 0.295 -215 1.2295
Origin +7 ∆ 1.40 0.413 -213 1.2413
11-2 An industrial engineer has developed a computer simulation model of a two-item inventory system.
The decision variables are the order quantity and the reorder point for each item. The response to be
minimized is the total inventory cost. The simulation model is used to produce the data shown in the
following table. Identify the experimental design. Find the path of steepest descent.
Item 1 Item 2
Order Reorder Order Reorder Total
Quantity (x1) Point (x2) Quantity (x3) Point (x4) Cost
100 25 250 40 625
140 45 250 40 670
140 25 300 40 663
140 25 250 80 654
100 45 300 40 648
100 45 250 80 634
100 25 300 80 692
140 45 300 80 686
120 35 275 60 680
120 35 275 60 674
120 35 275 60 681
The design is a 24-1
fractional factorial with generator I=ABCD, and three center points.
Response: Total Cost
Sum of Mean F
A 684.50 1 684.50 66.96 0.0038
C 1404.50 1 1404.50 137.40 0.0013
D 450.00 1 450.00 44.02 0.0070
AC 392.00 1 392.00 38.35 0.0085
AD 264.50 1 264.50 25.88 0.0147
CD 684.50 1 684.50 66.96 0.0038
Curvature 815.52 1 815.52 79.78 0.0030 significant
Residual 30.67 3 10.22
Pure Error 28.67 2 14.33
11-2

Intercept 659.00 1 1.13 655.40 662.60
A-Item 1 QTY 9.25 1 1.13 5.65 12.85 1.00
C-Item 2 QTY 13.25 1 1.13 9.65 16.85 1.00
D-Item 2 Reorder 7.50 1 1.13 3.90 11.10 1.00
AC -7.00 1 1.13 -10.60 -3.40 1.00
AD -5.75 1 1.13 -9.35 -2.15 1.00
CD 9.25 1 1.13 5.65 12.85 1.00
Center Point 19.33 1 2.16 12.44 26.22 1.00
Total Cost =
+659.00
+9.25 * A
+13.25 * C
+7.50 * D
-7.00 * A * C
-5.75 * A * D
+9.25 * C * D
Total Cost =
+175.00000
+5.17500 * Item 1 QTY
+1.10000 * Item 2 QTY
-2.98750 * Item 2 Reorder
-0.014000 * Item 1 QTY * Item 2 QTY
-0.014375 * Item 1 QTY * Item 2 Reorder
+0.018500 * Item 2 QTY * Item 2 Reorder
+0.019 * Item 2 QTY * Item 2 Reorder
The equation used to compute the path of steepest ascent is 431 5072513259659 x.x.x.yˆ +++= . Notice
that even though the model contains interaction, it is relatively common practice to ignore the interactions
in computing the path of steepest ascent. This means that the path constructed is only an approximation to
the path that would have been obtained if the interactions were considered, but it’s usually close enough to
give satisfactory results.
It is helpful to give a general method for finding the path of steepest ascent. Suppose we have a first-order
model in k variables, say
∑=
+=
k
i
ii xˆˆyˆ
1
0 ββ
The path of steepest ascent passes through the origin, x=0, and through the point on a hypersphere of
radius, R where is a maximum. Thus, the x’s must satisfy the constraintyˆ
∑=
=
k
i
i Rx
1
22
To find the set of x’s that maximize subject to this constraint, we maximizeyˆ
11-3

⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−−+= ∑∑ ==
k
i
i
k
i
ii RxxˆˆL
1
22
1
0 λββ
where λ is a LaGrange multiplier. From 0== ∂λ∂∂∂ /Lx/L i , we find
λ
β
2
i
i
ˆ
x =
It is customary to specify a basic step size in one of the variables, say ∆ xj, and then calculate 2 λ as
2 λ = jj x/ˆ ∆β . Then this value of 2 λ can be used to generate the remaining coordinates of a point on the
path of steepest ascent.
We demonstrate using the data from this problem. Suppose that we use -10 units in 1ξ as the basic step
size. Note that a decrease in 1ξ is called for, because we are looking for a path of steepest decent. Now
-10 units in 1ξ is equal to -10/20 = -0.5 units change in x1.
Thus, 2 λ = 11 x/ˆ ∆β = 9.25/(-0.5) = -18.50
Consequently,
7160
5018
2513
2
3
3 .
.
.ˆ
x −=
−
==
λ
β
∆
7050
5018
507
2
4
4 .
.
.ˆ
x −=
−
==
λ
β
∆
are the remaining coordinates of points along the path of steepest decent, in terms of the coded variables.
The path of steepest decent is shown below:
x1 x2 x3 x4 1ξ 2ξ 3ξ 4ξ
Origin 0 0 0 0 120 35 275 60
∆ -0.50 0 -0.716 -0.405 -10 0 -17.91 -8.11
Origin + ∆ -0.50 0 -0.716 -0.405 110 35 257.09 51.89
Origin +2 ∆ -1.00 0 -1.432 -0.810 100 35 239.18 43.78
11-3 Verify that the following design is a simplex. Fit the first-order model and find the path of steepest
ascent.
Position x1 x2 x3 y
1 0 2 -1 18.5
2 - 2 0 1 19.8
3 0 - 2 -1 17.4
4 2 0 1 22.5
11-4

1
2
3
4
x2
x1
x3
The graphical representation of the design identifies a tetrahedron; therefore, the design is a simplex.
Response: y
Sum of Mean F
Model 14.49 3 4.83
A 3.64 1 3.64
B 0.61 1 0.61
C 10.24 1 10.24
Pure Error 0.000 0
Cor Total 14.49 3
Std. Dev. R-Squared 1.0000
Mean 19.55 Adj R-Squared
C.V. Pred R-Squared N/A
PRESS N/A Adeq Precision 0.000
Case(s) with leverage of 1.0000: Pred R-Squared and PRESS statistic not defined
Intercept 19.55 1
A-x1 1.35 1 1.00
B-x2 0.55 1 1.00
C-x3 1.60 1 1.00
y =
+19.55
+1.35 * A
+0.55 * B
+1.60 * C
y =
+19.55000
+0.95459 * x1
+0.38891 * x2
+1.60000 * x3
11-5

The first order model is 321 6015503515519 x.x.x..yˆ +++= .
To find the path of steepest ascent, let the basic step size be 13 =x∆ . Then using the results obtained in the
previous problem, we obtain
λ
β
∆
2
3
3
ˆ
x = or 1.0 =
λ2
601.
which yields 6012 .=λ . Then the coordinates of points on the path of steepest ascent are defined by
600
601
960
2
1
1 .
.
.ˆ
x ===
λ
β
∆
240
601
240
2
2
2 .
.
.ˆ
x ===
λ
β
∆
Therefore, in the coded variables we have:
Coded Variables
x1 x2 x3
Origin 0 0 0
∆ 0.60 0.24 1.00
Origin + ∆ 0.60 0.24 1.00
Origin +2 ∆ 1.20 0.48 2.00
11-4 For the first-order model 321 02805160 x.x.x.yˆ +−+= find the path of steepest ascent. The
variables are coded as .11 ≤≤− ix
Let the basic step size be 13 =x∆ .
λ
β
∆
2
3
3
ˆ
x = or 1.0 =
λ2
02.
. Then 022 .=λ
750
02
501
2
1
1 .
.
.ˆ
x ===
λ
β
∆
400
02
80
2
2
2 .
.
.ˆ
x −=
−
==
λ
β
∆
Therefore, in the coded variables we have
Coded Variables
x1 x2 x3
Origin 0 0 0
∆ 0.75 -0.40 1.00
Origin + ∆ 0.75 -0.40 1.00
Origin +2 ∆ 1.50 -0.80 2.00
11-5 The region of experimentation for three factors are time ( 8040 1 ≤≤ T min), temperature
( °C), and pressure (300200 2 ≤≤ T 5020 ≤≤ P psig). A first-order model in coded variables has been fit
to yield data from a 23
design. The model is
11-6

321 5352530 x.x.xyˆ +++=
Is the point T1 = 85, T2 = 325, P=60 on the path of steepest ascent?
The coded variables are found with the following:
20
601
1
−
=
T
x
50
2502
2
−
=
T
x
15
351
3
−P
x
51 =T∆ 250
20
5
1 .x ==∆
λ
β
∆
2
1
1
ˆ
x = or 0.25 =
λ2
20
202 =λ
1250
20
52
2
2
2 .
.ˆ
x ===
λ
β
∆
1750
20
53
2
3
3 .
.ˆ
x ===
λ
β
∆
x1 x2 x3 T1 T2 P
Origin 0 0 0 60 250 35
∆ 0.25 0.125 0.175 5 6.25 2.625
Origin + ∆ 0.25 0.125 0.175 65 256.25 37.625
Origin +5 ∆ 1.25 0.625 0.875 85 281.25 48.125
The point T1=85, T2=325, and P=60 is not on the path of steepest ascent.
11-6 The region of experimentation for two factors are temperature ( °≤≤ 300100 T F) and catalyst feed
rate ( lb/h). A first order model in the usual3010 ≤≤ C ± 1 coded variables has been fit to a molecular
weight response, yielding the following model.
21 401252000 xxyˆ ++=
(a) Find the path of steepest ascent.
100
200
1
−
=
T
x
10
20
2
−
=
C
x
100=T∆ 1
100
100
1 ==x∆
λ
β
∆
2
1
1
ˆ
x = or
λ2
125
1 = 1252 =λ
320
125
40
2
2
2 .
ˆ
x ===
λ
β
∆
x1 x2 T C
Origin 0 0 200 20
∆ 1 0.32 100 3.2
11-7

Origin + ∆ 1 0.32 300 23.2
Origin +5 ∆ 5 1.60 700 36.0
(a) It is desired to move to a region where molecular weights are above 2500. Based on the information
you have from the experiment, in this region, about how may steps along the path of steepest ascent
might be required to move to the region of interest?
( )( ) ( )( ) 81374032012512211 ..ˆxˆxyˆ =+=+= β∆β∆∆
4633
8137
20002500
→=
−
= .
.
Steps#
11-7 The path of steepest ascent is usually computed assuming that the model is truly first-order.; that is,
there is no interaction. However, even if there is interaction, steepest ascent ignoring the interaction still
usually produces good results. To illustrate, suppose that we have fit the model
2121 38520 xxxxyˆ +−+=
using coded variables (-1 ≤ x1 ≤ +1)
(a) Draw the path of steepest ascent that you would obtain if the interaction were ignored.
Path of Steepest Ascent for
Main Effects Model
-5
-4
-3
-2
-1
0
0 1 2 3 4 5
X1
X2
(b) Draw the path of steepest ascent that you would obtain with the interaction included in the model.
Compare this with the path found in part (a).
11-8

Path of Steepest Ascent for
Full Model
-5
-4
-3
-2
-1
0
-2 -1 0 1 2 3
X1
X2
11-8 The data shown in the following table were collected in an experiment to optimize crystal growth as
a function of three variables x1, x2, and x3. Large values of y (yield in grams) are desirable. Fit a second
order model and analyze the fitted surface. Under what set of conditions is maximum growth achieved?
x1 x2 x3 y
-1 -1 -1 66
-1 -1 1 70
-1 1 -1 78
-1 1 1 60
1 -1 -1 80
1 -1 1 70
1 1 -1 100
1 1 1 75
-1.682 0 0 100
1.682 0 0 80
0 -1.682 0 68
0 1.682 0 63
0 0 -1.682 65
0 0 1.682 82
0 0 0 113
0 0 0 100
0 0 0 118
0 0 0 88
0 0 0 100
0 0 0 85
Response: Yield
ANOVA for Response Surface Quadratic Model
Sum of Mean F
A 22.08 1 22.08 0.12 0.7377
B 25.31 1 25.31 0.14 0.7200
C 30.50 1 30.50 0.16 0.6941
A2 204.55 1 204.55 1.10 0.3191
11-9

B2 2226.45 1 2226.45 11.96 0.0061
C2 1328.46 1 1328.46 7.14 0.0234
AB 66.12 1 66.12 0.36 0.5644
AC 55.13 1 55.13 0.30 0.5982
BC 171.13 1 171.13 0.92 0.3602
Residual 1860.95 10 186.09
Pure Error 859.33 5 171.87
Intercept 100.67 1 5.56 88.27 113.06
A-x1 1.27 1 3.69 -6.95 9.50 1.00
B-x2 1.36 1 3.69 -6.86 9.59 1.00
C-x3 -1.49 1 3.69 -9.72 6.73 1.00
A2 -3.77 1 3.59 -11.77 4.24 1.02
B2 -12.43 1 3.59 -20.44 -4.42 1.02
C2 -9.60 1 3.59 -17.61 -1.59 1.02
AB 2.87 1 4.82 -7.87 13.62 1.00
AC -2.63 1 4.82 -13.37 8.12 1.00
BC -4.63 1 4.82 -15.37 6.12 1.00
Yield =
+100.67
+1.27 * A
+1.36 * B
-1.49 * C
-3.77 * A2
-12.43 * B2
-9.60 * C2
+2.87 * A * B
-2.63 * A * C
-4.63 * B * C
Yield =
+100.66609
+1.27146 * x1
+1.36130 * x2
-1.49445 * x3
-3.76749 * x12
-12.42955 * x22
-9.60113 * x32
+2.87500 * x1 * x2
-2.62500 * x1 * x3
-4.62500 * x2 * x3
There are so many nonsignificant terms in this model that we should consider eliminating some of them. A
reasonable reduced model is shown below.
11-10

Response: Yield
ANOVA for Response Surface Reduced Quadratic Model
Sum of Mean F
B 25.31 1 25.31 0.16 0.6952
C 30.50 1 30.50 0.19 0.6673
B2 2115.31 1 2115.31 13.33 0.0024
C2 1239.17 1 1239.17 7.81 0.0136
Residual 2379.95 15 158.66
Pure Error 859.33 5 171.87
Intercept 97.58 1 4.36 88.29 106.88
B-x2 1.36 1 3.41 -5.90 8.63 1.00
C-x3 -1.49 1 3.41 -8.76 5.77 1.00
B2 -12.06 1 3.30 -19.09 -5.02 1.01
C2 -9.23 1 3.30 -16.26 -2.19 1.01
Yield =
+97.58
+1.36 * B
-1.49 * C
-12.06 * B2
-9.23 * C2
Yield =
+97.58260
+1.36130 * x2
-1.49445 * x3
-12.05546 * x22
-9.22703 * x32
The contour plot identifies a maximum near the center of the design space.
11-11

DESIGN-EXPERT Plot
Yield
X = B: x2
Y = C: x3
Design Points
Actual Factor
A: x1 = 0.00
Yield
B: x2
C:x3
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
80
80
80
85
85
85
90
95
6
Prediction97.682
95% Low 69.273
95% H igh126.090
SE m ean 4.35584
SE pred 13.3281
X 0.06
Y -0.08
11-9 The following data were collected by a chemical engineer. The response y is filtration time, x1 is
temperature, and x2 is pressure. Fit a second-order model.
x1 x2 y
-1 -1 54
-1 1 45
1 -1 32
1 1 47
-1.414 0 50
1.414 0 53
0 -1.414 47
0 1.414 51
0 0 41
0 0 39
0 0 44
0 0 42
0 0 40
Response: y
Sum of Mean F
A 13.11 1 13.11 0.51 0.4955
B 25.72 1 25.72 1.00 0.3467
A2 81.39 1 81.39 3.16 0.1132
AB 144.00 1 144.00 5.60 0.0455
Residual 205.78 8 25.72
Pure Error 14.80 4 3.70
Cor Total 470.00 12
11-12

Intercept 42.91 1 1.83 38.69 47.14
A-Temperature 1.28 1 1.79 -2.85 5.42 1.00
B-Pressure -1.79 1 1.79 -5.93 2.34 1.00
A2 3.39 1 1.91 -1.01 7.79 1.00
AB 6.00 1 2.54 0.15 11.85 1.00
Time =
+42.91
+1.28 * A
-1.79 * B
+3.39 * A2
+6.00 * A * B
Time =
+42.91304
-1.79289 * Pressure
+3.39130 * Temperature2
+6.00000 * Temperature * Pressure
The lack of fit test in the above analysis is significant. Also, the residual plot below identifies an outlier
which happens to be standard order number 8.
Residual
Normal%probability
-5.23112 -1.26772 2.69568 6.65909 10.6225
1
5
10
20
30
50
70
80
90
95
99
We chose to remove this run and re-analyze the data.
Response: y
Sum of Mean F
A 13.11 1 13.11 3.88 0.0895
11-13

B 132.63 1 132.63 39.25 0.0004
A2 155.27 1 155.27 45.95 0.0003
AB 144.00 1 144.00 42.61 0.0003
Residual 23.66 7 3.38
Pure Error 14.80 4 3.70
Cor Total 431.00 11
Intercept 40.68 1 0.73 38.95 42.40
A-Temperature 1.28 1 0.65 -0.26 2.82 1.00
B-Pressure -4.82 1 0.77 -6.64 -3.00 1.02
A2 4.88 1 0.72 3.18 6.59 1.02
AB 6.00 1 0.92 3.83 8.17 1.00
Time =
+40.68
+1.28 * A
-4.82 * B
+4.88 * A2
+6.00 * A * B
Time =
+40.67673
-4.82374 * Pressure
+6.00000 * Temperature * Pressure
The lack of fit test is satisfactory as well as the following normal plot of residuals:
11-14

Residual
Normal%probability
-1.67673 -0.42673 0.82327 2.07327 3.32327
1
5
10
20
30
50
70
80
90
95
99
(a) What operating conditions would you recommend if the objective is to minimize the filtration time?
Time
A: Tem perature
B:Pressure
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
34
36
38
40
42
44
46
46
48
50
52
5
Prediction 33.195
95% Low 27.885
95% H igh 38.506
SE m ean 1.29007
SE pred 2.24581
X -0.68
Y 1.00
(b) What operating conditions would you recommend if the objective is to operate the process at a mean
filtration time very close to 46?
11-15

Time
A: Tem perature
B:Pressure
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
34
36
38
40
42
44
46
46
48
50
52
5
There are two regions that enable a filtration time of 46. Either will suffice; however, higher temperatures
and pressures typically have higher operating costs. We chose the operating conditions at the lower
pressure and temperature as shown.
11-10 The hexagon design that follows is used in an experiment that has the objective of fitting a second-
order model.
x1 x2 y
1 0 68
0.5 0 75. 74
-0.5 0 75. 65
-1 0 60
-0.5 - 0 75. 63
0.5 - 0 75. 70
0 0 58
0 0 60
0 0 57
0 0 55
0 0 69
(a) Fit the second-order model.
Response: y
Sum of Mean F
A 85.33 1 85.33 3.30 0.1292
B 9.00 1 9.00 0.35 0.5811
A2 25.20 1 25.20 0.97 0.3692
B2 129.83 1 129.83 5.01 0.0753
AB 1.00 1 1.00 0.039 0.8519
Residual 129.47 5 25.89
11-16

Pure Error 118.80 4 29.70
Cor Total 374.73 10
Intercept 59.80 1 2.28 53.95 65.65
A-x1 5.33 1 2.94 -2.22 12.89 1.00
B-x2 1.73 1 2.94 -5.82 9.28 1.00
A2 4.20 1 4.26 -6.74 15.14 1.00
B2 9.53 1 4.26 -1.41 20.48 1.00
AB 1.15 1 5.88 -13.95 16.26 1.00
y =
+59.80
+5.33 * A
+1.73 * B
+4.20 * A2
+9.53 * B2
+1.15 * A * B
(a) Perform the canonical analysis. What type of surface has been found?
The full quadratic model is used in the following analysis because the reduced model is singular.
Solution
Variable Critical Value
X1 -0.627658
X2 -0.052829
Predicted Value at Solution 58.080492
Eigenvalues and Eigenvectors
Variable 9.5957 4.1382
X1 0.10640 0.99432
X2 0.99432 -0.10640
Since both eigenvalues are positive, the response is a minimum at the stationary point.
(c) What operating conditions on x1 and x2 lead to the stationary point?
The stationary point is (x1,x2) = (-0.62766, -0.05283)
(d) Where would you run this process if the objective is to obtain a response that is as close to 65 as
possible?
11-17

y
A: x1
B:x2
-1.00 -0.50 0.00 0.50 1.00
-0.87
-0.43
0.00
0.43
0.87
60
65
70
70
75
5
Any value of x1 and x2 that give a point on the contour with value of 65 would be satisfactory.
11-11 An experimenter has run a Box-Behnken design and has obtained the results below, where the
response variable is the viscosity of a polymer.
Level Temp.
Agitation
Rate Pressure x1 x2 x3
High 200 10.0 25 +1 +1 +1
Middle 175 7.5 20 0 0 0
Low 150 5.0 15 -1 -1 -1
Run x1 x2 x3 y1
1 -1 -1 0 535
2 1 -1 0 580
3 -1 1 0 596
4 1 1 0 563
5 -1 0 -1 645
6 1 0 -1 458
7 -1 0 1 350
8 1 0 1 600
9 0 -1 -1 595
10 0 1 -1 648
11 0 -1 1 532
12 0 1 1 656
13 0 0 0 653
14 0 0 0 599
15 0 0 0 620
(a) Fit the second-order model.
Response: Viscosity
Sum of Mean F
11-18

A 703.12 1 703.12 0.67 0.4491
B 6105.12 1 6105.12 5.85 0.0602
C 5408.00 1 5408.00 5.18 0.0719
A2 20769.23 1 20769.23 19.90 0.0066
B2 1404.00 1 1404.00 1.35 0.2985
C2 4719.00 1 4719.00 4.52 0.0868
AB 1521.00 1 1521.00 1.46 0.2814
AC 47742.25 1 47742.25 45.74 0.0011
BC 1260.25 1 1260.25 1.21 0.3219
Residual 5218.75 5 1043.75
Pure Error 1482.00 2 741.00
Cor Total 94871.33 14
Intercept 624.00 1 18.65 576.05 671.95
A-Temp 9.37 1 11.42 -19.99 38.74 1.00
B-Agitation Rate 27.62 1 11.42 -1.74 56.99 1.00
C-Pressure -26.00 1 11.42 -55.36 3.36 1.00
A2 -75.00 1 16.81 -118.22 -31.78 1.01
B2 19.50 1 16.81 -23.72 62.72 1.01
C2 -35.75 1 16.81 -78.97 7.47 1.01
AB -19.50 1 16.15 -61.02 22.02 1.00
AC 109.25 1 16.15 67.73 150.77 1.00
BC 17.75 1 16.15 -23.77 59.27 1.00
Viscosity =
+624.00
+9.37 * A
+27.62 * B
-26.00 * C
-75.00 * A2
+19.50 * B2
-35.75 * C2
-19.50 * A * B
+109.25 * A * C
+17.75 * B * C
Viscosity =
-629.50000
+27.23500 * Temp
-9.55000 * Agitation Rate
-111.60000 * Pressure
-0.12000 * Temp2
+3.12000 * Agitation Rate2
-1.43000 * Pressure2
-0.31200 * Temp * Agitation Rate
+0.87400 * Temp * Pressure
+1.42000 * Agitation Rate * Pressure
11-19

(b) Perform the canonical analysis. What type of surface has been found?
Solution
Variable Critical Value
X1 2.1849596
X2 -0.871371
X3 2.7586015
Predicted Value at Solution 586.34437
Eigevalues and Eigevectors
Variable 20.9229 2.5208 -114.694
X1 -0.02739 0.58118 0.81331
X2 0.99129 -0.08907 0.09703
X3 0.12883 0.80888 -0.57368
The system is a saddle point.
(c) What operating conditions on x1, x2, and x3 lead to the stationary point?
The stationary point is (x1, x2, x3) = (2.18496, -0.87167, 2.75860). This is outside the design region. It
would be necessary to either examine contour plots or use numerical optimization methods to find desired
operating conditions.
(d) What operating conditions would you recommend if it is important to obtain a viscosity that is as close
to 600 as possible?
DESIGN-EXPERT Plot
Viscosity
X = A: T em peratue
Y = C: Pressure
Design Points
Actual Factor
B: Agitation Rate = 7.50
Viscosity
C:Pressure
150.00 162.50 175.00 187.50 200.00
15.00
17.50
20.00
22.50
25.00
400
450
500
500
550
550
600
600
3
A: Temp
Any point on either of the contours showing a viscosity of 600 is satisfactory.
11-12 Consider the three-variable central composite design shown below. Analyze the data and draw
conclusions, assuming that we wish to maximize conversion (y1) with activity (y2) between 55 and 60.
Run
Time
(min)
Temperature
(°C)
Catalyst
(%)
Conversion (%)
y1
Activity
y2
1 -1.000 -1.000 -1.000 74.00 53.20
11-20

2 1.000 -1.000 -1.000 51.00 62.90
3 -1.000 1.000 -1.000 88.00 53.40
4 1.000 1.000 -1.000 70.00 62.60
5 -1.000 -1.000 1.000 71.00 57.30
6 1.000 -1.000 1.000 90.00 67.90
7 -1.000 1.000 1.000 66.00 59.80
8 1.000 1.000 1.000 97.00 67.80
9 0.000 0.000 0.000 81.00 59.20
10 0.000 0.000 0.000 75.00 60.40
11 0.000 0.000 0.000 76.00 59.10
12 0.000 0.000 0.000 83.00 60.60
13 -1.682 0.000 0.000 76.00 59.10
14 1.682 0.000 0.000 79.00 65.90
15 0.000 -1.682 0.000 85.00 60.00
16 0.000 1.682 0.000 97.00 60.70
17 0.000 0.000 -1.682 55.00 57.40
18 0.000 0.000 1.682 81.00 63.20
19 0.000 0.000 0.000 80.00 60.80
20 0.000 0.000 0.000 91.00 58.90
Quadratic models are developed for the Conversion and Activity response variables as follows:
Response: Conversion
Sum of Mean F
A 14.44 1 14.44 0.65 0.4391
B 222.96 1 222.96 10.02 0.0101
C 525.64 1 525.64 23.63 0.0007
A2
48.47 1 48.47 2.18 0.1707
B2
124.48 1 124.48 5.60 0.0396
C2
388.59 1 388.59 17.47 0.0019
AB 36.13 1 36.13 1.62 0.2314
AC 1035.13 1 1035.13 46.53 < 0.0001
BC 120.12 1 120.12 5.40 0.0425
Residual 222.47 10 22.25
Pure Error 166.00 5 33.20
Cor Total 287.28 19
Intercept 81.09 1 1.92 76.81 85.38
A-Time 1.03 1 1.28 -1.82 3.87 1.00
B-Temperature 4.04 1 1.28 1.20 6.88 1.00
C-Catalyst 6.20 1 1.28 3.36 9.05 1.00
11-21

A2 -1.83 1 1.24 -4.60 0.93 1.02
B2 2.94 1 1.24 0.17 5.71 1.02
C2 -5.19 1 1.24 -7.96 -2.42 1.02
AB 2.13 1 1.67 -1.59 5.84 1.00
AC 11.38 1 1.67 7.66 15.09 1.00
BC -3.87 1 1.67 -7.59 -0.16 1.00
Conversion =
+81.09
+1.03 * A
+4.04 * B
+6.20 * C
-1.83 * A2
+2.94 * B2
-5.19 * C2
+2.13 * A * B
+11.38 * A * C
-3.87 * B * C
Conversion =
+81.09128
+1.02845 * Time
+6.20396 * Catalyst
-1.83398 * Time2
-5.19274 * Catalyst2
+2.12500 * Time * Temperature
+11.37500 * Time * Catalyst
-3.87500 * Temperature * Catalyst
Response: Activity
Sum of Mean F
A 175.35 1 175.35 56.42 < 0.0001
B 0.89 1 0.89 0.28 0.6052
C 67.91 1 67.91 21.85 0.0009
A2
10.05 1 10.05 3.23 0.1024
B2
0.081 1 0.081 0.026 0.8753
C2
0.047 1 0.047 0.015 0.9046
AB 1.20 1 1.20 0.39 0.5480
AC 0.011 1 0.011 3.620E-003 0.9532
BC 0.78 1 0.78 0.25 0.6270
Residual 31.08 10 3.11
Cor Total 287.28 19
Intercept 59.85 1 0.72 58.25 61.45
11-22

A-Time 3.58 1 0.48 2.52 4.65 1.00
B-Temperature 0.25 1 0.48 -0.81 1.32 1.00
C-Catalyst 2.23 1 0.48 1.17 3.29 1.00
A2
0.83 1 0.46 -0.20 1.87 1.02
B2
0.075 1 0.46 -0.96 1.11 1.02
C2
0.057 1 0.46 -0.98 1.09 1.02
AB -0.39 1 0.62 -1.78 1.00 1.00
AC -0.038 1 0.62 -1.43 1.35 1.00
BC 0.31 1 0.62 -1.08 1.70 1.00
Conversion =
59.85+
+3.58 * A
+0.25 * B
+2.23 * C
+0.83 * A2
+0.075 * B2
0.057 * C2+
-0.39 * A * B
-0.038 * A * C
+0.31 * B * C
Final Equati al Factors:on in Terms of Actu
Conversion =
59.84984+
+3.58327 * Time
+2.22997 * Catalyst
+0.83491 * Time2
0.057094 * Catalyst2+
-0.38750 * Time * Temperature
-0.037500 * Time * Catalyst
+0.31250 * Temperature * Catalyst
ecause many of the terms are insignificant, the reduced quadratic model is fit as follows:B
eD sign Expert Output
Response: Activity
Ana e table [Partial sum of squares]lysis of varianc
Sum of Mean F
Source Squares F Square Value Prob > FD
A 175.35 1 175.35 82.34 < 0.0001
C 67.91 1 67.91 31.89 < 0.0001
A2
9.94 1 9.94 4.67 0.0463
Residual 34.07 16 2.13
Lack of Fit 0.42 1 2.773 1 3.78 0.0766 not significant
Cor Total 287.28 19
oefficient 5% CI 5% CIC Standard 9 9
11-23

Intercept 59.95 1 0.42 59.06 60.83
A-Time 3.58 1 0.39 2.75 4.42 1.00
C-Catalyst 2.23 1 0.39 1.39 3.07 1.00
A2
0.82 1 0.38 0.015 1.63 1.00
Activity =
59.95+
+3.58 * A
+2.23 * C
+0.82 * A2
Activity =
59.94802+
+3.58327 * Time
+2.22997 * Catalyst
+0.82300 * Time2
DESIGN-EXPERT Plot
Conversion
X = A: T im e
Y = C: Catalyst
Design Points
Actual Factor
B: T em perature = -1.00
Conversion
A: Tim e
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
5456
58606264
66
68
70
72
74
74
76
76
78
78
80
82
84
86
88
90
92
DESIGN-EXPERT Plot
Activity
X = A: T im e
Y = C: Catalyst
Design Points
Actual Factor
Activity
A: Tim e
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
56
58
60 62
64
66
DESIGN-EXPERT Plot
Overlay Plot
X = A: T im e
Y = C: Catalyst
Design Points
Actual Factor
Overlay Plot
A: Tim e
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Conv ersion: 82
Activ ity : 60
11-24

The contour plots visually describe the models while the overlay plots identifies the acceptable region for
the process.
11-13 A manufacturer of cutting tools has developed two empirical equations for tool life in hours (y1) and
for tool cost in dollars (y2). Both models are linear functions of steel hardness (x1) and manufacturing time
(x2). The two equations are
212
211
4323
2510
xxyˆ
xxyˆ
++=
++=
and both equations are valid over the range -1.5≤x1≤1.5. Unit tool cost must be below $27.50 and life
must exceed 12 hours for the product to be competitive. Is there a feasible set of operating conditions for
this process? Where would you recommend that the process be run?
The contour plots below graphically describe the two models. The overlay plot identifies the feasible
operating region for the process.
Life
A: Hardness
B:Time
-1.50 -0.75 0.00 0.75 1.50
-1.50
-0.75
0.00
0.75
1.50
2
4
6 8 10 12 14
16
18
20
Cost
A: Hardness
B:Time
-1.50 -0.75 0.00 0.75 1.50
-1.50
-0.75
0.00
0.75
1.50
14
16
18
20
22
24
26
28
30
32
27.5
11-25

Overlay Plot
A: Hardness
B:Time
-1.50 -0.75 0.00 0.75 1.50
-1.50
-0.75
0.00
0.75
1.50
Lif e: 12
C ost: 27.5
50274323
122510
21
21
.xx
xx
≤++
≥++
11-14 A central composite design is run in a chemical vapor deposition process, resulting in the
experimental data shown below. Four experimental units were processed simultaneously on each run of
the design, and the responses are the mean and variance of thickness, computed across the four units.
x1 x2 y 2
s
-1 -1 360.6 6.689
-1 1 445.2 14.230
1 -1 412.1 7.088
1 1 601.7 8.586
1.414 0 518.0 13.130
-1.414 0 411.4 6.644
0 1.414 497.6 7.649
0 -1.414 397.6 11.740
0 0 530.6 7.836
0 0 495.4 9.306
0 0 510.2 7.956
0 0 487.3 9.127
(a) Fit a model to the mean response. Analyze the residuals.
Response: Mean Thick
Sum of Mean F
A 22573.36 1 22573.36 38.19 0.0008
B 15261.91 1 15261.91 25.82 0.0023
A2 2795.58 1 2795.58 4.73 0.0726
B2 5550.74 1 5550.74 9.39 0.0221
AB 2756.25 1 2756.25 4.66 0.0741
Residual 3546.83 6 591.14
Pure Error 1084.79 3 361.60
11-26

Cor Total 51191.09 11
Intercept 505.88 1 12.16 476.13 535.62
A-x1 53.12 1 8.60 32.09 74.15 1.00
B-x2 43.68 1 8.60 22.64 64.71 1.00
A2 -20.90 1 9.61 -44.42 2.62 1.04
B2 -29.45 1 9.61 -52.97 -5.93 1.04
AB 26.25 1 12.16 -3.50 56.00 1.00
Mean Thick =
+505.88
+53.12 * A
+43.68 * B
-20.90 * A2
-29.45 * B2
+26.25 * A * B
Mean Thick =
+505.87500
+53.11940 * x1
+43.67767 * x2
-20.90000 * x12
-29.45000 * x22
+26.25000 * x1 * x2
Residual
Normal%probability
-24.3779 -12.1022 0.173533 12.4493 24.725
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-24.3779
-12.1022
0.173533
12.4493
24.725
384.98 433.38 481.78 530.17 578.57
A modest deviation from normality can be observed in the Normal Plot of Residuals; however, not enough
to be concerned.
11-27

(b) Fit a model to the variance response. Analyze the residuals.
Response: Var Thick
ANOVA for Response Surface 2FI Model
Sum of Mean F
A 41.46 1 41.46 67.79 < 0.0001
B 15.21 1 15.21 24.87 0.0011
AB 9.13 1 9.13 14.93 0.0048
Cor Total 70.69 11
Intercept 9.17 1 0.23 8.64 9.69
A-x1 2.28 1 0.28 1.64 2.91 1.00
B-x2 -1.38 1 0.28 -2.02 -0.74 1.00
AB -1.51 1 0.39 -2.41 -0.61 1.00
Var Thick =
+9.17
+2.28 * A
-1.38 * B
-1.51 * A * B
Var Thick =
+9.16508
+2.27645 * x1
-1.37882 * x2
-1.51075 * x1 * x2
11-28

Residual
Normal%probability
-1.32908 -0.810429 -0.291776 0.226878 0.745532
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-1.32908
-0.810429
-0.291776
0.226878
0.745532
5.95 8.04 10.14 12.23 14.33
The residual plots are not acceptable. A transformation should be considered. If not successful at
correcting the residual plots, further investigation into the two apparently unusual points should be made.
(c) Fit a model to the ln(s2
). Is this model superior to the one you found in part (b)?
Response: Var Thick Transform: Natural log Constant: 0
ANOVA for Response Surface 2FI Model
Sum of Mean F
A 0.46 1 0.46 74.99 < 0.0001
B 0.14 1 0.14 22.80 0.0014
AB 0.079 1 0.079 13.04 0.0069
Residual 0.049 8 6.081E-003
Lack of Fit 0.024 5 4.887E-003 0.61 0.7093 not significant
Pure Error 0.024 3 8.071E-003
Cor Total 0.72 11
Intercept 2.18 1 0.023 2.13 2.24
A-x1 0.24 1 0.028 0.18 0.30 1.00
B-x2 -0.13 1 0.028 -0.20 -0.068 1.00
AB -0.14 1 0.039 -0.23 -0.051 1.00
Ln(Var Thick) =
+2.18
+0.24 * A
-0.13 * B
-0.14 * A * B
11-29

Ln(Var Thick) =
+2.18376
+0.23874 * x1
-0.13165 * x2
-0.14079 * x1 * x2
Residual
Normal%probability
-0.125029 -0.070505 -0.0159805 0.0385439 0.0930684
1
5
10
20
30
50
70
80
90
95
99
Predicted
Residuals
-0.125029
-0.070505
-0.0159805
0.0385439
0.0930684
1.85 2.06 2.27 2.48 2.69
The residual plots are much improved following the natural log transformation; however, the two runs still
appear to be somewhat unusual and should be investigated further. They will be retained in the analysis.
(d) Suppose you want the mean thickness to be in the interval 450±25. Find a set of operating conditions
that achieve the objective and simultaneously minimize the variance.
Mean Thick
A: x1
B:x2
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
400
425
450
475
500
525
550
575
4
Ln(Var Thick)
A: x1
B:x2
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
2
2.1
2.2
2.3
2.4
2.5
2.6
4
11-30

Overlay Plot
A: x1
B:x2
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Mean Thick: 425
Mean Thick: 475
Ln(Var Thick): 2.000
4
The contour plots describe the two models while the overlay plot identifies the acceptable region for the
process.
(e) Discuss the variance minimization aspects of part (d). Have you minimized total process variance?
The within run variance has been minimized; however, the run-to-run variation has not been minimized in
the analysis. This may not be the most robust operating conditions for the process.
11-15 Verify that an orthogonal first-order design is also first-order rotatable.
To show that a first order orthogonal design is also first order rotatable, consider
)ˆ(Vx)ˆ(V)xˆˆ(V)yˆ(V
k
i
ii
k
i
ii ∑∑ ==
+=+=
1
2
0
1
0 ββββ
since all covariances between and are zero, due to design orthogonality. Furthermore, we have:βi βj
( ) n
)ˆ(V...)ˆ(V)ˆ(VˆV k
2
210
σ
ββββ ===== , so
∑=
+=
k
i
ix
nn
)yˆ(V
1
2
22
σσ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+= ∑=
k
i
ix
nn
)yˆ(V
1
2
22
1
σσ
which is a function of distance from the design center (i.e. x=0), and not direction. Thus the design must
be rotatable. Note that n is, in general, the number of points in the exterior portion of the design. If there
are nc centerpoints, then
)nn(
)ˆ(V
c+
=
2
0
σ
β .
11-31

11-16 Show that augmenting a 2k design with nc center points does not affect the estimates of the βi (i=1,
2, . . . , k), but that the estimate of the intercept β0 is the average of all 2k + nc observations.
In general, the X matrix for the 2k
design with nc center points and the y vector would be:
β0 β1 β2 . . . βk
1 1 1 1
1 1 1 1
1 1 1 1
1 0 0 0
1 0 0 0
1 0 0 0
− − −⎡ ⎤
⎢ ⎥− −⎢ ⎥
⎢
⎢
⎢
⎢ ⎥= − − −− − −− − −− − −− − −
⎢ ⎥
⎢
⎢
⎢
⎢ ⎥
⎢ ⎥
⎣ ⎦
X
⎥
⎥
⎥
⎥
⎥
⎥
The upper half of the matrix is the usual ± 1
notation of the 2k
design
The lower half of the matrix represents the
center points (nc rows)
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−−=
c
k
n
n
n
y
y
y
0
0
0
2
2
1
2
1
y 2k
+nc
observations
.
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡ +
=
k
k
c
k
n
2
02
002
XX'
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
kg
g
g
g
2
1
0
yX'
Grand total of
all 2k
+nc
observations
usual contrasts
from 2k
Therefore,
c
k
n
gˆ
+
=
2
0
0β , which is the average of all ( )c
k
n+2 observations, while k
i
i
gˆ
2
=β , which does
not depend on the number of center points, since in computing the contrasts gi, all observations at the
center are multiplied by zero.
11-17 The rotatable central composite design. It can be shown that a second-order design is rotatable if
if a or b (or both) are odd and if . Show that for the central
composite design these conditions lead to
∑ =
=
n
u
b
ju
a
iu xx
1
0 2
1
2
1
4
3 ju
n
u iu
n
u iu xxx ∑∑ ==
=
( ) 41/
fn=α for rotatability, where nf is the number of points in
the factorial portion.
The balance between +1 and -1 in the factorial columns and the orthogonality among certain column in the
X matrix for the central composite design will result in all odd moments being zero. To solve for α use the
following relations:
4
1
4
2α+=∑=
n
u
fiu nx , ∑=
=
n
u
fjuiu nxx
1
22
11-32

then
4
4
4
4
2
1
2
1
4
22
32
3
f
f
f
ff
ju
n
u iu
n
u iu
n
n
n
)n(n
xxx
=
=
=
=+
= ∑∑ ==
α
α
α
α
11-18 Verify that the central composite design shown below blocks orthogonally.
x1 x2 x3 x1 x2 x3 x1 x2 x3
0 0 0 0 0 0 -1.633 0 0
0 0 0 0 0 0 1.633 0 0
1 1 1 1 1 -1 0 -1.633 0
1 -1 -1 1 -1 1 0 1.633 0
-1 -1 1 -1 1 1 0 0 -1.633
-1 1 -1 -1 -1 -1 0 0 1.633
0 0 0
0 0 0
Note that each block is an orthogonal first order design, since the cross products of elements in different
columns add to zero for each block. To verify the second condition, choose a column, say column x2.
Now
∑=
=
k
u
u .x
1
2
2 33413 , and n=20
For blocks 1 and 2,
∑ =
m
mx 42
2 , nm=6
So
6
1
2
2
2
2
==
∑
∑
=
mn
u
u
m
m
n
x
x
20
6
33413
4
=
.
0.3 = 0.3
and condition 2 is satisfied by blocks 1 and 2. For block 3, we have
11-33

∑ =
m
m .x 33452
2 , nm = 8, so
n
n
x
x
m
n
u
u
m
m
=
∑
∑
=1
2
2
2
2
20
8
33413
3345
=
.
.
0.4 = 0.4
And condition 2 is satisfied by block 3. Similar results hold for the other columns.
11-19 Blocking in the central composite design. Consider a central composite design for k = 4 variables
in two blocks. Can a rotatable design always be found that blocks orthogonally?
To run a central composite design in two blocks, assign the nf factorial points and the n01 center points to
block 1 and the 2k
axial points plus n02 center points to block 2. Both blocks will be orthogonal first order
designs, so the first condition for orthogonal blocking is satisfied.
The second condition implies that
( )
( ) 2
1
2
2
2
2
1
c
cf
m
im
m
im
nk
nn
blockx
blockx
+
+
=
∑
∑
However, in block 1 and in block 2, so∑ =
m
fim nx2
∑ =
m
imx 22
2α
2
1
2 22 c
cff
nk
nnn
+
+
=
α
Which gives:
( )
( )
2
1
1
2
2
2
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
=
cf
cf
nn
nkn
α
Since 4
fn=α if the design is to be rotatable, then the design must satisfy
( )
( )
2
1
2
2
2
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
=
cf
cf
f
nn
nkn
n
11-34

It is not possible to find rotatable central composite designs which block orthogonally for all k. For
example, if k=3, the above condition cannot be satisfied. For k=2, there must be an equal number of center
points in each block, i.e. nc1 = nc2. For k=4, we must have nc1 = 4 and nc2 = 2.
11-20 How could a hexagon design be run in two orthogonal blocks?
The hexagonal design can be blocked as shown below. There are nc1 = nc2 = nc center points with nc even.
1 2
3
45
6
n
Put the points 1,3,and 5 in block 1 and 2,4,and 6 in block 2. Note that each block is a simplex.
11-21 Yield during the first four cycles of a chemical process is shown in the following table. The
variables are percent concentration (x1) at levels 30, 31, and 32 and temperature (x2) at 140, 142, and
144°F. Analyze by EVOP methods.
Conditions
Cycle (1) (2) (3) (4) (5)
1 60.7 59.8 60.2 64.2 57.5
2 59.1 62.8 62.5 64.6 58.3
3 56.6 59.1 59.0 62.3 61.1
4 60.5 59.8 64.5 61.0 60.1
Cycle: n=1 Phase 1
Calculation of Averages Calculation of Standard Deviation
Operating Conditions (1) (2) (3) (4) (5)
(i) Previous Cycle Sum Previous Sum S=
(ii) Previous Cycle Average Previous Average =
(iii) New Observation 60.7 59.8 60.2 64.2 57.5 New S=Range x fk,n
(iv) Differences Range=
(v) New Sums 60.7 59.8 60.2 64.2 57.5 New Sum S=
(vi) New Averages 60.7 59.8 60.2 64.2 57.5 New average S = New Sum S/(n-1)=
Calculation of Effects Calculation of Error Limits
( =−−+= 5243
2
1
yyyyA )
3.55
For New Average: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2
( =+−−= 5243
2
1
yyyyB )
-3.55
For New Effects: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2
11-35

( )=−+−= 5243
2
1
yyyyAB
-0.85
For CIM: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
.781
( )=−+++= 15243 4
2
1
yyyyyCIM
-0.22
Cycle: n=2 Phase 1
(i) Previous Cycle Sum 60.7 59.8 60.2 64.2 57.5 Previous Sum S=
(ii) Previous Cycle Average 60.7 59.8 60.2 64.2 57.5 Previous Average =
(iii) New Observation 59.1 62.8 62.5 64.6 58.3 New S=Range x fk,n=1.38
(iv) Differences 1.6 -3.0 -2.3 -0.4 -0.8 Range=4.6
(v) New Sums 119.8 122.6 122.7 128.8 115.8 New Sum S=1.38
(vi) New Averages 59.90 61.30 61.35 64.40 57.90 New average S = New Sum S/(n-1)=1.38
( =−−+= 5243
2
1
yyyyA )
3.28
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 1.95
( =+−−= 5243
2
1
yyyyB )
-3.23
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 1.95
( )=−+−= 5243
2
1
yyyyAB
0.18
For CIM: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
.781 1.74
( )=−+++= 15243 4
2
1
yyyyyCIM
1.07
Cycle: n=3 Phase 1
(i) Previous Cycle Sum 119.8 122.6 122.7 128.8 115.8 Previous Sum S=1.38
(ii) Previous Cycle Average 59.90 61.30 61.35 64.40 57.90 Previous Average =1.38
(iv) Differences 3.30 2.20 2.35 2.10 -3.20 Range=6.5
( =−−+= 5243
2
1
yyyyA )
2.37
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 2.11
( =+−−= 5243
2
1
yyyyB )
-2.37
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 2.11
( )=−+−= 5243
2
1
yyyyAB
-0.77
For CIM: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
.781 1.74
( )=−+++= 15243 4
2
1
yyyyyCIM
1.72
Cycle: n=4 Phase 1
11-36

(i) Previous Cycle Sum 176.4 181.7 181.7 191.1 176.9 Previous Sum S=3.66
(ii) Previous Cycle Average 58.80 60.57 60.57 63.70 58.97 Previous Average =1.83
(iv) Differences -1.70 0.77 -3.93 2.70 -1.13 Range=6.63
( =−−+= 5243
2
1
yyyyA )
2.48
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 2.04
( =+−−= 5243
2
1
yyyyB )
-1.31
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
2 2.04
( )=−+−= 5243
2
1
yyyyAB
-0.18
For CIM: =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
S
n
.781 1.82
( )=−+++= 15243 4
2
1
yyyyyCIM
1.46
From studying cycles 3 and 4, it is apparent that A (and possibly B) has a significant effect. A new phase
should be started following cycle 3 or 4.
11-22 Suppose that we approximate a response surface with a model of order d1, such as y=X1β1+ε, when
the true surface is described by a model of order d2>d1; that is E(y)= X1β1+ X2β2.
(a) Show that the regression coefficients are biased, that is, that E( )=ββ1 1+Aβ2, where A=(X’
1X1)-1
X’
1X2.
A is usually called the alias matrix.
[ ] ( )
( ) [ ]
( ) ( )
( ) ( )
21
221
1
11111
1
11
22111
1
11
1
1
11
1
1
111
Aββ
βXXXXβXXXX
βXβXXXX
yXXX
yXXXβ
+=
+=
+=
=
⎥⎦
⎤
⎢⎣
⎡=
−−
−
−
−
''''
''
''
''
E
EˆE
where ( ) 2
'
1
'
1 XXXXA
1
1
−
=
(a) If d1=1 and d2=2, and a full 2k
is used to fit the model, use the result in part (a) to determine the alias
structure.
In this situation, we have assumed the true surface to be first order, when it is really second order. If a full
factorial is used for k=2, then
X1 = X
β β β0 1 2
1 1 1
1 1 1
1 1 1
1 1 1
− −
−
−
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
2 =
β11
1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎢
β22
1
1
1
1
⎥
⎥
⎥
⎥
⎦
⎤
−
−
1
1
1
1
12β
and A =
1 1 0
0 0 0
0 0 0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
11-37

Then, E[ = E]β1
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡ ++
=
=
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
=
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
2
1
22110
12
22
11
2
1
0
2
1
0
000
000
011
β
β
βββ
β
β
β
β
β
β
β
β
β
ˆ
ˆ
ˆ
The pure quadratic terms bias the intercept.
(b) If d1=1, d2=2 and k=3, find the alias structure assuming that a 23-1
design is used to fit the model.
X1 = X
β β β β0 1 2 3
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
− −
− −
− −
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
2 =
β11
1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎢
β22
1
1
1
1
β33
1
1
1
1
β12
1
1
1
1
−
−
β13
1
1
1
1
−
−
β23
1
1
1
1
−
−
⎤
⎦
⎥
⎥
⎥
⎥
and A =
1 1 0 0 0 0
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 1 0 0
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Then, E[ = E]β1
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
+
+
+
+++
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
=
=
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
123
132
231
2222110
23
13
12
33
22
11
3
2
1
0
3
2
1
0
001000
010000
100000
000011
ββ
ββ
ββ
ββββ
β
β
β
β
β
β
β
β
β
β
β
β
β
β
ˆ
ˆ
ˆ
ˆ
(d) If d1=1, d2=2, k=3, and the simplex design in Problem 11-3 is used to fit the model, determine the alias
structure and compare the results with part (c).
X1 = X
β β β β0 1 2 3
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
− −
− −
− −
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
2 =
β11
0
2
0
2
⎡
⎣
⎢
⎢
⎢
⎢
β22
2
0
2
0
β33
1
1
1
1
β12
0
0
0
0
β13
0
2
0
2
−
−
β23
2
0
2
0
−
−
⎤
⎦
⎥
⎥
⎥
⎥
and A =
1 1 1 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
1 1 0 0 0 0
−
−
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Then, E[ = E]β1
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−+
−
+
+++
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
=
=
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
22113
232
131
2222110
23
13
12
33
22
11
3
2
1
0
3
2
1
0
000011
100000
010000
000111
βββ
ββ
ββ
ββββ
β
β
β
β
β
β
β
β
β
β
β
β
β
β
ˆ
ˆ
ˆ
ˆ
Notice that the alias structure is different from that found in the previous part for the 23-1
design. In
general, the A matrix will depend on which simplex design is used.
11-23 Suppose that you need to design an experiment to fit a quadratic model over the region
, i=1,2 subject to the constraint11 +≤≤− ix 121 ≤+ xx . If the constraint is violated, the process will not
work properly. You can afford to make no more than n=12 runs. Set up the following designs:
11-38

(a) An “inscribed” CCD with center points at 021 == xx
x1 x2
-0.5 -0.5
0.5 -0.5
-0.5 0.5
0.5 0.5
-0.707 0
0.707 0
0 -0.707
0 0.707
0 0
0 0
0 0
0 0
(a)* An “inscribed” CCD with center points at 25021 .xx −== so that a larger design could be fit within
the constrained region
x1 x2
-1 -1
0.5 -1
-1 0.5
0.5 0.5
-1.664 -0.25
1.164 -0.25
-0.25 -1.664
-0.25 1.164
-0.25 -0.25
-0.25 -0.25
-0.25 -0.25
-0.25 -0.25
(a) An “inscribed” 32
factorial with center points at 25.021 −= xx
x1 x2
-1 -1
-0.25 -1
0.5 -1
-1 -0.25
-0.25 -0.25
0.5 -0.25
-1 0.5
-0.25 0.5
0.5 0.5
-0.25 -0.25
-0.25 -0.25
-0.25 -0.25
(a) A D-optimal design.
11-39

x1 x2
-1 -1
1 -1
-1 1
1 0
0 1
0 0
-1 0
0 -1
0.5 0.5
-1 -1
1 -1
-1 1
(a) A modified D-optimal design that is identical to the one in part (c), but with all replicate runs at the
design center.
x1 x2
1 0
0 0
0 1
-1 -1
1 -1
-1 1
-1 0
0 -1
0.5 0.5
0 0
0 0
0 0
(a) Evaluate the 1
)( −
′XX criteria for each design.
(a) (a)* (b) (c) (d)
( ) 1−
′XX
0.5 0.00005248 0.007217 0.0001016 0.0002294
(a) Evaluate the D-efficiency for each design relative to the D-optimal design in part (c).
(a) (a)* (b) (c) (d)
D-efficiency 24.25% 111.64% 49.14% 100.00% 87.31%
(a) Which design would you prefer? Why?
The offset CCD, (a)*, is the preferred design based on the D-efficiency. Not only is it better than the D-
optimal design, (c), but it maintains the desirable design features of the CCD.
11-24 Consider a 23
design for fitting a first-order model.
11-40

(a) Evaluate the D-criterion 1
)( −
′XX for this design.
1
)( −
′XX = 2.441E-4
(b) Evaluate the A-criterion for this design.1
)( −
′XXtr
1
)( −
′XXtr = 0.5
(c) Find the maximum scaled prediction variance for this design. Is this design G-optimal?
( ) ( )( ) ( )( ) ( ) 4111
2
=′′== −
xXXx
x
x N
yˆNVar
v
σ
. Yes, this is a G-optimal design.
11-25 Repeat Problem 11-24 using a first order model with the two-factor interaction.
1
)( −
′XX = 4.768E-7
1
)( −
′XXtr = 0.875
( ) ( )( ) ( )( ) ( ) 7111
2
=′′== −
xXXx
x
x N
yˆNVar
v
σ
. Yes, this is a G-optimal design.
11-26 A chemical engineer wishes to fit a calibration curve for a new procedure used to measure the
concentration of a particular ingredient in a product manufactured in his facility. Twelve samples can be
prepared, having known concentration. The engineer’s interest is in building a model for the measured
concentrations. He suspects that a linear calibration curve will be adequate to model the measured
concentration as a function of the known concentrations; that is, where x is the actual concentration. Four
experimental designs are under consideration. Design 1 consists of 6 runs at known concentration 1 and 6
runs at known concentration 10. Design 2 consists of 4 runs at concentrations 1, 5.5, and 10. Design 3
consists of 3 runs at concentrations 1, 4, 7, and 10. Finally, design 4 consists of 3 runs at concentrations 1
and 10 and 6 runs at concentration 5.5.
(a) Plot the scaled variance of prediction for all four designs on the same graph over the concentration
range. Which design would be preferable, in your opinion?
11-41

Scaled Variance of Prediction
0
0.5
1
1.5
2
2.5
3
3.5
1 3 5 7 9
Design 4
Design 3
Design 2
Design 1
Because it has the lowest scaled variance of prediction at all points in the design space with the exception
of 5.5, Design 1 is preferred.
(b) For each design calculate the determinant of . Which design would be preferred according
to the “D” criterion?
1
)( −
′XX
Design 1
)( −
′XX
1 0.000343
2 0.000514
3 0.000617
4 0.000686
Design 1 would be preferred.
(c) Calculate the D-efficiency of each design relative to the “best” design that you found in part b.
Design D-efficiency
1 100.00%
2 81.65%
3 74.55%
4 70.71%
(a) For each design, calculate the average variance of prediction over the set of points given by x = 1, 1.5,
2, 2.5, . . ., 10. Which design would you prefer according to the V-criterion?
Average Variance of Prediction
Design Actual Coded
1 1.3704 0.1142
2 1.5556 0.1296
3 1.6664 0.1389
4 1.7407 0.1451
Design 1 is still preferred based on the V-criterion.
11-42

(e) Calculate the V-efficiency of each design relative to the best design you found in part (d).
Design V-efficiency
1 100.00%
2 88.10%
3 82.24%
4 78.72%
(f) What is the G-efficiency of each design?
Design G-efficiency
1 100.00%
2 80.00%
3 71.40%
4 66.70%
11-27 Rework Problem 11-26 assuming that the model the engineer wishes to fit is a quadratic.
Obviously, only designs 2, 3, and 4 can now be considered.
Scaled Variance of Prediction
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
1 3 5 7 9
Design 4
Design 3
Design 2
2
Based on the plot, the preferred design would depend on the region of interest. Design 4 would be
preferred if the center of the region was of interest; otherwise, Design 2 would be preferred.
Design 1
)( −
′XX
2 4.704E-07
3 6.351E-07
4 5.575E-07
Design 2 is preferred based on 1
)( −
′XX .
Design D-efficiency
2 100.00%
11-43

3 90.46%
4 94.49%
Average Variance of Prediction
Design Actual Coded
2 2.441 0.2034
3 2.393 0.1994
4 2.242 0.1869
Design 4 is preferred.
Design V-efficiency
2 91.89%
3 93.74%
4 100.00%
Design G-efficiency
2 100.00%
3 79.00%
4 75.00%
11-28 An experimenter wishes to run a three-component mixture experiment. The constraints are the
components proportions are as follows:
7.04.0
3.01.0
4.02.0
3
2
1
≤≤
≤≤
≤≤
x
x
x
(a) Set up an experiment to fit a quadratic mixture model. Use n=14 runs, with 4 replicates. Use the D-
criteria.
Std x1 x2 x3
1 0.2 0.3 0.5
2 0.3 0.3 0.4
3 0.3 0.15 0.55
4 0.2 0.1 0.7
5 0.4 0.2 0.4
6 0.4 0.1 0.5
7 0.2 0.2 0.6
8 0.275 0.25 0.475
9 0.35 0.175 0.475
10 0.3 0.1 0.6
11 0.2 0.3 0.5
12 0.3 0.3 0.4
13 0.2 0.1 0.7
14 0.4 0.1 0.5
(a) Draw the experimental design region.
11-44

A: x1
0.50
B: x2
0.40
C: x3
0.70
0.40 0.10
0.20
2
2
2
2
(c) Set up an experiment to fit a quadratic mixture model with n=12 runs, assuming that three of these
runs are replicated. Use the D-criterion.
Std x1 x2 x3
1 0.3 0.15 0.55
2 0.2 0.3 0.5
3 0.3 0.3 0.4
4 0.2 0.1 0.7
5 0.4 0.2 0.4
6 0.4 0.1 0.5
7 0.2 0.2 0.6
8 0.275 0.25 0.475
9 0.35 0.175 0.475
10 0.2 0.1 0.7
11 0.4 0.1 0.5
12 0.4 0.2 0.4
11-45

A: x1
0.50
B: x2
0.40
C: x3
0.70
0.40 0.10
0.20
2
2 2
(d) Comment on the two designs you have found.
The design points are the same for both designs except that the edge center on the x1-x3 edge is not
included in the second design. None of the replicates for either design are in the center of the experimental
region. The experimental runs are fairly uniformly spaced in the design region.
11-29 Myers and Montgomery (2002) describe a gasoline blending experiment involving three mixture
components. There are no constraints on the mixture proportions, and the following 10 run design is used.
Design Point x1 x2 x3 y(mpg)
1 1 0 0 24.5, 25.1
2 0 1 0 24.8, 23.9
3 0 0 1 22.7, 23.6
4 ½ ½ 0 25.1
5 ½ 0 ½ 24.3
6 0 ½ ½ 23.5
7 1/3 1/3 1/3 24.8, 24.1
8 2/3 1/6 1/6 24.2
9 1/6 2/3 1/6 23.9
10 1/6 1/6 2/3 23.7
(a) What type of design did the experimenters use?
A simplex centroid design was used.
(b) Fit a quadratic mixture model to the data. Is this model adequate?
Response: y
ANOVA for Mixture Quadratic Model
Sum of Mean F
Linear Mixture 3.92 2 1.96 9.06 0.0088
11-46

AB 0.15 1 0.15 0.69 0.4289
AC 0.081 1 0.081 0.38 0.5569
BC 0.077 1 0.077 0.36 0.5664
Cor Total 5.95 13
Component Estimate DF Error Low High
A-x1 24.74 1 0.32 24.00 25.49
B-x2 24.31 1 0.32 23.57 25.05
C-x3 23.18 1 0.32 22.43 23.92
AB 1.51 1 1.82 -2.68 5.70
AC 1.11 1 1.82 -3.08 5.30
BC -1.09 1 1.82 -5.28 3.10
Final Equation in Terms of Pseudo Components:
y =
+24.74 * A
+24.31 * B
+23.18 * C
+1.51 * A * B
+1.11 * A * C
-1.09 * B * C
Final Equation in Terms of Real Components:
y =
+24.74432 * x1
+24.31098 * x2
+23.17765 * x3
+1.51364 * x1 * x2
+1.11364 * x1 * x3
-1.08636 * x2 * x3
The quadratic terms appear to be insignificant. The analysis below is for the linear mixture model:
Response: y
ANOVA for Mixture Quadratic Model
Sum of Mean F
Linear Mixture 3.92 2 1.96 10.64 0.0027
Residual 2.03 11 0.18
Cor Total 5.95 13
11-47

Component Estimate DF Error Low High
A-x1 24.93 1 0.25 24.38 25.48
B-x2 24.35 1 0.25 23.80 24.90
C-x3 23.19 1 0.25 22.64 23.74
Adjusted Adjusted Approx t for H0
Component Effect DF Std Error Effect=0 Prob > |t|
A-x1 1.16 1 0.33 3.49 0.0051
B-x2 0.29 1 0.33 0.87 0.4021
C-x3 -1.45 1 0.33 -4.36 0.0011
Final Equation in Terms of Pseudo Components:
y =
+24.93 * A
+24.35 * B
+23.19 * C
Final Equation in Terms of Real Components:
y =
+24.93048 * x1
+24.35048 * x2
+23.19048 * x3
(c) Plot the response surface contours. What blend would you recommend to maximize the MPG?
A: x1
1.00
B: x2
1.00
C: x3
1.00
0.00 0.00
0.00
y
23.4
23.6
23.8
24
24.2
24.4
24.6
24.8
2
2 2
To maximize the miles per gallon, the recommended blend is x1 = 1, x2 = 0, and x3 = 0.
11-48

Chapter 12
Robust Parameter Design and Process Robustness Studies
Solutions
12-1 Reconsider the leaf spring experiment in Table 12-1. Suppose that the objective is to find a set of
conditions where the mean free height is as close as possible to 7.6 inches with a variance of free height as
small as possible. What conditions would you recommend to achieve these objectives?
A B C D E(-) E(+) y 2
s
- - - - 7.78,7.78, 7.81 7.50, 7,25, 7.12 7.54 0.090
+ - - + 8.15, 8.18, 7.88 7.88, 7.88, 7.44 7.90 0.071
- + - + 7.50, 7.56, 7.50 7.50, 7.56, 7.50 7.52 0.001
+ + - - 7.59, 7.56, 7.75 7.63, 7.75, 7.56 7.64 0.008
- - + + 7.54, 8.00, 7.88 7.32, 7.44, 7.44 7.60 0.074
+ - + - 7.69, 8.09, 8.06 7.56, 7.69, 7.62 7.79 0.053
- + + - 7.56, 7.52, 7.44 7.18, 7.18, 7.25 7.36 0.030
+ + + + 7.56, 7.81, 7.69 7.81, 7.50, 7.59 7.66 0.017
By overlaying the contour plots for Free Height Mean and the Free Height Variance, optimal solutions can
be found. To minimize the variance, factor B must be at the high level while factors A and D are adjusted
to assure a mean of 7.6. The two overlay plots below set factor D at both low and high levels. Therefore, a
mean as close as possible to 7.6 with minimum variance of 0.008 can be achieved at A = 0.78, B = +1, and
D = -1. This can also be achieved with A = +0.07, B = +1, and D = +1.
Free Height Mean
A: Furnace Temp
B:HeatingTime
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
7.5
7.55
7.6
7.65
7.7
7.75
7.8
7.85
-1.00 -0.50 0.00 0.50 1.00
0.001
0.0451273
0.0892545
0.133382
0.177509
B: Heating Time
FreeHeightVariance
One Factor Plot
12-1

Overlay Plot
A: Furnace Temp
B:HeatingTime
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Free Height Mean: 7.58
Free Height Variance: 0.009
Overlay Plot
A: Furnace Temp
B:HeatingTime
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Free Height Variance: 0.009
Factor D = -1 Factor D = +1
12-2 Consider the bottle filling experiment in Problem 6-18. Suppose that the percentage of
carbonation (A) is a noise variable ( 2
1zσ = in coded units).
(a) Fit the response model to these data. Is there a robust design problem?
The following is the analysis of variance with all terms in the model followed by a reduced model.
Because the noise factor A is significant, and the AB interaction is moderately significant, there is a robust
design problem.
Response: Fill Deviation
ANOVA for Response Surface Reduced Cubic Model
Sum of Mean F
Cor Total 300.05 3
A 36.00 1 36.00 57.60 < 0.0001
B 20.25 1 20.25 32.40 0.0005
C 12.25 1 12.25 19.60 0.0022
AB 2.25 1 2.25 3.60 0.0943
AC 0.25 1 0.25 0.40 0.5447
BC 1.00 1 1.00 1.60 0.2415
ABC 1.00 1 1.00 1.60 0.2415
Cor Total 78.00 15
Based on the above analysis, the AB interaction is removed from the model and used as error.
Response: Fill Deviation
Sum of Mean F
A 36.00 1 36.00 54.62 < 0.0001
B 20.25 1 20.25 30.72 0.0002
C 12.25 1 12.25 18.59 0.0012
12-2

AB 2.25 1 2.25 3.41 0.0917
Residual 7.25 11 0.66
Cor Total 78.00 15
Fill Deviation =
+1.00
+1.50 * A
+1.13 * B
+0.88 * C
+0.38 * A * B
(b) Find the mean model and either the variance model or the POE.
From the final equation shown in the above analysis, the mean model and corresponding contour plot is
shown below.
( )1 2, 1 1.13 0.88z 3E y z x x⎡ ⎤ = + +⎣ ⎦x
Fill Deviation
B: Pressure
C:Speed
-1.000 -0.500 0.000 0.500 1.000
-1.000
-0.500
0.000
0.500
1.000
-0.5
0
0.5
1
1.5
2
2.5
Contour and 3-D plots of the POE are shown below.
12-3

POE(Fill Deviation)
B: Pressure
C:Speed
-1.000 -0.500 0.000 0.500 1.000
-1.000
-0.500
0.000
0.500
1.000
1.5 1.6 1.7 1.8 1.9 2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
POE(FillDeviation)
-1.000
-0.500
0.000
0.500
1.000
-1.000
-0.500
0.000
0.500
1.000
B: Pressure
C: Speed
(c) Find a set of conditions that result in mean fill deviation as close to zero as possible with minimum
transmitted variance.
The overlay plot below identifies a an operating region for pressure and speed that in a mean fill deviation
as close to zero as possible with minimum transmitted variance.
Overlay Plot
B: Pressure
C:Speed
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Fill Deviation: -0.1
Fill Deviation: 0.1
POE(Fill Deviation): 1.5
12-3 Consider the experiment in Problem 11-12. Suppose that temperature is a noise variable
( in coded units). Fit response models for both responses. Is there a robust design problem with
respect to both responses? Find a set of conditions that maximize conversion with activity between 55 and
60 and that minimize variability transmitted from temperature.
2
1zσ =
The analysis and models as found in problem 11-12 are shown below for both responses. There is a robust
design problem with regards to the conversion response because of the significance of factor B,
temperature, and the BC interaction. However, temperature is not significant in the analysis of the second
response, activity.
12-4

Response: Conversion
Sum of Mean F
A 14.44 1 14.44 0.65 0.4391
B 222.96 1 222.96 10.02 0.0101
C 525.64 1 525.64 23.63 0.0007
A2
48.47 1 48.47 2.18 0.1707
B2
124.48 1 124.48 5.60 0.0396
C2
388.59 1 388.59 17.47 0.0019
AB 36.13 1 36.13 1.62 0.2314
AC 1035.13 1 1035.13 46.53 < 0.0001
BC 120.12 1 120.12 5.40 0.0425
Residual 222.47 10 22.25
Pure Error 166.00 5 33.20
Cor Total 287.28 19
Intercept 81.09 1 1.92 76.81 85.38
A-Time 1.03 1 1.28 -1.82 3.87 1.00
B-Temperature 4.04 1 1.28 1.20 6.88 1.00
C-Catalyst 6.20 1 1.28 3.36 9.05 1.00
A2 -1.83 1 1.24 -4.60 0.93 1.02
B2 2.94 1 1.24 0.17 5.71 1.02
C2 -5.19 1 1.24 -7.96 -2.42 1.02
AB 2.13 1 1.67 -1.59 5.84 1.00
AC 11.38 1 1.67 7.66 15.09 1.00
BC -3.87 1 1.67 -7.59 -0.16 1.00
Conversion =
+81.09
+1.03 * A
+4.04 * B
+6.20 * C
-1.83 * A2
+2.94 * B2
-5.19 * C2
+2.13 * A * B
+11.38 * A * C
-3.87 * B * C
Response: Activity
Sum of Mean F
A 175.35 1 175.35 82.34 < 0.0001
C 67.91 1 67.91 31.89 < 0.0001
A2
9.94 1 9.94 4.67 0.0463
Residual 34.07 16 2.13
12-5

Cor Total 287.28 19
Intercept 59.95 1 0.42 59.06 60.83
A-Time 3.58 1 0.39 2.75 4.42 1.00
C-Catalyst 2.23 1 0.39 1.39 3.07 1.00
A2
0.82 1 0.38 0.015 1.63 1.00
Activity =
+59.95
+3.58 * A
+2.23 * C
+0.82 * A2
The following contour plots of conversion, activity, and POE and the corresponding optimization plot
identify a region where conversion is maximized, activity is between 55 and 60, and the transmitted
variability from temperature is minimized. Factor A is set at 0.5 while C is set at 0.4.
Conversion
A: Time
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
65
70
70
75
75
80
85
90
666666
Activity
A: Time
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
56
58
60
62
64
66
666666
12-6

POE(Conversion)
A: Time
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
5
6
7
8
9
10
666666
Overlay Plot
A: Time
C:Catalyst
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Conversion: 80
POE(Conversion): 5
Activity: 60
666666
12-4 Reconsider the leaf spring experiment from Table 12-1. Suppose that factors A, B and C are
controllable variables, and that factors D and E are noise factors. Set up a crossed array design to
investigate this problem, assuming that all of the two-factor interactions involving the controllable
variables are thought to be important. What type of design have you obtained?
The following experimental design has a 23
inner array for the controllable variables and a 22
outer array
for the noise factors. A total of 32 runs are required.
Outer Array
Inner Array D -1 1 -1 1
A B C E -1 -1 1 1
-1 -1 -1
1 -1 -1
-1 1 -1
1 1 -1
-1 -1 1
1 -1 1
-1 1 1
1 1 1
12-5 Continuation of Problem 12-4. Reconsider the leaf spring experiment from Table 12-1.
Suppose that A, B and C are controllable factors and that factors D and E are noise factors. Show how a
combined array design can be employed to investigate this problem that allows all two-factor interactions
to be estimated and only require 16 runs. Compare this with the crossed array design from Problem 12-5.
Can you see how in general combined array designs that have fewer runs than crossed array designs?
The following experiment is a 25-1
fractional factorial experiment where the controllable factors are A, B,
and C and the noise factors are D and E. Only 16 runs are required versus the 32 runs required for the
crossed array design in problem 12-4.
12-7

A B C D E Free Height
- - - - +
+ - - - -
- + - - -
+ + - - +
- - + - -
+ - + - +
- + + - +
+ + + - -
- - - + -
+ - - + +
- + - + +
+ + - + -
- - + + +
+ - + + -
- + + + -
+ + + + +
12-6 Consider the connector pull-off force experiment shown in Table 12-2. What main effects and
interactions involving the controllable variables can be estimated with this design? Remember that all of
the controllable variables are quantitative factors.
The design in Table 12-2 contains a 34-2
inner array for the controllable variables. This is a resolution III
design which aliases the main effects with two factor interactions. The alias table below identifies the alias
structure for this design. Because of the partial aliasing in this design, it is difficult to interpret the
interactions.
Alias Matrix
[Est. Terms] Aliased Terms
[Intercept] = Intercept - BC - BD - CD
[A] = A - 0.5 * BC - 0.5 * BD - 0.5 * CD
[B] = B - 0.5 * AC - 0.5 * AD
[C] = C - 0.5 * AB - 0.5 * AD
[D] = D - 0.5 * AB - 0.5 * AC
[A2] = A2 + 0.5 * BC + 0.5 * BD + 0.5 * CD
[B2] = B2 + 0.5 * AC - 0.5 * AD + CD
[C2] = C2 - 0.5 * AB + 0.5 * AD + BD
[D2] = D2 + 0.5 * AB - 0.5 * AC + BC
12-7 Consider the connector pull-off force experiment shown in Table 12-2. Show how an experiment
can be designed for this problem that will allow a full quadratic model to be fit in the controllable variables
along all main effects of the noise variables and their interactions with the controllable variables. How
many runs will be required in this design? How does this compare with the design in Table 12-2?
There are several designs that can be employed to achieve the requirements stated above. Below is a small
central composite design with the axial points removed for the noise variables. Five center points are also
included which brings the total runs to 35. As shown in the alias analysis, the full quadratic model for the
controllable variables is achieved.
12-8

A B C D E F G
+1 +1 +1 -1 +1 +1 +1
+1 +1 -1 +1 -1 +1 -1
+1 +1 -1 +1 +1 -1 +1
+1 -1 +1 +1 -1 +1 +1
-1 +1 +1 -1 -1 +1 -1
+1 -1 -1 -1 +1 -1 -1
-1 +1 -1 +1 +1 -1 +1
+1 +1 +1 +1 -1 +1 -1
+1 -1 +1 -1 -1 -1 +1
-1 -1 -1 -1 +1 +1 -1
-1 +1 -1 +1 -1 -1 -1
+1 +1 +1 -1 +1 -1 -1
+1 -1 -1 +1 -1 -1 -1
-1 -1 +1 -1 -1 -1 +1
-1 +1 -1 -1 -1 +1 +1
+1 -1 -1 -1 -1 +1 +1
-1 +1 -1 -1 +1 +1 +1
+1 -1 -1 +1 +1 +1 +1
-1 -1 +1 +1 +1 -1 +1
-1 -1 +1 +1 +1 +1 -1
-1 +1 +1 +1 -1 -1 +1
-1 -1 -1 -1 -1 -1 -1
-2.17 0 0 0 0 0 0
2.17 0 0 0 0 0 0
0 -2.17 0 0 0 0 0
0 2.17 0 0 0 0 0
0 0 -2.17 0 0 0 0
0 0 2.17 0 0 0 0
0 0 0 -2.17 0 0 0
0 0 0 2.17 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Alias Matrix
[Est. Terms] Aliased Terms
[Intercept] = Intercept
[A] = A
[B] = B
[C] = C
[D] = D
[E] = E + 0.211 * EG + 0.789 * FG
[F] = F - EF - EG
[G] = G - EF - 0.158 * EG + 0.158 * FG
[A2
] = A2
[B2
] = B2
[C2
] = C2
[D2
] = D2
[E2
] = E2
+ F2
+ G2
[AB] = AB - 0.105 * EG - 0.895 * FG
[AC] = AC - 0.158 * EG + 0.158 * FG
[AD] = AD + 0.421 * EG + 0.579 * FG
[AE] = AE - 0.474 * EG + 0.474 * FG
[AF] = AF + EF + 1.05 * EG - 0.0526 * FG
12-9

[AG] = AG + EF + 1.05 * EG - 0.0526 * FG
[BC] = BC - 0.263 * EG + 0.263 * FG
[BD] = BD - EF - 0.158 * EG + 0.158 * FG
[BE] = BE - 0.368 * EG + 0.368 * FG
[BF] = BF + 1.11 * EG - 0.105 * FG
[BG] = BG + EF + 0.421 * EG - 0.421 * FG
[CD] = CD - 0.421 * EG + 0.421 * FG
[CE] = CE - EF + 0.158 * EG + 0.842 * FG
[CF] = CF - EF - 0.211 * EG + 0.211 * FG
[CG] = CG - 1.21 * EG + 0.211 * FG
[DE] = DE - 0.842 * EG - 0.158 * FG
[DF] = DF - 0.211 * EG + 0.211 * FG
[DG] = DG - EF + 0.263 * EG - 0.263 * FG
12-8 Consider the experiment in Problem 11-11. Suppose that pressure is a noise variable ( in
coded units). Fit the response model for the viscosity response. Find a set of conditions that result in
viscosity as close as possible to 600 and that minimize the variability transmitted from the noise variable
pressure.
2
1zσ =
Response: Viscosity
Sum of Mean F
A 703.12 1 703.12 0.60 0.4615
B 6105.12 1 6105.12 5.19 0.0522
C 5408.00 1 5408.00 4.60 0.0643
A2 21736.93 1 21736.93 18.49 0.0026
C2 5153.80 1 5153.80 4.38 0.0696
AC 47742.25 1 47742.25 40.61 0.0002
Residual 9404.00 8 1175.50
Pure Error 1482.00 2 741.00
Cor Total 94871.33 14
Viscosity =
+636.00
+9.37 * A
+27.62 * B
-26.00 * C
-76.50 * A2
-37.25 * C2
+109.25 * A * C
From the final equation shown in the above analysis, the mean model and corresponding contour plot is
shown below.
( ) 2
1 1 2, 636.00 9.37 27.62 76.50z 1E y z x x x⎡ ⎤ = + + −⎣ ⎦x
12-10

Viscosity
A: Temperatue
B:AgitationRate
-1.000 -0.500 0.000 0.500 1.000
-1.000
-0.500
0.000
0.500
1.000
540
560
560
580
580
600 600
620
640
660
333
525
550
575
600
625
650
675
Viscosity
-1.000
-0.500
0.000
0.500
1.000 -1.000
-0.500
0.000
0.500
1.000
A: Temperatue
B: Agitation Rate
Contour and 3-D plots of the POE are shown below.
POE(Viscosity)
A: Temperatue
B:AgitationRate
-1.000 -0.500 0.000 0.500 1.000
-1.000
-0.500
0.000
0.500
1.000
35 3536 3637 3738394041
42
43 333
34
36
38
40
42
44
POE(Viscosity)
-1.000
-0.500
0.000
0.500
1.000 -1.000
-0.500
0.000
0.500
1.000
A: Temperatue
B: Agitation Rate
The stacked contour plots below identify a region with viscosity between 590 and 610 while minimizing
the variability transmitted from the noise variable pressure. The conditions are in the region of factor A =
0.5 and factor B = -1.
12-11

Overlay Plot
A: Temperatue
B:AgitationRate
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Viscosity: 590
Viscosity: 590
Viscosity: 600
Viscosity: 600POE(Viscosity): 35
POE(Viscosity): 35
333
12-9 A variation of Example 12-1. In example 12-1 (which utilized data from Example 6-2) we found
that one of the process variables (B = pressure) was not important. Dropping this variable produced two
replicates of a 23
design. The data are shown below.
C D A(+) A(-) y 2
s
- - 45, 48 71, 65 57.75 121.19
+ - 68, 80 60, 65 68.25 72.25
- + 43, 45 100, 104 73.00 1124.67
+ + 75, 70 86, 96 81.75 134.92
Assume that C and D are controllable factors and that A is a noise factor.
(a) Fit a model to the mean response.
The following is the analysis of variance with all terms in the model:
Response: Mean
Sum of Mean F
Model 300.05 3 100.02
A 92.64 1 92.64
B 206.64 1 206.64
AB 0.77 1 0.77
Pure Error 0.000 0
Cor Total 300.05 3
Based on the above analysis, the AB interaction is removed from the model and used as error.
Response: Mean
Sum of Mean F
12-12

A 92.64 1 92.64 121.00 0.0577
B 206.64 1 206.64 269.90 0.0387
Cor Total 300.05 3
Intercept 70.19 1 0.44 64.63 75.75
A-Concentration 4.81 1 0.44 -0.75 10.37 1.00
B-Stir Rate 7.19 1 0.44 1.63 12.75 1.00
Mean =
+70.19
+4.81 * A
+7.19 * B
Mean =
+70.18750
+4.81250 * Concentration
+7.18750 * Stir Rate
The following is a contour plot of the mean model:
Mean
A: Concentration
B:StirRate
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
60
65
70
75
80
(b) Fit a model to the ln(s2
) response.
The following is the analysis of variance with all terms in the model:
Response: Variance Transform: Natural log Constant: 0
12-13

Sum of Mean F
Model 4.42 3 1.47
A 1.74 1 1.74
B 2.03 1 2.03
AB 0.64 1 0.64
Pure Error 0.000 0
Cor Total 4.42 3
Based on the above analysis, the AB interaction is removed from the model and applied to the residual
error.
Response: Variance Transform: Natural log Constant: 0
Sum of Mean F
A 1.74 1 1.74 2.71 0.3477
B 2.03 1 2.03 3.17 0.3260
Cor Total 4.42 3
Intercept 5.25 1 0.40 0.16 10.34
A-Concentration -0.66 1 0.40 -5.75 4.43 1.00
B-Stir Rate 0.71 1 0.40 -4.38 5.81 1.00
Ln(Variance) =
+5.25
-0.66 * A
+0.71 * B
Ln(Variance) =
+5.25185
-0.65945 * Concentration
+0.71311 * Stir Rate
The following is a contour plot of the variance model in the untransformed form:
12-14

Variance
A: Concentration
B:StirRate
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
100
150
200
250
300
350
400
450
500550
600650
(c) Find operating conditions that result in the mean filtration rate response exceeding 75 with minimum
variance.
The overlay plot shown below identifies the region required by the process:
Overlay Plot
A: Concentration
B:StirRate
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Mean: 75
Variance: 130
(d) Compare your results with those from Example 12-1 which used the transmission of error approach.
How similar are the two answers.
The results are very similar. Both require the Concentration to be held at the high level while the stirring
rate is held near the middle.
12-10 In an article (“Let’s All Beware the Latin Square,” Quality Engineering, Vol. 1, 1989, pp. 453-
465) J.S. Hunter illustrates some of the problems associated with 3k-p
fractional factorial designs. Factor A
12-15

is the amount of ethanol added to a standard fuel and factor B represents the air/fuel ratio. The response
variable is carbon monoxide (CO) emission in g/m2
. The design is shown below.
Design Observations
A B x1 x2 y y
0 0 -1 -1 66 62
1 0 0 -1 78 81
2 0 1 -1 90 94
0 1 -1 0 72 67
1 1 0 0 80 81
2 1 1 0 75 78
0 2 -1 1 68 66
1 2 0 1 66 69
2 2 1 1 60 58
Notice that we have used the notation system of 0, 1, and 2 to represent the low, medium, and high levels
for the factors. We have also used a “geometric notation” of -1, 0, and 1. Each run in the design is
replicated twice.
(a) Verify that the second-order model
21
2
2
2
121 0904540754578 xx.x.x.x.x..yˆ −−−−+=
is a reasonable model for this experiment. Sketch the CO concentration contours in the x1, x2 space.
In the computer output that follows, the “coded factors” model is in the -1, 0, +1 scale.
Response: CO Emis
Sum of Mean F
A 243.00 1 243.00 38.12 < 0.0001
B 588.00 1 588.00 92.24 < 0.0001
A2 81.00 1 81.00 12.71 0.0039
B2 64.00 1 64.00 10.04 0.0081
AB 648.00 1 648.00 101.65 < 0.0001
Residual 76.50 12 6.37
Pure Error 46.50 9 5.17
Intercept 78.50 1 1.33 75.60 81.40
A-Ethanol 4.50 1 0.73 2.91 6.09 1.00
B-Air/Fuel Ratio -7.00 1 0.73 -8.59 -5.41 1.00
A2 -4.50 1 1.26 -7.25 -1.75 1.00
B2 -4.00 1 1.26 -6.75 -1.25 1.00
AB -9.00 1 0.89 -10.94 -7.06 1.00
12-16

CO Emis =
+78.50
+4.50 * A
-7.00 * B
-4.50 * A2
-4.00 * B2
-9.00 * A * B
CO Emis
A: Ethanol
B:Air/FuelRatio
-1 -0.5 0 0.5 1
-1.00
-0.50
0.00
0.50
1.00
65
65
70
75
80
85
2 2 2
2 2 2
2 2 2
(b) Now suppose that instead of only two factors, we had used four factors in a 34-2
fractional factorial
design and obtained exactly the same data in part (a). The design would be as follows:
Design Observations
A B C D x1 x2 x3 x4 y y
0 0 0 0 -1 -1 -1 -1 66 62
1 0 1 1 0 -1 0 0 78 81
2 0 2 2 +1 -1 +1 +1 90 94
0 1 2 1 -1 0 +1 0 72 67
1 1 0 2 0 0 -1 +1 80 81
2 1 1 0 +1 0 0 -1 75 78
0 2 1 2 -1 +1 0 +1 68 66
1 2 2 0 0 +1 +1 -1 66 69
2 2 0 1 +1 +1 -1 0 60 58
(c) The design in part (b) allows the model
∑ ∑= =
+++=
4
1
4
1
2
0
i i
iiiii xxy εβββ
to be fitted. Suppose that the true model is
∑ ∑ ∑∑= = <
++++=
4
1
4
1
2
0
i i ji
jiijiiiii xxxxy εββββ
12-17

Show that if represents the least squares estimates of the coefficients in the fitted model, thenβj
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) 1323124444
1412243333
3414132222
24231111
231244
241233
34141322
242311
34141300
2
2
2
2
2
2
2
2
βββββ
βββββ
βββββ
ββββ
ββββ
ββββ
βββββ
ββββ
βββββ
+−−=
+−−=
+++=
−−=
+−=
+−=
++−=
+−=
−−−=
/Ê
/Ê
/Ê
/Ê
/Ê
/Ê
/Ê
/Ê
Ê
Does this help explain the strong effects for factors C and D observed graphically in part (b)?
Let X1 = and X
β0
1
1
1
1
1
1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
β1
1
0
1
1
0
1
1
0
1
−
−
−
β2
1
1
1
0
0
0
1
1
1
−
−
−
β3
1
0
1
1
1
0
1
1
0
−
−
−
β4
1
0
1
0
1
1
1
1
0
−
−
−
β11
1
0
1
1
0
1
1
0
1
β22
1
1
1
0
0
0
1
1
1
β33
1
0
1
1
1
0
1
1
0
β44
1
0
1
0
1
1
1
1
0
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
2 =
β12
1
0
1
0
0
0
1
0
1
−
−
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
β13
1
0
1
1
0
0
0
0
1
−
−
β14
1
0
1
0
0
1
1
0
0
−
−
β23
1
0
1
0
0
0
0
1
1
−
−
β24
1
0
1
0
0
0
1
1
0
−
−
β34
1
0
1
0
1
0
0
1
0
−
−
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
Then, = A =( )A X X X X=
−
1 1
1
1 2
' '
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−−
−−
−−−
−−
−−−
00210121
02101021
210021210
02121000
00210021
02100021
210021210
02121000
100110
12-18

⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
++−
−++
+++
+−
−−
−−
−−−
−−
−−−
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−−
−−
−−−
−−
−−−
+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
23131244
24141233
34141322
242311
23124
24123
3414132
24231
3414130
34
24
23
14
13
12
44
33
22
11
4
3
2
1
0
44
33
22
11
4
3
2
1
0
2121
2121
212121
2121
2121
2121
212121
2121
00210121
02101021
210021210
02121000
00210021
02100021
210021210
02121000
100110
ββββ
ββββ
ββββ
βββ
βββ
βββ
ββββ
βββ
ββββ
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
β
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
E
12-11 An experiment has been run in a process that applies a coating material to a wafer. Each run in the
experiment produced a wafer, and the coating thickness was measured several times at different locations
on the wafer. Then the mean y1, and standard deviation y2 of the thickness measurement was obtained. The
data [adapted from Box and Draper (1987)] are shown in the table below.
Run Speed Pressure Distance Mean (y1) Std Dev (y2)
1 -1.000 -1.000 -1.000 24.0 12.5
2 0.000 -1.000 -1.000 120.3 8.4
3 1.000 -1.000 -1.000 213.7 42.8
4 -1.000 0.000 -1.000 86.0 3.5
5 0.000 0.000 -1.000 136.6 80.4
6 1.000 0.000 -1.000 340.7 16.2
7 -1.000 1.000 -1.000 112.3 27.6
8 0.000 1.000 -1.000 256.3 4.6
9 1.000 1.000 -1.000 271.7 23.6
10 -1.000 -1.000 0.000 81.0 0.0
11 0.000 -1.000 0.000 101.7 17.7
12 1.000 -1.000 0.000 357.0 32.9
13 -1.000 0.000 0.000 171.3 15.0
14 0.000 0.000 0.000 372.0 0.0
15 1.000 0.000 0.000 501.7 92.5
16 -1.000 1.000 0.000 264.0 63.5
17 0.000 1.000 0.000 427.0 88.6
18 1.000 1.000 0.000 730.7 21.1
19 -1.000 -1.000 1.000 220.7 133.8
20 0.000 -1.000 1.000 239.7 23.5
21 1.000 -1.000 1.000 422.0 18.5
22 -1.000 0.000 1.000 199.0 29.4
23 0.000 0.000 1.000 485.3 44.7
24 1.000 0.000 1.000 673.7 158.2
25 -1.000 1.000 1.000 176.7 55.5
26 0.000 1.000 1.000 501.0 138.9
27 1.000 1.000 1.000 1010.0 142.4
(a) What type of design did the experimenters use? Is this a good choice of design for fitting a
quadratic model?
The design is a 33
. A better choice would be a 23
central composite design. The CCD gives more
information over the design region with fewer points.
(b) Build models of both responses.
The model for the mean is developed as follows:
Response: Mean
12-19

Sum of Mean F
A 5.640E+005 1 5.640E+005 185.16 < 0.0001
B 2.155E+005 1 2.155E+005 70.75 < 0.0001
C 3.111E+005 1 3.111E+005 102.14 < 0.0001
AB 52324.81 1 52324.81 17.18 0.0006
AC 68327.52 1 68327.52 22.43 0.0001
BC 22794.08 1 22794.08 7.48 0.0131
ABC 54830.16 1 54830.16 18.00 0.0004
Residual 57874.57 19 3046.03
Cor Total 1.347E+006 26
PRESS 1.271E+005 Adeq Precision 33.333
Intercept 314.67 1 10.62 292.44 336.90
A-Speed 177.01 1 13.01 149.78 204.24 1.00
B-Pressure 109.42 1 13.01 82.19 136.65 1.00
C-Distance 131.47 1 13.01 104.24 158.70 1.00
AB 66.03 1 15.93 32.69 99.38 1.00
AC 75.46 1 15.93 42.11 108.80 1.00
BC 43.58 1 15.93 10.24 76.93 1.00
ABC 82.79 1 19.51 41.95 123.63 1.00
Mean =
+314.67
+177.01 * A
+109.42 * B
+131.47 * C
+66.03 * A * B
+75.46 * A * C
+43.58 * B * C
+82.79 * A * B * C
Mean =
+314.67037
+177.01111 * Speed
+109.42222 * Pressure
+131.47222 * Distance
+66.03333 * Speed * Pressure
+75.45833 * Speed * Distance
+43.58333 * Pressure * Distance
+82.78750 * Speed * Pressure * Distance
The model for the Std. Dev. response is as follows. A square root transformation was applied to correct
problems with the normality assumption.
Response: Std. Dev. Transform: Square root Constant: 0
ANOVA for Response Surface Linear Model
Sum of Mean F
A 16.52 1 16.52 1.84 0.1878
12-20

B 26.32 1 26.32 2.94 0.1001
C 73.92 1 73.92 8.25 0.0086
Residual 206.17 23 8.96
Cor Total 322.92 26
Intercept 6.00 1 0.58 4.81 7.19
A-Speed 0.96 1 0.71 -0.50 2.42 1.00
B-Pressure 1.21 1 0.71 -0.25 2.67 1.00
C-Distance 2.03 1 0.71 0.57 3.49 1.00
Sqrt(Std. Dev.) =
+6.00
+0.96 * A
+1.21 * B
+2.03 * C
Sqrt(Std. Dev.) =
+6.00273
+0.95796 * Speed
+1.20916 * Pressure
+2.02643 * Distance
Because Factor A is insignificant, it is removed from the model. The reduced linear model analysis is
shown below:
Response: Std. Dev. Transform: Square root Constant: 0
ANOVA for Response Surface Reduced Linear Model
Sum of Mean F
B 26.32 1 26.32 2.84 0.1051
C 73.92 1 73.92 7.97 0.0094
Residual 222.68 24 9.28
Cor Total 322.92 26
Intercept 6.00 1 0.59 4.79 7.21
B-Pressure 1.21 1 0.72 -0.27 2.69 1.00
C-Distance 2.03 1 0.72 0.54 3.51 1.00
Sqrt(Std. Dev.) =
12-21

+6.00
+1.21 * B
+2.03 * C
Sqrt(Std. Dev.) =
+6.00273
+1.20916 * Pressure
+2.02643 * Distance
The following contour plots graphically represent the two models:
DESIGN-EXPERT Plot
M ean
X = B: Pressure
Y = C: Distance
Design Points
Actual Factor
A: Speed = 1.00
Mean
B: Press ure
C:Distance
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
DESIGN-EXPERT Plot
Sqrt(Std. Dev.)
X = B: Pressure
Y = C: Distance
Design Points
Actual Factor
A: Speed = 1.00
Std. Dev.
B: Press ure
C:Distance
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
(c) Find a set of optimum conditions that result in the mean as large as possible with the standard
deviation less than 60.
The overlay plot identifies a region that meets the criteria of the mean as large as possible with the standard
deviation less than 60. The optimum conditions in coded terms are approximately Speed = 1.0, Pressure =
0.75 and Distance = 0.25.
DESIGN-EXPERT Plot
Overlay Plot
X = B: Pressure
Y = C: Distance
Design Points
Actual Factor
A: Speed = 1.00
Overlay Plot
B: Press ure
C:Distance
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Mean: 700
Std. D ev .: 60
12-22

12-12 Suppose that there are four controllable variables and two noise variables. It is necessary to
estimate the main effects and two-factor interactions of all of the controllable variables, the main effects of
the noise variables, and the two-factor interactions between all controllable and noise factors. If all factors
are at two levels, what is the minimum number of runs that can be used to estimate all of the model
parameters using a combined array design? Use a D-optimal algorithm to find a design.
Twenty-one runs are required for the model, with five additional runs for lack of fit, and five as replicates
for a total of 31 runs as follows.
Std A B C D E F
1 +1 +1 -1 +1 +1 +1
2 -1 +1 -1 +1 -1 -1
3 +1 -1 -1 +1 -1 -1
4 +1 +1 -1 -1 -1 +1
5 -1 +1 -1 -1 +1 +1
6 -1 +1 +1 +1 +1 +1
7 +1 +1 -1 -1 +1 -1
8 -1 -1 +1 +1 -1 -1
9 -1 +1 +1 -1 +1 -1
10 -1 +1 +1 -1 -1 +1
11 +1 -1 +1 +1 +1 +1
12 +1 +1 +1 +1 -1 +1
13 +1 -1 -1 -1 +1 +1
14 +1 +1 +1 -1 +1 +1
15 -1 -1 -1 -1 -1 -1
16 +1 +1 +1 +1 +1 -1
17 -1 -1 -1 +1 -1 +1
18 -1 -1 -1 +1 +1 -1
19 -1 -1 +1 -1 +1 +1
20 +1 -1 +1 -1 +1 -1
21 +1 -1 +1 -1 -1 +1
22 +1 +1 +1 -1 -1 -1
23 +1 -1 -1 -1 -1 -1
24 -1 +1 -1 -1 -1 -1
25 +1 +1 -1 -1 -1 -1
26 -1 -1 +1 -1 -1 -1
27 +1 +1 +1 +1 -1 +1
28 -1 -1 -1 +1 -1 +1
29 +1 +1 +1 +1 +1 -1
30 -1 -1 -1 +1 +1 -1
31 -1 +1 -1 -1 +1 +1
12-13 Suppose that there are four controllable variables and two noise variables. It is necessary to fit a
complete quadratic model in the controllable variables, the main effects of the noise variables, and the two-
factor interactions between all controllable and noise factors. Set up a combined array design for this by
modifying a central composite design.
The following design is a half fraction central composite design with the axial points removed from the
noise factors. The total number of runs is forty-eight which includes eight center points.
Std A B C D E F
1 -1 -1 -1 -1 -1 -1
2 +1 -1 -1 -1 -1 +1
3 -1 +1 -1 -1 -1 +1
4 +1 +1 -1 -1 -1 -1
5 -1 -1 +1 -1 -1 +1
6 +1 -1 +1 -1 -1 -1
7 -1 +1 +1 -1 -1 -1
8 +1 +1 +1 -1 -1 +1
9 -1 -1 -1 +1 -1 +1
10 +1 -1 -1 +1 -1 -1
11 -1 +1 -1 +1 -1 -1
12 +1 +1 -1 +1 -1 +1
13 -1 -1 +1 +1 -1 -1
12-23

14 +1 -1 +1 +1 -1 +1
15 -1 +1 +1 +1 -1 +1
16 +1 +1 +1 +1 -1 -1
17 -1 -1 -1 -1 +1 +1
18 +1 -1 -1 -1 +1 -1
19 -1 +1 -1 -1 +1 -1
20 +1 +1 -1 -1 +1 +1
21 -1 -1 +1 -1 +1 -1
22 +1 -1 +1 -1 +1 +1
23 -1 +1 +1 -1 +1 +1
24 +1 +1 +1 -1 +1 -1
25 -1 -1 -1 +1 +1 -1
26 +1 -1 -1 +1 +1 +1
27 -1 +1 -1 +1 +1 +1
28 +1 +1 -1 +1 +1 -1
29 -1 -1 +1 +1 +1 +1
30 +1 -1 +1 +1 +1 -1
31 -1 +1 +1 +1 +1 -1
32 +1 +1 +1 +1 +1 +1
33 -2.378 0 0 0 0 0
34 +2.378 0 0 0 0 0
35 0 -2.378 0 0 0 0
36 0 +2.378 0 0 0 0
37 0 0 -2.378 0 0 0
38 0 0 +2.378 0 0 0
39 0 0 0 -2.378 0 0
40 0 0 0 +2.378 0 0
41 0 0 0 0 0 0
42 0 0 0 0 0 0
43 0 0 0 0 0 0
44 0 0 0 0 0 0
45 0 0 0 0 0 0
46 0 0 0 0 0 0
47 0 0 0 0 0 0
48 0 0 0 0 0 0
12-14 Reconsider the situation in Problem 12-13. Could a modified small composite design be used for
this problem? Are there any disadvantages associated with the use of the small composite design?
The axial points for the noise factors were removed in following small central composite design. Five
center points are included. The small central composite design does have aliasing with noise factor E
aliased with the AD interaction and noise factor F aliased with the BC interaction. These aliases are
corrected by leaving the axial points for the noise factors in the design.
Std A B C D E F
1 +1 +1 +1 +1 -1 -1
2 +1 +1 +1 -1 +1 -1
3 +1 +1 -1 +1 -1 +1
4 +1 -1 +1 -1 +1 +1
5 -1 +1 -1 +1 +1 +1
6 +1 -1 +1 +1 -1 +1
7 -1 +1 +1 -1 -1 -1
8 +1 +1 -1 -1 +1 +1
9 +1 -1 -1 +1 -1 -1
10 -1 -1 +1 -1 -1 +1
11 -1 +1 -1 -1 -1 +1
12 +1 -1 -1 -1 +1 -1
13 -1 -1 -1 +1 +1 -1
14 -1 -1 +1 +1 +1 +1
15 -1 +1 +1 +1 +1 -1
16 -1 -1 -1 -1 -1 -1
17 -2 0 0 0 0 0
18 +2 0 0 0 0 0
19 0 -2 0 0 0 0
20 0 +2 0 0 0 0
21 0 0 -2 0 0 0
12-24

22 0 0 +2 0 0 0
23 0 0 0 -2 0 0
24 0 0 0 +2 0 0
25 0 0 0 0 0 0
26 0 0 0 0 0 0
27 0 0 0 0 0 0
28 0 0 0 0 0 0
29 0 0 0 0 0 0
12-15 Reconsider the situation in Problem 12-13. What is the minimum number of runs that can be used
to estimate all of the model parameters using a combined array design? Use a D-optimal algorithm to find
a reasonable design for this problem.
The following design is a 36 run D-optimal design with five runs included for lack of fit and five as
replicates.
Std A B C D E F
1 +1 +1 +1 -1 -1 -1
2 -1 +1 -1 -1 +1 +1
3 -1 +1 +1 +1 -1 +1
4 +1 +1 -1 +1 -1 -1
5 -1 -1 +1 +1 -1 -1
6 -1 +1 -1 -1 -1 -1
7 +1 -1 -1 +1 +1 -1
8 +1 -1 +1 -1 +1 -1
9 +1 +1 -1 +1 +1 +1
10 +1 -1 -1 -1 -1 -1
11 +1 -1 +1 +1 -1 +1
12 -1 +1 -1 +1 +1 -1
13 +1 +1 +1 -1 +1 +1
14 +1 +1 -1 -1 -1 +1
15 +1 +1 +1 +1 +1 -1
16 -1 -1 -1 +1 -1 +1
17 0 -1 -1 -1 +1 -1
18 0 -1 +1 -1 -1 +1
19 0 +1 0 0 0 0
20 0 0 0 -1 0 0
21 0 0 +1 0 0 0
22 -1 +1 +1 -1 +1 -1
23 -1 -1 +1 0 +1 +1
24 +1 +1 -1 -1 +1 -1
25 0 -1 +1 +1 +1 +1
26 +1 -1 -1 -1 +1 +1
27 -1 -1 -1 0 -1 -1
28 +1 -1 0 +1 -1 -1
29 -1 -1 0 +1 +1 -1
30 +1 -1 -1 0 -1 +1
31 -1 0 -1 +1 +1 +1
32 +1 +1 +1 +1 +1 -1
33 +1 -1 +1 -1 +1 -1
34 -1 +1 +1 +1 -1 +1
35 +1 +1 +1 -1 -1 -1
36 +1 +1 -1 +1 +1 +1
12-16 An experiment was run in a wave soldering process. There are five controllable variables and
three noise variables. The response variable is the number of solder defects per million opportunities. The
experimental design employed was the crossed array shown below.
Outer Array
F -1 1 1 -1
Inner Array G -1 1 -1 1
A B C D E H -1 -1 1 1
12-25

1 1 1 -1 -1 194 197 193 275
1 1 -1 1 1 136 136 132 136
1 -1 1 -1 1 185 261 264 264
1 -1 -1 1 -1 47 125 127 42
-1 1 1 1 -1 295 216 204 293
-1 1 -1 -1 1 234 159 231 157
-1 -1 1 1 1 328 326 247 322
-1 -1 -1 -1 -1 186 187 105 104
(a) What types of designs were used for the inner and outer arrays?
The inner array is a 25-2
fractional factorial design with a defining relation of I = -ACD = -BCE = ABDE.
The outer array is a 23-1
fractional factorial design with a defining relation of I = -FGH.
(b) Develop models for the mean and variance of solder defects. What set of operating conditions would
you recommend?
A B C D E y 2
s
1 1 1 -1 -1 214.75 1616.25
1 1 -1 1 1 135.00 4.00
1 -1 1 -1 1 243.50 1523.00
1 -1 -1 1 -1 85.25 2218.92
-1 1 1 1 -1 252.00 2376.67
-1 1 -1 -1 1 195.25 1852.25
-1 -1 1 1 1 305.75 1540.25
-1 -1 -1 -1 -1 145.50 2241.67
The following analysis identifies factors A, C, and E as being significant for the solder defects mean model.
DESIGN-EXPERT Plot
Solder Def ects Mean
A: A
B: B
C: C
D: D
E: E
Half Normal plot
HalfNormal%probability
|Effect|
0.00 28.44 56.87 85.31 113.75
0
20
40
60
70
80
85
90
95
97
99
A
C
E
Response: Solder Defects Mean
Sum of Mean F
A 6050.00 1 6050.00 97.78 0.0006
12-26

C 25878.13 1 25878.13 418.23 < 0.0001
E 4140.50 1 4140.50 66.92 0.0012
Residual 247.50 4 61.88
The "Model F-value" of 194.31implies the model is not significant relative to the noise. There is a
Solder Defects Mean =
+197.13
-27.50 * A
+56.88 * C
+22.75 * E
Although the natural log transformation is often utilized for variance response, a power transformation
actually performed better for this problem per the Box-Cox plot below. The analysis for the solder defect
variance follows.
DESIGN-EXPERT Plot
(Solder Def ects Variance)^2.04
Lambda
Current = 2.04
Best = 2.04
Low C.I. = 1.63
High C.I. = 2.4
Recommend transf orm:
Power
(Lambda = 2.04)
Lambda
Ln(ResidualSS)
9.93
18.11
26.30
34.48
42.66
-3 -2 -1 0 1 2 3
DESIGN-EXPERT Plot
A: A
B: B
C: C
D: D
E: E
Half Normal plot
|Effect|
0.00 864648.81 1729297.632593946.443458595.25
0
20
40
60
70
80
85
90
95
97
99
A
B
E
AB
Response: Solder Defects VarianceTransform: Power Lambda: 2.04 Constant: 0
Sum of Mean F
A 1.023E+013 1 1.023E+013 293.08 0.0004
B 1.979E+012 1 1.979E+012 56.70 0.0049
E 2.392E+013 1 2.392E+013 685.33 0.0001
AB 9.289E+012 1 9.289E+012 266.11 0.0005
Residual 1.047E+011 3 3.491E+010
Cor Total 4.553E+013 7
Std. Dev. 1.868E+005 R-Squared 0.9977
Mean 4.461E+006 Adj R-Squared 0.9946
12-27

PRESS 7.447E+011 Adeq Precision 53.318
(Solder Defects Variance)2.04 =
+4.461E+006
-1.131E+006 * A
-4.974E+005 * B
-1.729E+006 * E
-1.078E+006 * A * B
The contour plots of the mean and variance models are shown below along with the overlay plot.
Assuming that we wish to minimize both solder defects mean and variance, a solution is shown in the
overlay plot with factors A = +1, B = +1, C = -1, D = 0, and E near -1.
DESIGN-EXPERT Plot
Solder Def ects Mean
X = A: A
Y = E: E
Actual Factors
B: B = 0.00
C: C = -1.00
D: D = 0.00
Solder Defects Mean
A: A
E:E
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
100
110
120
130
140
150
160
170
180
DESIGN-EXPERT Plot
X = A: A
Y = E: E
Actual Factors
B: B = 1.00
C: C = 0.00
D: D = 0.00
Solder Defects Variance
A: A
E:E
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
600
800
1000
1200
1400
16001800
2000
2200
DESIGN-EXPERT Plot
Ov erlay Plot
X = A: A
Y = E: E
Actual Factors
B: B = 1.00
C: C = -1.00
D: D = 0.00
Overlay Plot
A: A
E:E
-1.00 -0.50 0.00 0.50 1.00
-1.00
-0.50
0.00
0.50
1.00
Solder Defects Mean: 110
Solder Defects Variance: 1600
12-17 Reconsider the wave soldering experiment in Problem 12-16. Find a combined array design for
this experiment that requires fewer runs.
12-28

The following experiment is a 28-4
, resolution IV design with the defining relation I = BCDE = ACDF =
ABCG = ABDH. Only 16 runs are required.
A B C D E F G H
-1 -1 -1 -1 -1 -1 -1 -1
+1 -1 -1 -1 -1 +1 +1 +1
-1 +1 -1 -1 +1 -1 +1 +1
+1 +1 -1 -1 +1 +1 -1 -1
-1 -1 +1 -1 +1 +1 +1 -1
+1 -1 +1 -1 +1 -1 -1 +1
-1 +1 +1 -1 -1 +1 -1 +1
+1 +1 +1 -1 -1 -1 +1 -1
-1 -1 -1 +1 +1 +1 -1 +1
+1 -1 -1 +1 +1 -1 +1 -1
-1 +1 -1 +1 -1 +1 +1 -1
+1 +1 -1 +1 -1 -1 -1 +1
-1 -1 +1 +1 -1 -1 +1 +1
+1 -1 +1 +1 -1 +1 -1 -1
-1 +1 +1 +1 +1 -1 -1 -1
+1 +1 +1 +1 +1 +1 +1 +1
12-18 Reconsider the wave soldering experiment in Problem 12-17. Suppose that it was necessary to fit
a complete quadratic model in the controllable variables, all main effects of the noise variables, and all
controllable variable-noise variable interactions. What design would you recommend?
The following experiment is a small central composite design with five center points; the axial points for
the noise factors have been removed. A total of 45 runs are required.
A B C D E F G H
+1 +1 +1 -1 -1 +1 +1 +1
-1 +1 +1 -1 +1 +1 +1 -1
+1 +1 -1 -1 +1 +1 -1 -1
+1 -1 -1 +1 +1 -1 +1 -1
-1 -1 +1 +1 +1 +1 -1 -1
-1 +1 +1 +1 -1 +1 -1 -1
-1 +1 +1 +1 +1 +1 -1 -1
+1 +1 +1 +1 +1 -1 +1 -1
+1 +1 -1 +1 -1 +1 -1 +1
+1 -1 +1 +1 -1 -1 +1 -1
+1 +1 +1 -1 +1 -1 -1 +1
-1 +1 +1 -1 -1 -1 +1 +1
+1 -1 -1 -1 -1 +1 -1 -1
+1 -1 +1 -1 -1 -1 -1 +1
-1 +1 -1 -1 +1 +1 +1 -1
+1 -1 -1 +1 +1 +1 -1 +1
-1 -1 -1 -1 -1 +1 +1 -1
-1 -1 +1 +1 +1 -1 -1 +1
-1 +1 -1 -1 +1 -1 +1 +1
-1 -1 +1 +1 -1 +1 +1 +1
+1 +1 -1 +1 -1 -1 +1 -1
-1 -1 +1 -1 +1 -1 -1 -1
+1 +1 +1 -1 -1 +1 -1 -1
+1 -1 -1 -1 +1 +1 +1 +1
+1 -1 +1 -1 -1 +1 +1 +1
-1 +1 -1 +1 +1 +1 +1 +1
-1 -1 -1 -1 +1 -1 +1 +1
+1 -1 +1 +1 +1 -1 +1 -1
-1 +1 -1 +1 -1 -1 -1 +1
-1 -1 -1 -1 -1 -1 -1 -1
-2.34 0 0 0 0 0 0 0
2.34 0 0 0 0 0 0 0
12-29

0 -2.34 0 0 0 0 0 0
0 2.34 0 0 0 0 0 0
0 0 -2.34 0 0 0 0 0
0 0 2.34 0 0 0 0 0
0 0 0 -2.34 0 0 0 0
0 0 0 2.34 0 0 0 0
0 0 0 0 -2.34 0 0 0
0 0 0 0 2.34 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
12-30

Chapter 13
Experiments with Random Factors
Solutions
13-1 A textile mill has a large number of looms. Each loom is supposed to provide the same output of
cloth per minute. To investigate this assumption, five looms are chosen at random and their output is noted
at different times. The following data are obtained:
Loom Output (lb/min)
1 14.0 14.1 14.2 14.0 14.1
2 13.9 13.8 13.9 14.0 14.0
3 14.1 14.2 14.1 14.0 13.9
4 13.6 13.8 14.0 13.9 13.7
5 13.8 13.6 13.9 13.8 14.0
(a) Explain why this is a random effects experiment. Are the looms equal in output? Use α = 0.05.
The looms used in the experiment are a random sample of all the looms in the manufacturing area. The
following is the analysis of variance for the data:
Minitab Output
ANOVA: Output versus Loom
Loom random 5 1 2 3 4 5
Analysis of Variance for Output
Source DF SS MS F P
Loom 4 0.34160 0.08540 5.77 0.003
Error 20 0.29600 0.01480
Total 24 0.63760
Source Variance Error Expected Mean Square for Each Term
component term (using restricted model)
1 Loom 0.01412 2 (2) + 5(1)
2 Error 0.01480 (2)
(b) Estimate the variability between looms.
014120
5
01480085402
.
..
n
MSMS
ˆ EModel
=
−
=
−
=τσ
(c) Estimate the experimental error variance.
014802
.MSˆ E ==σ
(d) Find a 95 percent confidence interval for ( )222
σσσ ττ + .
2, 1,
1 1 1 0.08540 1
1 1 0.1288
5 0.01480 3.51
Model
E a n a
MS
L
n MS Fα − −
⎡ ⎤ ⎡ ⎤
= − = × −⎢ ⎥ ⎢ ⎥
⎣ ⎦⎢ ⎥⎣ ⎦
=
12-1

1 2, 1,
2
2 2
2
2 2
1 1 1 0.08540
1 8,56 1 9.6787
5 0.01480
1 1
0.1141 0.9064
Model
E a n a
MS
U
n MS F
L U
L U
α
τ
τ
τ
τ
σ
σ σ
σ
σ σ
− − −
⎡ ⎤ ⎡ ⎤
= − = × −⎢ ⎥ ⎢ ⎥
⎣ ⎦⎢ ⎥⎣ ⎦
≤ ≤
+ ++
≤ ≤
+
=
(e) Analyze the residuals from this experiment. Do you think that the analysis of variance assumptions are
satisfied?
There is nothing unusual about the residual plots; therefore, the analysis of variance assumptions are
satisfied.
0.20.10.0-0.1-0.2
2
1
0
-1
-2
NormalScore
Residual
(response is Output)
14.114.013.913.8
0.2
0.1
0.0
-0.1
-0.2
Fitted Value
Residual
12-2

54321
0.2
0.1
0.0
-0.1
-0.2
Loom
Residual
Residuals Versus Loom
13-2 A manufacturer suspects that the batches of raw material furnished by her supplier differ
significantly in calcium content. There are a large number of batches currently in the warehouse. Five of
these are randomly selected for study. A chemist makes five determinations on each batch and obtains the
following data:
Batch 1 Batch 2 Batch 3 Batch 4 Batch 5
23.46 23.59 23.51 23.28 23.29
23.48 23.46 23.64 23.40 23.46
23.56 23.42 23.46 23.37 23.37
23.39 23.49 23.52 23.46 23.32
23.40 23.50 23.49 23.39 23.38
(a) Is there significant variation in calcium content from batch to batch? Use α = 0.05.
Yes, as shown in the Minitab Output below, there is a difference.
Minitab Output
ANOVA: Calcium versus Batch
Analysis of Variance for Calcium
Source DF SS MS F P
Batch 4 0.096976 0.024244 5.54 0.004
Error 20 0.087600 0.004380
Total 24 0.184576
1 Batch 0.00397 2 (2) + 5(1)
2 Error 0.00438 (2)
(b) Estimate the components of variance.
. .
.στ
2 024244 004380
5
0 00397=
−
=
−
=
MS MS
n
Model E
12-3

.σ 2
0 004380= =MSE
(c) Find a 95 percent confidence interval for ( )222
σσσ ττ + .
9027010350
11
27691
11
115401
11
22
2
22
2
121
12
..
U
U
L
L
.
FMS
MS
n
U
.
FMS
MS
n
L
an,a,E
Model
an,a,E
Model
≤
+
≤
+
≤
+
≤
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−=
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−=
−−−
−−
σσ
σ
σσ
σ
τ
τ
τ
τ
α
α
(d) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
There are five residuals that stand out in the normal probability plot. From the Residual vs. Batch plot, we
see that one point per batch appears to stand out. A natural log transformation was applied to the data but
did not change the results of the residual analysis. Further investigation should probably be performed to
determine if these points are outliers.
0.10.0-0.1
2
1
0
-1
-2
NormalScore
Residual
(response is Calcium)
12-4

54321
0.1
0.0
-0.1
Batch
Residual
Residuals Versus Batch
(response is Calcium)
13-3 Several ovens in a metal working shop are used to heat metal specimens. All the ovens are supposed
to operate at the same temperature, although it is suspected that this may not be true. Three ovens are
selected at random and their temperatures on successive heats are noted. The data collected are as follows:
Oven Temperature
1 491.50 498.30 498.10 493.50 493.60
2 488.50 484.65 479.90 477.35
3 490.10 484.80 488.25 473.00 471.85 478.65
(a) Is there significant variation in temperature between ovens? Use α = 0.05.
The analysis of variance shown below identifies significant variation in temperature between the ovens.
Minitab Output
General Linear Model: Temperature versus Oven
12-5

Oven random 3 1 2 3
Analysis of Variance for Temperat, using Adjusted SS for Tests
Oven 2 594.53 594.53 297.27 8.62 0.005
Error 12 413.81 413.81 34.48
Total 14 1008.34
Expected Mean Squares, using Adjusted SS
Source Expected Mean Square for Each Term
1 Oven (2) + 4.9333(1)
2 Error (2)
Error Terms for Tests, using Adjusted SS
Source Error DF Error MS Synthesis of Error MS
1 Oven 12.00 34.48 (2)
Variance Components, using Adjusted SS
Source Estimated Value
Oven 53.27
Error 34.48
(b) Estimate the components of variance.
n
a
n
n
n
i
i
i
0
2
1
1
1
2
15
25 16 36
15
4 93=
−
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= −
+ +⎡
⎣
⎢
⎤
⎦
⎥ =
∑
∑
∑ .
. .
.
.στ
2 297 27 34 48
4 93
5330=
−
=
−
=
MS MS
n
Model E
.σ 2
34 48= =MSE
(c) Analyze the residuals from this experiment. Draw conclusions about model adequacy.
There is a funnel shaped appearance in the plot of residuals versus predicted value indicating a possible
non-constant variance. There is also some indication of non-constant variance in the plot of residuals
versus oven. The inequality of variance problem is not severe.
100-10
2
1
0
-1
-2
NormalScore
Residual
(response is Temperat)
12-6

495490485480
10
0
-10
Fitted Value
Residual
321
10
0
-10
Oven
Residual
Residuals Versus Oven
13-4 An article in the Journal of the Electrochemical Society (Vol. 139, No. 2, 1992, pp. 524-532)
describes an experiment to investigate the low-pressure vapor deposition of polysilicon. The experiment
was carried out in a large-capacity reactor at Sematech in Austin, Texas. The reactor has several wafer
positions, and four of these positions are selected at random. The response variable is film thickness
uniformity. Three replicates of the experiments were run, and the data are as follows:
Wafer Position Uniformity
1 2.76 5.67 4.49
2 1.43 1.70 2.19
3 2.34 1.97 1.47
4 0.94 1.36 1.65
(a) Is there a difference in the wafer positions? Use α = 0.05.
Yes, there is a difference.
12-7

Minitab Output
ANOVA: Uniformity versus Wafer Position
Wafer Po fixed 4 1 2 3 4
Analysis of Variance for Uniformi
Source DF SS MS F P
Wafer Po 3 16.2198 5.4066 8.29 0.008
Error 8 5.2175 0.6522
Total 11 21.4373
1 Wafer Po 2 (2) + 3Q[1]
2 Error 0.6522 (2)
(b) Estimate the variability due to wafer positions.
. .
.
σ
σ
τ
τ
2
2 54066 0 6522
3
15844
=
−
=
−
=
MS MS
n
Treatment E
(c) Estimate the random error component.
.σ 2
0 6522=
(d) Analyze the residuals from this experiment and comment on model adequacy.
Variability in film thickness seems to depend on wafer position. These observations also show up as
outliers on the normal probability plot. Wafer position number 1 appears to have greater variation in
uniformity than the other positions.
10-1
2
1
0
-1
-2
NormalScore
Residual
(response is Uniformi)
12-8

4321
1
0
-1
Fitted Value
Residual
4321
1
0
-1
Wafer Po
Residual
Residuals Versus Wafer Po
13-5 Consider the vapor deposition experiment described in Problem 13-4.
(a) Estimate the total variability in the uniformity response.
23702652205848122
...ˆˆ =+=+σστ
(b) How much of the total variability in the uniformity response is due to the difference between positions
in the reactor?
708450
23702
58481
22
2
.
.
.
ˆˆ
ˆ
==
+ τ
τ
σσ
σ
(c) To what level could the variability in the uniformity response be reduced, if the position-to-position
variability in the reactor could be eliminated? Do you believe this is a significant reduction?
12-9

The variability would be reduced from 2.2370 to which is a reduction of approximately:652202
.ˆ =σ
%
.
..
71
23702
6522023702
=
−
13-6 An article in the Journal of Quality Technology (Vol. 13, No. 2, 1981, pp. 111-114) describes and
experiment that investigates the effects of four bleaching chemicals on pulp brightness. These four
chemicals were selected at random from a large population of potential bleaching agents. The data are as
follows:
Chemical Pulp Brightness
1 77.199 74.466 92.746 76.208 82.876
2 80.522 79.306 81.914 80.346 73.385
3 79.417 78.017 91.596 80.802 80.626
4 78.001 78.358 77.544 77.364 77.386
(a) Is there a difference in the chemical types? Use α = 0.05.
The computer output shows that the null hypothesis cannot be rejected. Therefore, there is no evidence that
there is a difference in chemical types.
Minitab Output
ANOVA: Brightness versus Chemical
Chemical random 4 1 2 3 4
Analysis of Variance for Brightne
Source DF SS MS F P
Chemical 3 53.98 17.99 0.75 0.538
Error 16 383.99 24.00
Total 19 437.97
1 Chemical -1.201 2 (2) + 5(1)
2 Error 23.999 (2)
(b) Estimate the variability due to chemical types.
. .
.
σ
σ
τ
τ
2
2
17 994 23 999
5
1201
=
−
=
−
= −
MS MS
n
Treatment E
which agrees with the Minitab output.
Because the variance component cannot be negative, this likely means that the variability due to chemical
types is zero.
(c) Estimate the variability due to random error.
.σ 2
23999=
(d) Analyze the residuals from this experiment and comment on model adequacy.
12-10

Two data points appear to be outliers in the normal probability plot of effects. These outliers belong to
chemical types 1 and 3 and should be investigated. There seems to be much less variability in brightness
with chemical type 4.
1050-5
2
1
0
-1
-2
NormalScore
Residual
(response is Brightne)
8281807978
10
5
0
-5
Fitted Value
Residual
12-11

4321
10
5
0
-5
Chemical
Residual
Residuals Versus Chemical
13-7 Consider the one-way balanced, random effects method. Develop a procedure for finding a 100(1-
α) percent confidence interval for .σ σ στ
2 2 2
/ ( )+
We know that P L U≤ ≤
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= −
σ
σ
ατ
2
2
1
P L U+ ≤ + ≤ +
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= −1 1
2
2
2
2
σ
σ
σ
σ
ατ
1
P L U+ ≤
+
≤ +
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= −1 1
2 2
2
σ σ
σ
ατ
1
P
L
L
U
U1 1
1
2
2 2+
≥
+
≥
+
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= −
σ
σ σ
α
τ
(a) What is the probability of accepting H0 if is four times the error variance ?στ
2
σ 2
( ) 6421
45
11 2
2
2
2
.
n
==+=+=
σ
σ
σ
σ
λ τ
υ1 1 4= − =a υ2 25 5 20= − = − =N a β ≈ 0 035. , from the OC curve.
(b) If the difference between looms is large enough to increase the standard deviation of an observation by
20 percent, we wish to detect this with a probability of at least 0.80. What sample size should be used?
υ1 1 4= − =a υ2 25 5 20= − = − =N a α = 0 05. P accept( ) ≤ 0 2.
( )[ ] ( )( )[ ] n..nP.n 440112001011101011 22
+=−++=−++=λ
12-12

n υ2 λ P(accept)
5 20 1.79 0.6
10 45 2.32 0.3
14 65 2.67 0.2
13-9 An experiment was performed to investigate the capability of a measurement system. Ten parts
were randomly selected, and two randomly selected operators measured each part three times. The tests
were made in random order, and the data below resulted.
Operator 1 Operator 2
Measurements MeasurementsPart
Number 1 2 3 1 2 3
1 50 49 50 50 48 51
2 52 52 51 51 51 51
3 53 50 50 54 52 51
4 49 51 50 48 50 51
5 48 49 48 48 49 48
6 52 50 50 52 50 50
7 51 51 51 51 50 50
8 52 50 49 53 48 50
9 50 51 50 51 48 49
10 47 46 49 46 47 48
(a) Analyze the data from this experiment.
Minitab Output
ANOVA: Measurement versus Part, Operator
Part random 10 1 2 3 4 5 6 7
8 9 10
Operator random 2 1 2
Analysis of Variance for Measurem
Source DF SS MS F P
Part 9 99.017 11.002 18.28 0.000
Operator 1 0.417 0.417 0.69 0.427
Part*Operator 9 5.417 0.602 0.40 0.927
Error 40 60.000 1.500
Total 59 164.850
1 Part 1.73333 3 (4) + 3(3) + 6(1)
2 Operator -0.00617 3 (4) + 3(3) + 30(2)
3 Part*Operator -0.29938 4 (4) + 3(3)
4 Error 1.50000 (4)
(b) Find point estimates of the variance components using the analysis of variance method.
12-13

σ 2
= MSE .σ 2
15=
στβ
2
=
−MS MS
n
AB E . .
στβ
2 0 6018519 15000000
3
0=
−
< , assume =0στβ
2
σβ
2
=
−MS MS
an
B AB
( )
73331
32
60185190001852112
.
..
ˆ =
−
=βσ
στ
2
=
−MS MS
bn
A AB
( )
0
310
6018519041666702
<
−
=
..
ˆ τσ , assume =0στ
2
All estimates agree with the Minitab output.
13-10 An article by Hoof and Berman (“Statistical Analysis of Power Module Thermal Test Equipment
Performance”, IEEE Transactions on Components, Hybrids, and Manufacturing Technology Vol. 11, pp.
516-520, 1988) describes an experiment conducted to investigate the capability of measurements on
thermal impedance (Cº/W x 100) on a power module for an induction motor starter. There are 10 parts,
three operators, and three replicates. The data are shown in the following table.
Inspector 1 Inspector 2 Inspector 3Part
Number Test 1 Test 2 Test 3 Test 1 Test 2 Test 3 Test 1 Test 2 Test 3
1 37 38 37 41 41 40 41 42 41
2 42 41 43 42 42 42 43 42 43
3 30 31 31 31 31 31 29 30 28
4 42 43 42 43 43 43 42 42 42
5 28 30 29 29 30 29 31 29 29
6 42 42 43 45 45 45 44 46 45
7 25 26 27 28 28 30 29 27 27
8 40 40 40 43 42 42 43 43 41
9 25 25 25 27 29 28 26 26 26
10 35 34 34 35 35 34 35 34 35
(a) Analyze the data from this experiment, assuming both parts and operators are random effects.
Minitab Output
ANOVA: Impedance versus Inspector, Part
Inspecto random 3 1 2 3
Part random 10 1 2 3 4 5 6 7
8 9 10
Analysis of Variance for Impedanc
Source DF SS MS F P
Inspecto 2 39.27 19.63 7.28 0.005
Part 9 3935.96 437.33 162.27 0.000
Inspecto*Part 18 48.51 2.70 5.27 0.000
Error 60 30.67 0.51
Total 89 4054.40
1 Inspecto 0.5646 3 (4) + 3(3) + 30(1)
2 Part 48.2926 3 (4) + 3(3) + 9(2)
3 Inspecto*Part 0.7280 4 (4) + 3(3)
4 Error 0.5111 (4)
12-14

(b) Estimate the variance components using the analysis of variance method.
σ 2
= MSE
2
ˆ 0.51σ =
στβ
2
=
−MS MS
n
AB E 2 2.70 0.51
ˆ 0.73
3
τβσ
−
= =
σβ
2
=
−MS MS
an
B AB
( )
2 437.33 2.70
ˆ 48.29
3 3
βσ
−
= =
στ
2
=
−MS MS
bn
A AB
( )
2 19.63 2.70
ˆ 0.56
10 3
τσ
−
= =
All estimates agree with the Minitab output.
13-11 Reconsider the data in Problem 5-6. Suppose that both factors, machines and operators, are chosen
at random.
Machine
Operator 1 2 3 4
1 109 110 108 110
110 115 109 108
2 110 110 111 114
112 111 109 112
3 116 112 114 120
114 115 119 117
The following Minitab output contains the analysis of variance and the variance component estimates:
Minitab Output
ANOVA: Strength versus Operator, Machine
Operator random 3 1 2 3
Machine random 4 1 2 3 4
Analysis of Variance for Strength
Source DF SS MS F P
Operator 2 160.333 80.167 10.77 0.010
Machine 3 12.458 4.153 0.56 0.662
Operator*Machine 6 44.667 7.444 1.96 0.151
Error 12 45.500 3.792
Total 23 262.958
1 Operator 9.0903 3 (4) + 2(3) + 8(1)
2 Machine -0.5486 3 (4) + 2(3) + 6(2)
3 Operator*Machine 1.8264 4 (4) + 2(3)
4 Error 3.7917 (4)
(b) Find point estimates of the variance components using the analysis of variance method.
12-15

σ 2
= MSE .σ 2
379167=
στβ
2
=
−MS MS
n
AB E . .
.στβ
2 7 44444 379167
2
182639=
−
=
σβ
2
=
−MS MS
an
B AB . .
( )
σβ
2 415278 7 44444
3 2
0=
−
< , assume σβ
2
0=
στ
2
=
−MS MS
bn
A AB . .
( )
.στ
2 8016667 7 44444
4 2
9 09028=
−
=
These results agree with the Minitab variance component analysis.
13-12 Reconsider the data in Problem 5-13. Suppose that both factors are random.
Column Factor
Row Factor 1 2 3 4
1 36 39 36 32
2 18 20 22 20
3 30 37 33 34
Minitab Output
General Linear Model: Response versus Row, Column
Row random 3 1 2 3
Column random 4 1 2 3 4
Analysis of Variance for Response, using Adjusted SS for Tests
Row 2 580.500 580.500 290.250 60.40 **
Column 3 28.917 28.917 9.639 2.01 **
Row*Column 6 28.833 28.833 4.806 **
Error 0 0.000 0.000 0.000
Total 11 638.250
** Denominator of F-test is zero.
1 Row (4) + (3) + 4.0000(1)
2 Column (4) + (3) + 3.0000(2)
3 Row*Column (4) + (3)
4 Error (4)
1 Row * 4.806 (3)
2 Column * 4.806 (3)
3 Row*Column * * (4)
Row 71.3611
Column 1.6111
Row*Column 4.8056
Error 0.0000
12-16

(b) Estimate the variance components.
Because the experiment is unreplicated and the interaction term was included in the model, there is no
estimate of MSE, and therefore, no estimate of .2
σ
στβ
2
=
−MS MS
n
AB E
80564
1
0805642
.
.
ˆ =
−
=τβσ
σβ
2
=
−MS MS
an
B AB
( )
61111
13
80564638992
.
..
ˆ =
−
=βσ
στ
2
=
−MS MS
bn
A AB
( )
361171
14
8056425002902
.
..
ˆ =
−
=τσ
These estimates agree with the Minitab output.
13-13 Suppose that in Problem 5-11 the furnace positions were randomly selected, resulting in a mixed
model experiment. Reanalyze the data from this experiment under this new assumption. Estimate the
appropriate model components.
Temperature (°C)
Position 800 825 850
570 1063 565
1 565 1080 510
583 1043 590
528 988 526
2 547 1026 538
521 1004 532
The following analysis assumes a restricted model:
Minitab Output
ANOVA: Density versus Position, Temperature
Position random 2 1 2
Temperat fixed 3 800 825 850
Analysis of Variance for Density
Source DF SS MS F P
Position 1 7160 7160 16.00 0.002
Temperat 2 945342 472671 1155.52 0.001
Position*Temperat 2 818 409 0.91 0.427
Error 12 5371 448
Total 17 958691
1 Position 745.83 4 (4) + 9(1)
2 Temperat 3 (4) + 3(3) + 6Q[2]
3 Position*Temperat -12.83 4 (4) + 3(3)
4 Error 447.56 (4)
σ 2
= MSE 564472
.ˆ =σ
στβ
2
=
−MS MS
n
AB E
0
3
4484092
<
−
=τβσˆ assume στβ
2
0=
12-17

bn
MSMS
ˆ EA −
=2
τσ
( )
83745
33
44871602
.ˆ =
−
=τσ
These results agree with the Minitab output.
13-14 Reanalyze the measurement systems experiment in Problem 12-9, assuming that operators are a
fixed factor. Estimate the appropriate model components.
Minitab Output
Part random 10 1 2 3 4 5 6 7
8 9 10
Operator fixed 2 1 2
Source DF SS MS F P
Part 9 99.017 11.002 7.33 0.000
Operator 1 0.417 0.417 0.69 0.427
Part*Operator 9 5.417 0.602 0.40 0.927
Error 40 60.000 1.500
Total 59 164.850
1 Part 1.5836 4 (4) + 6(1)
2 Operator 3 (4) + 3(3) + 30Q[2]
3 Part*Operator -0.2994 4 (4) + 3(3)
4 Error 1.5000 (4)
σ 2
= MSE 500012
.ˆ =σ
n
MSMS
ˆ EAB −
=2
τβσ 0
3
500016018502
<
−
=
..
ˆτβσ assume στβ
2
0=
bn
MSMS
ˆ EA −
=2
τσ
( )
583641
32
50000100185112
.
..
ˆ =
−
=τσ
13-15 Reanalyze the measurement system experiment in Problem 13-10, assuming that operators are a
fixed factor. Estimate the appropriate model components.
Minitab Output
ANOVA: Impedance versus Inspector, Part
Inspecto fixed 3 1 2 3
Part random 10 1 2 3 4 5 6 7
8 9 10
Analysis of Variance for Impedanc
Source DF SS MS F P
Inspecto 2 39.27 19.63 7.28 0.005
Part 9 3935.96 437.33 855.64 0.000
Inspecto*Part 18 48.51 2.70 5.27 0.000
12-18

Error 60 30.67 0.51
Total 89 4054.40
1 Inspecto 3 (4) + 3(3) + 30Q[1]
2 Part 48.5353 4 (4) + 9(2)
3 Inspecto*Part 0.7280 4 (4) + 3(3)
4 Error 0.5111 (4)
σ 2
= MSE
2
ˆ 0.51σ =
στβ
2
=
−MS MS
n
AB E 2 2.70 0.51
ˆ 0.73
3
τβσ
−
= =
2
ˆ B EMS MS
an
βσ
−
=
( )
2 437.33 0.51
ˆ 48.54
3 3
βσ
−
= =
13-16 In problem 5-6, suppose that there are only four machines of interest, but the operators were selected
at random.
(a) What type of model is appropriate?
A mixed model is appropriate.
(b) Perform the analysis and estimate the model components.
Minitab Output
ANOVA: Strength versus Operator, Machine
Machine fixed 4 1 2 3 4
Source DF SS MS F P
Operator 2 160.333 80.167 21.14 0.000
Machine 3 12.458 4.153 0.56 0.662
Operator*Machine 6 44.667 7.444 1.96 0.151
Error 12 45.500 3.792
Total 23 262.958
1 Operator 9.547 4 (4) + 8(1)
2 Machine 3 (4) + 2(3) + 6Q[2]
3 Operator*Machine 1.826 4 (4) + 2(3)
4 Error 3.792 (4)
σ 2
= MSE 79232
.ˆ =σ
στβ
2
=
−MS MS
n
AB E
8261
2
792344472
.
..
ˆ =
−
=τβσ
bn
MSMS
ˆ EA −
=2
τσ
( )
5479
24
7923167802
.
..
ˆ =
−
=τσ
12-19

13-17 By application of the expectation operator, develop the expected mean squares for the two-factor
factorial, mixed model. Use the restricted model assumptions. Check your results with the expected mean
squares given in Equation 13-23 to see that they agree.
The sums of squares may be written as
( )∑=
−=
a
i
.....iA yybnSS
1
2
, ( )∑=
−=
b
j
....j.B yyanSS
1
2
( )∑∑= =
+−−=
a
i
b
j
....j...i.ijAB yyyynSS
1 1
2
, ( )∑∑∑= = =
−=
a
i
b
j
n
k
...ijkE yySS
1 1 1
2
Using the model , we may find that( ) ijkijjiijky ετββτµ ++++=
( )
( )
.......
.ijijji.ij
.j.j.j.
..i.ii..i
y
y
y
y
εβµ
ετββτµ
εβµ
εβττµ
++=
++++=
++=
+++=
Using the assumptions for the restricted form of the mixed model, τ. = 0, ( ) 0=j.τβ , which imply that
( ) 0=..τβ . Substituting these expressions into the sums of squares yields
( )( )
( )
( ) ( )( )
( )∑∑∑
∑∑
∑
∑
= = =
= =
=
=
−=
+−−+−=
−+=
−++=
a
i
b
j
n
k
.ijijkE
a
i
b
j
....j...i.ij.iijAB
b
j
....j.jB
a
i
.....i.iA
SS
)nSS
anSS
bnSS
1 1 1
2
1 1
2
1
2
1
2
εε
εεεετβτβ
εεβ
εετβτ
Using the assumption that ( ) 0=ijkE ε , V , andijk( )ε = 0 ( ) 0=⋅ 'k'j'iijkE εε , we may divide each sum of
squares by its degrees of freedom and take the expectation to produce
12-20

( )
( )
( )( )
( )
( )
( )
( )( )
( ) ( )( )
( ) 2
1 1
22
1
22
1
22
11
1
1
σ
βττβσ
βσ
βττσ
=
−⎥
⎦
⎤
⎢
⎣
⎡
−−
+=
⎥
⎦
⎤
⎢
⎣
⎡
−
+=
+⎥
⎦
⎤
⎢
⎣
⎡
−
+=
∑∑
∑
∑
= =
=
=
E
a
i
b
j
.iijAB
b
j
jB
a
i
.iiA
MSE
E
ba
n
MSE
b
an
MSE
E
a
bn
MSE
Note that and are the results given in Table 8-3. We need to simplify and
. Consider
( )BMSE ( EMSE ) )
)
( AMSE
( ABMSE ( )AMSE
( ) ( ) ( ) ( )
( )
( )
( ) ∑
∑
∑ ∑
=
=
= =
−
++=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡ −
+
−
+=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
=++
−
+=
a
i
iA
a
i
iA
a
i
a
i
.iiA
a
bn
nMSE
b
a
a
a
a
bn
MSE
ctscrossproduEE
a
bn
MSE
1
222
2
1
22
1 1
222
1
1
1
0
1
τσσ
στσ
τβτσ
τβ
τβ
since ( is)ijτβ ⎟
⎠
⎞
⎜
⎝
⎛ − 21
0 τβσ
a
a
,NID . Consider ( )ABMSE
( )
( )( )
( ) ( )( )
( )
( )( )
( ) 22
2
1 1
2
1 1
22
11
11
11
τβ
τβ
σσ
σσ
βττβσ
nMSE
a
a
b
b
ba
n
MSE
E
ba
n
MSE
AB
a
i
b
j
AB
a
i
b
j
.iijAB
+=
⎟
⎠
⎞
⎜
⎝
⎛ −
⎟
⎠
⎞
⎜
⎝
⎛ −
−−
+=
−
−−
+=
∑∑
∑∑
= =
= =
Thus and agree with Equation 13-23..( )AMSE ( ABMSE )
13-18 Consider the three-factor factorial design in Example 13-6. Propose appropriate test statistics for all
main effects and interactions. Repeat for the case where A and B are fixed and C is random.
If all three factors are random there are no exact tests on main effects. We could use the following:
BCAC
ABCC
BCAB
ABCB
ACAB
ABCA
MSMS
MSMS
F:C
MSMS
MSMS
F:B
MSMS
MSMS
F:A
+
+
=
+
+
=
+
+
=
12-21

If A and B are fixed and C is random, the expected mean squares are (assuming the restricted for m of the
model):
F F R R
a b c n
Factor i j k l E(MS)
τi 0 b c n
( )∑ −
++
1
2
22
a
bcnbn iτ
σσ τγ
βj a 0 c n
( )∑ −
++
1
2
22
b
acnan
jβ
σσ βγ
γ k a b 1 n σ σγ
2 2
+ abn
( )ijτβ 0 0 c n
( )
( )( )∑∑ −−
++
11
2
22
ba
cnn
jiτβ
σσ τβγ
( )ikτγ 0 b 1 n σ στγ
2 2
+ bn
( )jkβγ a 0 1 n σ σβγ
2 2
+ an
( )ijkτβγ 0 0 1 n σ στβγ
2 2
+ n
( )lijkε 1 1 1 1 σ 2
These are exact tests for all effects.
13-19 Consider the experiment in Example 13-7. Analyze the data for the case where A, B, and C are
random.
Minitab Output
ANOVA: Drop versus Temp, Operator, Gauge
Temp random 3 60 75 90
Gauge random 3 1 2 3
Analysis of Variance for Drop
Source DF SS MS F P
Temp 2 1023.36 511.68 2.30 0.171 x
Operator 3 423.82 141.27 0.63 0.616 x
Gauge 2 7.19 3.60 0.06 0.938 x
Temp*Operator 6 1211.97 202.00 14.59 0.000
Temp*Gauge 4 137.89 34.47 2.49 0.099
Operator*Gauge 6 209.47 34.91 2.52 0.081
Temp*Operator*Gauge 12 166.11 13.84 0.65 0.788
Error 36 770.50 21.40
Total 71 3950.32
x Not an exact F-test.
1 Temp 12.044 * (8) + 2(7) + 8(5) + 6(4) + 24(1)
2 Operator -4.544 * (8) + 2(7) + 6(6) + 6(4) + 18(2)
3 Gauge -2.164 * (8) + 2(7) + 6(6) + 8(5) + 24(3)
4 Temp*Operator 31.359 7 (8) + 2(7) + 6(4)
5 Temp*Gauge 2.579 7 (8) + 2(7) + 8(5)
6 Operator*Gauge 3.512 7 (8) + 2(7) + 6(6)
7 Temp*Operator*Gauge -3.780 8 (8) + 2(7)
8 Error 21.403 (8)
12-22

* Synthesized Test.
Error Terms for Synthesized Tests
1 Temp 6.97 222.63 (4) + (5) - (7)
2 Operator 7.09 223.06 (4) + (6) - (7)
3 Gauge 5.98 55.54 (5) + (6) - (7)
Since all three factors are random there are no exact tests on main effects. Minitab uses an approximate F
test for the these factors.
13-20 Derive the expected mean squares shown in Table 13-11.
F R R R
a b c n
Factor i j k l E(MS)
τi 0 b c n
( )∑ −
++++
1
2
2222
a
bcncnbnn iτ
σσσσ τβτγτβγ
βj a 1 c n σ σβγ β
2 2
+ +an acnσ 2
γ k a b 1 n σ σβγ γ
2 2
+ +an abnσ 2
( )ijτβ 0 1 c n σ σ στβγ τβ
2 2
+ +n cn 2
( )ikτγ 0 b 1 n σ σ στβγ τγ
2 2
+ +n bn 2
( )jkβγ a 1 1 n σ σβγ
2 2
+ an
( )ijkτβγ 0 1 1 n σ στβγ
2 2
+ n
εijkl 1 1 1 1 σ 2
13-21 Consider a four-factor factorial experiment where factor A is at a levels, factor B is at b levels, factor
C is at c levels, factor D is at d levels, and there are n replicates. Write down the sums of squares, the
degrees of freedom, and the expected mean squares for the following cases. Do exact tests exist for all
effects? If not, propose test statistics for those effects that cannot be directly tested. Assume the restricted
model on all cases. You may use a computer package such as Minitab.
The four factor model is:
( ) ( ) ( ) ( ) ( ) ( ) +++++++++++= kljljkilikijlkjiijklhy γδβδβγτδτγτβδγβτµ
( ) ( ) ( ) ( ) ( ) ijklhijklikljklijlijk ετβγδτγδβγδτβδτβγ +++++
To simplify the expected mean square derivations, let capital Latin letters represent the factor effects or
variance components. For example, A
bcdn
a
i
=
−
∑τ 2
1
, or .B acdn= σβ
2
(a) A, B, C, and D are fixed factors.
F F F F R
a b c d n
Factor i j k l h E(MS)
τi 0 b c d n σ2
+ A
βj a 0 c d n σ2
+ B
12-23

γ k a b 0 d n σ2
+ C
δl a b c 0 n σ2
+ D
( )τβ ij 0 0 c d n σ2
+ AB
( )τγ ik 0 b 0 d n σ2
+ AC
( )τδ il 0 b c 0 n σ2
+ AD
( )βγ jk a 0 0 d n σ2
+ BC
( )βδ jl a 0 c 0 n σ2
+ BD
( )γδ kl a b 0 0 n σ2
+ CD
( )τβγ ijk 0 0 0 d n σ2
+ ABC
( )τβδ ijl 0 0 c 0 n σ2
+ ABD
( )βγδ jkl a 0 0 0 n σ2
+ BCD
( )τγδ ikl 0 b 0 0 n σ2
+ ACD
( )τβγδ ijkl 0 0 0 0 n σ2
+ ABCD
ε( )ijkl h 1 1 1 1 1 σ2
There are exact tests for all effects. The results can also be generated in Minitab as follows:
Minitab Output
ANOVA: y versus A, B, C, D
A fixed 2 H L
B fixed 2 H L
C fixed 2 H L
D fixed 2 H L
Analysis of Variance for y
Source DF SS MS F P
A 1 6.13 6.13 0.49 0.492
B 1 0.13 0.13 0.01 0.921
C 1 1.13 1.13 0.09 0.767
D 1 0.13 0.13 0.01 0.921
A*B 1 3.13 3.13 0.25 0.622
A*C 1 3.13 3.13 0.25 0.622
A*D 1 3.13 3.13 0.25 0.622
B*C 1 3.13 3.13 0.25 0.622
B*D 1 3.13 3.13 0.25 0.622
C*D 1 3.13 3.13 0.25 0.622
A*B*C 1 3.13 3.13 0.25 0.622
A*B*D 1 28.13 28.13 2.27 0.151
A*C*D 1 3.13 3.13 0.25 0.622
B*C*D 1 3.13 3.13 0.25 0.622
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
1 A 16 (16) + 16Q[1]
2 B 16 (16) + 16Q[2]
3 C 16 (16) + 16Q[3]
4 D 16 (16) + 16Q[4]
5 A*B 16 (16) + 8Q[5]
6 A*C 16 (16) + 8Q[6]
7 A*D 16 (16) + 8Q[7]
8 B*C 16 (16) + 8Q[8]
9 B*D 16 (16) + 8Q[9]
10 C*D 16 (16) + 8Q[10]
11 A*B*C 16 (16) + 4Q[11]
12 A*B*D 16 (16) + 4Q[12]
13 A*C*D 16 (16) + 4Q[13]
14 B*C*D 16 (16) + 4Q[14]
12-24

15 A*B*C*D 16 (16) + 2Q[15]
16 Error 12.38 (16)
(b) A, B, C, and D are random factors.
R R R R R
a b c d n
τi 1 b c d n σ2
+ + + + + + + +ABCD ACD ABD ABC AD AC AB A
βj a 1 c d n σ2
+ + + + + + + +ABCD BCD ABD ABC BD BC AB B
γ k a b 1 d n σ2
+ + + + + + + +ABCD ACD BCD ABC AB BC CD C
δl a b c 1 n σ2
+ + + + + + + +ABCD ACD BCD ABD BD AD CD D
( )τβ ij 1 1 c d n σ2
+ + + +ABCD ABC ABD AB
( )τγ ik 1 b 1 d n σ2
+ + + +ABCD ABC ACD AC
( )τδ il 1 b c 1 n σ2
+ + + +ABCD ABD ACD AD
( )βγ jk a 1 1 d n σ2
+ + + +ABCD ABC BCD BC
( )βδ jl a 1 c 1 n σ2
+ + + +ABCD ABD BCD BD
( )γδ kl a b 1 1 n σ2
+ + + +ABCD ACD BCD CD
( )τβγ ijk 1 1 1 d n σ2
+ +ABCD ABC
( )τβδ ijl 1 1 c 1 n σ2
+ +ABCD ABD
( )βγδ jkl a 1 1 1 n σ2
+ +ABCD BCD
( )τγδ ikl 1 b 1 1 n σ2
+ +ABCD ACD
+ ABCD
ε( )ijkl h 1 1 1 1 1 σ2
No exact tests exist on main effects or two-factor interactions. For main effects use statistics such as:
A F
MS MS MS MS
MS MS MS MS
A ABC ABD ACD
AB AC AD ABCD
: =
+ + +
+ + +
For testing two-factor interactions use statistics such as: AB F
MS MS
MS MS
AB ABCD
ABC ABD
: =
+
+
The results can also be generated in Minitab as follows:
Minitab Output
A random 2 H L
B random 2 H L
C random 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 **
B 1 0.13 0.13 **
C 1 1.13 1.13 0.36 0.843 x
D 1 0.13 0.13 **
A*B 1 3.13 3.13 0.11 0.796 x
A*C 1 3.13 3.13 1.00 0.667 x
A*D 1 3.13 3.13 0.11 0.796 x
B*C 1 3.13 3.13 1.00 0.667 x
12-25

B*D 1 3.13 3.13 0.11 0.796 x
C*D 1 3.13 3.13 1.00 0.667 x
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 1.00 0.500
B*C*D 1 3.13 3.13 1.00 0.500
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
1 A 1.7500 * (16) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6)
+ 8(5) + 16(1)
2 B 1.3750 * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8)
+ 8(5) + 16(2)
3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8)
+ 8(6) + 16(3)
4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9)
+ 8(7) + 16(4)
5 A*B -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5)
6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6)
7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7)
8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8)
9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9)
10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10)
11 A*B*C 0.0000 15 (16) + 2(15) + 4(11)
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D 0.0000 15 (16) + 2(15) + 4(13)
14 B*C*D 0.0000 15 (16) + 2(15) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15)
2 B 0.56 * (5) + (8) + (9) - (11) - (12) - (14) + (15)
3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15)
4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15)
5 A*B 0.98 28.13 (11) + (12) - (15)
6 A*C 0.33 3.13 (11) + (13) - (15)
7 A*D 0.98 28.13 (12) + (13) - (15)
8 B*C 0.33 3.13 (11) + (14) - (15)
9 B*D 0.98 28.13 (12) + (14) - (15)
10 C*D 0.33 3.13 (13) + (14) - (15)
(c) A is fixed and B, C, and D are random.
F R R R R
a b c d n
τi 0 b c d n σ2
+ + + + + + + +ABCD ACD ABD ABC AD AC AB A
βj a 1 c d n σ 2
+ + + +BCD ABD BC B
γ k a b 1 d n σ2
+ + + +BCD BC CD C
δl a b c 1 n σ2
+ + + +BCD BD CD D
( )τβ ij 0 1 c d n σ2
( )τγ ik 0 b 1 d n σ2
+ + + +ABCD ABC ACD AC
( )τδ il 0 b c 1 n σ2
+ + + +ABCD ABD ACD AD
12-26

( )βγ jk a 1 1 d n σ2
+ +BCD BC
( )βδ jl a 1 c 1 n σ2
+ +BCD BD
( )γδ kl a b 1 1 n σ2
+ +BCD CD
( )τβγ ijk 0 1 1 d n σ2
+ +ABCD ABC
( )τβδ ijl 0 1 c 1 n σ2
+ +ABCD ABD
( )βγδ jkl a 1 1 1 n σ2
+ BCD
( )τγδ ikl 0 b 1 1 n σ2
+ +ABCD ACD
+ ABCD
ε( )ijkl h 1 1 1 1 1 σ2
No exact tests exist on main effects or two-factor interactions involving the fixed factor A. To test the fixed
factor A use
A F
MS MS MS MS
MS MS MS MS
A ABC ABD ACD
AB AC AD ABCD
: =
+ + +
+ + +
Random main effects could be tested by, for example: D F
MS MS
MS MS
D ABCD
ABC ABD
: =
+
+
For testing two-factor interactions involving A use: AB F
MS MS
MS MS
AB ABCD
ABC ABD
: =
+
+
Minitab Output
A fixed 2 H L
B random 2 H L
C random 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 **
B 1 0.13 0.13 0.04 0.907 x
C 1 1.13 1.13 0.36 0.761 x
D 1 0.13 0.13 0.04 0.907 x
A*B 1 3.13 3.13 0.11 0.796 x
A*C 1 3.13 3.13 1.00 0.667 x
A*D 1 3.13 3.13 0.11 0.796 x
B*C 1 3.13 3.13 1.00 0.500
B*D 1 3.13 3.13 1.00 0.500
C*D 1 3.13 3.13 1.00 0.500
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 1.00 0.500
B*C*D 1 3.13 3.13 0.25 0.622
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
12-27

1 A * (16) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6)
+ 8(5) + 16Q[1]
2 B -0.1875 * (16) + 4(14) + 8(9) + 8(8) + 16(2)
3 C -0.1250 * (16) + 4(14) + 8(10) + 8(8) + 16(3)
4 D -0.1875 * (16) + 4(14) + 8(10) + 8(9) + 16(4)
5 A*B -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5)
6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6)
7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7)
8 B*C 0.0000 14 (16) + 4(14) + 8(8)
9 B*D 0.0000 14 (16) + 4(14) + 8(9)
10 C*D 0.0000 14 (16) + 4(14) + 8(10)
11 A*B*C 0.0000 15 (16) + 2(15) + 4(11)
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D 0.0000 15 (16) + 2(15) + 4(13)
14 B*C*D -2.3125 16 (16) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15)
2 B 0.33 3.13 (8) + (9) - (14)
3 C 0.33 3.13 (8) + (10) - (14)
4 D 0.33 3.13 (9) + (10) - (14)
5 A*B 0.98 28.13 (11) + (12) - (15)
6 A*C 0.33 3.13 (11) + (13) - (15)
7 A*D 0.98 28.13 (12) + (13) - (15)
(d) A and B are fixed and C and D are random.
F F R R R
a b c d n
τi 0 b c d n σ2
+ + + +ACD AD AC A
βj a 0 c d n σ2
+ + + +BCD BC BD B
γ k a b 1 d n σ2
+ +CD C
δl a b c 1 n σ2
+ +CD D
( )τβ ij 0 0 c d n σ2
( )τγ ik 0 b 1 d n σ2
+ +ACD AC
( )τδ il 0 b c 1 n σ2
+ +ACD AD
( )βγ jk a 0 1 d n σ2
+ +BCD BC
( )βδ jl a 0 c 1 n σ2
+ +BCD BD
( )γδ kl a b 1 1 n σ2
+ CD
( )τβγ ijk 0 0 1 d n σ2
+ +ABCD ABC
( )τβδ ijl 0 0 c 1 n σ2
+ +ABCD ABD
( )βγδ jkl a 0 1 1 n σ2
+ BCD
( )τγδ ikl 0 b 1 1 n σ2
+ ACD
+ ABCD
ε( )ijkl h 1 1 1 1 1 σ2
There are no exact tests on the fixed factors A and B, or their two-factor interaction AB. The appropriate
test statistics are:
12-28

A F
MS MS
MS MS
B F
MS MS
MS MS
A AC
AC AD
B BC
BC BD
:
:
= D
D
+
+
=
+
+
AB F
MS MS
MS MS
AB ABCD
ABC ABD
: =
+
+
Minitab Output
A fixed 2 H L
B fixed 2 H L
C random 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 1.96 0.604 x
B 1 0.13 0.13 0.04 0.907 x
C 1 1.13 1.13 0.36 0.656
D 1 0.13 0.13 0.04 0.874
A*B 1 3.13 3.13 0.11 0.796 x
A*C 1 3.13 3.13 1.00 0.500
A*D 1 3.13 3.13 1.00 0.500
B*C 1 3.13 3.13 1.00 0.500
B*D 1 3.13 3.13 1.00 0.500
C*D 1 3.13 3.13 0.25 0.622
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 0.25 0.622
B*C*D 1 3.13 3.13 0.25 0.622
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
1 A * (16) + 4(13) + 8(7) + 8(6) + 16Q[1]
2 B * (16) + 4(14) + 8(9) + 8(8) + 16Q[2]
3 C -0.1250 10 (16) + 8(10) + 16(3)
4 D -0.1875 10 (16) + 8(10) + 16(4)
5 A*B * (16) + 2(15) + 4(12) + 4(11) + 8Q[5]
6 A*C 0.0000 13 (16) + 4(13) + 8(6)
7 A*D 0.0000 13 (16) + 4(13) + 8(7)
8 B*C 0.0000 14 (16) + 4(14) + 8(8)
9 B*D 0.0000 14 (16) + 4(14) + 8(9)
10 C*D -1.1563 16 (16) + 8(10)
11 A*B*C 0.0000 15 (16) + 2(15) + 4(11)
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D -2.3125 16 (16) + 4(13)
14 B*C*D -2.3125 16 (16) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
1 A 0.33 3.13 (6) + (7) - (13)
12-29

2 B 0.33 3.13 (8) + (9) - (14)
5 A*B 0.98 28.13 (11) + (12) - (15)
(e) A, B and C are fixed and D is random.
F F F R R
a b c d n
τi 0 b c d n σ2
+ +AD A
βj a 0 c d n σ2
+ +BD B
γ k a b 0 d n σ2
+ +CD C
δl a b c 1 n σ2
+ D
( )τβ ij 0 0 c d n σ2
+ +ABD AB
( )τγ ik 0 b 0 d n σ2
+ +ACD AC
( )τδ il 0 b c 1 n σ2
+ AD
( )βγ jk a 0 0 d n σ2
+ +BCD BC
( )βδ jl a 0 c 1 n σ2
+ BD
( )γδ kl a b 0 1 n σ2
+ CD
( )τβγ ijk 0 0 0 d n σ2
+ +ABCD ABC
( )τβδ ijl 0 0 c 1 n σ2
+ ABD
( )βγδ jkl a 0 0 1 n σ2
+ BCD
( )τγδ ikl 0 b 0 1 n σ2
+ ACD
+ ABCD
ε( )ijkl h 1 1 1 1 1 σ2
There are exact tests for all effects. The results can also be generated in Minitab as follows:
Minitab Output
A fixed 2 H L
B fixed 2 H L
C fixed 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 1.96 0.395
B 1 0.13 0.13 0.04 0.874
C 1 1.13 1.13 0.36 0.656
D 1 0.13 0.13 0.01 0.921
A*B 1 3.13 3.13 0.11 0.795
A*C 1 3.13 3.13 1.00 0.500
A*D 1 3.13 3.13 0.25 0.622
B*C 1 3.13 3.13 1.00 0.500
B*D 1 3.13 3.13 0.25 0.622
C*D 1 3.13 3.13 0.25 0.622
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 2.27 0.151
A*C*D 1 3.13 3.13 0.25 0.622
B*C*D 1 3.13 3.13 0.25 0.622
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
12-30

1 A 7 (16) + 8(7) + 16Q[1]
2 B 9 (16) + 8(9) + 16Q[2]
3 C 10 (16) + 8(10) + 16Q[3]
4 D -0.7656 16 (16) + 16(4)
5 A*B 12 (16) + 4(12) + 8Q[5]
6 A*C 13 (16) + 4(13) + 8Q[6]
7 A*D -1.1563 16 (16) + 8(7)
8 B*C 14 (16) + 4(14) + 8Q[8]
9 B*D -1.1563 16 (16) + 8(9)
10 C*D -1.1563 16 (16) + 8(10)
11 A*B*C 15 (16) + 2(15) + 4Q[11]
12 A*B*D 3.9375 16 (16) + 4(12)
13 A*C*D -2.3125 16 (16) + 4(13)
14 B*C*D -2.3125 16 (16) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
13-22 Reconsider cases (c), (d) and (e) of Problem 13-21. Obtain the expected mean squares assuming the
unrestricted model. You may use a computer package such as Minitab. Compare your results with those
for the restricted model.
A is fixed and B, C, and D are random.
Minitab Output
A fixed 2 H L
B random 2 H L
C random 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 **
B 1 0.13 0.13 **
C 1 1.13 1.13 0.36 0.843 x
D 1 0.13 0.13 **
A*B 1 3.13 3.13 0.11 0.796 x
A*C 1 3.13 3.13 1.00 0.667 x
A*D 1 3.13 3.13 0.11 0.796 x
B*C 1 3.13 3.13 1.00 0.667 x
B*D 1 3.13 3.13 0.11 0.796 x
C*D 1 3.13 3.13 1.00 0.667 x
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 1.00 0.500
B*C*D 1 3.13 3.13 1.00 0.500
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
component term (using unrestricted model)
1 A * (16) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6)
+ 8(5) + Q[1]
2 B 1.3750 * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8)
+ 8(5) + 16(2)
3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8)
+ 8(6) + 16(3)
4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9)
+ 8(7) + 16(4)
12-31

5 A*B -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5)
6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6)
7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7)
8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8)
9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9)
10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10)
11 A*B*C 0.0000 15 (16) + 2(15) + 4(11)
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D 0.0000 15 (16) + 2(15) + 4(13)
14 B*C*D 0.0000 15 (16) + 2(15) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15)
2 B 0.56 * (5) + (8) + (9) - (11) - (12) - (14) + (15)
3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15)
4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15)
5 A*B 0.98 28.13 (11) + (12) - (15)
6 A*C 0.33 3.13 (11) + (13) - (15)
7 A*D 0.98 28.13 (12) + (13) - (15)
8 B*C 0.33 3.13 (11) + (14) - (15)
9 B*D 0.98 28.13 (12) + (14) - (15)
10 C*D 0.33 3.13 (13) + (14) - (15)
A and B are fixed and C and D are random.
Minitab Output
A fixed 2 H L
B fixed 2 H L
C random 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 1.96 0.604 x
B 1 0.13 0.13 0.04 0.907 x
C 1 1.13 1.13 0.36 0.843 x
D 1 0.13 0.13 **
A*B 1 3.13 3.13 0.11 0.796 x
A*C 1 3.13 3.13 1.00 0.667 x
A*D 1 3.13 3.13 0.11 0.796 x
B*C 1 3.13 3.13 1.00 0.667 x
B*D 1 3.13 3.13 0.11 0.796 x
C*D 1 3.13 3.13 1.00 0.667 x
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 1.00 0.500
B*C*D 1 3.13 3.13 1.00 0.500
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
1 A * (16) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6)
+ Q[1,5]
2 B * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8)
12-32

+ Q[2,5]
3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8)
+ 8(6) + 16(3)
4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9)
+ 8(7) + 16(4)
5 A*B * (16) + 2(15) + 4(12) + 4(11) + Q[5]
6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6)
7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7)
8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8)
9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9)
10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10)
11 A*B*C 0.0000 15 (16) + 2(15) + 4(11)
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D 0.0000 15 (16) + 2(15) + 4(13)
14 B*C*D 0.0000 15 (16) + 2(15) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
1 A 0.33 3.13 (6) + (7) - (13)
2 B 0.33 3.13 (8) + (9) - (14)
3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15)
4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15)
5 A*B 0.98 28.13 (11) + (12) - (15)
6 A*C 0.33 3.13 (11) + (13) - (15)
7 A*D 0.98 28.13 (12) + (13) - (15)
8 B*C 0.33 3.13 (11) + (14) - (15)
9 B*D 0.98 28.13 (12) + (14) - (15)
10 C*D 0.33 3.13 (13) + (14) - (15)
(e) A, B and C are fixed and D is random.
Minitab Output
A fixed 2 H L
B fixed 2 H L
C fixed 2 H L
D random 2 H L
Source DF SS MS F P
A 1 6.13 6.13 1.96 0.395
B 1 0.13 0.13 0.04 0.874
C 1 1.13 1.13 0.36 0.656
D 1 0.13 0.13 **
A*B 1 3.13 3.13 0.11 0.795
A*C 1 3.13 3.13 1.00 0.500
A*D 1 3.13 3.13 0.11 0.796 x
B*C 1 3.13 3.13 1.00 0.500
B*D 1 3.13 3.13 0.11 0.796 x
C*D 1 3.13 3.13 1.00 0.667 x
A*B*C 1 3.13 3.13 1.00 0.500
A*B*D 1 28.13 28.13 9.00 0.205
A*C*D 1 3.13 3.13 1.00 0.500
B*C*D 1 3.13 3.13 1.00 0.500
A*B*C*D 1 3.13 3.13 0.25 0.622
Error 16 198.00 12.38
Total 31 264.88
12-33

1 A 7 (16) + 2(15) + 4(13) + 4(12) + 8(7) + Q[1,5,6,11]
2 B 9 (16) + 2(15) + 4(14) + 4(12) + 8(9) + Q[2,5,8,11]
3 C 10 (16) + 2(15) + 4(14) + 4(13) + 8(10) + Q[3,6,8,11]
4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9)
+ 8(7) + 16(4)
5 A*B 12 (16) + 2(15) + 4(12) + Q[5,11]
6 A*C 13 (16) + 2(15) + 4(13) + Q[6,11]
7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7)
8 B*C 14 (16) + 2(15) + 4(14) + Q[8,11]
9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9)
10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10)
11 A*B*C 15 (16) + 2(15) + Q[11]
12 A*B*D 6.2500 15 (16) + 2(15) + 4(12)
13 A*C*D 0.0000 15 (16) + 2(15) + 4(13)
14 B*C*D 0.0000 15 (16) + 2(15) + 4(14)
15 A*B*C*D -4.6250 16 (16) + 2(15)
16 Error 12.3750 (16)
* Synthesized Test.
4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15)
7 A*D 0.98 28.13 (12) + (13) - (15)
9 B*D 0.98 28.13 (12) + (14) - (15)
10 C*D 0.33 3.13 (13) + (14) - (15)
13-23 In Problem 5-17, assume that the three operators were selected at random. Analyze the data under
these conditions and draw conclusions. Estimate the variance components.
Minitab Output
ANOVA: Score versus Cycle Time, Operator, Temperature
Cycle Ti fixed 3 40 50 60
Temperat fixed 2 300 350
Analysis of Variance for Score
Source DF SS MS F P
Cycle Ti 2 436.000 218.000 2.45 0.202
Operator 2 261.333 130.667 39.86 0.000
Temperat 1 50.074 50.074 8.89 0.096
Cycle Ti*Operator 4 355.667 88.917 27.13 0.000
Cycle Ti*Temperat 2 78.815 39.407 3.41 0.137
Operator*Temperat 2 11.259 5.630 1.72 0.194
Cycle Ti*Operator*Temperat 4 46.185 11.546 3.52 0.016
Error 36 118.000 3.278
Total 53 1357.333
1 Cycle Ti 4 (8) + 6(4) + 18Q[1]
2 Operator 7.0772 8 (8) + 18(2)
3 Temperat 6 (8) + 9(6) + 27Q[3]
4 Cycle Ti*Operator 14.2731 8 (8) + 6(4)
5 Cycle Ti*Temperat 7 (8) + 3(7) + 9Q[5]
6 Operator*Temperat 0.2613 8 (8) + 9(6)
7 Cycle Ti*Operator*Temperat 2.7562 8 (8) + 3(7)
8 Error 3.2778 (8)
The following calculations agree with the Minitab results:
12-34

σ 2
= MSE 2777832
.ˆ =σ
στβγ
2
=
−MS MS
n
ABC E
75622
3
2777783546296112
.
..
ˆ =
−
=τβγσ
σβγ
2
=
−MS MS
an
BC E
( )
2731514
32
277778391667882
.
..
ˆ =
−
=βγσ
στγ
2
=
−MS MS
bn
AC E
( )
261320
33
277778362963052
.
..
ˆ =
−
=τγσ
σγ
2
=
−MS MS
abn
C E
( )( )
077167
332
2777783666671302
.
..
ˆ =
−
=γσ
13-24 Consider the three-factor model
( ) ( ) ijkjkijkjiijky εβγτβγβτµ ++++++=
Assuming that all the factors are random, develop the analysis of variance table, including the expected
mean squares. Propose appropriate test statistics for all effects.
Source DF E(MS)
A a-1 σ σ στβ τ
2 2
+ +c bc 2
B b-1 σ σ σ στβ βγ β
2 2 2
+ + +c a ac 2
2
C c-1 σ σ σβγ γ
2 2
+ +a ab
AB (a-1)(b-1) σ στβ
2 2
+ c
BC (b-1)(c-1) σ σβγ
2 2
+ a
Error (AC + ABC) b(a-1)(c-1) σ 2
Total abc-1
There are exact tests for all effects except B. To test B, use the statistic F
MS MS
MS MS
B E
AB BC
=
+
+
13-25 The three-factor model for a single replicate is
yijk i j k ij jk ik ijk ijk= + + + + + + + +µ τ β γ τβ βγ τγ τβγ ε( ) ( ) ( ) ( )
If all the factors are random, can any effects be tested? If the three-factor interaction and the ( )τβ ij
interaction do not exist, can all the remaining effects be tested.
The expected mean squares are found by referring to Table 12-9, deleting the line for the error term ε( )ijk l
and setting n=1. The three-factor interaction now cannot be tested; however, exact tests exist for the two-
factor interactions and approximate F tests can be conducted for the main effects. For example, to test the
main effect of A, use
F
MS MS
MS MS
A AB
AB AC
=
+
+
C
12-35

If ( )τβγ ijk and ( )τβ ij can be eliminated, the model becomes
( ) ( ) ( ) ( ) ijkijkikjkijkjiijky ετβγτγβγτβγβτµ ++++++++=
For this model, the analysis of variance is
Source DF E(MS)
A a-1 σ σ στγ τ
2 2
+ +b bc 2
B b-1 σ σ σβγ β
2 2
+ +a ac 2
2
C c-1 σ σ σ σβγ τγ γ
2 2 2
+ + +a b ab
AC (a-1)(c-1) σ στγ
2 2
+ b
BC (b-1)(c-1) σ σβγ
2 2
+ a
Error (AB + ABC) c(a-1)(b-1) σ 2
Total abc-1
There are exact tests for all effect except C. To test the main effect of C, use the statistic:
F
MS MS
MS MS
C E
BC AC
=
+
+
13-26 In Problem 5-6, assume that both machines and operators were chosen randomly. Determine the
power of the test for detecting a machine effect such that , where is the variance component
for the machine factor. Are two replicates sufficient?
σ σβ
2
= 2
σβ
2
λ
σ
σ σ
β
τβ
= +
+
1
2
2 2
an
n
If , then an estimate of , and an estimate of , from the analysis
of variance table. Then
σ σβ
2
= 2
σ σβ
2 2
379= = . σ στβ
2 2
7 45= =n .
( )( )( ) 491222
457
79323
1 ..
.
.
==+=λ
and the other OC curve parameters are υ1 3= and υ2 6= . This results in β ≈ 075. approximately, with
α = 0 05. , or β ≈ 0 9. with α = 0 01. . Two replicates does not seem sufficient.
13-27 In the two-factor mixed model analysis of variance, show that Cov ( ) ( )[ ] ( )2
1 τβστβτβ a, j'iij −= for
i≠i'.
12-36

Since (constant) we have , which implies that( )∑=
=
a
i
ij
1
0τβ ( ) 0
1
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
∑=
a
i
ijV τβ
( ) ( ) ( )[ ]
( )
( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]
( ) ( )[ ] 2
2
2
1
1
011
02
22
1
0
2
2
τβ
τβ
τβ
στββτ
βττβσ
τβτβσ
τβτβτβ
⎟
⎠
⎞
⎜
⎝
⎛
−=
=−+−
=
−
+⎥
⎦
⎤
⎢
⎣
⎡ −
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+∑=
a
,Cov
,Covaaa
,Cov
!a!
!a
a
a
a
,Cov
a
V
j'iij
j'iij
j'iij
j'iij
a
i
ij
13-28 Show that the method of analysis of variance always produces unbiased point estimates of the
variance component in any random or mixed model.
Let g be the vector of mean squares from the analysis of variance, chosen so that E(g) does not contain any
fixed effects. Let be the vector of variance components such that , where A is a matrix of
constants. Now in the analysis of variance method of variance component estimation, we equate observed
and expected mean squares, i.e.
σ2
E( )g A= σ2
gAssA=g -122
=⇒ ˆ
Since always exists then,-1
A
( ) ( ) ( ) ( ) 22-1-1-12
ssAA=gAgA=s == EEE
Thus is an unbiased estimator of σ . This and other properties of the analysis of variance method are
discussed by Searle (1971a).
σ2 2
13-29 Invoking the usual normality assumptions, find an expression for the probability that a negative
estimate of a variance component will be obtained by the analysis of variance method. Using this result,
write a statement giving the probability that στ
2
0< in a one-factor analysis of variance. Comment on the
usefulness of this probability statement.
Suppose σ 2 1
=
−MS MS
c
2
, where for i=1,2 are two mean squares and c is a constant. The
probability that (negative) is
MSi
02
<τσˆ
{ } { }
( )
( )
( )
( )
( )
( )⎭
⎬
⎫
⎩
⎨
⎧
<=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
<=
⎭
⎬
⎫
⎩
⎨
⎧
<=<−=<
2
1
2
1
2
2
1
1
2
1
21
2
100
MSE
MSE
FP
MSE
MSE
MSE
MS
MSE
MS
P
MS
MS
PMSMSPˆP v,uσ
where u is the number of degrees of freedom for and v is the number of degrees of freedom for .
For the one-way model, this equation reduces to
MS1 MS2
12-37

{ }P P F
n
P F
nka N a a N a, ,σ
σ
σ στ
2
1
2
2 2 10
1
1
< = <
+
⎧
⎨
⎪
⎩⎪
⎫
⎬
⎪
⎭⎪
= <
+
⎧
⎨
⎩
⎫
⎬
⎭
− − − −
where 2
2
σ
στ
=k . Using arbitrary values for some of the parameters in this equation will give an
experimenter some idea of the probability of obtaining a negative estimate of .02
<τσˆ
13-30 Analyze the data in Problem 13-9, assuming that the operators are fixed, using both the unrestricted
and restricted forms of the mixed models. Compare the results obtained from the two models.
The restricted model is as follows:
Minitab Output
Part random 10 1 2 3 4 5 6 7
8 9 10
Source DF SS MS F P
Part 9 99.017 11.002 7.33 0.000
Operator 1 0.417 0.417 0.69 0.427
Part*Operator 9 5.417 0.602 0.40 0.927
Error 40 60.000 1.500
Total 59 164.850
1 Part 1.5836 4 (4) + 6(1)
2 Operator 3 (4) + 3(3) + 30Q[2]
3 Part*Operator -0.2994 4 (4) + 3(3)
4 Error 1.5000 (4)
The second approach is the unrestricted mixed model.
Minitab Output
Part random 10 1 2 3 4 5 6 7
8 9 10
Source DF SS MS F P
Part 9 99.017 11.002 18.28 0.000
Operator 1 0.417 0.417 0.69 0.427
Part*Operator 9 5.417 0.602 0.40 0.927
Error 40 60.000 1.500
Total 59 164.850
1 Part 1.7333 3 (4) + 3(3) + 6(1)
2 Operator 3 (4) + 3(3) + Q[2]
3 Part*Operator -0.2994 4 (4) + 3(3)
4 Error 1.5000 (4)
12-38

Source Sum of
Squares
DF Mean
Square
E(MS) F-test F
A 0.416667 a-1=1 0.416667 σ σ
τ
τβ
2 2
2
1
1
+ +
−
=
∑
n bn
a
i
i
a
AB
A
MS
MS
F = 0.692
B 99.016667 b-1=9 11.00185 σ σ στβ β
2 2
+ +n an 2
AB
B
MS
MS
F = 18.28
AB 5.416667 (a-1)(b-1)=9 0.60185 σ στβ
2 2
+ n
E
AB
MS
MS
F = 0.401
Error 60.000000 40 1.50000 σ 2
Total 164.85000 nabc-1=59
In the unrestricted model, the F-test for B is different. The F-test for B in the unrestricted model should
generally be more conservative, since MSAB will generally be larger than MSE. However, this is not the
case with this particular experiment.
13-31 Consider the two-factor mixed model. Show that the standard error of the fixed factor mean (e.g. A)
is [ ] 21
bn/MSAB .
The standard error is often used in Duncan’s Multiple Range test. Duncan’s Multiple Range Test requires
the variance of the difference in two means, say
( )..m..i yyV −
where rows are fixed and columns are random. Now, assuming all model parameters to be independent, we
have the following:
( ) ( ) ( ) ∑∑∑∑∑∑ = == ===
−+−+−=−
b
j
n
k
mjk
b
j
n
k
ijk
b
j
mj
b
j
ijmi..m..i
bnbnbb
yy
1 11 111
1111
εετβτβττ
and
( )
( )
bn
n
bn
bn
bn
bn
b
b
b
b
yyV ..m..i
22
2
2
2
2
2
2
2
2
21111 τβ
τβτβ
σσ
σσσσ
+
=⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=−
Since estimates , we would useMSAB σ στβ
2
+ n 2
2MS
bn
AB
as the standard error to test the difference. However, the table of ranges for Duncan’s Multiple Range test
already include the constant 2.
13-32 Consider the variance components in the random model from Problem 13-9.
(a) Find an exact 95 percent confidence interval on σ2.
12-39

f MS f MSE E
f
E E
fE E
χ
σ
χα α2
2
2
1 2
2
, ,
≤ ≤
−
( )( ) ( )( )
4324
5140
3459
5140 2
.
.
.
.
≤≤ σ
1011 2 4562
. .≤ ≤σ
(b) Find approximate 95 percent confidence intervals on the other variance components using the
Satterthwaite method.
στβ
2
and are negative, and the Satterthwaithe method does not apply. The confidence interval on
is
στ
2
σβ
2
σβ
2
=
−MS MS
an
B AB
( )
73331
32
60185190001852112
.
..
ˆ =
−
=βσ
( )
( ) ( )( )
( )
( ) ( )( )
018268
91
60185190
9
0018521
6018519000185211
111
22
2
22
2
.
..
..
ba
MS
b
MS
MSMS
r
ABB
ABB
=
+
−
=
−−
+
−
−
=
r rO
r r, ,
σ
χ
σ
σ
χα
β
β
α
2
2
2
2
2
1 2
2≤ ≤
−
( )( ) ( )( )
189502
73331018268
5575217
73331018268 2
.
..
.
..
≤≤ βσ
0 79157 6 347592
. .≤ ≤σβ
13-33 Use the experiment described in Problem 5-6 and assume that both factor are random. Find an exact
95 percent confidence interval on σ2. Construct approximate 95 percent confidence interval on the other
variance components using the Satterthwaite method.
σ 2
= MSE .σ 2
379167=
f MS f MSE E
f
E E
fE E
χ
σ
χα α2
2
2
1 2
2
, ,
≤ ≤
−
( )( ) ( )( )
404
79167312
3423
79167312 2
.
.
.
.
≤≤ σ
19494 10 34092
. .≤ ≤σ
Satterthwaite Method:
στβ
2
=
−MS MS
n
AB E . .
.στβ
2 7 44444 379167
2
182639=
−
=
( )
( )( )
( )
( )( ) ( )
29402
12
791673
32
444447
791673444447
11
22
2
22
2
.
..
..
df
MS
ba
MS
MSMS
r
E
EAB
EAB
=
+
−
=
+
−−
−
=
r r
r r, ,
σ
χ
σ
σ
χ
β
α
β
β
α
2
2
2
2
2
1 2
2≤ ≤
−
12-40

( )( ) ( )( )
099980
82639129402
959187
82639129402 2
.
..
.
..
≤≤ βσ
052640 41905772
. .≤ ≤σβ
σβ
2
0< , this variance component does not have a confidence interval using Satterthwaite’s Method.
στ
2
=
−MS MS
bn
A AB
( )
090289
24
44444716667802
.
..
ˆ =
−
=τσ
( )
( ) ( )( )
( )
( ) ( )( )
641081
32
444447
2
1666780
4444471666780
111
22
2
22
2
.
..
..
ba
MS
a
MS
MSMS
r
ABA
ABA
=
+
−
=
−−
+
−
−
=
r r
r r, ,
σ
χ
σ
σ
χ
τ
α
τ
τ
α
2
2
2
2
2
1 2
2≤ ≤
−
( . )( . )
.
( . )( . )
.
164108 9 09028
653295
164108 9 09028
0 03205
2
≤ ≤στ
2 28348 465456372
. .≤ ≤στ
13-34 Consider the three-factor experiment in Problem 5-17 and assume that operators were selected at
random. Find an approximate 95 percent confidence interval on the operator variance component.
σγ
2
=
−MS MS
abn
C E
( )( )
077167
332
2777783666671302
.
..
ˆ =
−
=γσ
( )
( )
( )
( ) ( )
900851
36
277783
2
66667130
27778366667130
1
22
2
22
2
.
..
..
df
MS
c
MS
MSMS
r
E
EC
EC
=
+
−
=
+
−
−
=
r r
r r, ,
σ
χ
σ
σ
χ
γ
α
γ
γ
α
2
2
2
2
2
1 2
2≤ ≤
−
( )( ) ( )( )
045040
077167900851
154679
077167900851 2
.
..
.
..
≤≤ γσ
146948 4298 665322
. .≤ ≤σγ
13-35 Rework Problem 13-32 using the modified large-sample approach described in Section 13-7.2.
Compare the two sets of confidence intervals obtained and discuss.
σ σβO
B AMS MS
an
2 2
= =
− B
( )
73331
32
60185190001852112
.
..
ˆ O =
−
=σ
12-41

( ) ( ) ( ) ( ) 363660
183
7027118346809011831
702711
3700
1
1
9
1
1
1
468090
881
1
1
1
1
2222
1
2
1
2
2
995995
1
9050
1
.
.
....
F
HFGF
G
.
.
.
F
H
.
.F
G
ji
jiji
i
f,f,
f,f,f,f,
ij
,.,,.
,,.
=
−−−
=
−−−
=
=−=−=−=
=−=−=
∞
∞
α
αα
χ
( ) ( ) ( ) ( ) ( ) ( )(
832750
6018500018511
6
1
6
1
363660601850
6
1
702710018511
6
1
468090 2
2
22
2
2
2111
22
2
2
1
22
1
2
1
.V
.......V
MSMSccGMScHMScGV
L
L
ABBABBL
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=
++=
)
L VL= − = − =. . .σβ
2
17333 083275 082075
13-36 Rework Problem 13-34 using the modified large-sample method described in Section 13-7.2.
Compare this confidence interval with he one obtained previously and discuss.
σγ
2
=
−MS MS
abn
C E
( )( )
077167
332
2777783666671302
.
..
ˆ =
−
=γσ
( ) ( ) ( ) ( ) 745420
882
54493088261538018821
5449301
647280
1
1
36
1
1
1
615380
602
1
1
1
1
2222
1
2
1
2
2
36953695
1
3050
1
.
.
....
F
HFGF
G
.
.
.
F
H
.
.F
G
ji
jiji
f,f,
f,f,f,f,
ij
,.,,.
,,.
=
−−−
=
−−−
=
=−=−=−=
=−=−=
∞
∞
α
αα
χ
( ) ( ) ( ) ( ) ( ) ( )(
9511220
27778366667130
18
1
18
1
745420277783
18
1
54493066667130
18
1
615380 2
2
22
2
2
2111
22
2
2
1
22
1
2
1
.V
.......V
MSMSccGMScHMScGV
L
L
ABBABBL
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=
++=
)
L VL= − = − =. . .σγ
2
7 07716 20 95112 2 49992
12-42

Chapter 14
Nested and Split-Plot Designs
Solutions
In this chapter we have not shown residual plots and other diagnostics to conserve space. A complete
analysis would, of course, include these model adequacy checking procedures.
14-1 A rocket propellant manufacturer is studying the burning rate of propellant from three production
processes. Four batches of propellant are randomly selected from the output of each process and three
determinations of burning rate are made on each batch. The results follow. Analyze the data and draw
conclusions.
Process 1 Process 2 Process 3
Batch 1 2 3 4 1 2 3 4 1 2 3 4
25 19 15 15 19 23 18 35 14 35 38 25
30 28 17 16 17 24 21 27 15 21 54 29
26 20 14 13 14 21 17 25 20 24 50 33
Minitab Output
ANOVA: Burn Rate versus Process, Batch
Process fixed 3 1 2 3
Batch(Process) random 4 1 2 3 4
Analysis of Variance for Burn Rat
Source DF SS MS F P
Process 2 676.06 338.03 1.46 0.281
Batch(Process) 9 2077.58 230.84 12.20 0.000
Error 24 454.00 18.92
Total 35 3207.64
1 Process 2 (3) + 3(2) + 12Q[1]
2 Batch(Process) 70.64 3 (3) + 3(2)
3 Error 18.92 (3)
There is no significant effect on mean burning rate among the different processes; however, different
batches from the same process have significantly different burning rates.
14-2 The surface finish of metal parts made on four machines is being studied. An experiment is
conducted in which each machine is run by three different operators and two specimens from each operator
are collected and tested. Because of the location of the machines, different operators are used on each
machine, and the operators are chosen at random. The data are shown in the following table. Analyze the
data and draw conclusions.
Machine 1 Machine 2 Machine 3 Machine 4
Operator 1 2 3 1 2 3 1 2 3 1 2 3
79 94 46 92 85 76 88 53 46 36 40 62
62 74 57 99 79 68 75 56 57 53 56 47
Minitab Output
ANOVA: Finish versus Machine, Operator
13-1

Machine fixed 4 1 2 3 4
Operator(Machine) random 3 1 2 3
Analysis of Variance for Finish
Source DF SS MS F P
Machine 3 3617.67 1205.89 3.42 0.073
Operator(Machine) 8 2817.67 352.21 4.17 0.013
Error 12 1014.00 84.50
Total 23 7449.33
1 Machine 2 (3) + 2(2) + 6Q[1]
2 Operator(Machine) 133.85 3 (3) + 2(2)
3 Error 84.50 (3)
There is a slight effect on surface finish due to the different processes; however, the different operators
running the same machine have significantly different surface finish.
14-3 A manufacturing engineer is studying the dimensional variability of a particular component that is
produced on three machines. Each machine has two spindles, and four components are randomly selected
from each spindle. These results follow. Analyze the data, assuming that machines and spindles are fixed
factors.
Machine 1 Machine 2 Machine 3
Spindle 1 2 1 2 1 2
12 8 14 12 14 16
9 9 15 10 10 15
11 10 13 11 12 15
12 8 14 13 11 14
Minitab Output
ANOVA: Variability versus Machine, Spindle
Machine fixed 3 1 2 3
Spindle(Machine) fixed 2 1 2
Analysis of Variance for Variabil
Source DF SS MS F P
Machine 2 55.750 27.875 18.93 0.000
Spindle(Machine) 3 43.750 14.583 9.91 0.000
Error 18 26.500 1.472
Total 23 126.000
There is a significant effect on dimensional variability due to the machine and spindle factors.
14-4 To simplify production scheduling, an industrial engineer is studying the possibility of assigning one
time standard to a particular class of jobs, believing that differences between jobs is negligible. To see if
this simplification is possible, six jobs are randomly selected. Each job is given to a different group of
three operators. Each operator completes the job twice at different times during the week, and the
following results were obtained. What are your conclusions about the use of a common time standard for
all jobs in this class? What value would you use for the standard?
Job Operator 1 Operator 2 Operator 3
1 158.3 159.4 159.2 159.6 158.9 157.8
2 154.6 154.9 157.7 156.8 154.8 156.3
13-2

3 162.5 162.6 161.0 158.9 160.5 159.5
4 160.0 158.7 157.5 158.9 161.1 158.5
5 156.3 158.1 158.3 156.9 157.7 156.9
6 163.7 161.0 162.3 160.3 162.6 161.8
Minitab Output
ANOVA: Time versus Job, Operator
Job random 6 1 2 3 4 5 6
Operator(Job) random 3 1 2 3
Analysis of Variance for Time
Source DF SS MS F P
Job 5 148.111 29.622 27.89 0.000
Operator(Job) 12 12.743 1.062 0.69 0.738
Error 18 27.575 1.532
Total 35 188.430
1 Job 4.7601 2 (3) + 2(2) + 6(1)
2 Operator(Job) -0.2350 3 (3) + 2(2)
3 Error 1.5319 (3)
The jobs differ significantly; the use of a common time standard would likely not be a good idea.
14-5 Consider the three-stage nested design shown in Figure 13-5 to investigate alloy hardness. Using
the data that follow, analyze the design, assuming that alloy chemistry and heats are fixed factors and
ingots are random.
Alloy Chemistry
1 2
Heats 1 2 3 1 2 3
Ingots 1 2 1 2 1 2 1 2 1 2 1 2
40 27 95 69 65 78 22 23 83 75 61 35
63 30 67 47 54 45 10 39 62 64 77 42
Minitab Output
ANOVA: Hardness versus Alloy, Heat, Ingot
Alloy fixed 2 1 2
Heat(Alloy) fixed 3 1 2 3
Ingot(Alloy Heat) random 2 1 2
Analysis of Variance for Hardness
Source DF SS MS F P
Alloy 1 315.4 315.4 0.85 0.392
Heat(Alloy) 4 6453.8 1613.5 4.35 0.055
Ingot(Alloy Heat) 6 2226.3 371.0 2.08 0.132
Error 12 2141.5 178.5
Total 23 11137.0
1 Alloy 3 (4) + 2(3) + 12Q[1]
2 Heat(Alloy) 3 (4) + 2(3) + 4Q[2]
3 Ingot(Alloy Heat) 96.29 4 (4) + 2(3)
4 Error 178.46 (4)
Alloy hardness differs significantly due to the different heats within each alloy.
13-3

14-6 Reanalyze the experiment in Problem 14-5 using the unrestricted form of the mixed model.
Comment on any differences you observe between the restricted and unrestricted model results. You may
use a computer software package.
Minitab Output
ANOVA: Hardness versus Alloy, Heat, Ingot
Alloy fixed 2 1 2
Heat(Alloy) fixed 3 1 2 3
Ingot(Alloy Heat) random 2 1 2
Analysis of Variance for Hardness
Source DF SS MS F P
Alloy 1 315.4 315.4 0.85 0.392
Heat(Alloy) 4 6453.8 1613.5 4.35 0.055
Ingot(Alloy Heat) 6 2226.3 371.0 2.08 0.132
Error 12 2141.5 178.5
Total 23 11137.0
1 Alloy 3 (4) + 2(3) + Q[1,2]
2 Heat(Alloy) 3 (4) + 2(3) + Q[2]
3 Ingot(Alloy Heat) 96.29 4 (4) + 2(3)
4 Error 178.46 (4)
14-7 Derive the expected means squares for a balanced three-stage nested design, assuming that A is
fixed and that B and C are random. Obtain formulas for estimating the variance components.
The expected mean squares can be generated in Minitab as follows:
Minitab Output
ANOVA: y versus A, B, C
A fixed 2 -1 1
B(A) random 2 -1 1
C(A B) random 2 -1 1
Source DF SS MS F P
A 1 0.250 0.250 0.06 0.831
B(A) 2 8.500 4.250 0.35 0.726
C(A B) 4 49.000 12.250 2.13 0.168
Error 8 46.000 5.750
Total 15 103.750
1 A 2 (4) + 2(3) + 4(2) + 8Q[1]
2 B(A) -2.000 3 (4) + 2(3) + 4(2)
3 C(A B) 3.250 4 (4) + 2(3)
4 Error 5.750 (4)
14-8 Repeat Problem 14-7 assuming the unrestricted form of the mixed model. You may use a computer
software package. Comment on any differences you observe between the restricted and unrestricted model
analysis and conclusions.
Minitab Output
13-4

A fixed 2 -1 1
B(A) random 2 -1 1
Source DF SS MS F P
A 1 0.250 0.250 0.06 0.831
B(A) 2 8.500 4.250 0.35 0.726
C(A B) 4 49.000 12.250 2.13 0.168
Error 8 46.000 5.750
Total 15 103.750
1 A 2 (4) + 2(3) + 4(2) + Q[1]
2 B(A) -2.000 3 (4) + 2(3) + 4(2)
3 C(A B) 3.250 4 (4) + 2(3)
4 Error 5.750 (4)
In this case there is no difference in results between the restricted and unrestricted models.
14-9 Derive the expected means squares for a balanced three-stage nested design if all three factors are
random. Obtain formulas for estimating the variance components. Assume the restricted form of the
mixed model.
The expected mean squares can be generated in Minitab as follows:
Minitab Output
A random 2 -1 1
B(A) random 2 -1 1
Source DF SS MS F P
A 1 0.250 0.250 0.06 0.831
B(A) 2 8.500 4.250 0.35 0.726
C(A B) 4 49.000 12.250 2.13 0.168
Error 8 46.000 5.750
Total 15 103.750
1 A -0.5000 2 (4) + 2(3) + 4(2) + 8(1)
2 B(A) -2.0000 3 (4) + 2(3) + 4(2)
3 C(A B) 3.2500 4 (4) + 2(3)
4 Error 5.7500 (4)
14-10 Verify the expected mean squares given in Table 14-1.
F F R
a b n
Factor i j l E(MS)
τi 0 b n
∑−
+ 22
1
i
a
bn
τσ
13-5

( )ijβ 1 0 n
( ) ( )∑∑−
+ 22
1
ij
ba
n
βσ
( )lijkε 1 1 1 σ 2
R R R
a b n
Factor i j l E(MS)
τi 1 b n 222
τβ σσσ bnn ++
( )ijβ 1 1 n 22
βσσ n+
( )lijkε 1 1 1 σ 2
F R R
a b n
Factor i j l E(MS)
τi 0 b n
∑−
++ 222
1
i
a
bn
n τσσ β
( )ijβ 1 1 n 22
βσσ n+
( )lijkε 1 1 1 σ 2
14-11 Unbalanced designs. Consider an unbalanced two-stage nested design with bj levels of B under the
ith level of A and nij replicates in the ijth cell.
(a) Write down the least squares normal equations for this situation. Solve the normal equations.
The least squares normal equations are:
( )∑∑∑ = ==
=++=
a
i
b
j
...ijij
a
i
i.i..
i
yˆnˆnˆn
1 11
βτµµ
( )∑=
=++=
ib
j
..iijiji.i.ii yˆnˆnˆn
1
βτµτ , for i a= 1 2, ,...,
( ) ( ) .ijijijiijijij yˆnˆnˆn =++= βτµβ , for i a= 1 2, ,..., and j bi= 1 2, ,...,
There are 1+a+b equations in 1+a+b unknowns. However, there are a+1linear dependencies in these
equations, and consequently, a+1 side conditions are needed to solve them. Any convenient set of a+1
linearly independent equations can be used. The easiest set is , , for i=1,2,…,a. Using these
conditions we get
µ = 0 τi = 0
µ = 0 , ,τi = 0 ( ) .βj i ijy=
as the solution to the normal equations. See Searle (1971) for a full discussion.
(b) Construct the analysis of variance table for the unbalanced two-stage nested design.
The analysis of variance table is
Source SS DF
13-6

A
..
...
a
i .i
..i
n
y
n
y 2
1
2
−∑−
a-1
B ∑∑∑ −= =
−
a
i .i
..i
a
i
b
j ij
.ij
n
y
n
yi
1
2
1 1
2
b.-a
Error ∑∑∑∑∑ = == = =
−
a
i
b
j ij
.ij
a
i
b
j
n
k
ijk
ii ij
n
y
y
1 1
2
1 1 1
2
n..-b
Total
..
...
a
i
b
j
n
k
ijk
n
y
y
i ij 2
1 1 1
2
−∑∑∑= = =
n..-1
(c) Analyze the following data, using the results in part (b).
Factor A 1 2
Factor B 1 2 1 2 3
6 -3 5 2 1
4 1 7 4 0
8 9 3 -3
6
Note that a=2, b1=2, b2=3, b.=b1+b2=5, n11=3, n12=2, n21=4, n22=3 and n23=3
Source SS DF MS
A 0.13 1 0.13
B 153.78 3 51.26
Error 35.42 10 3.54
Total 189.33 14
The analysis can also be performed in Minitab as follows. The adjusted sum of squares is utilized by
Minitab’s general linear model routine.
Minitab Output
General Linear Model: y versus A, B
A fixed 2 1 2
B(A) fixed 5 1 2 1 2 3
Analysis of Variance for y, using Adjusted SS for Tests
A 1 0.133 0.898 0.898 0.25 0.625
B(A) 3 153.783 153.783 51.261 14.47 0.001
Error 10 35.417 35.417 3.542
Total 14 189.333
14-12 Variance components in the unbalanced two-stage nested design. Consider the model
( ) ( )ijkijiijky εβτµ +++=
⎪
⎩
⎪
⎨
⎧
=
=
=
ijn,...,,k
b,...,,j
a,...,,i
21
21
21
where A and B are random factors. Show that
13-7

( )
( )( )
( ) 2
2
0
2
2
2
2
1
2
σ
σσ
σσσ
β
τβ
=
+=
++=
E
AB
A
MSE
cMSE
ccMSE
where
1
1
1
2
2
1 1
2
1
2
1
1 1
2
0
−
−
=
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
∑
∑ ∑∑∑
∑ ∑
=
= ==
= =
a
N
n
N
c
a
N
n
n
n
c
ab
n
n
N
c
a
i
.i
a
i
a
i
b
j
ij
b
j .i
ij
a
i
b
j .i
ij
ii
i
1=
See “Variance Component Estimation in the 2-way Nested Classification,” by S.R. Searle, Annals of
Mathematical Statistics, Vol. 32, pp. 1161-1166, 1961. A good discussion of variance component
estimation from unbalanced data is in Searle (1971a).
14-13 A process engineer is testing the yield of a product manufactured on three machines. Each machine
can be operated at two power settings. Furthermore, a machine has three stations on which the product is
formed. An experiment is conducted in which each machine is tested at both power settings, and three
observations on yield are taken from each station. The runs are made in random order, and the results
follow. Analyze this experiment, assuming all three factors are fixed.
Machine 1 Machine 2 Machine 3
Station 1 2 3 1 2 3 1 2 3
Power Setting 1 34.1 33.7 36.2 32.1 33.1 32.8 32.9 33.8 33.6
30.3 34.9 36.8 33.5 34.7 35.1 33.0 33.4 32.8
31.6 35.0 37.1 34.0 33.9 34.3 33.1 32.8 31.7
Power Setting 2 24.3 28.1 25.7 24.1 24.1 26.0 24.2 23.2 24.7
26.3 29.3 26.1 25.0 25.1 27.1 26.1 27.4 22.0
27.1 28.6 24.9 26.3 27.9 23.9 25.3 28.0 24.8
The linear model is ( ) ( ) ( ) ( )lijk)j(ikjkijjiijkly ετγγτββτµ ++++++=
Minitab Output
ANOVA: Yield versus Machine, Power, Station
Power fixed 2 1 2
Station(Machine) fixed 3 1 2 3
Analysis of Variance for Yield
Source DF SS MS F P
Machine 2 21.143 10.572 6.46 0.004
13-8

Power 1 853.631 853.631 521.80 0.000
Station(Machine) 6 32.583 5.431 3.32 0.011
Machine*Power 2 0.616 0.308 0.19 0.829
Power*Station(Machine) 6 28.941 4.824 2.95 0.019
Error 36 58.893 1.636
Total 53 995.808
1 Machine 6 (6) + 18Q[1]
2 Power 6 (6) + 27Q[2]
3 Station(Machine) 6 (6) + 6Q[3]
4 Machine*Power 6 (6) + 9Q[4]
5 Power*Station(Machine) 6 (6) + 3Q[5]
6 Error 1.636 (6)
14-14 Suppose that in Problem 14-13 a large number of power settings could have been used and that the
two selected for the experiment were chosen randomly. Obtain the expected mean squares for this
situation and modify the previous analysis appropriately.
The analysis of variance and the expected mean squares can be obtained from Minitab as follows:
Minitab Output
Power random 2 1 2
Source DF SS MS F P
Machine 2 21.143 10.572 34.33 0.028
Power 1 853.631 853.631 521.80 0.000
Station(Machine) 6 32.583 5.431 1.13 0.445
Machine*Power 2 0.616 0.308 0.19 0.829
Error 36 58.893 1.636
Total 53 995.808
1 Machine 4 (6) + 9(4) + 18Q[1]
2 Power 31.5554 6 (6) + 27(2)
3 Station(Machine) 5 (6) + 3(5) + 6Q[3]
4 Machine*Power -0.1476 6 (6) + 9(4)
5 Power*Station(Machine) 1.0625 6 (6) + 3(5)
6 Error 1.6359 (6)
14-15 Reanalyze the experiment in Problem 14-14 assuming the unrestricted form of the mixed model.
You may use a computer software program to do this. Comment on any differences between the restricted
and unrestricted model analysis and conclusions.
Minitab Output
Power random 2 1 2
Source DF SS MS F P
Machine 2 21.143 10.572 34.33 0.028
Power 1 853.631 853.631 2771.86 0.000
13-9

Station(Machine) 6 32.583 5.431 1.13 0.445
Machine*Power 2 0.616 0.308 0.06 0.939
Error 36 58.893 1.636
Total 53 995.808
1 Machine 4 (6) + 3(5) + 9(4) + Q[1,3]
2 Power 31.6046 4 (6) + 3(5) + 9(4) + 27(2)
3 Station(Machine) 5 (6) + 3(5) + Q[3]
4 Machine*Power -0.5017 5 (6) + 3(5) + 9(4)
5 Power*Station(Machine) 1.0625 6 (6) + 3(5)
6 Error 1.6359 (6)
There are differences between several of the expected mean squares. However, the conclusions that could
be drawn do not differ in any meaningful way from the restricted model analysis.
14-16 A structural engineer is studying the strength of aluminum alloy purchased from three vendors.
Each vendor submits the alloy in standard-sized bars of 1.0, 1.5, or 2.0 inches. The processing of different
sizes of bar stock from a common ingot involves different forging techniques, and so this factor may be
important. Furthermore, the bar stock if forged from ingots made in different heats. Each vendor submits
two tests specimens of each size bar stock from the three heats. The resulting strength data follow.
Analyze the data, assuming that vendors and bar size are fixed and heats are random.
Vendor 1 Vendor 2 Vendor 3
Heat 1 2 3 1 2 3 1 2 3
Bar Size: 1 inch 1.230 1.346 1.235 1.301 1.346 1.315 1.247 1.275 1.324
1.259 1.400 1.206 1.263 1.392 1.320 1.296 1.268 1.315
1 1/2 inch 1.316 1.329 1.250 1.274 1.384 1.346 1.273 1.260 1.392
1.300 1.362 1.239 1.268 1.375 1.357 1.264 1.265 1.364
2 inch 1.287 1.346 1.273 1.247 1.362 1.336 1.301 1.280 1.319
1.292 1.382 1.215 1.215 1.328 1.342 1.262 1.271 1.323
( ) ( ) ( ) ( )lijkjikjkijjiijkl )(y ετγγτββτµ ++++++=
Minitab Output
ANOVA: Strength versus Vendor, Bar Size, Heat
Vendor fixed 3 1 2 3
Heat(Vendor) random 3 1 2 3
Bar Size fixed 3 1.0 1.5 2.0
Source DF SS MS F P
Vendor 2 0.0088486 0.0044243 0.26 0.776
Heat(Vendor) 6 0.1002093 0.0167016 41.32 0.000
Bar Size 2 0.0025263 0.0012631 1.37 0.290
Vendor*Bar Size 4 0.0023754 0.0005939 0.65 0.640
Bar Size*Heat(Vendor) 12 0.0110303 0.0009192 2.27 0.037
Error 27 0.0109135 0.0004042
Total 53 0.1359034
1 Vendor 2 (6) + 6(2) + 18Q[1]
2 Heat(Vendor) 0.00272 6 (6) + 6(2)
3 Bar Size 5 (6) + 2(5) + 18Q[3]
4 Vendor*Bar Size 5 (6) + 2(5) + 6Q[4]
5 Bar Size*Heat(Vendor) 0.00026 6 (6) + 2(5)
6 Error 0.00040 (6)
13-10

14-17 Reanalyze the experiment in Problem 14-16 assuming the unrestricted form of the mixed model.
You may use a computer software program to do this. Comment on any differences between the restricted
and unrestricted model analysis and conclusions.
Minitab Output
Bar Size fixed 3 1.0 1.5 2.0
Source DF SS MS F P
Vendor 2 0.0088486 0.0044243 0.26 0.776
Heat(Vendor) 6 0.1002093 0.0167016 18.17 0.000
Bar Size 2 0.0025263 0.0012631 1.37 0.290
Vendor*Bar Size 4 0.0023754 0.0005939 0.65 0.640
Error 27 0.0109135 0.0004042
Total 53 0.1359034
1 Vendor 2 (6) + 2(5) + 6(2) + Q[1,4]
2 Heat(Vendor) 0.00263 5 (6) + 2(5) + 6(2)
3 Bar Size 5 (6) + 2(5) + Q[3,4]
4 Vendor*Bar Size 5 (6) + 2(5) + Q[4]
6 Error 0.00040 (6)
There are some differences in the expected mean squares. However, the conclusions do not differ from
those of the restricted model analysis.
14-18 Suppose that in Problem 14-16 the bar stock may be purchased in many sizes and that the three sizes
are actually used in experiment were selected randomly. Obtain the expected mean squares for this
situation and modify the previous analysis appropriately. Use the restricted form of the mixed model.
Minitab Output
Bar Size random 3 1.0 1.5 2.0
Source DF SS MS F P
Vendor 2 0.0088486 0.0044243 0.27 0.772 x
Heat(Vendor) 6 0.1002093 0.0167016 18.17 0.000
Bar Size 2 0.0025263 0.0012631 1.37 0.290
Vendor*Bar Size 4 0.0023754 0.0005939 0.65 0.640
Error 27 0.0109135 0.0004042
Total 53 0.1359034
1 Vendor * (6) + 2(5) + 6(4) + 6(2) + 18Q[1]
2 Heat(Vendor) 0.00263 5 (6) + 2(5) + 6(2)
3 Bar Size 0.00002 5 (6) + 2(5) + 18(3)
4 Vendor*Bar Size -0.00005 5 (6) + 2(5) + 6(4)
13-11

6 Error 0.00040 (6)
* Synthesized Test.
1 Vendor 5.75 0.0163762 (2) + (4) - (5)
Notice that a Satterthwaite type test is used for vendor.
14-19 Steel in normalized by heating above the critical temperature, soaking, and then air cooling. This
process increases the strength of the steel, refines the grain, and homogenizes the structure. An experiment
is performed to determine the effect of temperature and heat treatment time on the strength of normalized
steel. Two temperatures and three times are selected. The experiment is performed by heating the oven to
a randomly selected temperature and inserting three specimens. After 10 minutes one specimen is
removed, after 20 minutes the second specimen is removed, and after 30 minutes the final specimen is
removed. Then the temperature is changed to the other level and the process is repeated. Four shifts are
required to collect the data, which are shown below. Analyze the data and draw conclusions, assume both
factors are fixed.
Temperature (F)
Shift Time(minutes) 1500 1600
1 10 63 89
20 54 91
30 61 62
2 10 50 80
20 52 72
30 59 69
3 10 48 73
20 74 81
30 71 69
4 10 54 88
20 48 92
30 59 64
This is a split-plot design. Shifts correspond to blocks, temperature is the whole plot treatment, and time is
the subtreatments (in the subplot or split-plot part of the design). The expected mean squares and analysis
of variance are shown below. The following Minitab Output has been modified to display the results of the
split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and
the statistical tests are not, in general, correct. Notice that the Error term in the analysis of variance is
actually the three factor interaction.
Minitab Output
ANOVA: Strength versus Shift, Temperature, Time
Shift random 4 1 2 3 4
Temperat fixed 2 1500 1600
Time fixed 3 10 20 30
Standard Split Plot
Source DF SS MS F P F P
Shift 3 145.46 48.49 1.19 0.390
Temperat 1 2340.38 2340.38 29.20 0.012 29.21 0.012
Shift*Temperat 3 240.46 80.15 1.97 0.220
Time 2 159.25 79.63 1.00 0.422 1.00 0.422
Shift*Time 6 478.42 79.74 1.96 0.217
Temperat*Time 2 795.25 397.63 9.76 0.013 9.76 0.013
Error 6 244.42 40.74
Total 23 4403.63
13-12

1 Shift 1.292 7 (7) + 6(1)
2 Temperat 3 (7) + 3(3) + 12Q[2]
3 Shift*Temperat 13.139 7 (7) + 3(3)
4 Time 5 (7) + 2(5) + 8Q[4]
5 Shift*Time 19.500 7 (7) + 2(5)
6 Temperat*Time 7 (7) + 4Q[6]
7 Error 40.736 (7)
14-20 An experiment is designed to study pigment dispersion in paint. Four different mixes of a particular
pigment are studied. The procedure consists of preparing a particular mix and then applying that mix to a
panel by three application methods (brushing, spraying, and rolling). The response measured is the
percentage reflectance of the pigment. Three days are required to run the experiment, and the data
obtained follow. Analyze the data and draw conclusions, assuming that mixes and application methods are
fixed.
Mix
Day App Method 1 2 3 4
1 1 64.5 66.3 74.1 66.5
2 68.3 69.5 73.8 70.0
3 70.3 73.1 78.0 72.3
2 1 65.2 65.0 73.8 64.8
2 69.2 70.3 74.5 68.3
3 71.2 72.8 79.1 71.5
3 1 66.2 66.5 72.3 67.7
2 69.0 69.0 75.4 68.6
3 70.8 74.2 80.1 72.4
This is a split plot design. Days correspond to blocks, mix is the whole plot treatment, and method is the
subtreatment (in the subplot or split plot part of the design). The following Minitab Output has been
modified to display the results of the split-plot analysis. Minitab will calculate the sums of squares
correctly, but the expected mean squares and the statistical tests are not, in general, correct. Notice that the
Error term in the analysis of variance is actually the three factor interaction.
Minitab Output
ANOVA: Reflectance versus Day, Mix, Method
Day random 3 1 2 3
Mix fixed 4 1 2 3 4
Method fixed 3 1 2 3
Analysis of Variance for Reflecta
Standard Split Plot
Day 2 2.042 1.021 1.39 0.285
Mix 3 307.479 102.493 135.77 0. 000 135.75 0.000
Day*Mix 6 4.529 0.755 1.03 0.451
Method 2 222.095 111.047 226.24 0.000 226.16 0.000
Day*Method 4 1.963 0.491 0.67 0.625
Mix*Method 6 10.036 1.673 2.28 0.105 2.28 0.105
Error 12 8.786 0.732
Total 35 556.930
1 Day 0.02406 7 (7) + 12(1)
2 Mix 3 (7) + 3(3) + 9Q[2]
3 Day*Mix 0.00759 7 (7) + 3(3)
4 Method 5 (7) + 4(5) + 12Q[4]
5 Day*Method -0.06032 7 (7) + 4(5)
6 Mix*Method 7 (7) + 3Q[6]
7 Error 0.73213 (7)
13-13

14-21 Repeat Problem 14-20, assuming that the mixes are random and the application methods are fixed.
The F-tests are the same as those in Problem 13-20. The following Minitab Output has been edited to
display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the
expected mean squares and the statistical tests are not, in general, correct. Again, the Error term in the
analysis of variance is actually the three factor interaction.
Minitab Output
ANOVA: Reflectance versus Day, Mix, Method
Day random 3 1 2 3
Mix random 4 1 2 3 4
Method fixed 3 1 2 3
Analysis of Variance for Reflecta
Standard Split Plot
Day 2 2.042 1.021 1.35 0.328
Mix 3 307.479 102.493 135.77 0.000 135.75 0.000
Day*Mix 6 4.529 0.755 1.03 0.451
Method 2 222.095 111.047 77.58 0.001 x 226.16 0.000
Day*Method 4 1.963 0.491 0.67 0.625
Mix*Method 6 10.036 1.673 2.28 0.105 2.28 0.105
Error 12 8.786 0.732
Total 35 556.930
1 Day 0.0222 3 (7) + 3(3) + 12(1)
2 Mix 11.3042 3 (7) + 3(3) + 9(2)
3 Day*Mix 0.0076 7 (7) + 3(3)
4 Method * (7) + 3(6) + 4(5) + 12Q[4]
5 Day*Method -0.0603 7 (7) + 4(5)
6 Mix*Method 0.3135 7 (7) + 3(6)
7 Error 0.7321 (7)
* Synthesized Test.
4 Method 3.59 1.431 (5) + (6) - (7)
14-22 Consider the split-split-plot design described in example 14-3. Suppose that this experiment is
conducted as described and that the data shown below are obtained. Analyze and draw conclusions.
Technician
1 2 3
Blocks Dose Strengths 1 2 3 1 2 3 1 2 3
Wall Thickness
1 1 95 71 108 96 70 108 95 70 100
2 104 82 115 99 84 100 102 81 106
3 101 85 117 95 83 105 105 84 113
4 108 85 116 97 85 109 107 87 115
2 1 95 78 110 100 72 104 92 69 101
2 106 84 109 101 79 102 100 76 104
3 103 86 116 99 80 108 101 80 109
4 109 84 110 112 86 109 108 86 113
3 1 96 70 107 94 66 100 90 73 98
2 105 81 106 100 84 101 97 75 100
3 106 88 112 104 87 109 100 82 104
4 113 90 117 121 90 117 110 91 112
4 1 90 68 109 98 68 106 98 72 101
2 100 84 112 102 81 103 102 78 105
13-14

3 102 85 115 100 85 110 105 80 110
4 114 88 118 118 85 116 110 95 120
Using the computer output, the F-ratios were calculated by hand using the expected mean squares found in
Table 14-18. The following Minitab Output has been edited to display the results of the split-plot analysis.
Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests
are not, in general, correct. Notice that the Error term in the analysis of variance is actually the four factor
interaction.
Minitab Output
ANOVA: Time versus Day, Tech, Dose, Thick
Tech fixed 3 1 2 3
Dose fixed 3 1 2 3
Thick fixed 4 1 2 3 4
Standard Split Plot
Day 3 48.41 16.14 3.38 0.029
Tech 2 248.35 124.17 4.62 0.061 4.62 0.061
Day*Tech 6 161.15 26.86 5.62 0.000
Dose 2 20570.06 10285.03 550.44 0.000 550.30 0.000
Day*Dose 6 112.11 18.69 3.91 0.004
Tech*Dose 4 125.94 31.49 3.32 0.048 3.32 0.048
Day*Tech*Dose 12 113.89 9.49 1.99 0.056
Thick 3 3806.91 1268.97 36.47 0.000 36.48 0.000
Day*Thick 9 313.12 34.79 7.28 0.000
Tech*Thick 6 126.49 21.08 2.26 0.084 2.26 0.084
Day*Tech*Thick 18 167.57 9.31 1.95 0.044
Dose*Thick 6 402.28 67.05 17.13 0.000 17.15 0.000
Day*Dose*Thick 18 70.44 3.91 0.82 0.668
Tech*Dose*Thick 12 205.89 17.16 3.59 0.001 3.59 0.001
Error 36 172.06 4.78
Total 143 26644.66
1 Day 0.3155 15 (15) + 36(1)
2 Tech 3 (15) + 12(3) + 48Q[2]
3 Day*Tech 1.8400 15 (15) + 12(3)
4 Dose 5 (15) + 12(5) + 48Q[4]
5 Day*Dose 1.1588 15 (15) + 12(5)
6 Tech*Dose 7 (15) + 4(7) + 16Q[6]
7 Day*Tech*Dose 1.1779 15 (15) + 4(7)
8 Thick 9 (15) + 9(9) + 36Q[8]
9 Day*Thick 3.3346 15 (15) + 9(9)
10 Tech*Thick 11 (15) + 3(11) + 12Q[10]
11 Day*Tech*Thick 1.5100 15 (15) + 3(11)
12 Dose*Thick 13 (15) + 3(13) + 12Q[12]
13 Day*Dose*Thick -0.2886 15 (15) + 3(13)
14 Tech*Dose*Thick 15 (15) + 4Q[14]
15 Error 4.7793 (15)
14-23 Rework Problem 14-22, assuming that the dosage strengths are chosen at random. Use the restricted
form of the mixed model.
The following Minitab Output has been edited to display the results of the split-plot analysis. Minitab will
calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, in
general, correct. Again, the Error term in the analysis of variance is actually the four factor interaction.
Minitab Output
ANOVA: Time versus Day, Tech, Dose, Thick
13-15

Tech fixed 3 1 2 3
Dose random 3 1 2 3
Thick fixed 4 1 2 3 4
Standard Split Plot
Day 3 48.41 16.14 0.86 0.509
Tech 2 248.35 124.17 2.54 0.155 4.62 0.061
Day*Tech 6 161.15 26.86 2.83 0.059
Dose 2 20570.06 10285.03 550.44 0.000 550.30 0.000
Day*Dose 6 112.11 18.69 3.91 0.004
Tech*Dose 4 125.94 31.49 3.32 0.048 3.32 0.048
Day*Tech*Dose 12 113.89 9.49 1.99 0.056
Thick 3 3806.91 1268.97 12.96 0.001 x 36.48 0.000
Day*Thick 9 313.12 34.79 8.89 0.000
Tech*Thick 6 126.49 21.08 0.97 0.475 x 2.26 0.084
Day*Tech*Thick 18 167.57 9.31 1.95 0.044
Dose*Thick 6 402.28 67.05 17.13 0.000 17.15 0.000
Day*Dose*Thick 18 70.44 3.91 0.82 0.668
Tech*Dose*Thick 12 205.89 17.16 3.59 0.001 3.59 0.001
Error 36 172.06 4.78
Total 143 26644.66
1 Day -0.071 5 (15) + 12(5) + 36(1)
2 Tech * (15) + 4(7) + 16(6) + 12(3) + 48Q[2]
3 Day*Tech 1.447 7 (15) + 4(7) + 12(3)
4 Dose 213.882 5 (15) + 12(5) + 48(4)
5 Day*Dose 1.159 15 (15) + 12(5)
6 Tech*Dose 1.375 7 (15) + 4(7) + 16(6)
7 Day*Tech*Dose 1.178 15 (15) + 4(7)
8 Thick * (15) + 3(13) + 12(12) + 9(9) + 36Q[8]
9 Day*Thick 3.431 13 (15) + 3(13) + 9(9)
10 Tech*Thick * (15) + 4(14) + 3(11) + 12Q[10]
11 Day*Tech*Thick 1.510 15 (15) + 3(11)
12 Dose*Thick 5.261 13 (15) + 3(13) + 12(12)
13 Day*Dose*Thick -0.289 15 (15) + 3(13)
14 Tech*Dose*Thick 3.095 15 (15) + 4(14)
15 Error 4.779 (15)
* Synthesized Test.
2 Tech 6.35 48.85 (3) + (6) - (7)
8 Thick 10.84 97.92 (9) + (12) - (13)
10 Tech*Thick 15.69 21.69 (11) + (14) - (15)
There are no exact tests on technicians βj , dosage strengths γ k , wall thickness δh , or the technician x
wall thickness interaction . The approximate F-tests are as follows:( )jhβδ
H0: =0β j
2912
4863185926
4919174124
.
..
..
MSMS
MSMS
F
BCAB
ABCB
=
+
+
=
+
+
=
( ) ( ) 3152
12
4919
2
174124
4919174124
122
22
2
22
2
.
..
..
MSMS
MSMS
p
ABCB
ABCB
=
+
+
=
+
+
=
13-16

( ) ( ) 2489
4
48631
6
85926
4863185926
46
22
2
22
2
.
..
..
MSMS
MSMS
q
BCAB
BCAB
=
+
+
=
+
+
=
Do not reject H0: =0β j
H0: γ k =0
039101
7913404667
914302810285
.
..
..
MSMS
MSMS
F
ADCD
ACDC
=
+
+
=
+
+
=
( ) ( ) 0022
18
9143
2
02810285
914302810285
182
22
2
22
2
.
..
..
MSMS
MSMS
p
ACDC
ACDC
=
+
+
=
+
+
=
( ) ( ) 73611
9
79134
6
04667
7913404667
96
22
2
22
2
.
..
..
MSMS
MSMS
q
ADCD
ADCD
=
+
+
=
+
+
=
Reject H0: γ k =0
H0: δh =0
49912
7913404667
91439701268
.
..
..
MSMS
MSMS
F
ADCD
ACDD
=
+
+
=
+
+
=
( ) ( ) 0193
18
9143
3
9701268
91439701268
183
22
2
22
2
.
..
..
MSMS
MSMS
p
ACDD
ACDD
=
+
+
=
+
+
=
( ) ( ) 73611
9
79134
6
04667
7913404667
96
22
2
22
2
.
..
..
MSMS
MSMS
q
ADCD
ADCD
=
+
+
=
+
+
=
Reject H0: δh =0
H0: =0( )jhβδ
9770
309915717
779408121
.
..
..
MSMS
MSMS
F
ABDBCD
ABCDBD
=
+
+
=
+
+
=
F<1, Do not reject H0: =0( )jhβδ
14-24 Suppose that in Problem 14-22 four technicians had been used. Assuming that all the factors are
fixed, how many blocks should be run to obtain an adequate number of degrees of freedom on the test for
differences among technicians?
The number of degrees of freedom for the test is (a-1)(4-1)=3(a-1), where a is the number of blocks used.
13-17

Number of Blocks (a) DF for test
2 3
3 6
4 9
5 12
At least three blocks should be run, but four would give a better test.
14-25 Consider the experiment described in Example 14-4. Demonstrate how the order in which the
treatments combinations are run would be determined if this experiment were run as (a) a split-split-plot,
(b) a split-plot, (c) a factorial design in a randomized block, and (d) a completely randomized factorial
design.
(a) Randomization for the split-split plot design is described in Example 14-4.
(b) In the split-plot, within a block, the technicians would be the main treatment and within a block-
technician plot, the 12 combinations of dosage strength and wall thickness would be run in random
order. The design would be a two-factor factorial in a split-plot.
(c) To run the design in a randomized block, the 36 combinations of technician, dosage strength, and wall
thickness would be run in random order within each block. The design would be a three factor
factorial in a randomized block.
(d) The blocks would be considered as replicates, and all 144 observations would be 4 replicates of a three
factor factorial.
14-26 An article in Quality Engineering (“Quality Quandaries: Two-Level Factorials Run as Split-Plot
Experiments”, Bisgaard, et al, Vol. 8, No. 4, pp. 705-708, 1996) describes a 25
factorial experiment on a
plasma process focused on making paper more susceptible to ink. Four of the factors (A-D) are difficult to
change from run-to-run, so the experimenters set up the reactor at the eight sets of conditions specific by
the low and high levels of these factors, and then processed the two paper types (factor E) together. The
placement of the paper specimens in the reactors (right versus left) was randomized. This produces a split-
plot design with A-D as the whole-plot factors and factor E as the subplot factor. The data from this
experiment are shown below. Analyze the data from this experiment and draw conclusions.
Standard
Order
Run
Number
A =
Pressure
B =
Power
C =
Gas Flow
D =
Gas Type
E =
Paper
Type
y
Contact
Angle
1 23 -1 -1 -1 Oxygen E1 48.6
2 3 +1 -1 -1 Oxygen E1 41.2
3 11 -1 +1 -1 Oxygen E1 55.8
4 29 +1 +1 -1 Oxygen E1 53.5
5 1 -1 -1 +1 Oxygen E1 37.6
6 15 +1 -1 +1 Oxygen E1 47.2
13-18

7 27 -1 +1 +1 Oxygen E1 47.2
8 25 +1 +1 +1 Oxygen E1 48.7
9 19 -1 -1 -1 SiCl4 E1 5
10 5 +1 -1 -1 SiCl4 E1 56.8
11 9 -1 +1 -1 SiCl4 E1 25.6
12 31 +1 +1 -1 SiCl4 E1 41.8
13 13 -1 -1 +1 SiCl4 E1 13.3
14 7 +1 -1 +1 SiCl4 E1 47.5
15 21 -1 +1 +1 SiCl4 E1 11.3
16 17 +1 +1 +1 SiCl4 E1 49.5
17 24 -1 -1 -1 Oxygen E2 57
18 4 +1 -1 -1 Oxygen E2 38.2
19 12 -1 +1 -1 Oxygen E2 62.9
20 30 +1 +1 -1 Oxygen E2 51.3
21 2 -1 -1 +1 Oxygen E2 43.5
22 16 +1 -1 +1 Oxygen E2 44.8
23 28 -1 +1 +1 Oxygen E2 54.6
24 26 +1 +1 +1 Oxygen E2 44.4
25 20 -1 -1 -1 SiCl4 E2 18.1
26 6 +1 -1 -1 SiCl4 E2 56.2
27 10 -1 +1 -1 SiCl4 E2 33
28 32 +1 +1 -1 SiCl4 E2 37.8
29 14 -1 -1 +1 SiCl4 E2 23.7
30 8 +1 -1 +1 SiCl4 E2 43.2
31 22 -1 +1 +1 SiCl4 E2 23.9
32 18 +1 +1 +1 SiCl4 E2 48.2
Half normal probability plots of the effects for both the whole plot with factors A, B, C, D, and their
corresponding interactions, as well as the sub-plot with factor E and all interactions involving E, are shown
below. The analysis of variance is not shown because of the known errors in the calculations; however, the
models are also shown below.
DESIGN-EXPERT Plot
Contact Angle
A: Pressure
B: Power
C: Gas Flow
D: Gas Ty pe
E: Paper Ty pe
Half Normal plot
|Effect|
0.00 4.14 8.28 12.42 16.56
0
20
40
60
70
80
85
90
95
97
99
A
D
AD
DESIGN-EXPERT Plot
Contact Angle
A: Pressure
B: Power
C: Gas Flow
D: Gas Ty pe
E: Paper Ty pe
Half Normal plot
|Effect|
0.00 1.48 2.95 4.43 5.90
0
20
40
60
70
80
85
90
95
97
99
E
AE
Response: Contact Angle
Contact Angle =
+40.98
+5.91 * A
-7.55 * D
+1.57 * E
+8.28 * A * D
-2.95 * A * E
13-19

Gas Type Oxygen
Paper Type E1
Contact Angle =
+46.96250
+0.58125 * Pressure
Gas Type SiCl4
Paper Type E1
Contact Angle =
+31.86250
+17.14375 * Pressure
Gas Type Oxygen
Paper Type E2
Contact Angle =
+50.10000
-5.31875 * Pressure
Gas Type SiCl4
Paper Type E2
Contact Angle =
+35.00000
14-27 Reconsider the experiment in problem 14-26. This is a rather large experiment, so suppose that the
experimenter had used a 25-1
design instead. Set up the 25-1
design in a split-plot, using the principle
fraction. Then select the response data using the information from the full factorial. Analyze the data and
draw conclusions. Do they agree with the results of Problem 14-26?
Standard
Order
Run
Number
A =
Pressure
B =
Power
C =
Gas Flow
D =
Gas Type
E =
Paper
Type
y
Contact
Angle
1 12 -1 -1 -1 Oxygen E2 57
2 2 +1 -1 -1 Oxygen E1 41.2
3 6 -1 +1 -1 Oxygen E1 55.8
4 15 +1 +1 -1 Oxygen E2 51.3
5 1 -1 -1 +1 Oxygen E1 37.6
6 8 +1 -1 +1 Oxygen E2 44.8
7 14 -1 +1 +1 Oxygen E2 54.6
8 13 +1 +1 +1 Oxygen E1 48.7
9 10 -1 -1 -1 SiCl4 E1 5
10 3 +1 -1 -1 SiCl4 E2 56.2
11 5 -1 +1 -1 SiCl4 E2 33
12 16 +1 +1 -1 SiCl4 E1 41.8
13 7 -1 -1 +1 SiCl4 E2 23.7
14 4 +1 -1 +1 SiCl4 E1 47.5
15 11 -1 +1 +1 SiCl4 E1 11.3
16 9 +1 +1 +1 SiCl4 E2 48.2
Similar results are found with the half fraction other than the AE interaction is no longer significant and the
effect for factor E is larger. The half normal probability plot of effects for the whole and sub-plots are
shown below. The resulting model is also shown.
13-20

DESIGN-EXPERT Plot
Contact Angle
A: Pressure
B: Power
C: Gas Flow
D: Gas Ty pe
E: Paper Ty pe
Half Normal plot
|Effect|
0.00 4.37 8.73 13.10 17.46
0
20
40
60
70
80
85
90
95
97
99
A
D
AD
DESIGN-EXPERT Plot
Contact Angle
A: Pressure
B: Power
C: Gas Flow
D: Gas Ty pe
E: Paper Ty pe
Half Normal plot
|Effect|
0.00 2.50 4.99 7.49 9.99
0
20
40
60
70
80
85
90
95
97
99
E
Response: Contact Angle
Contact Angle =
+41.11
+6.36 * A
-7.77 * D
+4.99 * E
+8.73 * A * D
Gas Type Oxygen
Paper Type E1
Contact Angle =
+43.88125
-2.37500 * Pressure
Gas Type SiCl4
Paper Type E1
Contact Angle =
+28.34375
Gas Type Oxygen
Paper Type E2
Contact Angle =
+53.86875
-2.37500 * Pressure
Gas Type SiCl4
Paper Type E2
Contact Angle =
+38.33125
13-21

Chapter 15
Other Design and Analysis Topics
Solutions
15-1 Reconsider the experiment in Problem 5-22. Use the Box-Cox procedure to determine if a
transformation on the response is appropriate (or useful) in the analysis of the data from this experiment.
DESIGN-EXPERT Plot
Crack Growth
Lam bda
Current = 1
Best = 0.11
Low C.I. = -0.44
High C.I. = 0.56
Recom m end transform :
Log
(Lam bda = 0)
Lam bda
Ln(ResidualSS)
1.10
2.23
3.36
4.49
5.62
-3 -2 -1 0 1 2 3
With the value of lambda near zero, and since the confidence interval does not include one, a natural log
transformation would be appropriate.
15-2 In example 6-3 we selected a log transformation for the drill advance rate response. Use the Box-
Cox procedure to demonstrate that this is an appropriate data transformation.
DESIGN-EXPERT Plot
Advance Rate
Lam bda
Current = 1
Best = -0.23
Low C.I. = -0.79
High C.I. = 0.32
Log
(Lam bda = 0)
Lam bda
Ln(ResidualSS)
1.05
2.50
3.95
5.40
6.85
-3 -2 -1 0 1 2 3
14-1

Because the value of lambda is very close to zero, and the confidence interval does not include one, the
natural log was the correct transformation chosen for this analysis.
15-3 Reconsider the smelting process experiment in Problem 8-23, where a 26-3
fractional factorial design
was used to study the weight of packing material stuck to carbon anodes after baking. Each of the eight
runs in the design was replicated three times and both the average weight and the range of the weights at
each test combination were treated as response variables. Is there any indication that that a transformation
is required for either response?
DESIGN-EXPERT Plot
Weight
Lam bda
Current = 1
Best = 1.33
Low C.I. = -0.71
High C.I. = 4.29
None
(Lam bda = 1)
Lam bda
Ln(ResidualSS)
7.89
8.68
9.47
10.26
11.05
-3 -2 -1 0 1 2 3
DESIGN-EXPERT Plot
Range
Lam bda
Current = 1
Best = 0.58
Low C.I. = -1.74
High C.I. = 2.92
None
(Lam bda = 1)
Lam bda
Ln(ResidualSS)
9.29
10.23
11.17
12.12
13.06
-3 -2 -1 0 1 2 3
There is no indication that a transformation is required for either response.
15-4 In Problem 8-25 a replicated fractional factorial design was used to study substrate camber in
semiconductor manufacturing. Both the mean and standard deviation of the camber measurements were
used as response variables. Is there any indication that a transformation is required for either response?
14-2

DESIGN-EXPERT Plot
Cam ber Avg
Lam bda
Current = 1
Best = -0.03
Low C.I. = -0.79
High C.I. = 0.74
Log
(Lam bda = 0)
Lam bda
Ln(ResidualSS)
8.76
9.62
10.49
11.35
12.22
-3 -2 -1 0 1 2 3
DESIGN-EXPERT Plot
Cam ber StDev
Lam bda
Current = 1
Best = 0.57
Low C.I. = -0.03
High C.I. = 1.16
None
(Lam bda = 1)
Lam bda
Ln(ResidualSS)
6.40
8.05
9.70
11.35
13.00
-3 -2 -1 0 1 2 3
The Box-Cox plot for the Camber Average suggests a natural log transformation should be applied. This
decision is based on the confidence interval for lambda not including one and the point estimate of lambda
being very close to zero. With a lambda of approximately 0.5, a square root transformation could be
considered for the Camber Standard Deviation; however, the confidence interval indicates that no
transformation is needed.
15-5 Reconsider the photoresist experiment in Problem 8-26. Use the variance of the resist thickness at
each test combination as the response variable. Is there any indication that a transformation is required?
DESIGN-EXPERT Plot
T hick StDev
Lam bda
Current = 1
Best = -0.04
Low C.I. = -0.77
High C.I. = 0.76
Log
(Lam bda = 0)
Lam bda
Ln(ResidualSS)
7.31
7.97
8.62
9.28
9.93
-3 -2 -1 0 1 2 3
With the point estimate of lambda near zero, and the confidence interval for lambda not inclusive of one, a
natural log transformation would be appropriate.
14-3

15-6 In the grill defects experiment described in Problem 8-30 a variation of the square root
transformation was employed in the analysis of the data. Use the Box-Cox method to determine if this is
the appropriate transformation.
DESIGN-EXPERT Plot
c
Lam bda
Current = 1
Best = -0.06
Low C.I. = -0.69
High C.I. = 0.74
Log
(Lam bda = 0)
k = 0.56
(used to m ake
response values
positive)
Lam bda
Ln(ResidualSS)
3.15
5.55
7.95
10.35
12.75
-3 -2 -1 0 1 2 3
Because the confidence interval for the minimum lambda does not include one, the decision to use a
transformation is correct. Because the lambda point estimate is close to zero, the natural log transformation
would be appropriate. This is a stronger transformation than the square root.
15-7 In the central composite design of Problem 11-14, two responses were obtained, the mean and
variance of an oxide thickness. Use the Box-Cox method to investigate the potential usefulness of
transformation for both of these responses. Is the log transformation suggested in part (c) of that problem
appropriate?
DESIGN-EXPERT Plot
M ean T hick
Lam bda
Current = 1
Best = -0.2
Low C.I. = -3.58
High C.I. = 3.18
None
(Lam bda = 1)
Lam bda
Ln(ResidualSS)
8.09
8.25
8.41
8.57
8.73
-3 -2 -1 0 1 2 3
DESIGN-EXPERT Plot
Var T hick
Lam bda
Current = 1
Best = -0.47
Low C.I. = -2.85
High C.I. = 1.51
None
(Lam bda = 1)
Lam bda
Ln(ResidualSS)
1.32
1.65
1.97
2.30
2.63
-3 -2 -1 0 1 2 3
The Box-Cox plot for the Mean Thickness model suggests that a natural log transformation could be
applied; however, the confidence interval for lambda includes one. Therefore, a transformation would
14-4

have a minimal effect. The natural log transformation applied to the Variance of Thickness model appears
to be acceptable; however, again the confidence interval for lambda includes one.
15-8 In the 33
factorial design of Problem 12-12 one of the responses is a standard deviation. Use the
Box-Cox method to investigate the usefulness of transformations for this response. Would your answer
change if we used the variance of the response?
DESIGN-EXPERT Plot
Std. Dev.
Lam bda
Current = 1
Best = 0.29
Low C.I. = 0.01
High C.I. = 0.61
Square Root
(Lam bda = 0.5)
k = 1.582
(used to m ake
response values
positive)
Lam bda
Ln(ResidualSS)
9.94
13.03
16.13
19.22
22.32
-3 -2 -1 0 1 2 3
Because the confidence interval for lambda does not include one, a transformation should be applied. The
natural log transformation should not be considered due to zero not being included in the confidence
interval. The square root transformation appears to be acceptable. However, notice that the value of zero
is very close to the lower confidence limit, and the minimizing value of lambda is between 0 and 0.5. It is
likely that either the natural log or the square root transformation would work reasonably well.
15-9 Problem 12-10 suggests using the ln(s2
) as the response (refer to part b). Does the Box-Cox method
indicate that a transformation is appropriate?
14-5

DESIGN-EXPERT Plot
Variance
Lam bda
Current = 1
Best = -1.17
Low C.I. = -1.53
High C.I. = -0.72
Inverse
(Lam bda = -1)
Lam bda
Ln(ResidualSS)
3.85
7.28
10.70
14.13
17.56
-3 -2 -1 0 1 2 3
Because the confidence interval for lambda does not include one, a transformation should be applied. The
confidence interval does not include zero; therefore, the natural log transformation is inappropriate. With
the point estimate of lambda at –1.17, the reciprocal transformation is appropriate.
15-10 Myers, Montgomery and Vining (2002) describe an experiment to study spermatozoa survival. The
design factors are the amount of sodium citrate, the amount of glycerol, and equilibrium time, each at two
levels. The response variable is the number of spermatozoa that survive out of fifty that were tested at each
set of conditions. The data are in the following table. Analyze the data from this experiment with
logistical regression.
Sodium
Citrate Glycerol
Equilibriu
m Time
Number
Survived
- - - 34
+ - - 20
- + - 8
+ + - 21
- - + 30
+ - + 20
- + + 10
+ + + 25
Minitab Output
Binary Logistic Regression: Number Survi, Freq versus Sodium Citra, Glycerol, .
Link Function: Logit
Response Information
Variable Value Count
Number Survived Success 168
Failure 232
Freq Total 400
Logistic Regression Table
Odds 95% CI
Predictor Coef SE Coef Z P Ratio Lower Upper
Constant -0.376962 0.110113 -3.42 0.001
Sodium Citrate 0.0932642 0.110103 0.85 0.397 1.10 0.88 1.36
14-6

Glycerol -0.463247 0.110078 -4.21 0.000 0.63 0.51 0.78
Equilbrium Time 0.0259045 0.109167 0.24 0.812 1.03 0.83 1.27
AB 0.585116 0.110066 5.32 0.000 1.80 1.45 2.23
AC 0.0543714 0.109317 0.50 0.619 1.06 0.85 1.31
BC 0.112190 0.108845 1.03 0.303 1.12 0.90 1.38
Log-Likelihood = -248.028
Test that all slopes are zero: G = 48.178, DF = 6, P-Value = 0.000
Goodness-of-Fit Tests
Method Chi-Square DF P
Pearson 0.113790 1 0.736
Deviance 0.113865 1 0.736
Hosmer-Lemeshow 0.113790 6 1.000
This analysis shows that Glycerol (B) and the Sodium Citrate x Glycerol (AB) interaction have an effect on
the survival rate of spermatozoa.
15-11 A soft drink distributor is studying the effectiveness of delivery methods. Three different types of
hand trucks have been developed, and an experiment is performed in the company’s methods engineering
laboratory. The variable of interest is the delivery time in minutes (y); however, delivery time is also
strongly related to the case volume delivered (x). Each hand truck is used four times and the data that
follow are obtained. Analyze the data and draw the appropriate conclusions. Use α=0.05.
Hand Truck Type
1 1 2 2 3 3
y x y x y x
27 24 25 26 40 38
44 40 35 32 22 26
33 35 46 42 53 50
41 40 26 25 18 20
From the analysis performed in Minitab, hand truck does not have a statistically significant effect on
delivery time. Volume, as expected, does have a significant effect.
Minitab Output
General Linear Model: Time versus Truck
Truck fixed 3 1 2 3
Volume 1 1232.07 1217.55 1217.55 232.20 0.000
Truck 2 11.65 11.65 5.82 1.11 0.375
Error 8 41.95 41.95 5.24
Total 11 1285.67
Constant -4.747 2.638 -1.80 0.110
Volume 1.17326 0.07699 15.24 0.000
15-12 Compute the adjusted treatment means and the standard errors of the adjusted treatment means for
the data in Problem 15-11.
adj ( )...i.i.i xxˆyy −−= β
14-7

adj ( ) 3934
12
398
4
139
1731
4
145
1 ..y . =⎟
⎠
⎞
⎜
⎝
⎛
−−=
adj ( ) 2535
12
398
4
125
1731
4
132
2 ..y . =⎟
⎠
⎞
⎜
⎝
⎛
−−=
adj ( ) 8632
12
398
4
134
1731
4
133
3 ..y . =⎟
⎠
⎞
⎜
⎝
⎛
−−=
( ) 2
1
2
1
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
xx
...i
Ey.adj
E
xx
n
MSS .i
( ) 1511
50884
17337534
4
1
245
2
1
2
1
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 1541
50884
17332531
4
1
245
2
1
2
2
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 1451
50884
17335033
4
1
245
2
1
2
3
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
The solutions can also be obtained with Minitab as follows:
Minitab Output
Least Squares Means for Time
Truck Mean SE Mean
1 34.39 1.151
2 35.25 1.154
3 32.86 1.145
15-13 The sums of squares and products for a single-factor analysis of covariance follow. Complete the
analysis and draw appropriate conclusions. Use α = 0.05.
Source of Degrees of Sums of Squares and Products
Variation Freedom x xy x
Treatment 3 1500 1000 650
Error 12 6000 1200 550
Total 15 7500 2200 1200
Sums of Squares & Products Adjusted
Source df x xy y y df MS F0
Treatment 3 1500 1000 650 - -
Error 12 6000 1200 550 310 11 28.18
Total 15 7500 2200 1200 559.67 14
Adjusted Treat. 244.67 3 81.56 2.89
Treatments differ only at 10%.
15-14 Find the standard errors of the adjusted treatment means in Example 15-5.
14-8

From Example 14-4 y1 40 38. .= , adj y2 4142. .= , adj y3 37 78. .=
( ) 72310
60195
13242025
5
1
542
2
1
2
1
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 74390
60195
13240026
5
1
542
2
1
2
2
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 78710
60195
13242021
5
1
542
2
1
2
3
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
15-15 Four different formulations of an industrial glue are being tested. The tensile strength of the glue
when it is applied to join parts is also related to the application thickness. Five observations on strength (y)
in pounds and thickness (x) in 0.01 inches are obtained for each formulation. The data are shown in the
following table. Analyze these data and draw appropriate conclusions.
Glue Formulation
1 1 2 2 3 3 4 4
y x y x y x y x
46.5 13 48.7 12 46.3 15 44.7 16
45.9 14 49.0 10 47.1 14 43.0 15
49.8 12 50.1 11 48.9 11 51.0 10
46.1 12 48.5 12 48.2 11 48.1 12
44.3 14 45.2 14 50.3 10 48.6 11
From the analysis performed in Minitab, glue formulation does not have a statistically significant effect on
strength. As expected, glue thickness does affect strength.
Minitab Output
General Linear Model: Strength versus Glue
Glue fixed 4 1 2 3 4
Thick 1 68.852 59.566 59.566 42.62 0.000
Glue 3 1.771 1.771 0.590 0.42 0.740
Error 15 20.962 20.962 1.397
Total 19 91.585
Constant 60.089 1.944 30.91 0.000
Thick -1.0099 0.1547 -6.53 0.000
Unusual Observations for Strength
Obs Strength Fit SE Fit Residual St Resid
3 49.8000 47.5299 0.5508 2.2701 2.17R
1 Thick (3) + Q[1]
2 Glue (3) + Q[2]
14-9

3 Error (3)
1 Thick 15.00 1.397 (3)
2 Glue 15.00 1.397 (3)
Error 1.397
15-16 Compute the adjusted treatment means and their standard errors using the data in Problem 15-15.
adj ( )...i.i.i xxˆyy −−= β
adj ( )( ) 0847451200130099152461 .....y . =−−−=
adj ( )( ) 6447451280110099130482 .....y . =−−−=
adj ( )( ) 9147451220120099116483 .....y . =−−−=
adj ( )( ) 4347451280120099108474 .....y . =−−−=
( ) 2
1
2
1
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
xx
...i
Ey.adj
E
xx
n
MSS .i
( ) 53600
4058
45120013
5
1
401
2
1
2
1
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 53860
4058
45128011
5
1
401
2
1
2
2
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 53060
4058
45122012
5
1
401
2
1
2
3
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
( ) 53190
4058
45128012
5
1
401
2
1
2
4
.
.
..
.S .y.adj =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧ −
+=
The adjusted treatment means can also be generated in Minitab as follows:
Minitab Output
Least Squares Means for Strength
Glue Mean SE Mean
1 47.08 0.5355
2 47.64 0.5382
3 47.91 0.5301
4 47.43 0.5314
15-17 An engineer is studying the effect of cutting speed on the rate of metal removal in a machining
operation. However, the rate of metal removal is also related to the hardness of the test specimen. Five
observations are taken at each cutting speed. The amount of metal removed (y) and the hardness of the
specimen (x) are shown in the following table. Analyze the data using and analysis of covariance. Use
α=0.05.
14-10

Cutting Speed (rpm)
1000 1000 1200 1200 1400 1400
y x y x y x
68 120 112 165 118 175
90 140 94 140 82 132
98 150 65 120 73 124
77 125 74 125 92 141
88 136 85 133 80 130
As shown in the analysis performed in Minitab, there is no difference in the rate of removal between the
three cutting speeds. As expected, the hardness does have an impact on rate of removal.
Minitab Output
General Linear Model: Removal versus Speed
Speed fixed 3 1000 1200 1400
Analysis of Variance for Removal, using Adjusted SS for Tests
Hardness 1 3075.7 3019.3 3019.3 347.96 0.000
Speed 2 2.4 2.4 1.2 0.14 0.872
Error 11 95.5 95.5 8.7
Total 14 3173.6
Constant -41.656 6.907 -6.03 0.000
Hardness 0.93426 0.05008 18.65 0.000
Speed
1000 0.478 1.085 0.44 0.668
1200 0.036 1.076 0.03 0.974
Unusual Observations for Removal
Obs Removal Fit SE Fit Residual St Resid
8 65.000 70.491 1.558 -5.491 -2.20R
1 Hardness (3) + Q[1]
2 Speed (3) + Q[2]
3 Error (3)
1 Hardness 11.00 8.7 (3)
2 Speed 11.00 8.7 (3)
Error 8.677
Means for Covariates
Covariate Mean StDev
Hardness 137.1 15.94
Least Squares Means for Removal
Speed Mean SE Mean
1000 86.88 1.325
1200 86.44 1.318
14-11

1400 85.89 1.328
15-18 Show that in a single factor analysis of covariance with a single covariate a 100(1-α) percent
confidence interval on the ith adjusted treatment mean is
( ) ( )
( ) 2
1
2
112
1
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+±−− −−
xx
...i
Ena,...i.i
E
xx
n
MStxxˆy αβ
Using this formula, calculate a 95 percent confidence interval on the adjusted mean of machine 1 in
Example 14-4.
The 100(1-α) percent interval on the ith adjusted treatment mean would be
( ) ( ) .iyadjna,...i.i Stxxˆy 112 −−±−− αβ
since ( ...i.i xx )ˆy −− β is an estimator of the ith adjusted treatment mean. The standard error of the adjusted
treatment mean is found as follows:
( ) ( )[ ] ( ) ( ) ( )ββ ˆVxxyVxxˆyVy.adjV ...i.i...i.i.i
2
−+=−−=
Since the { }yi. and are independent. From regression analysis, we haveβ ( )
xxE
ˆV
2
σ
β = . Therefore,
( ) ( ) ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+=
−
+=
xx
...i
xx
...i
.i
E
xx
nE
xx
n
y.adjV
2
2
222
1
σ
σσ
Replacing by its estimator MSσ 2
E, yields
( ) ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+=
xx
...i
E.i
E
xx
n
MSy.adjVˆ
2
1
or
( ) ( ) 2
1
2
1
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+=
xx
...i
E.i
E
xx
n
MSy.adjS
Substitution of this result into ( ) .1)1(,2....
ˆ
iyadjnaii Stxxy −−±−− αβ will produce the desired confidence
interval. A 95% confidence interval on the mean of machine 1 would be found as follows:
( ) 3840.xxˆyy.adj ...i.i.i =−−= β
( ) 72310.y.adjS .i =
( )[ ]
( )( )[ ]
[ ]5913840
723102023840
723103840 110250
..
...
.t. ,.
±
±
±
Therefore, 96417938 1 .. ≤≤ µ , where µ1 denotes the true adjusted mean of treatment one.
14-12

15-19 Show that in a single-factor analysis of covariance with a single covariate, the standard error of the
difference between any two adjusted treatment means is
( ) 2
1
2
2
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+=−
xx
...i
EyAdjyAdj
E
xx
n
MSS .j.i
( ) ( )[ ]...j.j...i.i.j.i xxˆyxxˆyy.adjy.adj −−−−−=− ββ
( )......
ˆ.. jijiji xxyyyadjyadj −−−=− β
The variance of this statistic is
( )[ ] ( ) ( ) ( ) ( )ββ ˆVxxyVyVxxˆyyV .j.i.j.i.j.i.j.i
2
−++=−−−
( ) ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+=
−
++=
xx
.j.i
xx
.j.i
E
xx
nE
xx
nn
2
2
2222
2
σ
σσσ
Replacing by its estimator MSσ 2
E, , and taking the square root yields the standard error
( ) 2
1
2
2
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+=−
xx
...i
EyAdjyAdj
E
xx
n
MSS .j.i
15-20 Discuss how the operating characteristic curves for the analysis of variance can be used in the
analysis of covariance.
To use the operating characteristic curves, fixed effects case, we would use as the parameter Φ2
,
2
2
2
σ
τ
Φ
n
a i∑=
The test has a-1 degrees of freedom in the numerator and a(n-1)-1 degrees of freedom in the denominator.
14-13

Solutions. Design and Analysis of Experiments. Montgomery

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