DOE L-02 C-01 Introduction C02 Basic Statistical Methods.pptx

BITS Pilani
Pilani Campus
BITS Pilani
Design of Experiments
Lecture-2
1
Varinder Singh @Goa.

DOE L-02 30Jul’23
Today’ssession
2
Chapter # Topic Dates
1 Introduction to Designed Experiments
2 Basic Statistical Methods
3 Analysis of Variance
4 Experiments with Blocking Factors
5 Factorial Experiments
6 Two-Level Factorial Designs
7 Blocking and Confounding Systems for Two-Level Factorials
8 Two-Level Fractional Factorial Designs
9 Other Topics on Factorial and Fractional Factorial Designs
10 Regression Modeling
11 Response Surface Methodology
12 Robust Design
13 Random Effects Models
14 Experiments with Nested Factors and Hard-to-Change Factors
Not in previous semester’s Course Handout. Syllabus yet to be revised.
Design and Analysis of Experiments-
Douglas C. Montgomery, Eighth
edition, 2013, Wiley India.

DOE L-01 30Jul’23 Introduction
1. Trial and error/Best guess.
2. One factor at a time (OFAT).
3. Factorial- Full and Fractional.
Threeapproaches…1/2
3

BITS Pilani, K K Birla Goa Campus
4
Designed Experiments
designed experiment is a test or series of tests in which purposeful
changes are made to the input variables of a process so that we may observe and identify
corresponding changes in the output response.
The objectives of the experiment may include
1. Determining which variables are most influential on the response, y.
2. Determining where to set the influential x’s so that y is near the nominal requirement.
3. Determining where to set the influential x’s so that variability in y is small.
4. Determining where to set the influential x’s so that the effects of the uncontrollable variables
are minimized.
If a process is in statistical control but still has poor capability, then to improve process capability
it will be necessary to reduce variability. Designed experiments may offer a more effective
way to do this than SPC.

5
EXPERIMENTAL DESIGN
FUNDAMENTALS
An experiment is a planned method of inquiry conducted to support or refute a hypothesized
belief or to discover new information about a product, process, or service. It is an active
method of obtaining information, as opposed to passive observation. It involves inducing a
response to determine the effects of certain controlled parameters.
Examples of controllable
parameters, often called factors, are the cutting tool, the temperature setting of an oven,
the selected vendor, the amount of a certain additive, or the type of decor in a restaurant. The
factors may be quantitative or qualitative (discrete).
response variable is generally the resulting effect or output of system .
a treatment is a certain combination of factor levels whose effect on the response variable is of interest.
The variation in experimental units that have been exposed to the same treatment is
attributed to experimental error. This variability is due to uncontrollable factors, or noise
factors.

6
Effect of one factor at a time
Experiments with two or more variables usually avoid changing only one variable at a
time—that is, keeping the other variables constant. There are reasons for this.
Figure (a) shows the design points for such an approach, with value of factor A is
fixed, the two design points are ( 0 , - 1 ) and (0,1).
Figure (b) shows the response
function value at the design points. Note that to maximize the response function in this
experiment, we would move in the direction of the arrow shown along the major axis of factor A.
This, however, does not lead to the proper point for maximizing the
response function.
Another drawback to the one-variable-at-a-time approach is that it cannot detect
interactions

7
FACTORIAL EXPERIMENT
2 level Experiment
Each factor can be set at two levels, denoted by - 1 and 1. The point (0,0) represents the current setting of the
factor levels. In this notation, the first coordinate represents the level of factor A, and the second
coordinate represents the level of factor B. A level of - 1 represents a value below the current
setting of 0, and a level of 1 represents a value above the current setting. Thus, four new design
points, (—1, —1), (1, —1), (—1,1), and (1,1), can be investigated.
to maximize the response function, we will consider the
direction in which the gradient is steepest.
Our next experiment could be conducted in a
region with the center at (2,2) with respect to the current
origin. These same principles apply
to more than two variables as well.

8
Interaction effects
Interactions exist when the nature of the relationship between the response variable and a certain factor is
influenced by the level of some other factor.
factor A stays the same regardless of the
level of factor B.
The rate of change of the response as a function of factor A's level
changes as B's level changes from -1 to 1
Interactions depict the joint relationship of factors on the response function. Such effects
should be accounted for in multifactor designs because they better approximate real-world
events.
For example, a new motivational training program for employees (factor B) might
impact productivity (response variable) differently for different departments (factor A)
within the organization.

9
Another way to represent interaction is
through contour plots

Replication:
Replication involves a repetition of the experiment under similar conditions (that is, similar treatments). allows us to
obtain an estimate of the experimental error, the variation in the experimental units under identically controlled
conditions.
The experimental error forms a basis for determining whether differences in the statistics found from the
observations are significant.
Randomization
Randomization means that the treatments should be assigned to the experimental units in such a way
that every unit has an equal chance of being assigned to any treatment. Such a process
eliminates bias and ensures that no particular treatment is favored.
Random assignments of the treatments and the order in which the experimental trials are run ensures that the
observations, and therefore the experimental errors, will be independent of each other.
An experimental unit is the quantity of material (in manufacturing) to which one trial of a single treatment is
applied.
Features of experimentation
(experimentstoensurereplication,randomizationandifneededblocikng)
10

Blocking
Using this concept,
variability between blocks is eliminated from the
experimental error, which leads to an increase in the precision of the experiment.
Variability of the response function within a block can be attributed to the differences in the treatments
because the impact of other extraneous variables has been minimized. Treatments are assigned at
random to the units within each block.
Features of experimentation
11

1. Randomization.
2. Blocking.
3. Replication.
 Randomization: Accuracy.
 Blocking: Impact of variables.
 Replication: Precision.
Three main DOEs-
1. Completely Randomized design.
2. Randomized Block design.
3. Matched-Pair design.
ThreeprinciplesofDOE
12
Accuracy and Precision

What factors to vary?
No. of experiments?
Sequence of experiments?
Designs-
1. Full factorial designs.
2. Fractional factorial designs (Screening designs).
3. Response surface designs.
4. Mixture designs.
5. Taguchi array designs.
6. Split plot designs.
Different design in different stage of Experimentation.
DOE:Designs
13

DOE L-01 30Jul’23 C-02 Basic Statistical Methods
FactorsandLevels…1/2
14
Speed (10, 20) Surface finish
One Factor (Speed), Two Levels (10 and 20).
Two Factors (Speed and Feed rate), Two (10 and 20)/Three Levels (60, 70 and 80).
Speed (10, 20)
Feed rate (60, 70, 80) Surface finish

FactorsandLevels….2/2
15
Four Factors & Two Levels of each
1. Sunlight (4, 8 hrs)
2. Plant Food (2, 20 g)
3. Water (100 ml, 500ml)
4. Brand of plant (A, B)

BITS Pilani
Pilani Campus
BITS Pilani
Basic Statistical Methods
(Comparative Experiments)
Chapter-2
16

17
Sampling Distribution
• Every time we take a random sample and calculate a statistic,
the value of the statistic changes (remember, a statistic is a
random variable).
• If we continue to take random samples and calculate a given
statistic over time, we will build up a distribution of values for the
statistic. This distribution is referred to as a sampling
distribution.
• A sampling distribution is a distribution that describes the
chance fluctuations of a statistic calculated from a random
sample.

18
Sampling Distribution of the Mean
• The probability distribution of is called the
sampling distribution of the mean.
• The distribution of , for a given sample size n,
describes the variability of sample averages around the
population mean μ.
X
X

19
• If a random sample of size n is taken from a normal
population having mean μx and variance , then
is a random variable which is also normally
distributed with mean μx and variance .
• Further,
is a standard normal random variable.
2
x
 X
n
x
2

n
X
Z x





20
Original Population
80 85 90 95 100 105 110 115 120
X
n(100,5)
n(100,3.54)
5/sqrt(2)=3.54
1
2
3
4
Original population

21
• Example: A manufacturer of steel rods claims that
the length of his bars follows a normal distribution
with a mean of 30 cm and a standard deviation of
0.5 cm.
(a) Assuming that the claim is true, what is the probability that
a given bar will exceed 30.1 cm?
(b) Assuming the claim is true, what is the probability that the
mean of 10 randomly chosen bars will exceed 30.1 cm?
(c) Assuming the claim is true, what is the probability that the
mean of 100 randomly chosen bars will exceed 30.1 cm?

22
• Example: A manufacturer of steel rods claims that the length
of his bars follows a normal distribution with a mean of 30 cm
and a standard deviation of 0.5 cm.
(a) Assuming that the claim is true, what is the probability that a
given bar will exceed 30.1 cm? (z=30.1-30)/0.5=0.2 p=0.42)
(b) Assuming the claim is true, what is the probability that the mean of
10 randomly chosen bars will exceed 30.1 cm?
(z=30.1-30)/(0.5/sqrt(10)=0.63 p=0.26)
(c) Assuming the claim is true, what is the probability that the mean of
100 randomly chosen bars will exceed 30.1 cm?
(z=30.1-30)/(0.5/sqrt(100)=2 p=0.02)

23
Steel Bar Lengths (cm)
28.5 29 29.5 30 30.5 31 31.5
X
28.5 29 29.5 30 30.5 31 31.5
X(10)
28.5 29 29.5 30 30.5 31 31.5
X(100)
.42
.02
. 26

27
Central Limit Theorem
As Sample
Size Gets
Large
Enough
Sampling
Distribution
Becomes almost
normal regardless
of shape of
population
X
X

28
n =30
X = 1.8
n = 4
X = 5
When The Population is Not Normal
Population Distribution
Sampling Distributions
 = 50
 = 10
X
X
50

X
Variation

x

=


x =
n
_
_
Central Tendency

29
Central Limit Theorem
• Rule of thumb: normal approximation for
will be good if n > 30. If n < 30, the
approximation is only good if the
population from which you are sampling
is not too different from normal.
X

30
t-Distribution
• So far, we have been assuming that we
knew the value of σ. This may be true if
one has a large amount of experience with
a certain process.
• However, it is often true that one is
estimating σ along with μ from the same set
of data.
1
)
(
ˆ
2





n
X
X
s i


31
t-Distribution
• To allow for such a situation, we will consider the t statistic:
which follows a t-distribution.
n
S
X
T x



S
n
 standard error of the mean

32
t-Distribution
t-Distribution
-4 -3 -2 -1 0 1 2 3 4
t
t(n=) = Z
t(n=6)
t(n=3)

33
t-Distribution
• If is the mean of a random sample of size n
then
is a random variable following the t- distribution with
parameter ν = n – 1, where ν is degrees of freedom.
X
1
)
( 2
2




n
X
X
S i
n
S
X
t
/




34
t-Distribution
• The t-distribution has been tabularized.
• tα represents the t-value that has an area of α to the
right of it.
• Note, due to symmetry, t1-α = -tα
t-Distribution
-4 -3 -2 -1 0 1 2 3 4
t(n=3) t.05
t.80 t.20
t.95

35

36
Example: t-Distribution
• The resistivity of batches of electrolyte follow a normal
distribution. We sample 5 batches and get the following
readings: 1400, 1450, 1375, 1500, 1550.
• Does this data support or refute a population average of
1400?
1455

X 72

S

37
t-Distribution
-4 -3 -2 -1 0 1 2 3 4
t(n=5)
71
.
1
5
/
72
1400
1455
/





n
S
X
t

1.71
Refute
Refute
Support
t=2.78
p=0.025
Example: t-Distribution

A cement company wants to know whether the
average bond strength of Modified mortar
(more water) is the same or different from that
of Unmodified mortar (standard water).
The company tested 10 samples of Modified
mortar and 10 samples of Unmodified mortar.
Bond strengths recorded by the tests are-
Comparativeexperiment-1:Problem
Textbook:p-26.
38
Cement Formulation
(Modified, Unmodified)
Strength
One Factor, Two Levels
Sample
no.
Modified
Mortar
Sample
no.
Unmodified
Mortar
1 16.9 1 16.6
2 16.4 2 16.8
3 17.2 3 17.4
4 16.4 4 17.1
5 16.5 5 17.0
6 17.0 6 16.9
7 17.0 7 17.3
8 17.2 8 17.0
9 16.6 9 17.1
10 16.6 10 17.3
Goal of the experiment-
 Average bond strength of Modified mortar
(more water) is the same or different from
that of Unmodified mortar (standard water).

Comparativeexperiment-1:Solution
Textbook:p-26,28,40.
39
Modified . Unmodified
Is average bond strength of Modified mortar the same as that of Unmodified mortar?
Ho: Modified = Unmodified. Average strength of mortar…
H1: Modified ≠ Unmodified.
Alpha =0.05. Level of Significance.
p-value= 0.042.
Excel function- 0.042 =T.Test(Array1,Array2,2Tail,EqualVariance)
p-value<0.05. Hence reject Ho. .
That is, average bond strengths of two mortars are different.
Sample
no.
Modified
Mortar
Sample
no.
Unmodified
Mortar
1 16.9 1 16.6
2 16.4 2 16.8
3 17.2 3 17.4
4 16.4 4 17.1
5 16.5 5 17.0
6 17.0 6 16.9
7 17.0 7 17.3
8 17.2 8 17.0
9 16.6 9 17.1
10 16.6 10 17.3
Sample
no.
Modified
Mortar
Sample
no.
Unmodified
Mortar
1 16.9 1 16.6
2 16.4 2 16.8
3 17.2 3 17.4
4 16.4 4 17.1
5 16.5 5 17.0
6 17.0 6 16.9
7 17.0 7 17.3
8 17.2 8 17.0
9 16.6 9 17.1
10 16.6 10 17.3
Mean 16.764 Mean 17.042
Var.S 0.100 Var.S 0.061
Stdev.S 0.316 Stdev.S 0.248
n 10 n 10
Bond strengths recorded by
the tests are-

BITS Pilani
Pilani Campus
BITS Pilani
Previousproblemwas-
• Two-samplet-test.

BITS Pilani
Pilani Campus
BITS Pilani
Nextproblem:
• FirstasTwosamplet-test,sameasthepreviousproblem.
• ThenasTwo-pairedsamplet-test

The purchase department of a company has suggested its
Metrology lab to use hardness measurement Tips of a
new supplier.
The lab wants to know whether the hardness measured
by the two Tips (old and new) is the same.
The lab tested 10 samples of Tip1 and 10 samples of Tip2.
The hardness measurements recorded are-
Comparativeexperiment-2:Problem
Textbook:p-53.
42
Sample
no.
Tip1
Sample
no.
Tip2
1 7 1 6
2 3 2 3
3 3 3 5
4 4 4 3
5 8 5 8
6 3 6 2
7 2 7 4
8 9 8 9
9 5 9 4
10 4 10 5
Design of the Experiment-
1. Take 20 samples to be tested.
2. Randomly assign 10 samples to Tip1 and 10
samples to Tip2.
3. The sequence in which the samples are
tested is also random.
Completely Randomized Design.

Comparativeexperiment-2:Solution
Textbook:p-53.
43
Sample
no.
Tip1
Sample
no.
Tip2
1 7 1 6
2 3 2 3
3 3 3 5
4 4 4 3
5 8 5 8
6 3 6 2
7 2 7 4
8 9 8 9
9 5 9 4
10 4 10 5
Sample
no.
Tip1
Sample
no.
Tip2
1 7 1 6
2 3 2 3
3 3 3 5
4 4 4 3
5 8 5 8
6 3 6 2
7 2 7 4
8 9 8 9
9 5 9 4
10 4 10 5
Mean 4.800 Mean 4.900
Var.S 5.733 Var.S 4.989
Stdev.S 2.394 Stdev.S 2.234
n 10 n 10
Is average Hardness same of Tip1 and Tip2?
Ho: Tip1 = Tip2. Average Hardness…
H1: Tip1 ≠ Tip2.
Alpha =0.05. Level of Significance.
p-value= 0.9241.
Excel function- 0.9241 =T.Test(Array1,Array2,2-Tail,EqualVariance).
p-value>0.05. Hence, do not reject Ho. Or, Ho is ok.
In English: There is no difference in the Hardness measured by Tip1 and Tip2.

Remaining chapter will be done on the next session.
44

DOE L-02 C-01 Introduction C02 Basic Statistical Methods.pptx

More Related Content

Similar to DOE L-02 C-01 Introduction C02 Basic Statistical Methods.pptx (20)

Recently uploaded (20)

DOE L-02 C-01 Introduction C02 Basic Statistical Methods.pptx