Nonparametric methods and chi square tests (1)

Nonparametric Methods and
Chi-Square Tests (1)

Slide 1

Shakeel Nouman
M.Phil Statistics

Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical
Officer

14 Nonparametric Methods and Chi-Square Tests (1)
• Using Statistics
• The Sign Test
• The Runs Test - A Test for Randomness
• The Mann-Whitney U Test
• The Wilcoxon Signed-Rank Test

• The Kruskal-Wallis Test - A Nonparametric
Alternative to One-Way ANOVA

Officer

14 Nonparametric Methods and Chi-Square Tests (2)
• The Friedman Test for a Randomized

•
•
•
•

•

Block Design
The Spearman Rank Correlation
Coefficient
A Chi-Square Test for Goodness of Fit
Contingency Table Analysis - A ChiSquare Test for Independence
A Chi-Square Test for Equality of
Proportions
Summary and Review of Terms

Officer

14-1 Using Statistics
(Parametric Tests)

Slide 4

• Parametric Methods
Inferences based on assumptions about the nature of the
population distribution
» Usually: population is normal
Types of tests
» z-test or t-test
» Comparing two population means or proportions
» Testing value of population mean or proportion
» ANOVA
» Testing equality of several population means

Officer

Nonparametric Tests

Slide 5

• Nonparametric Tests
Distribution-free methods making no assumptions about the
Types of tests

» Sign tests
» Sign Test: Comparing paired observations
» McNemar Test: Comparing qualitative variables
» Cox and Stuart Test: Detecting trend
» Runs tests
» Runs Test: Detecting randomness
» Wald-Wolfowitz Test: Comparing two distributions

Officer

Nonparametric Tests
(Continued)

Slide 6

Ranks tests
•
•
•

Mann-Whitney U Test: Comparing two populations
Wilcoxon Signed-Rank Test: Paired comparisons
Comparing several populations: ANOVA with ranks
– Kruskal-Wallis Test
– Friedman Test: Repeated measures

Spearman Rank Correlation Coefficient
Chi-Square Tests
•
•
•

Goodness of Fit
Testing for independence: Contingency Table Analysis
Equality of Proportions

Officer

Nonparametric Tests
(Continued)
•
•

•

Slide 7

Deal with enumerative (frequency counts) data.
Do not deal with specific population parameters,
such as the mean or standard deviation.
Do not require assumptions about specific
population distributions (in particular, the
normality assumption).

Officer

14-2 Sign Test

Slide 8

• Comparing paired observations
Paired observations: X and Y
p = P(X>Y)

» Two-tailed test H0: p = 0.50

H1: p0.50
» Right-tailed test H0: p  0.50
H1: p0.50
» Left-tailed test H0: p  0.50
H1: p 0.50
» Test statistic: T = Number of + signs

Officer

Sign Test Decision Rule

Slide 9

• Small Sample: Binomial Test
For a two-tailed test, find a critical point

corresponding as closely as possible to /2
(C1) and define C2 as n-C1. Reject null
hypothesis if T  C1or T  C2.
For a right-tailed test, reject H0 if T  C, where
C is the value of the binomial distribution with
parameters n and p = 0.50 such that the sum of
the probabilities of all values less than or equal
to C is as close as possible to the chosen level
of significance, .
For a left-tailed test, reject H0 if T  C, where C
is defined as above.

Officer

Example 14-1
CEO Before After
1
3
4
1
2
5
5
0
3
2
3
1
4
2
4
1
5
4
4
0
6
2
3
1
7
1
2
1
8
5
4
-1
9
4
5
1
10
5
4
-1
11
3
4
1
12
2
5
1
13
2
5
1
14
2
3
1
15
1
2
1
16
3
2
-1
17
4
5
1

Sign
+

+
+
+
+
+
+
+
+
+
+
+

n = 15
T = 12
  0.025
C1=3 C2 = 15-3 = 12
H0 rejected, since
T  C2

Slide 10

Cumulative
Binomial
Probabilities
(n=15, p=0.5)
x
F(x)
0
0.00003
1
0.00049
2
0.00369
3
0.01758
4
0.05923
5
0.15088
6
0.30362
7
0.50000
8
0.69638
9
0.84912
10
0.94077
11
0.98242
12
0.99631
13
0.99951
14
0.99997
15
1.00000

C1

Officer

Example 14-1- Using the
Template

Slide 11

H0: p = 0.5
H1: p  0.5
Test Statistic: T = 12
p-value = 0.0352.
For a = 0.05, the null hypothesis
is rejected since 0.0352 < 0.05.

Thus one can conclude that there
is a change in attitude toward a
CEO following the award of an
MBA degree.

Officer

14-3 The Runs Test - A Test for
Randomness

Slide 12

A run is a sequence of like elements that are preceded and followed by different
elements or no element at all.
Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E
Case 2: SSSSSSSSSS|EEEEEEEEEE
Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E

: R = 20 Apparently nonrandom
: R = 12 Perhaps random

A two-tailed hypothesis test for randomness:
H0: Observations are generated randomly
H1: Observations are not generated randomly
Test Statistic:
R=Number of Runs
Reject H0 at level if R C1 or R C2, as given in Table 8, with total tail
probability P(R C1) + P(R C2) = 
Officer

Runs Test: Examples
Table 8:
(n1,n2) 11

12

(10,10)

0.586 0.758 0.872 0.949 0.981 0.996 0.999 1.000 1.000 1.000

.
.
.

13

Number of Runs (r)
14
15
16
17

Slide 13

18

19

20

Case 1: n1 = 10 n2 = 10 R= 20 p-value
0
Case 2: n1 = 10 n2 = 10 R = 2 p-value 
0
Case 3: n1 = 10 n2 = 10 R= 12
p-value R 
P
F(11)]
= (2)(1-0.586) = (2)(0.414) = 0.828
H0 not rejected

Officer

Large-Sample Runs Test: Using
the Normal Approximation

Slide 14

The mean of the normal distribution of the number of runs:
2n n
E ( R) 
1
n n
1

2

1

2

The standard deviation:

 
R

2n n ( 2n n  n  n )
( n  n ) ( n  n  1)
1

2

1

2

1

2

2

1

2

1

2

The standard normal test statistic:
R  E ( R)
z



R

Officer

Large-Sample Runs Test:
Example 14-2

Slide 15

Example 14-2: n1 = 27 n2 = 26 R = 15
2n n
( 2)( 27)( 26)
E ( R)  1 2  1 
 1  26.49  1  27.49
n n
( 27  26)
1 2
2n n ( 2n n  n  n )
1 2 1 2 1 2  ( 2)( 27)( 26)(( 2)( 27)( 26)  27  26))
 
R
( n  n ) 2 ( n  n  1)
( 27  26) 2 ( 27  26  1)
1 2
1 2
1896804

 12.986  3.604
146068
R  E ( R ) 15  27.49
z

 3.47

3.604
R

p - value = 2(1 - .9997) = 0.0006

H0 should be rejected at any common level of significance.
Officer

Example 14-2 – Using the
Template

Slide 16

Note: The
computed p-value
using the template
is 0.0005 as
compared to the
manually
computed value of
0.0006. The value
of 0.0005 is more
accurate.
Reject the null
hypothesis that
the residuals are
random.

Officer

Using the Runs Test to Compare Two
Population Distributions (Means): the
Wald-Wolfowitz Test

Slide 17

The null and alternative hypotheses for the Wald-Wolfowitz test:
H0: The two populations have the same distribution
H1: The two populations have different distributions
The test statistic:
R = Number of Runs in the sequence of samples, when
the data from both samples have been sorted

Example 14-3:
Salesperson A:
Salesperson B:

35 44
17 23

39
13

50
24

48
33

29
21

60
18

75
16

49
32

66

Officer

The Wald-Wolfowitz Test:
Example 14-3

Sales
35
44
39
48
60
75
49
66
17
23
13
24
33
21
18
16
32

Sales
Person
A
A
A
A
A
A
A
A
B
B
B
B
B
B
B
B
B

Sales
(Sorted)
13
16
17
21
24
29
32
33
35
39
44
48
49
50
60
66
75

Sales
Person
(Sorted)
B
B
B
B
B
A
B
B
A
A
A
A
A
A
A
A
A

Runs

1
2
3

Slide 18

n1 = 10 n2 = 9 R= 4
p-value PR  4
H0 may be rejected

Table
Number of Runs (r)
(n1,n2) 2
3
4
5
.
.
.
(9,10) 0.000 0.000 0.002 0.004 ...

4

Officer

Ranks Tests

Slide 19

• Ranks tests
 Mann-Whitney U Test: Comparing two populations
 Wilcoxon Signed-Rank Test: Paired comparisons
 Comparing several populations: ANOVA with ranks
• Kruskal-Wallis Test
• Friedman Test: Repeated measures

Officer

14-4 The Mann-Whitney U Test
(Comparing Two Populations)

Slide 20

The null and alternative hypotheses:
H0: The distributions of two populations are
identical
H1: The two population distributions are not
identical
The Mann-Whitney U statistic:
n1 ( n1  1)
U  n1 n2 
 R1
2

R 1   Ranks from sample 1

where n1 is the sample size from population 1 and n2 is
the sample size from population 2.
n1 n2
n1 n2 (n1  n2  1)
E [U ] 
U 
2
12
U  E [U ]
The large - sample test statistic: z 



U
Officer

The Mann-Whitney U Test:
Example 14-4
Model
A
A
A
A
A
A
B
B
B
B
B
B

Time
35
38
40
42
41
36
29
27
30
33
39
37

Rank
5
8
10
12
11
6
2
1
3
4
9
7

Rank
Sum

U  n1 n 2 

n1 ( n1  1)
2
(6)(6 + 1)

= (6)(6) +
5

52

26

Slide 21

 R1

 52

2

Cumulative Distribution Function of the
Mann-Whitney U Statistic
n2=6
n1=6
u
.
.
.
4
0.0130
P(u
5)
5
0.0206
6
0.0325
.
.
.

Officer

Example 14-5: Large-Sample
Mann-Whitney U Test
Score Rank
Score Program Rank Sum
85
1
20.0 20.0
87
1
21.0 41.0
92
1
27.0 68.0
98
1
30.0 98.0
90
1
26.0 124.0
88
1
23.0 147.0
75
1
17.0 164.0
72
1
13.5 177.5
60
1
6.5 184.0
93
1
28.0 212.0
88
1
23.0 235.0
89
1
25.0 260.0
96
1
29.0 289.0
73
1
15.0 304.0
62
1
8.5 312.5

Score Rank
Score Program Rank Sum
65
2
10.0 10.0
57
2
4.0 14.0
74
2
16.0 30.0
43
2
2.0 32.0
39
2
1.0 33.0
88
2
23.0 56.0
62
2
8.5 64.5
69
2
11.0 75.5
70
2
12.0 87.5
72
2
13.5 101.0
59
2
5.0 106.0
60
2
6.5 112.5
80
2
18.0 130.5
83
2
19.0 149.5
50
2
3.0 152.5

U  n1n2 

Slide 22

n1 ( n1  1)

 R1
2
(15)(15  1)
 (15)(15) 
 312 .5  32 .5
2
n1n2
(15)(15)
E [U ] 
=
= 112.5
2
2
n1n2 ( n1  n2  1)
U 
12
(15)(15)(15  15  1)

 24 .109
12
U  E [U ] 32 .5  112 .5
z 

 3.32
U
24 .109

Since the test statistic is z = -3.32,
the p-value 0.0005,Tests (1) HShakeelrejected. Statistics Govt. College University Lahore, Statistical
Nonparametric Methods and Chi-Square and By 0 is Nouman M.Phil
Officer

Mann-Whitney U Test – Using the
Template

Since the test
statistic is z = 3.32, the p-value
 0.0005, and H0
is rejected.
That is, the LC
(Learning
Curve) program
is more effective.
Officer

14-5 The Wilcoxon SignedRanks Test (Paired Ranks)

Slide 24

H0: The median difference between populations are 1 and 2 is zero
H1: The median difference between populations are 1 and 2 is not
zero
Find the difference between the ranks for each pair, D = x1 -x2, and then rank
the absolute values of the differences.
The Wilcoxon T statistic is the smaller of the sums of the positive ranks and
the sum of the negative ranks: T  min  (  ),  (  )
For small samples, a left-tailed test is used, using the values in Appendix C,
Table 10.

E[T ] 

n ( n  1)
4

The large-sample test statistic:

n ( n  1)( 2 n  1)

T 

24
z

T  E[T ]

T

Officer

Example 14-6
Sold Sold
(1)
(2)

Rank
Rank Rank
D=x1-x2 ABS(D) ABS(D) (D>0) (D<0)

56
48
100
85
22
44
35
28
52
77
89
10
65
90
70
33

16
-22
40
15
14
4
-10
21
-8
7
-1
0
-20
29
30
7

40
70
60
70
8
40
45
7
60
70
90
10
85
61
40
26

16
22
40
15
14
4
10
21
8
7
1
*
20
29
30
7

9.0
12.0
15.0
8.0
7.0
2.0
6.0
11.0
5.0
3.5
1.0
*
10.0
13.0
14.0
3.5

9.0
0.0
15.0
8.0
7.0
2.0
0.0
11.0
0.0
3.5
0.0
*
0.0
13.0
14.0
3.5

0
12
0
0
0
0
6
0
5
0
1
*
10
0
0
0

Sum:

86

Slide 25

T=34

n=15
P=0.05 30
P=0.02525
P=0.01 20
P=0.00516

34

H0 is not rejected (Note the
arithmetic error in the text
for store 13)

Officer

Example 14-7
Hourly
Messages

149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149

Rank
D=x1-x2 ABS(D) ABS(D)
2
-5
-26
29
-44
-37
-9
18
28
36
-20
11
-39
21
49
16
-40
-31
6
-47
15
31
-10
17
33

2
5
26
29
44
37
9
18
28
36
20
11
39
21
49
16
40
31
6
47
15
31
10
17
33

Rank
(D>0)

Rank
(D<0)

1.0
2.0
13.0
15.0
23.0
20.0
4.0
10.0
14.0
19.0
11.0
6.0
21.0
12.0
25.0
8.0
22.0
16.5
3.0
24.0
7.0
16.5
5.0
9.0
18.0

1.0
0.0
0.0
15.0
0.0
0.0
0.0
10.0
14.0
19.0
0.0
6.0
0.0
12.0
25.0
8.0
0.0
0.0
3.0
0.0
7.0
16.5
0.0
9.0
18.0

0.0
2.0
13.0
0.0
23.0
20.0
4.0
0.0
0.0
0.0
11.0
0.0
21.0
0.0
0.0
0.0
22.0
16.5
0.0
24.0
0.0
0.0
5.0
0.0
0.0

Sum:

151
144
123
178
105
112
140
167
177
185
129
160
110
170
198
165
109
118
155
102
164
180
139
166
82

Md0

163.5

161.5

Slide 26

E[ T ] 

n ( n  1)

T 


(25)(25 + 1)

=
4
4
n ( n  1)( 2 n  1)

= 162.5

24
25( 25  1)(( 2 )( 25)  1)
24



33150

 37 .165
24
The large - sample test statistic:
z 

T  E[ T ]

T



163.5  162 .5

 0.027

37 .165
H 0 cannot be rejected
Officer

Example 14-7 using the
Template

Slide 27

Note 1: You should enter
the claimed value of the
mean (median) in every
used row of the second
column of data. In this case
it is 149.
Note 2: In order for the
large sample
approximations to be
computed you will need to
change n > 25 to n >= 25 in
cells M13 and M14.

Officer

14-6 The Kruskal-Wallis Test - A
Nonparametric Alternative to One-Way
ANOVA

Slide 28

The Kruskal-Wallis hypothesis test:
H0: All k populations have the same distribution
H1: Not all k populations have the same distribution
The Kruskal-Wallis test statistic:
2
12  k Rj 
H
 3(n  1)
n(n  1)   n j 
 j 1 

2
If each nj > 5, then H is approximately distributed as a  .

Officer

Example 14-8: The KruskalWallis Test
SoftwareTime Rank Group RankSum
1
45
14
1
90
1
38
10
2
56
1
56
16
3
25
1
60
17
1
47
15
1
65
18
2
30
8
2
40
11
2
28
7
2
44
13
2
25
5
2
42
12
3
22
4
3
19
3
3
15
1
3
31
9
3
27
6
3
17
2

Slide 29

 k R2 
12
j

H 
 j 1   3( n  1)
n ( n  1) 
nj 
12
 902 562 252 




  3(18  1)
18(18  1)  6
6
6 
 12   11861  57



 342   6 
 12 .3625

2(2,0.005)=10.5966, so H0 is
rejected.

Officer

Example 14-8: The Kruskal-Wallis
Test – Using the Template

Officer

Further Analysis (Pairwise
Comparisons of Average Ranks)

Slide 31

If the null hypothesis in the Kruskal-Wallis test is rejected, then
we may wish, in addition, compare each pair of populations to
determine which are different and which are the same.
The pairwise comparison test statistic:
D  Ri  R j
where R i is the mean of the ranks of the observations from
population i.
The critical point for the paired comparisons:
 n(n  1)  1 1 
2
C KW  (   , k 1 ) 
  ni  n j 

 12 
Reject if D > C KW
Officer

Pairwise Comparisons: Example
14-8

C KW

Slide 32

Critical Point:
n(n  1)  1 1 
2
 (   ,k 1 ) 
 12  ni  n j 



18(18  1)  1 1
 ( 9.21034)
  
12
 6 6
 87.49823  9.35

90
 15
6
56
R2   9.33
6
25
R3   4.17
6
R1 

D1,2  15  9.33  5.67
D1,3  15  4.17  10.83 ***
D2,3  9.33  4.17  516
.

Officer

14-7 The Friedman Test for a
Randomized Block Design

Slide 33

The Friedman test is a nonparametric version of the randomized block design
ANOVA. Sometimes this design is referred to as a two-way ANOVA with one
item per cell because it is possible to view the blocks as one factor and the
treatment levels as the other factor. The test is based on ranks.

The Friedman hypothesis test:
H0: The distributions of the k treatment populations are
identical
H1: Not all k distribution are identical
The Friedman test statistic:
12
 
 R  3n(k  1)
nk (k  1)
2

k

j 1

2

j

The degrees of freedom for the chi-square distribution is (k – 1).
Officer

Example 14-10 – using the
Template

Slide 34

Note: The p-value
is small relative to
a significance level
of a = 0.05, so one
should conclude
that there is
evidence that not
all three lowbudget cruise lines
are equally
preferred by the
frequent cruiser
population

Officer

14-8 The Spearman Rank
Correlation Coefficient

Slide 35

The Spearman Rank Correlation Coefficient is the simple correlation coefficient
calculated from variables converted to ranks from their original values.

The Spearman Rank Correlation Coefficient (assuming no ties):
n 2
6  di
rs  1  i 1
where d = R(x ) - R(y )
2
i
i
i
n ( n  1)
Null and alternative hypotheses:
H 0:  s = 0
H1:  s  0
Critical values for small sample tests from Appendix C, Table 11
Large sample test statistic:
z = rs ( n  1)
Officer

14-1 Using Statistics
(Parametric Tests)

Slide 38

• Parametric Methods
Inferences based on assumptions about the nature of the
» Usually: population is normal
Types of tests
» z-test or t-test
» Comparing two population means or proportions
» Testing value of population mean or proportion
» ANOVA
» Testing equality of several population means

Officer

Nonparametric Tests

Slide 39

Distribution-free methods making no assumptions about the
Types of tests

» Sign tests
» Sign Test: Comparing paired observations
» McNemar Test: Comparing qualitative variables
» Cox and Stuart Test: Detecting trend
» Runs tests
» Runs Test: Detecting randomness
» Wald-Wolfowitz Test: Comparing two distributions

Officer

Nonparametric Tests
(Continued)

Slide 40

Ranks tests
•
•
•

Mann-Whitney U Test: Comparing two populations
Wilcoxon Signed-Rank Test: Paired comparisons
Comparing several populations: ANOVA with ranks
– Kruskal-Wallis Test
– Friedman Test: Repeated measures

Spearman Rank Correlation Coefficient
Chi-Square Tests
•
•
•

Goodness of Fit
Testing for independence: Contingency Table Analysis
Equality of Proportions

Officer

Nonparametric Tests
(Continued)
•
•
•

Slide 41

Deal with enumerative (frequency counts) data.
Do not deal with specific population
parameters, such as the mean or standard
deviation.
Do not require assumptions about specific
population distributions (in particular, the
normality assumption).

Officer

14-2 Sign Test

Slide 42

• Comparing paired observations
Paired observations: X and Y
p = P(X>Y)

» Two-tailed test H0: p = 0.50

H1: p0.50
» Right-tailed test H0: p  0.50
H1: p0.50
» Left-tailed test H0: p  0.50
H1: p 0.50
» Test statistic: T = Number of + signs

Officer

Sign Test Decision Rule

Slide 43

• Small Sample: Binomial Test
For a two-tailed test, find a critical point

corresponding as closely as possible to /2
(C1) and define C2 as n-C1. Reject null
hypothesis if T  C1or T  C2.
For a right-tailed test, reject H0 if T  C, where
C is the value of the binomial distribution with
parameters n and p = 0.50 such that the sum of
the probabilities of all values less than or equal
to C is as close as possible to the chosen level
of significance, .
For a left-tailed test, reject H0 if T  C, where C
is defined as above.

Officer

Example 14-1
CEO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

Before After
3
4
5
5
2
3
2
4
4
4
2
3
1
2
5
4
4
5
5
4
3
4
2
5
2
5
2
3
1
2
3
2
4
5

Sign
1
0
1
1
0
1
1
-1
1
-1
1
1
1
1
1
-1
1

+
+
+
+
+
+
+
+
+
+
+
+

n = 15
T = 12
  0.025
C1=3 C2 = 15-3 = 12 C1
H0 rejected, since
T  C2

Slide 44

Cumulative
Binomial
Probabilities
(n=15, p=0.5)
x
F(x)
0
0.00003
1
0.00049
2
0.00369
3
0.01758
4
0.05923
5
0.15088
6
0.30362
7
0.50000
8
0.69638
9
0.84912
10
0.94077
11
0.98242
12
0.99631
13
0.99951
14
0.99997
15
1.00000

Officer

Example 14-1- Using the
Template

Slide 45

H0: p = 0.5
H1: p  5
Test Statistic: T = 12
p-value = 0.0352.
For  = 0.05, the null
hypothesis
is rejected since 0.0352 <
0.05.
Thus one can conclude
that there
is a change in attitude
toward a
CEO following the award
of an
MBA degree.
Officer

14-3 The Runs Test - A Test for
Randomness

Slide 46

A run is a sequence of like elements that are preceded and
followed by different elements or no element at all.
Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E
Case 2: SSSSSSSSSS|EEEEEEEEEE
Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E

: R = 12 Perhaps random

A two-tailed hypothesis test for randomness:
H0: Observations are generated randomly
H1: Observations are not generated randomly
Test Statistic:
R=Number of Runs
Reject H0 at level  if R  C1 or R  C2, as given in Table
8, with total tail probability P(R  C1) + P(R  C2) = 
Officer

Runs Test: Examples
Table 8:
(n1,n2)

11

12

(10,10)

0.586

0.758

.
.
.

Number of Runs (r)
13
14
0.872

0.949

Slide 47

15

16

17

18

19

20

0.981

0.996

0.999

1.000

1.000

1.000

Case 1: n1 = 10 n2 = 10 R= 20 p-value0
Case 2: n1 = 10 n2 = 10 R = 2 p-value 0
Case 3: n1 = 10 n2 = 10 R= 12
p-value PR  11F(11)]
= (2)(1-0.586) = (2)(0.414) = 0.828
H0 not rejected

Officer

Large-Sample Runs Test: Using
the Normal Approximation

Slide 48

The mean of the normal distribution of the number of runs:
2n n
E ( R) 
1
n n
1

2

1

2

The standard deviation:

 
R

2n n ( 2n n  n  n )
( n  n ) ( n  n  1)
1

2

1

2

1

2

2

1

2

1

2

The standard normal test statistic:
R  E ( R)
z



R

Officer

Example 14-2

Slide 49

Example 14-2: n1 = 27 n2 = 26 R = 15
2n n
( 2)( 27)( 26)
E ( R)  1 2  1 
 1  26.49  1  27.49
n n
( 27  26)
1 2
2n n ( 2n n  n  n )
1 2 1 2 1 2  ( 2)( 27)( 26)(( 2)( 27)( 26)  27  26))
 
R
( n  n ) 2 ( n  n  1)
( 27  26) 2 ( 27  26  1)
1 2
1 2
1896804

 12.986  3.604
146068
R  E ( R ) 15  27.49
z

 3.47

3.604
R

p - value = 2(1 - .9997) = 0.0006

H0 should be rejected at any common level of significance.
Officer

Example 14-2 – Using the
Template

Slide 50

Note: The
computed pvalue using the
template is
0.0005 as
compared to the
manually
computed value
of 0.0006. The
value of 0.0005
is more
accurate.
Reject the null
hypothesis that
the residuals are
random.
Officer

Using the Runs Test to Compare Two
Population Distributions (Means): the
Wald-Wolfowitz Test

Slide 51

The null and alternative hypotheses for the Wald-Wolfowitz test:
H0: The two populations have the same distribution
H1: The two populations have different distributions
The test statistic:
R = Number of Runs in the sequence of samples, when
the data from both samples have been sorted

Example 14-3:
Salesperson A:
Salesperson B:

35
17

44
23

39
13

50
24

48
33

29
21

60
18

75
16

49
32

66

Officer

The Wald-Wolfowitz Test:
Example 14-3
Sales
35
44
39
48
60
75
49
66
17
23
13
24
33
21
18
16
32

Sales
Person
A
A
A
A
A
A
A
A
B
B
B
B
B
B
B
B
B

Sales
(Sorted)
13
16
17
21
24
29
32
33
35
39
44
48
49
50
60
66
75

Sales
Person
(Sorted)
B
B
B
B
B
A
B
B
A
A
A
A
A
A
A
A
A

Runs

1
2
3

Slide 52

n1 = 10 n2 = 9 R= 4
p-value PR  4
H0 may be rejected

Table
Number of Runs (r)
(n1,n2) 2
3
4
5
.
.
.
(9,10) 0.000 0.000 0.002 0.004 ...

4

Officer

Ranks Tests

Slide 53

• Ranks tests

 Mann-Whitney U Test: Comparing two populations
 Wilcoxon Signed-Rank Test: Paired comparisons
 Comparing several populations: ANOVA with ranks
• Kruskal-Wallis Test
• Friedman Test: Repeated measures

Officer

14-4 The Mann-Whitney U Test
(Comparing Two Populations)

Slide 54

H0: The distributions of two populations are identical
H1: The two population distributions are not identical
The Mann-Whitney U statistic:
n1 ( n1  1)
U  n1 n2 
 R1
R 1   Ranks from sample 1
2
where n1 is the sample size from population 1 and n2 is the
sample size from population 2.
n1 n2
n1 n2 (n1  n2  1)
E [U ] 
U 
2
12
U  E [U ]
The large - sample test statistic: z 

U

Officer

The Mann-Whitney U Test:
Example 14-4
Model
A
A
A
A
A
A
B
B
B
B
B
B

Time
35
38
40
42
41
36
29
27
30
33
39
37

Rank
5
8
10
12
11
6
2
1
3
4
9
7

Rank
Sum

U  n1 n 2 

n1 ( n1  1)
2
(6)(6 + 1)

= (6)(6) +
5

52

26

Slide 55

 R1

 52

2

Cumulative Distribution Function of the
Mann-Whitney U Statistic
n2=6
n1=6
u
.
.
.
4
0.0130
Pu5
5
0.0206
6
0.0325
.
.
.

Officer

Mann-Whitney U Test
Score
Score Program Rank
85
1
20.0
87
1
21.0
92
1
27.0
98
1
30.0
90
1
26.0
88
1
23.0
75
1
17.0
72
1
13.5
60
1
6.5
93
1
28.0
88
1
23.0
89
1
25.0
96
1
29.0
73
1
15.0
62
1
8.5

Rank
Sum
20.0
41.0
68.0
98.0
124.0
147.0
164.0
177.5
184.0
212.0
235.0
260.0
289.0
304.0
312.5

Score
Score Program Rank
65
2
10.0
57
2
4.0
74
2
16.0
43
2
2.0
39
2
1.0
88
2
23.0
62
2
8.5
69
2
11.0
70
2
12.0
72
2
13.5
59
2
5.0
60
2
6.5
80
2
18.0
83
2
19.0
50
2
3.0

Rank
Sum
10.0
14.0
30.0
32.0
33.0
56.0
64.5
75.5
87.5
101.0
106.0
112.5
130.5
149.5
152.5

Since the test statistic is z = -3.32,
the p-value  0.0005, and H0 is rejected.

U  n1n2 

Slide 56

n1 ( n1  1)

 R1
2
(15)(15  1)
 (15)(15) 
 312 .5  32 .5
2
n1n2
(15)(15)
E [U ] 
=
= 112.5
2
2
n1n2 ( n1  n2  1)
U 
12
(15)(15)(15  15  1)

 24 .109
12
U  E [U ]
32 .5  112 .5
z 

 3.32
U
24 .109

Officer

Mann-Whitney U Test – Using the
Template

Since the test
statistic is z = 3.32, the p-value 
0.0005, and H0 is
rejected.
That is, the LC
(Learning Curve)
program is more
effective.

Officer

14-5 The Wilcoxon SignedRanks Test (Paired Ranks)

Slide 58

H0: The median difference between populations are 1 and 2 is zero
H1: The median difference between populations are 1 and 2 is not
zero
Find the difference between the ranks for each pair, D = x1 -x2, and then rank
the absolute values of the differences.
The Wilcoxon T statistic is the smaller of the sums of the positive ranks and
the sum of the negative ranks:  (  ),  (  )
T  min





For small samples, a left-tailed test is used, using the values in Appendix C,
Table 10.

E[T ] 

n ( n  1)
4

The large-sample test statistic: z 

T 
T  E[T ]

n ( n  1)( 2 n  1)
24

T

Officer

Example 14-6
Sold Sold
(1)
(2)
56
48
100
85
22
44
35
28
52
77
89
10
65
90
70
33

40
70
60
70
8
40
45
7
60
70
90
10
85
61
40
26

Rank Rank Rank
D=x1-x2 ABS(D) ABS(D)(D>0) (D<0)
16
-22
40
15
14
4
-10
21
-8
7
-1
0
-20
29
30
7

16
22
40
15
14
4
10
21
8
7
1
*
20
29
30
7

9.0
12.0
15.0
8.0
7.0
2.0
6.0
11.0
5.0
3.5
1.0
*
10.0
13.0
14.0
3.5

9.0
0.0
15.0
8.0
7.0
2.0
0.0
11.0
0.0
3.5
0.0
*
0.0
13.0
14.0
3.5

0
12
0
0
0
0
6
0
5
0
1
*
10
0
0
0

Sum:

86

Slide 59

T=34
P=0.05
P=0.025
P=0.01
P=0.005

n=15
30
25
20
16

34

H0 is not rejected (Note the arithmetic
error in the text for store 13)

Officer

Example 14-7
Hourly
Messages
151
144
123
178
105
112
140
167
177
185
129
160
110
170
198
165
109
118
155
102
164
180
139
166
82

Md0
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149
149

Rank Rank
D=x1-x2 ABS(D) ABS(D) (D>0)
2
-5
-26
29
-44
-37
-9
18
28
36
-20
11
-39
21
49
16
-40
-31
6
-47
15
31
-10
17
33

2
5
26
29
44
37
9
18
28
36
20
11
39
21
49
16
40
31
6
47
15
31
10
17
33

Rank
(D<0)

1.0
2.0
13.0
15.0
23.0
20.0
4.0
10.0
14.0
19.0
11.0
6.0
21.0
12.0
25.0
8.0
22.0
16.5
3.0
24.0
7.0
16.5
5.0
9.0
18.0

1.0
0.0
0.0
15.0
0.0
0.0
0.0
10.0
14.0
19.0
0.0
6.0
0.0
12.0
25.0
8.0
0.0
0.0
3.0
0.0
7.0
16.5
0.0
9.0
18.0

0.0
2.0
13.0
0.0
23.0
20.0
4.0
0.0
0.0
0.0
11.0
0.0
21.0
0.0
0.0
0.0
22.0
16.5
0.0
24.0
0.0
0.0
5.0
0.0
0.0

Sum:

163.5

161.5

Slide 60

E[ T ] 

n ( n  1)

(25)(25 + 1)
=

T 


= 162.5

4
4
n ( n  1)( 2 n  1)
24
25( 25  1)(( 2 )( 25)  1)
24



33150

 37 .165
24
The large - sample test statistic:
z 

T  E[ T ]

T



163.5  162 .5

 0.027

37 .165
H 0 cannot be rejected

Officer

Example 14-7 using the
Template

Slide 61

Note 1: You should enter
the claimed value of the
mean (median) in every
used row of the second
column of data. In this
case it is 149.
Note 2: In order for the
large sample
approximations to be
computed you will need
to change n > 25 to n >=
25 in cells M13 and M14.

Officer

14-6 The Kruskal-Wallis Test - A
Nonparametric Alternative to One-Way
ANOVA

Slide 62

The Kruskal-Wallis hypothesis test:
H0: All k populations have the same distribution
H1: Not all k populations have the same distribution

The Kruskal-Wallis test statistic:
2
12  k Rj 
H
  n   3(n  1)
n(n  1)  j 1 j 
If each nj > 5, then H is approximately distributed as a 2.

Officer

Example 14-8: The KruskalWallis Test
SoftwareTime Rank Group RankSum
1
45
14
1
90
1
38
10
2
56
1
56
16
3
25
1
60
17
1
47
15
1
65
18
2
30
8
2
40
11
2
28
7
2
44
13
2
25
5
2
42
12
3
22
4
3
19
3
3
15
1
3
31
9
3
27
6
3
17
2

Slide 63

 k R2 
12
j

H 
 j 1   3( n  1)
n ( n  1) 
nj 
12
 902 562 252 




  3(18  1)
18(18  1)  6
6
6 
 12   11861  57



 342   6 
 12 .3625

2(2,0.005)=10.5966, so H0 is
rejected.

Officer

Further Analysis (Pairwise
Comparisons of Average Ranks)

Slide 65

If the null hypothesis in the Kruskal-Wallis test is rejected,
then we may wish, in addition, compare each pair of
populations to determine which are different and which are
the same.
The pairwise comparison test statistic:
D  Ri  R j
where R i is the mean of the ranks of the observations from
population i.
The critical point for the paired comparisons:
 n(n  1)  1 1 
2
C KW  (   , k 1 ) 
  ni  n j 

 12 
Reject if D > C KW
Officer

Pairwise Comparisons: Example
14-8

C KW

Slide 66

Critical Point:
n(n  1)  1 1 
2
 (   ,k 1 ) 
 12  ni  n j 



18(18  1)  1 1
 ( 9.21034)
  
12
 6 6
 87.49823  9.35

90
 15
6
56
R2   9.33
6
25
R3   4.17
6
R1 

D1,2  15  9.33  5.67
D1,3  15  4.17  10.83 ***
D2,3  9.33  4.17  516
.

Officer

14-7 The Friedman Test for a
Randomized Block Design

Slide 67

The Friedman test is a nonparametric version of the randomized block design
ANOVA. Sometimes this design is referred to as a two-way ANOVA with one
item per cell because it is possible to view the blocks as one factor and the
treatment levels as the other factor. The test is based on ranks.

The Friedman hypothesis test:
H0: The distributions of the k treatment populations are identical
H1: Not all k distribution are identical
The Friedman test statistic:

12
 
 R  3n(k  1)
nk (k  1)
The degrees of freedom for the chi-square distribution is (k – 1).
2

k

j 1

2

j

Officer

Example 14-10 – using the
Template

Slide 68

Note: The pvalue is small
relative to a
significance level
of  = 0.05, so
one should
conclude that
there is evidence
that not all three
low-budget cruise
lines are equally
preferred by the
frequent cruiser
population

Officer

14-8 The Spearman Rank
Correlation Coefficient

Slide 69

The Spearman Rank Correlation Coefficient is the simple correlation coefficient
calculated from variables converted to ranks from their original values.

The Spearman Rank Correlation Coefficient (assuming no ties):
n 2
6  di
rs  1  i 1
where d = R(x ) - R(y )
2
i
i
i
n ( n  1)
H 0:  s = 0
H1:  s  0
Critical values for small sample tests from Appendix C, Table 11
Large sample test statistic:
z = rs ( n  1)
Officer

Spearman Rank Correlation
Coefficient: Example 14-11
MMI S&P100
220
151
218
150
216
148
217
149
215
147
213
146
219
152
236
165
237
162
235
161

R-MMI R-S&P Diff Diffsq
7
6
1
1
5
5
0
0
3
3
0
0
4
4
0
0
2
2
0
0
1
1
0
0
6
7
-1
1
9
10
-1
1
10
9
1
1
8
8
0
0
Sum:

Slide 70

Table 11: =0.005
n
.
.
.
7
-----8
0.881
9
0.833
10
0.794
11
0.818
.
.
.

4

n 
 di
6
64
4
1
 1  i
1
1
9758794H 


99
nn 1
11 1

rejected

Officer

Using the Template

Slide 71

Note: The pvalues in the
range J15:J17
will appear only
if the sample
size is large (n
> 30)

Officer

14-9 A Chi-Square Test for
Goodness of Fit


Slide 72

Steps in a chi-square analysis:
 Formulate null and alternative hypotheses
 Compute frequencies of occurrence that would be
expected if the null hypothesis were true - expected cell
counts
 Note actual, observed cell counts
 Use differences between expected and actual cell counts
to find chi-square statistic:

(Oi  Ei ) 2
2  
Ei
i 1
k

 Compare chi-statistic with critical values from the chisquare distribution (with k-1 degrees of freedom) to test
the null hypothesis
Officer

Example 14-12: Goodness-of-Fit
Test for the Multinomial
Distribution

Slide 73

H0: The probabilities of occurrence of events E1, E2...,Ek are given by
p1,p2,...,pk
H1: The probabilities of the k events are not as specified in the null
hypothesis

Assuming equal probabilities, p1= p2 = p3 = p4 =0.25 and n=80
Preference
Tan
Brown Maroon Black
Total
Observed
12
40
8
20
80
Expected(np)
20
20
20
20
80
(O-E)
-8
20
-12
0
0
k ( Oi  E i )
2
  
i 1
Ei

2


( 8 )
20

2


( 20 )

2


( 12 )

20

20

2


( 0)

2

20

 30.4  

2

 11.3449

( 0.01, 3)

H 0 is rejected at the 0.01 level.
Officer

Distribution using the Template

Note: the pvalue is
0.0000, so we
can reject the
null
hypothesis at
any a level.

Officer

Goodness-of-Fit for the Normal
Distribution: Example 14-13

p(z<-1)
p(-1<z<-0.44)
p(-0.44<z<0)
p(0<z<0.44)
p(0.44<z<14)
p(z>1)

= 0.1587
= 0.1713
= 0.1700
= 0.1700
= 0.1713
= 0.1587

Partitioning the Standard Normal Distribution
0.1700

0.1700
0.4

0.1713

0.1713
0.3

f(z)

1. Use the table of the standard
normal distribution to determine an
appropriate partition of the standard
normal distribution which gives
ranges with approximately equal
percentages.

Slide 75

0.2

0.1587

0.1587

0.1

z

0.0
-5

-1

0

1

5

-0.44 0.44

2. Given z boundaries, x boundaries can be determined from the inverse
standard normal transformation: x =  + z = 125 + 40z.
3. Compare with the critical value of the 2 distribution with k-3 degrees of
freedom.
Officer

Example 14-13: Solution

i

Oi

Ei

1
2
3
4
5
6

14 15.87
20 17.13
16 17.00
19 17.00
16 17.13
15 15.87

Slide 76

Oi - Ei (Oi - Ei)2 (Oi - Ei)2/ Ei

-1.87
2.87
-1.00
2.00
-1.13
-0.87

3.49690
8.23691
1.00000
4.00000
1.27690
0.75690
2
:

0.22035
0.48085
0.05882
0.23529
0.07454
0.04769
1.11755

2(0.10,k-3)= 6.5139 > 1.11755 H0 is not rejected at the 0.10 level

Officer

Example 14-13: Solution using
the Template

Slide 77

Note: p-value =
0.8002 > 0.01
H0 is not
rejected at the
0.10 level

Officer

14-9 Contingency Table
Analysis:
A Chi-Square Test for
Independence

Slide 78

First Classification Category
Second
Classification
Category
1
2
3
4
5
Column
Total

1
O11
O21
O31
O41
O51

2
O12
O22
O32
O42
O52

3
O13
O23
O33
O43
O53

4
O14
O24
O34
O44
O54

5
O15
O25
O35
O45
O55

C1

C2

C3

C4

C5

Row
Total
R1
R2
R3
R4
R5
n

Officer

Contingency Table Analysis:
A Chi-Square Test for Independence

Slide 79

A and B are independent if:P(AUB) = P(A)P(B).
If the first and second classification categories are independent:Eij = (Ri)(Cj)/
H0: The two classification variables are independent of each
other
H1: The two classification variables are not independent

( Oij  Eij ) 2
  
Eij
i 1 j 1

r
Chi-square test statistic for independence:
2

c

Degrees of freedom: df=(r-1)(c-1)
Expected cell count:

Ri C j
Eij 
n

Officer

Example 14-14
Industry Type

Profit

Service Nonservice
(Expected) (Expected) Total
42
18
60

(Expected)

Loss

6

(Expected)

(40*48/100)=19.2

Total
ij
11
12
21
22

(60*48/100)=28.8

48

O
42
18
6
34

E
28.8
31.2
19.2
20.8

O-E
13.2
-13.2
-13.2
13.2

(O-E)2
174.24
174.24
174.24
174.24
2:

Slide 80

2(0.01,(2-1)(21))=6.63490

H0 is rejected at
the 0.01 level
(60*52/100)=31.2
and
34
40
(40*52/100)=20.8
it is concluded
52
100
that the two
variables
2
(O-E) /E
Yates corrected  for a 2x2 table:
are not
6.0500
2
independent.
5.5846
O  E  0.5

29.0865

2

9.0750
8.3769



2


 

ij

ij
Eij



Officer

Contingency Table Analysis: Slide 81
Example 14-14 using the Template
Note: When
the
contingenc
y table is a
2x2, one
should use
the Yates
correction.

Since p-value = 0.000, H0 is rejected at the 0.01 level and it is concluded that the two
variables are not independent.
Officer

14-11 Chi-Square Test for
Equality
of Proportions

Slide 82

Tests of equality of proportions across several populations
are also called tests of homogeneity.
In general, when we compare c populations (or r populations if they are arranged as
rows rather than columns in the table), then the Null and alternative hypotheses:
H0: p1 = p2 = p3 = … = pc
H1: Not all pi, I = 1, 2, …, c, are equal

Chi-square test statistic for equal proportions:

( Oij  Eij ) 2
2  
Eij
i 1 j 1
r

c

Degrees of freedom: df = (r-1)(c-1)


Ri C j
Eij 
n

Officer

Equality
of Proportions - Extension

Slide 83

The Median Test
Here, the Null and alternative hypotheses are:
H0: The c populations have the same median
H1: Not all c populations have the same median

Officer

Chi-Square Test for the Median:
Example 14-16 Using the
Template

Slide 84

Note: The template was
used to help compute
the test statistic and the
p-value for the median
test. First you must
manually compute the
number of values that
are above the grand
median and the number
that is less than or
equal to the grand
median. Use these
values in the template.
See Table 14-16 in the
text.

Since the p-value = 0.6703 is very large there is no evidence to reject the null hypothesis.
Officer

MMI S&P100
220
151
218
150
216
148
217
149
215
147
213
146
219
152
236
165
237
162
235
161

R-MMI R-S&P Diff Diffsq
7
6
1
1
5
5
0
0
3
3
0
0
4
4
0
0
2
2
0
0
1
1
0
0
6
7
-1
1
9
10
-1
1
10
9
1
1
8
8
0
0
Sum:

Slide 85

Table 11: =0.005
n
.
.
.
7
-----8
0.881
9
0.833
10
0.794
11
0.818
.
.
.

4

n 
 di
6
64
4

1
 1  i
1
1
9758794H  rejected
 1

99
nn
11 1
Officer

Using the Template

Slide 86

Note: The pvalues in the
range J15:J17
will appear only
if the sample
size is large (n
> 30)

Officer

14-9 A Chi-Square Test for
Goodness of Fit


Slide 87

Steps in a chi-square analysis:
 Formulate null and alternative hypotheses
 Compute frequencies of occurrence that would be
expected if the null hypothesis were true - expected cell
counts
 Note actual, observed cell counts
 Use differences between expected and actual cell counts
to find chi-square statistic:

(Oi  Ei ) 2
 
Ei
i 1
k

2

 Compare chi-statistic with critical values from the chisquare distribution (with k-1 degrees of freedom) to test
the null hypothesis
Officer

Distribution

Slide 88

H0: The probabilities of occurrence of events E1, E2...,Ek are
given by
p1,p2,...,pk
H1: The probabilities of the k events are not as specified in the
null
hypothesis
Assuming equal probabilities, p1= p2 = p3 = p4 =0.25 and n=80
Preference
Tan
Brown Maroon Black
Total
Observed
12
40
8
20
80
Expected(np)
20
20
20
20
80
(O-E)
-8
20
-12
0
0
k ( Oi  E i )
2
  
i 1
Ei

2


( 8 )
20

2


( 20 )

2


( 12 )

20

20

2


( 0)

2

20

 30.4  

2

 11.3449

( 0.01, 3)

H 0 is rejected at the 0.01 level.
Officer

Distribution using the Template

Note: the pvalue is
0.0000, so
we can
reject the
null
hypothesis
at any 
level.

Officer

Goodness-of-Fit for the Normal
Distribution: Example 14-13

p(z<-1)
p(-1<z<-0.44)
p(-0.44<z<0)
p(0<z<0.44)
p(0.44<z<14)
p(z>1)

= 0.1587
= 0.1713
= 0.1700
= 0.1700
= 0.1713
= 0.1587

Partitioning the Standard Normal Distribution
0.1700

0.1700
0.4

0.1713

0.1713
0.3

f(z)

1. Use the table of the standard
normal distribution to determine an
appropriate partition of the standard
normal distribution which gives
ranges with approximately equal
percentages.

Slide 90

0.2

0.1587

0.1587

0.1

z

0.0
-5

-1

0

1

5

-0.44 0.44

2. Given z boundaries, x boundaries can be determined from the inverse
standard normal transformation: x =  + z = 125 + 40z.
3. Compare with the critical value of the 2 distribution with k-3 degrees of
freedom.
Officer

Example 14-13: Solution

i
1
2
3
4
5
6

Oi
14
20
16
19
16
15

Ei

15.87
17.13
17.00
17.00
17.13
15.87

Slide 91

Oi - Ei (Oi - Ei)2 (Oi - Ei)2/ Ei
-1.87
2.87
-1.00
2.00
-1.13
-0.87

3.49690
8.23691
1.00000
4.00000
1.27690
0.75690

2
:

0.22035
0.48085
0.05882
0.23529
0.07454
0.04769

1.11755

2(0.10,k-3)= 6.5139 > 1.11755 H0 is not rejected at the 0.10 level

Officer

Example 14-13: Solution using
the Template

Slide 92

Note: p-value =
0.8002 > 0.01
H0 is not
rejected at the
0.10 level

Officer

14-9 Contingency Table Analysis:
A Chi-Square Test for Independence

Slide 93

First Classification Category
Second
Classification
Category
1
2
3
4
5
Column
Total

1
O11
O21
O31
O41
O51

2
O12
O22
O32
O42
O52

3
O13
O23
O33
O43
O53

4
O14
O24
O34
O44
O54

5
O15
O25
O35
O45
O55

C1

C2

C3

C4

C5

Row
Total
R1
R2
R3
R4
R5
n

Officer

A Chi-Square Test for
Independence

Slide 94

A and B are independent if:P(AUB) = P(A)P(B).
If the first and second classification categories are independent:Eij = (Ri)(Cj)/
H0: The two classification variables are independent of each other
H1: The two classification variables are not independent
Chi-square test statistic for independence:

( Oij  Eij ) 2
2  
Eij
i 1 j 1
r

Degrees of freedom: df=(r-1)(c-1)

c

Ri C j
Eij 
n

Officer

Example 14-14
2(0.01,(2-1)(21))=6.63490

Industry Type

Profit

Service Nonservice
(Expected) (Expected) Total
42
18
60

(Expected)

6

34

(Expected)

(40*48/100)=19.2

(40*52/100)=20.8

Total

48

52

H0 is rejected at the
0.01 level and
it is concluded that
the two variables
are not independent.

(60*52/100)=31.2

Loss

ij
11
12
21
22

(60*48/100)=28.8

O
42
18
6
34

E
28.8
31.2
19.2
20.8

(O-E)2
174.24
174.24
174.24
174.24

O-E
13.2
-13.2
-13.2
13.2
2:

29.0865

(O-E)2/E
6.0500
5.5846
9.0750
8.3769

Slide 95

40
100

2
Yates corrected  for a 2x2 table:
2
Oij  Eij  0.5
2
  
Eij





Officer

Contingency Table Analysis: Slide 96
Example 14-14 using the Template
Note: When
the
contingenc
y table is a
2x2, one
should use
the Yates
correction.

Since p-value = 0.000, H0 is rejected at the 0.01 level and it is concluded that the two
variables are not independent.
Officer

Equality
of Proportions

Slide 97

Tests of equality of proportions across several populations
are also called tests of homogeneity.
In general, when we compare c populations (or r populations if they are arranged as
rows rather than columns in the table), then the Null and alternative hypotheses:
H0: p1 = p2 = p3 = … = pc
H1: Not all pi, I = 1, 2, …, c, are equal

Chi-square test statistic for equal proportions:

( Oij  Eij ) 2
2  
Eij
i 1 j 1
r

c

Degrees of freedom: df = (r-1)(c-1)


Ri C j
Eij 
n

Officer

14-11 Chi-Square Test for Equality
of Proportions - Extension

Slide 98

The Median Test
Here, the Null and alternative hypotheses are:
H0: The c populations have the same median
H1: Not all c populations have the same median

Officer

Chi-Square Test for the Median:
Example 14-16 Using the Template

Slide 99

Note: The template was
used to help compute
the test statistic and the
p-value for the median
test. First you must
manually compute the
number of values that
are above the grand
median and the number
that is less than or
equal to the grand
median. Use these
values in the template.
See Table 14-16 in the
text.

Since the p-value = 0.6703 is very large there is no evidence to reject the null hypothesis.
Officer

Name
Religion
Domicile
Contact #
E.Mail

M.Phil (Statistics)

Shakeel Nouman
Christian
Punjab (Lahore)
0332-4462527. 0321-9898767
sn_gcu@yahoo.com
sn_gcu@hotmail.com

Slide 100

GC University, .
(Degree awarded by GC University)

M.Sc (Statistics)
Statitical Officer
(BS-17)
(Economics & Marketing
Division)

GC University, .
(Degree awarded by GC University)

Livestock Production Research Institute
Bahadurnagar (Okara), Livestock & Dairy Development
Department, Govt. of Punjab

Officer

Nonparametric methods and chi square tests (1)

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