This chapter discusses chi-square tests and nonparametric tests. It begins by introducing contingency tables and how they are used to classify sample observations according to multiple characteristics. Examples are provided to demonstrate how to set up contingency tables and calculate expected frequencies. The chapter then explains how to perform chi-square tests to analyze differences between two or more proportions, test independence between categorical variables, and compare population medians using the Wilcoxon rank-sum test. Decision rules for each test are outlined. Worked examples are provided to demonstrate applying these statistical tests and interpreting the results.

1. Statistics for Managers
Using Microsoft® Excel
4th Edition
Chapter 11
Chi-Square Tests and
Nonparametric Tests
Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc.
Chap 11-1

2. Chapter Goals
After completing this chapter, you should be
able to:
Perform a χ2 test for the difference between two
proportions
 Use a χ2 test for differences in more than two proportions
 Perform a χ2 test of independence
 Apply and interpret the Wilcoxon rank sum test for the
difference between two medians
 Perform nonparametric analysis of variance using the
Kruskal-Wallis rank test for one-way ANOVA
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-2
Prentice-Hall, Inc.


3. Contingency Tables
Contingency Tables

Useful in situations involving multiple population
proportions

Used to classify sample observations according
to two or more characteristics

Also called a cross-classification table.
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-3

4. Contingency Table Example
Left-Handed vs. Gender
Dominant Hand: Left vs. Right
Gender: Male vs. Female
 2 categories for each variable, so
called a 2 x 2 table
 Suppose we examine a sample of
size 300
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-4

5. Contingency Table Example
(continued)
Sample results organized in a contingency table:
Hand Preference
sample size = n = 300:
180 Males, 24 were
left handed
Left
Right
Female
12
108
120
Male
24
156
180
36
120 Females, 12
were left handed
Gender
264
300
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-5

6. χ2 Test for the Difference
Between Two Proportions
H0: p1 = p2 (Proportion of females who are left
handed is equal to the proportion of
males who are left handed)
H1: p1 ≠ p2 (The two proportions are not the same –
Hand preference is not independent
of gender)

If H0 is true, then the proportion of left-handed females should be
the same as the proportion of left-handed males

The two proportions above should be the same as the proportion of
left-handed people overall
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-6

7. The Chi-Square Test Statistic
The Chi-square test statistic is:
( fo − fe )2
χ2 = ∑
fe
all cells

where:
fo = observed frequency in a particular cell
fe = expected frequency in a particular cell if H0 is true
χ2 for the 2 x 2 case has 1 degree of freedom
Statistics for Managers Using the contingency table has expected
(Assumed: each cell in
Microsoft Excel, 4e © of at least 5)
frequency 2004
Chap 11-7
Prentice-Hall, Inc.

8. Decision Rule
The χ2 test statistic approximately follows a chisquared distribution with one degree of freedom
Decision Rule:
If χ2 > χ2U, reject H0,
otherwise, do not
reject H0
α
0
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Do not
reject H0
Reject H0
χ2
χ2U
Chap 11-8

9. Computing the
Average Proportion
X1 + X 2 X
The average
p=
=
proportion is:
n1 + n2
n
120 Females, 12
were left handed
180 Males, 24 were
left handed
Here:
12 + 24
36
p=
=
= 0.12
120 + 180 300
i.e., the proportion
Statistics for Managers Using of left handers overall is 12%
Microsoft Excel, 4e © 2004
Chap 11-9
Prentice-Hall, Inc.

10. Finding Expected Frequencies

To obtain the expected frequency for left handed
females, multiply the average proportion left handed (p)
by the total number of females

To obtain the expected frequency for left handed males,
multiply the average proportion left handed (p) by the
total number of males
If the two proportions are equal, then
P(Left Handed | Female) = P(Left Handed | Male) = .12
i.e., we would expect (.12)(120) = 14.4 females to be left handed
Statistics for Managers (.12)(180) = 21.6 males to be left handed
Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-10

11. Observed vs. Expected
Frequencies
Hand Preference
Gender
Left
Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180
264
300
36
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-11

12. The Chi-Square Test Statistic
Hand Preference
Gender
Left
Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180
36
264
300
The test statistic is:
( fo − fe )2
χ = ∑
fe
all cells
2
Statistics12 − 14.4)2 (108 − 105.6)2 (24 − 21.6)2 (156 − 158.4)2
for Managers Using
(
=
+
+
+
= 0.6848
Microsoft Excel, 4e © 2004
14.4
105.6
21.6
158.4
Chap 11-12
Prentice-Hall, Inc.

13. Decision Rule
2
The test statistic is χ 2 = 0.6848 , χ U with 1 d.f. = 3.841
Decision Rule:
If χ2 > 3.841, reject H0,
otherwise, do not reject H0
α
0
Do not
reject H0
Reject H0
χ2U=3.841
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
χ2
Here,
χ2 = 0.6848 < χ2U = 3.841,
so we do not reject H0
and conclude that there is
not sufficient evidence
that the two proportions
are different at α = .05
Chap 11-13

14. χ2 Test for the Differences in
More Than Two Proportions

Extend the χ2 test to the case with more than
two independent populations:
H0: p1 = p2 = … = pc
H1: Not all of the pj are equal (j = 1, 2, …, c)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-14

15. The Chi-Square Test Statistic
The Chi-square test statistic is:
( fo − fe )2
χ2 = ∑
fe
all cells

where:
fo = observed frequency in a particular cell of the 2 x c table
fe = expected frequency in a particular cell if H0 is true
χ2 for the 2 x c case has (2-1)(c-1) = c - 1 degrees of freedom
Statistics for Managers Usingcontingency table has expected
(Assumed: each cell in the
Microsoft Excel, 4e at least 1)
frequency of © 2004
Chap 11-15
Prentice-Hall, Inc.

16. Computing the
Overall Proportion
The overall
proportion is:

X1 + X 2 +  + Xc X
p=
=
n1 + n2 +  + nc
n
Expected cell frequencies for the c categories
are calculated as in the 2 x 2 case, and the
decision rule is the same:
Decision Rule:
If χ2 > χ2U, reject H0,
otherwise, do not
Statistics for H
reject Managers Using
0
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Where χ2U is from the
chi-squared distribution
with c – 1 degrees of
freedom
Chap 11-16

17. χ2 Test of Independence

Similar to the χ2 test for equality of more than
two proportions, but extends the concept to
contingency tables with r rows and c columns
H0: The two categorical variables are independent
(i.e., there is no relationship between them)
H1: The two categorical variables are dependent
(i.e., there is a relationship between them)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-17

18. χ2 Test of Independence
(continued)
The Chi-square test statistic is:
( fo − fe )2
χ2 = ∑
fe
all cells

where:
fo = observed frequency in a particular cell of the r x c table
fe = expected frequency in a particular cell if H0 is true
χ2 for the r x c case has (r-1)(c-1) degrees of freedom
Statistics for Managers in the contingency table has expected
(Assumed: each cell Using
frequency of © 2004
Microsoft Excel, 4eat least 1)
Chap 11-18
Prentice-Hall, Inc.

19. Expected Cell Frequencies

Expected cell frequencies:
row total × column total
fe =
n
Where:
row total = sum of all frequencies in the row
column total = sum of all frequencies in the column
n = overall sample size
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-19
Prentice-Hall, Inc.

20. Decision Rule

The decision rule is
If χ2 > χ2U, reject H0,
otherwise, do not reject H0
Where χ2U is from the chi-squared distribution
with (r – 1)(c – 1) degrees of freedom
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-20

21. Example

The meal plan selected by 200 students is shown below:
Number of meals per week
20/week 10/week
Class
Standing
Fresh.
24
32
none
Total
14
70
Soph.
22
26
12
60
Junior
10
14
6
30
Senior
14
16
Statistics for Managers Using 88
Total
70
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
10
40
42
200
Chap 11-21

22. Example
(continued)

The hypothesis to be tested is:
H0: Meal plan and class standing are independent
(i.e., there is no relationship between them)
H1: Meal plan and class standing are dependent
(i.e., there is a relationship between them)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-22

23. Example:
Expected Cell Frequencies
(continued)
Observed:
Class
Standing
Number of meals
per week
20/wk 10/wk none
Expected cell
frequencies if H0 is true:
Total
Fresh.
24
32
14
70
Soph.
22
26
12
60
Junior
10
14
6
30
Senior
14
16
10
40
Class
Standing
Total
70
88
42
200
Fresh.
24.5
30.8
14.7
70
Soph.
21.0
26.4
12.6
60
Junior
10.5
13.2
6.3
30
Senior
14.0
17.6
8.4
40
70
88
42
200
Example for one cell:
fe =
row total × column total
n
Statistics for Managers Using
30 × 70
=
= 10 5
Microsoft Excel, .4e © 2004
200
Prentice-Hall, Inc.
Total
Number of meals
per week
20/wk 10/wk none
Total
Chap 11-23

24. Example: The Test Statistic
(continued)

The test statistic value is:
( fo − fe )2
χ2 = ∑
fe
all cells
(24 − 24.5)2 (32 − 30.8)2
(10 − 8.4)2
=
+
++
= 0.709
24.5
30.8
8. 4
χ2U = 12.592 for α = .05 from the chi-squared
distribution with (4 – 1)(3 – 1) = 6 degrees of
freedom
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-24
Prentice-Hall, Inc.

25. Example:
Decision and Interpretation
(continued)
2
The test statistic is χ 2 = 0.709 , χ U with 6 d.f. = 12.592
Decision Rule:
If χ2 > 12.592, reject H0,
otherwise, do not reject H0
α
0
Do not
reject H0
Reject H0
χ2U=12.592
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
χ2
Here,
χ2 = 0.709 < χ2U = 12.592,
so do not reject H0
Conclusion: there is not
sufficient evidence that meal
plan and class standing are
related at α = .05
Chap 11-25

26. Wilcoxon Rank-Sum Test for
Differences in 2 Medians

Test two independent population medians

Populations need not be normally distributed

Distribution free procedure

Used when only rank data are available

Must use normal approximation if either of the
sample sizes is larger than 10
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-26

27. Wilcoxon Rank-Sum Test:
Small Samples

Can use when both n1 , n2 ≤ 10

Assign ranks to the combined n1 + n2 sample
observations




If unequal sample sizes, let n1 refer to smaller-sized
sample
Smallest value rank = 1, largest value rank = n1 + n2
Assign average rank for ties
Sum the ranks for each sample: T1 and T2
Statistics for Managers Using
 Obtain test statistic, T (from smaller sample)
1
Microsoft Excel, 4e © 2004
Chap 11-27
Prentice-Hall, Inc.

28. Checking the Rankings


The sum of the rankings must satisfy the
formula below
Can use this to verify the sums T1 and T2
n(n + 1)
T1 + T2 =
2
where n = n1 + n2
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-28

29. Wilcoxon Rank-Sum Test:
Hypothesis and Decision Rule
M1 = median of population 1; M2 = median of population 2
Test statistic = T1 (Sum of ranks from smaller sample)
Two-Tail Test
H0: M1 = M2
H1: M1 ≠ M2
Reject Do Not
Reject
T1L
Reject
T1U
Left-Tail Test
H0: M1 ≥ M2
H1: M1 < M2
Reject
Do Not Reject
T1L
Reject H0 T1 T1L
Statisticsif for<Managers Using H0 if T1 < T1L
Reject
Microsoftif Excel, 4e © 2004
or T1 > T1U
Prentice-Hall, Inc.
Right-Tail Test
H0: M1 ≤ M2
H1: M1 > M2
Do Not Reject
Reject
T1U
Reject H0 if T1 > T1U
Chap 11-29

30. Wilcoxon Rank-Sum Test:
Small Sample Example
Sample data is collected on the capacity rates
(% of capacity) for two factories.
Are the median operating rates for two factories
the same?

For factory A, the rates are 71, 82, 77, 94, 88

For factory B, the rates are 85, 82, 92, 97
Test for equality of the sample medians
Statistics for Managers Using
at the 0.05 significance level
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-30

31. Wilcoxon Rank-Sum Test:
Small Sample Example
(continued)
Ranked
Capacity
values:
Capacity
Factory A
Rank
Factory B
Factory A
1
77
Tie in 3rd and
4th places
71
2
82
Factory B
3.5
82
3.5
85
5
88
6
92
94
97
Statistics for Managers Using
Microsoft Excel, 4e © 2004 Rank Sums:
Prentice-Hall, Inc.
7
8
9
20.5
24.5
Chap 11-31

32. Wilcoxon Rank-Sum Test:
Small Sample Example
(continued)
Factory B has the smaller sample size, so
the test statistic is the sum of the
Factory B ranks:
T1 = 24.5
The sample sizes are:
n1 = 4 (factory B)
n2 = 5 (factory A)
Statistics for Managers Using
Microsoft Excel, 4e © 2004 of significance is α = .05
The level
Chap 11-32
Prentice-Hall, Inc.

33. Wilcoxon Rank-Sum Test:
Small Sample Example
(continued)

Lower and
Upper
Critical
Values for
T1 from
Appendix
table E.8:
α
n2
n1
OneTailed
TwoTailed
4
5
.05
.10
12, 28
19, 36
.025
.05
11, 29
17, 38
.01
.02
10, 30
16, 39
.005
.01
--, --
15, 40
4
5
6
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
T1L = 11 and T1U = 29
Chap 11-33

34. Wilcoxon Rank-Sum Test:
Small Sample Solution
(continued)

α = .05
n1 = 4 , n2 = 5
Test Statistic (Sum of
ranks from smaller sample):
Two-Tail Test
H0: M1 = M2
H1: M1 ≠ M2

T1 = 24.5
Reject Do Not
Reject
T1L=11
Reject
T1U=29
Decision:
Do not reject at α = 0.05
Conclusion:
There is not enough evidence to
Reject H0 T1 T1L=11
Statisticsif for<Managers Using
prove that the medians are not
Microsoftif Excel, =29 © 2004 equal.
or T1 > T1U 4e
Chap 11-34
Prentice-Hall, Inc.

35. Wilcoxon Rank-Sum Test
(Large Sample)

For large samples, the test statistic T1 is
approximately normal with mean μT and
standard deviation σ T :
1
1
n1(n + 1)
μT1 =
2

n1 n2 (n + 1)
σ T1 =
12
Must use the normal approximation if either n1
or n2 > 10
 Assign n to be the
Statistics for Managers Usingsmaller of the two sample sizes
1
Microsoft Excel, 4e the2004
 Can use © normal approximation for small samples
Chap 11-35
Prentice-Hall, Inc.

36. Wilcoxon Rank-Sum Test
(Large Sample)
(continued)

The Z test statistic is
Z=

T1 − μT1
σ T1
Where Z approximately follows a
standardized normal distribution
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-36

37. Wilcoxon Rank-Sum Test:
Normal Approximation Example
Use the setting of the prior example:
The sample sizes were:
n1 = 4 (factory B)
n2 = 5 (factory A)
The level of significance was α = .05
The test statistic was T1 = 24.5
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-37
Prentice-Hall, Inc.

38. Wilcoxon Rank-Sum Test:
Normal Approximation Example
(continued)
μT1 =
σ T1 =

n1(n + 1) 4(9 + 1)
=
= 20
2
2
n1 n2 (n + 1)
=
12
4 (5) (9 + 1)
= 2.739
12
The test statistic is
T1 − μT1 24.5 − 20
Z=
=
= 1.64
σ T1
2.739
Z = 1.64 is not greater than the critical Z value of 1.96
(for for Managers Using
Statisticsα = .05) so we do not reject H0 – there is not
sufficient evidence that
Microsoft Excel, 4e © 2004 the medians are not equal

Prentice-Hall, Inc.
Chap 11-38

39. Kruskal-Wallis Rank Test



Tests the equality of more than 2 population
medians
Use when the normality assumption for oneway ANOVA is violated
Assumptions:
The samples are random and independent
 variables have a continuous distribution
 the data can be ranked
 populations have the same variability
 populations have the same shape
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-39
Prentice-Hall, Inc.


40. Kruskal-Wallis Test Procedure

Obtain relative rankings for each value


In event of tie, each of the tied values gets the
average rank
Sum the rankings for data from each of the c
groups

Compute the H test statistic
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-40

41. Kruskal - Wallis Test Procedure
(continued)

The Kruskal - Wallis H test statistic:
(with c – 1 degrees of freedom)
c T2 
 12
j
H=
∑ n  − 3(n + 1)
 n(n + 1) j=1 j 


where:
n = sum of sample sizes in all samples
c = Number of samples
Tj = Sum of ranks in the jth sample
nj = Size of the
Managers Using jth sample
Statistics for
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-41

42. Kruskal-Wallis Test Procedure
(continued)

Complete the test by comparing the
calculated H value to a critical χ2 value from
the chi-square distribution with c – 1
degrees of freedom

Decision rule

α
0
χ2
Do not
Reject H
reject H
Statistics for Managers Using
χ2U
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Reject H0 if test statistic H > χ2U

Otherwise do not reject H0
0
0
Chap 11-42

43. Kruskal-Wallis Example

Do different departments have different class
sizes?
Class size
(Math, M)
Class size
(English, E)
Class size
(Biology, B)
23
45
54
78
66
55
60
72
45
70
30
40
18
34
44
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-43

44. Kruskal-Wallis Example

Do different departments have different class
sizes?
Class size
Class size
Class size
Ranking
Ranking
(Math, M)
(English, E)
(Biology, B)
23
41
54
78
66
2
6
9
15
12
55
60
72
45
70
Σ = 44
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
10
11
14
8
13
Σ = 56
30
40
18
34
44
Ranking
3
5
1
4
7
Σ = 20
Chap 11-44

45. Kruskal-Wallis Example
(continued)
H0 : MedianM = MedianE = MedianH
HA : Not all population Medians are equal

The H statistic is
c R2 
 12
j
H= 
∑ n  − 3(n + 1)
 n(n + 1) j=1 j 



 44 2 56 2 20 2 
12

=
 5 + 5 + 5  − 3(15 + 1) = 6.72

15(15 + 1) 


Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 11-45
Prentice-Hall, Inc.

46. Kruskal-Wallis Example
(continued)

Compare H = 6.72 to the critical value from the
chi-square distribution for 5 – 1 = 4 degrees of
freedom and α = .05:
χ = 9.4877
2
U
2
χU = 9.4877
Since H = 6.72 <
do not reject H0
There is not sufficient evidence to reject
Statistics for Managers Using
that the 2004
Microsoft Excel, 4e ©population medians are all equal
Chap 11-46
Prentice-Hall, Inc.

47. Chapter Summary




Developed and applied the χ2 test for the
difference between two proportions
Developed and applied the χ2 test for
differences in more than two proportions
Examined the χ2 test for independence
Used the Wilcoxon rank sum test for two
population medians


Small Samples
Large sample Z approximation
Applied the Kruskal - Wallis H-test for multiple
Statistics for Managers Using
population medians

Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 11-47