Basic Business Statistics
10th Edition
Chapter 12
Chi-Square Tests and
Nonparametric Tests
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-1
Learning Objectives
In this chapter, you learn:
 How and when to use the chi-square test for
contingency tables
 How to use the Marascuilo procedure for
determining pairwise differences when evaluating
more than two proportions
 How to use the chi-square test to evaluate the
goodness of fit of a set of data to a specific
probability distribution
 How and when to use nonparametric tests
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-2
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.
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-3
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-4
Contingency Table Example
(continued)
Sample results organized in a contingency table:
Hand Preference
sample size = n = 300:
Gender
Left
Right
120 Females, 12
were left handed
Female
12
108
120
180 Males, 24 were
left handed
Male
24
156
180
36
264
300
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-5
2 Test for the Difference
Between Two Proportions
H0: π1 = π2 (Proportion of females who are left
handed is equal to the proportion of
males who are left handed)
H1: π1 ≠ π2 (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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-6
The Chi-Square Test Statistic
The Chi-square test statistic is:
2
(
f

f
)
2   o e
fe
all cells

where:
fo = observed frequency in a particular cell
fe = expected frequency in a particular cell if H0 is true
2 for the 2 x 2 case has 1 degree of freedom
(Assumed: each cell in the contingency table has expected
frequency of at least 5)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-7
Decision Rule
The 2 test statistic approximately follows a chisquared distribution with one degree of freedom
Decision Rule:
If 2 > 2U, reject H0,
otherwise, do not
reject H0

0
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Do not
reject H0
Reject H0
2
2U
Chap 12-8
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 of left handers overall is 0.12, that is, 12%
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-9
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
(.12)(180) = 21.6 males to be left handed
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-10
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
36
264
300
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-11
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
(12  14.4) 2 (108  105.6) 2 (24  21.6) 2 (156  158.4) 2




 0.7576
14.4
105.6
21.6
158.4
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-12
Decision Rule
The test statistic is χ 2  0.7576 , χU2 with 1 d.f.  3.841
Decision Rule:
If 2 > 3.841, reject H0,
otherwise, do not reject H0

0
Do not
reject H0
Reject H0
2U=3.841
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
2
Here,
2 = 0.7576 < 2U = 3.841,
so we do not reject H0
and conclude that there is
not sufficient evidence
that the two proportions
are different at  = 0.05
Chap 12-13
2 Test for Differences Among
More Than Two Proportions

Extend the 2 test to the case with more than
two independent populations:
H0: π1 = π2 = … = πc
H1: Not all of the πj are equal (j = 1, 2, …, c)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-14
The Chi-Square Test Statistic
The Chi-square test statistic is:
2
(
f

f
)
2   o e
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
2 for the 2 x c case has (2-1)(c-1) = c - 1 degrees of freedom
(Assumed: each cell in the contingency table has expected
frequency of at least 1)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-15
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 2 > 2U, reject H0,
otherwise, do not
reject H0
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Where 2U is from the
chi-squared distribution
with c – 1 degrees of
freedom
Chap 12-16
The Marascuilo Procedure




Used when the null hypothesis of equal
proportions is rejected
Enables you to make comparisons between all
pairs
Start with the observed differences, pj – pj’, for
all pairs (for j ≠ j’) . . .
. . .then compare the absolute difference to a
calculated critical range
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-17
The Marascuilo Procedure
(continued)

Critical Range for the Marascuilo Procedure:
Critical range  χ


2
U
p j (1  p j ) p j' (1  p j' )

nj
n j'
(Note: the critical range is different for each pairwise comparison)
A particular pair of proportions is significantly
different if
|pj – pj’| > critical range for j and j’
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-18
Marascuilo Procedure Example
A University is thinking of switching to a trimester academic
calendar. A random sample of 100 administrators, 50 students,
and 50 faculty members were surveyed
Opinion
Administrators
Students
Faculty
Favor
63
20
37
Oppose
37
30
13
100
50
50
Totals
Using a 1% level of significance, which groups have a
different attitude?
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-19
Marascuilo Procedure: Solution
Excel Output:
compare
Marascuilo Procedure
Sample
Sample
Absolute Std. Error Critical
Group Proportion
Size
Comparison Difference of Difference Range Results
1
0.63
100 1 to 2
0.23 0.08444525 0.256 Means are not different
2
0.4
50 1 to 3
0.11 0.07860662 0.239 Means are not different
3
0.74
50 2 to 3
0.34 0.09299462 0.282 Means are different
Other Data
Level of significance
d.f
Q Statistic
0.01
2
3.034856
Chi-sq Critical Value
9.21
At 1% level of significance, there is evidence of a difference
in attitude between students and faculty
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-20
2 Test of Independence

Similar to the 2 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)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-21
2 Test of Independence
(continued)
The Chi-square test statistic is:
2
(
f

f
)
2   o e
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
2 for the r x c case has (r-1)(c-1) degrees of freedom
(Assumed: each cell in the contingency table has expected
frequency of at least 1)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-22
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-23
Decision Rule

The decision rule is
If 2 > 2U, reject H0,
otherwise, do not reject H0
Where 2U is from the chi-squared distribution
with (r – 1)(c – 1) degrees of freedom
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-24
Example

The meal plan selected by 200 students is shown below:
Number of meals per week
Class
none
Standing 20/week 10/week
Total
Fresh.
24
32
14
70
Soph.
22
26
12
60
Junior
10
14
6
30
Senior
14
16
10
40
Total
70
88
42
200
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-25
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)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-26
Example:
Expected Cell Frequencies
(continued)
Observed:
Class
Standing
Number of meals
per week
Expected cell
frequencies if H0 is true:
20/wk
10/wk
none
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
Example for one cell:
row total  column total
fe 
n
30  70

 10.5
200
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Number of meals
per week
20/wk
10/wk
none
Total
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
Total
Chap 12-27
Example: The Test Statistic
(continued)

The test statistic value is:
2
(
f

f
)
2   o e
fe
all cells
(24  24.5)2 (32  30.8)2
(10  8.4)2



 0.709
24.5
30.8
8 .4
2U = 12.592 for  = 0.05 from the chi-squared
distribution with (4 – 1)(3 – 1) = 6 degrees of
freedom
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-28
Example:
Decision and Interpretation
(continued)
The test statistic is  2  0.709 , U2 with 6 d.f.  12.592
Decision Rule:
If 2 > 12.592, reject H0,
otherwise, do not reject H0

0
Do not
reject H0
Reject H0
2U=12.592
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
2
Here,
2 = 0.709 < 2U = 12.592,
so do not reject H0
Conclusion: there is not
sufficient evidence that meal
plan and class standing are
related at  = 0.05
Chap 12-29
McNemar Test (Related Samples)

Used to determine if there is a difference
between proportions of two related samples

Uses a test statistic the follows the normal
distribution
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-30
McNemar Test (Related Samples)
(continued)

Consider a 2 X 2 contingency table:
Condition 2
Condition 1
Yes
No
Totals
Yes
A
B
A+B
No
C
D
C+D
A+C
B+D
n
Totals
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-31
McNemar Test (Related Samples)
(continued)

p1 
The sample proportions of interest are
A B
 proportion of respondent s who answer yes to condition 1
n
A C
p2 
 proportion of respondent s who answer yes to condition 2
n

Test
H0: π1 = π2
(the two population proportions are equal)
H1: π1 ≠ π2
(the two population proportions are not equal)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-32
McNemar Test (Related Samples)
(continued)

The test statistic for the McNemar test:
BC
Z
BC
where the test statistic Z is approximately
normally distributed
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-33
Chi-Square Test for a Variance or
Standard Deviation

A χ2 test statistic is used to test whether or not
the population variance or standard deviation is
equal to a specified value:
2
(n
1)S
χ2 
σ2
Where
n = sample size
S2 = sample variance
σ2 = hypothesized population variance
χ2 follows a chi-square distribution with n – 1 d.f.
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-34
Chi-Square Goodness-of-Fit Test

Does sample data conform to a hypothesized
distribution?
 Examples:


Are technical support calls equal across all days of
the week? (i.e., do calls follow a uniform
distribution?)
Do measurements from a production process
follow a normal distribution?
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-35
Chi-Square Goodness-of-Fit Test
(continued)

Are technical support calls equal across all days of the
week? (i.e., do calls follow a uniform distribution?)
 Sample data for 10 days per day of week:
Sum of calls for this day:
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
290
250
238
257
265
230
192
 = 1722
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-36
Logic of Goodness-of-Fit Test
 If calls are uniformly distributed, the 1722 calls
would be expected to be equally divided across
the 7 days:
1722
 246 expected calls per day if uniform
7
 Chi-Square Goodness-of-Fit Test: test to see if
the sample results are consistent with the
expected results
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-37
Observed vs. Expected
Frequencies
Observed
fo
Expected
fe
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
290
250
238
257
265
230
192
246
246
246
246
246
246
246
TOTAL
1722
1722
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-38
Chi-Square Test Statistic
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform

The test statistic is
(fo  fe )
 
fe
k
2
2
(where df  k  p  1)
where:
k = number of categories
fo = observed frequency
fe = expected frequency
p = number of parameters estimated from the data
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-39
The Rejection Region
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform
2
(f

f
)
2   o e
fe
k

Reject H0 if
 
2
k – 2 degrees of freedom,
since p = 1 here (the
mean was estimated)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
2
α

0
2
Do not
reject H0
2
Reject H0

Chap 12-40
Chi-Square Test Statistic
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform
2
2
2
(290

246)
(250

246)
(192

246)
2 

 ... 
 23.05
246
246
246
k – 2 = 5 (k = 7 days of the week) so
use 5 degrees of freedom:
2.05 = 11.0705
Conclusion:
2 = 23.05 > 2 = 11.0705 so
reject H0 and conclude that the
distribution is not uniform
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
 = .05
0
Do not
reject H0
Reject H0
2.05 = 11.0705
2
Chap 12-41
Normal Distribution Example

Do measurements from a production
process follow a normal distribution with
μ = 50 and σ = 15?

Process:



Get sample data
Group sample results into classes (cells)
(Expected cell frequency must be at least
5 for each cell)
Compare actual cell frequencies with expected cell
frequencies
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-42
Normal Distribution Example
(continued)

Sample data and values grouped into classes:
150 Sample
Measurements
80
65
36
66
50
38
57
77
59
…etc…
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Class
Frequency
less than 30
10
30 but < 40
21
40 but < 50
33
50 but < 60
41
60 but < 70
26
70 but < 80
10
80 but < 90
7
90 or over
2
TOTAL
150
Chap 12-43
Normal Distribution Example
(continued)

What are the expected frequencies for these classes for
a normal distribution with μ = 50 and σ = 15?
Class
Frequency
less than 30
10
30 but < 40
21
40 but < 50
33
50 but < 60
41
60 but < 70
26
70 but < 80
10
80 but < 90
7
90 or over
2
TOTAL
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Expected
Frequency
?
150
Chap 12-44
Expected Frequencies
Value
P(X < value)
Expected
frequency
Expected frequencies
in a sample of size
n=150, from a normal
distribution with
μ=50, σ=15
less than 30
0.09121
13.68
30 but < 40
0.16128
24.19
40 but < 50
0.24751
37.13
50 but < 60
0.24751
37.13
Example:
60 but < 70
0.16128
24.19
70 but < 80
0.06846
10.27
30  50 

P(x  30)  P z 

15


80 but < 90
0.01892
2.84
90 or over
0.00383
0.57
1.00000
150.00
TOTAL
 P(z  1.3333)
 .0912
(.0912)(15 0)  13.68
Combine class groups so no class has
expected frequency <1
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-45
The Test Statistic
(observed, fo)
Expected
Frequency, fe
less than 30
10
13.68
30 but < 40
40 but < 50
21
33
24.19
37.13
50 but < 60
41
37.13
60 but < 70
26
24.19
70 but < 80
10
10.27
80 or over
9
3.41
150
150.00
Frequency
Class
TOTAL
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
The test statistic is
2
(f

f
)
2   o e
fe
k

Reject H0 if
 
2
2
α
(with k – p – 1
degrees of freedom)
Chap 12-46
The Rejection Region
H0: The distribution of values is normal
with μ = 50 and σ = 15
H1: The distribution of calls does not
have this distribution
(fo  fe )2 (10  13.68) 2
(9  3.41)2
χ 

 ... 
 11.580
fe
13.68
3.41
k
2
8 classes so use 8 – 2 – 1 = 5 d.f.:
2.05 = 11.0705
=.05
Conclusion:
2 = 11.580 > 2 = 11.0705 so reject H0:
There is sufficient evidence that the data
are not normal with μ = 50 and σ = 15
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
0
2
Do not
reject H0
Reject H0
2.05 = 11.0705
Chap 12-47
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-48
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

Obtain test statistic, T1 (from smaller sample)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-49
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-50
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
Left-Tail Test
H0: M1  M2
H1: M1  M2
Reject
T1U
Reject H0 if T1 < T1L
Do Not Reject
T1L
Reject H0 if T1 < T1L
Right-Tail Test
H0: M1  M2
H1: M1  M2
Do Not Reject
Reject
T1U
Reject H0 if T1 > T1U
or if T1 > T1U
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-51
Wilcoxon Rank-Sum Test:
Small Sample Example
Sample data are 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 population medians
at the 0.05 significance level
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-52
Wilcoxon Rank-Sum Test:
Small Sample Example
(continued)
Ranked
Capacity
values:
Tie in 3rd and
4th places
Capacity
Factory A
Rank
Factory B
Factory A
71
1
77
2
82
3.5
82
3.5
85
5
88
6
92
94
7
8
97
Rank Sums:
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Factory B
9
20.5
24.5
Chap 12-53
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)
The level of significance is  = .05
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-54
Wilcoxon Rank-Sum Test:
Small Sample Example
(continued)

Lower and
Upper
Critical
Values for
T1 from
Appendix
table E.8:

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
T1L = 11 and T1U = 29
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-55
Wilcoxon Rank-Sum Test:
Small Sample Solution
(continued)


 = .05
n1 = 4 , n2 = 5
Two-Tail Test
H0: M1 = M2
H1: M1  M2
Reject Do Not
Reject
T1L=11
Reject
T1U=29
Reject H0 if T1 < T1L=11
or if T1 > T1U=29
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Test Statistic (Sum of
ranks from smaller sample):
T1 = 24.5
Decision:
Do not reject at  = 0.05
Conclusion:
There is not enough evidence to
prove that the medians are not
equal.
Chap 12-56
Wilcoxon Rank-Sum Test
(Large Sample)

For large samples, the test statistic T1 is
approximately normal with mean μ T 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 n1 to be the smaller of the two sample sizes

Can use the normal approximation for small samples
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-57
Wilcoxon Rank-Sum Test
(Large Sample)
(continued)

The Z test statistic is
n1(n  1)
T1  μT1 T1 
2
Z

σ T1
n1n2 (n  1)
12

Where Z approximately follows a
standardized normal distribution
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-58
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-59
Wilcoxon Rank-Sum Test:
Normal Approximation Example
(continued)
μT1 
σ T1 

n1(n  1) 4(9  1)

 20
2
2
n1 n2 (n  1)
4 (5) (9  1)

 2.739
12
12
The test statistic is
Z

T1  μT1
σ T1
24.5  20

 1.64
2.739
Z = 1.64 is not greater than the critical Z value of 1.96
(for α = .05) so we do not reject H0 – there is not
sufficient evidence that the medians are not equal
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-60
Wilcoxon Signed Ranks Test

A nonparametric test for two related populations

Steps:
1. For each of n sample items, compute the difference,
Di, between two measurements
2. Ignore + and – signs and find the absolute values, |Di|
3. Omit zero differences, so sample size is n’
4. Assign ranks Ri from 1 to n’ (give average rank to
ties)
5. Reassign + and – signs to the ranks Ri
6. Compute the Wilcoxon test statistic W as the sum of
the positive ranks
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-61
Wilcoxon Signed Ranks
Test Statistic

The Wilcoxon signed ranks test statistic is the
sum of the positive ranks:
n'
W  R
()
i
i 1

For small samples (n’ < 20), use Table E.9 for
the critical value of W
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-62
Wilcoxon Signed Ranks
Test Statistic

For samples of n’ > 20, W is approximately
normally distributed with
n' (n' 1)
μW 
4
σW
n' (n' 1)(2n' 1)

24
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-63
Wilcoxon Signed Ranks Test

The large sample Wilcoxon signed ranks Z
test statistic is
Z

n' (n' 1)
W
4
n' (n' 1)(2n' 1)
24
The appropriate hypothesis test is
H0: MD ≤ 0
H1: MD > 0
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-64
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-65
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-66
Kruskal-Wallis Test Procedure
(continued)

The Kruskal-Wallis H-test statistic:
(with c – 1 degrees of freedom)
c T2 
 12
j
H
  3(n  1)

 n(n  1) j1 n j 
where:
n = sum of sample sizes in all samples
c = Number of samples
Tj = Sum of ranks in the jth sample
nj = Number of values in the jth sample (j = 1, 2, … , c)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-67
Kruskal-Wallis Test Procedure
(continued)

Complete the test by comparing the
calculated H value to a critical 2 value from
the chi-square distribution with c – 1
degrees of freedom


0
Do not
reject H0
2U
Reject H0
Decision rule

2
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.

Reject H0 if test statistic H > 2U
Otherwise do not reject H0
Chap 12-68
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-69
Kruskal-Wallis Example
(continued)

Do different departments have different class
sizes?
Class size
Class size
Ranking
Ranking
(Math, M)
(English, E)
23
41
54
78
66
2
6
9
15
12
 = 44
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
55
60
72
45
70
10
11
14
8
13
 = 56
Class size
(Biology, B)
Ranking
30
40
18
34
44
3
5
1
4
7
 = 20
Chap 12-70
Kruskal-Wallis Example
(continued)
H0 : MedianM  MedianE  MedianB
H1 : Not all population Medians are equal

The H statistic is
c T2 
 12
j
H 
  3(n  1)

 n(n  1) j1 n j 

 44 2 56 2 20 2 
12

  3(15  1)  6.72



5
5 
15(15  1)  5
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-71
Kruskal-Wallis Example
(continued)

Compare H = 6.72 to the critical value from the
chi-square distribution for 3 – 1 = 2 degrees of
freedom and  = 0.05:
χ  5.991
2
U
2

Since H = 6.72 > U  5.991 ,
reject H0
There is sufficient evidence to reject that
the population medians are all equal
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-72
Friedman Rank Test

Use the Friedman rank test to determine
whether c groups (i.e., treatment levels) have
been selected from populations having equal
medians
H0: M.1 = M.2 = . . . = M.c
H1: Not all M.j are equal (j = 1, 2, …, c)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-73
Friedman Rank Test
(continued)

Friedman rank test for differences among c
medians:
c
12
2
FR 
R.j  3r(c  1)

rc(c  1) j1
where
R.j2 = the square of the total ranks for group j
r = the number of blocks
c = the number of groups
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-74
Friedman Rank Test
(continued)

The Friedman rank test statistic is approximated
by a chi-square distribution with c – 1 d.f.
Reject H0 if
FR  
2
U
Otherwise do not reject H0
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-75
Chapter Summary




Developed and applied the 2 test for the
difference between two proportions
Developed and applied the 2 test for
differences in more than two proportions
Examined the 2 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
population medians
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 12-76