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Chapter 12: Analysis of Categorical Data
1
Chapter 12
Analysis of Categorical Data
LEARNING OBJECTIVES
This chapter presents several nonparametric statistics that can be used to analyze data
enabling you to:
1.
Understand the chi-square goodness-of-fit test and how to use it.
2.
Analyze data using the chi-square test of independence.
CHAPTER TEACHING STRATEGY
Chapter 12 is a chapter containing the two most prevalent chi-square tests: chisquare goodness-of-fit and chi-square test of independence. These two techniques are
important because they give the statistician a tool that is particularly useful for analyzing
nominal data (even though independent variable categories can sometimes have ordinal
or higher categories). It should be emphasized that there are many instances in business
research where the resulting data gathered are merely categorical identification. For
example, in segmenting the market place (consumers or industrial users), information is
gathered regarding gender, income level, geographical location, political affiliation,
religious preference, ethnicity, occupation, size of company, type of industry, etc. On
these variables, the measurement is often a tallying of the frequency of occurrence of
individuals, items, or companies in each category. The subject of the research is given no
"score" or "measurement" other than a 0/1 for being a member or not of a given category.
These two chi-square tests are perfectly tailored to analyze such data.
The chi-square goodness-of-fit test examines the categories of one variable to
determine if the distribution of observed occurrences matches some expected or
theoretical distribution of occurrences. It can be used to determine if some standard or
previously known distribution of proportions is the same as some observed distribution of
Chapter 12: Analysis of Categorical Data
2
proportions. It can also be used to validate the theoretical distribution of occurrences of
phenomena such as random arrivals which are often assumed to be Poisson distributed.
You will note that the degrees of freedom which are k - 1 for a given set of expected
values or for the uniform distribution change to k - 2 for an expected Poisson distribution
and to k - 3 for an expected normal distribution. To conduct a chi-square goodness-of-fit
test to analyze an expected Poisson distribution, the value of lambda must be estimated
from the observed data. This causes the loss of an additional degree of freedom. With
the normal distribution, both the mean and standard deviation of the expected distribution
are estimated from the observed values causing the loss of two additional degrees of
freedom from the k - 1 value.
The chi-square test of independence is used to compare the observed frequencies
along the categories of two independent variables to expected values to determine if the
two variables are independent or not. Of course, if the variables are not independent,
they are dependent or related. This allows business researchers to reach some
conclusions about such questions as is smoking independent of gender or is type of
housing preferred independent of geographic region. The chi-square test of independence
is often used as a tool for preliminary analysis of data gathered in exploratory research
where the researcher has little idea of what variables seem to be related to what variables,
and the data are nominal. This test is particularly useful with demographic type data.
A word of warning is appropriate here. When an expected frequency is small, the
observed chi-square value can be inordinately large thus yielding an increased possibility
of committing a Type I error. The research on this problem has yielded varying results
with some authors indicating that expected values as low as two or three are acceptable
and other researchers demanding that expected values be ten or more. In this text, we
have settled on the fairly widespread accepted criterion of five or more.
CHAPTER OUTLINE
16.1
Chi-Square Goodness-of-Fit Test
Testing a Population Proportion Using the Chi-square Goodness-of-Fit
Test as an Alternative Technique to the z Test
16.2
Contingency Analysis: Chi-Square Test of Independence
KEY TERMS
Categorical Data
Chi-Square Distribution
Chi-Square Goodness-of-Fit Test
Chi-Square Test of Independence
Contingency Analysis
Contingency Table
Chapter 12: Analysis of Categorical Data
3
SOLUTIONS TO CHAPTER 16
12.1
( f0  fe )2
f0
68
42
33
22
10
8
f0
53
37
32
28
18
15
fe
3.309
0.595
0.030
1.636
6.400
6.125
Ho: The observed distribution is the same
as the expected distribution.
Ha: The observed distribution is not the same
as the expected distribution.
Observed  2  
( f0  fe )2
= 18.095
fe
df = k - 1 = 6 - 1 = 5,  = .05
2.05,5 = 11.07
Since the observed 2 = 18.095 > 2.05,5 = 11.07, the decision is to reject the null
hypothesis.
The observed frequencies are not distributed the same as the expected
frequencies.
12.2
f0
19
17
14
18
19
21
18
18
fo = 144
fe
18
18
18
18
18
18
18
18
fe = 144
( f0  fe )2
f0
0.056
0.056
0.889
0.000
0.056
0.500
0.000
0.000
1.557
Chapter 12: Analysis of Categorical Data
4
Ho: The observed frequencies are uniformly distributed.
Ha: The observed frequencies are not uniformly distributed.
x
f
k
0

144
= 18
8
In this uniform distribution, each fe = 18
df = k – 1 = 8 – 1 = 7,  = .01
2.01,7 = 18.4753
Observed  2  
( f0  fe )2
= 1.557
fe
Since the observed 2 = 1.557 < 2.01,7 = 18.4753, the decision is to fail to reject
the null hypothesis
There is no reason to conclude that the frequencies are not uniformly
distributed.
12.3
Number
f0
0
1
2
3
28
17
11
5
(Number)(f0)
0
17
22
15
54
Ho: The frequency distribution is Poisson.
Ha: The frequency distribution is not Poisson.
=
54
=0.9
61
Number
0
1
2
>3
Expected
Probability
.4066
.3659
.1647
.0628
Expected
Frequency
24.803
22.312
10.047
3.831
Since fe for > 3 is less than 5, collapse categories 2 and >3:
Chapter 12: Analysis of Categorical Data
Number
fo
fe
0
1
>2
28
17
16
61
24.803
22.312
13.878
60.993
df = k - 2 = 3 - 2 = 1,
5
( f0  fe )2
f0
0.412
1.265
0.324
2.001
= .05
2.05,1 = 3.84146
Calculated  2  
( f0  fe )2
= 2.001
fe
Since the observed 2 = 2.001 < 2.05,1 = 3.84146, the decision is to fail to reject
the null hypothesis.
There is insufficient evidence to reject the distribution as Poisson distributed.
The conclusion is that the distribution is Poisson distributed.
12.4
Category
f(observed)
10-20
6
20-30
14
30-40
29
40-50
38
50-60
25
60-70
10
70-80
7
n = f = 129
x
s=
 fm  5,715
 f 129

fM 2 
Midpt.
fm
15
90
25
350
35
1,015
45
1,710
55
1,375
65
650
75
525
fm = 5,715
= 44.3
( fM ) 2
n 1
fm2
1,350
8,750
35,525
76,950
75,625
42,250
39,375
2
fm = 279,825
n

(5,715) 2
129
= 14.43
128
279,825 
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.
Chapter 12: Analysis of Categorical Data
For Category 10 - 20
Prob
10  44.3
= -2.38
14.43
20  44.3
z =
= -1.68
14.43
z =
.4913
- .4535
Expected prob.:
For Category 20-30
.0378
Prob
for x = 20,
z = -1.68
30  44.3
z=
= -0.99
14.43
.4535
-.3389
Expected prob:
For Category 30 - 40
.1146
Prob
for x = 30,
z = -0.99
40  44.3
z =
= -0.30
14.43
.3389
-.1179
Expected prob:
For Category 40 - 50
.2210
Prob
for x = 40,
z = -0.30
50  44.3
z =
= 0.40
14.43
.1179
+.1554
Expected prob:
.2733
For Category 50 - 60
Prob
60  44.3
= 1.09
14.43
for x = 50,
z = 0.40
.3621
z =
6
-.1554
Expected prob: .2067
Chapter 12: Analysis of Categorical Data
For Category 60 - 70
Prob
70  44.3
= 1.78
14.43
for x = 60, z = 1.09
.4625
z =
7
-.3621
Expected prob: .1004
For Category 70 - 80
Prob
80  44.3
= 2.47
14.43
for x = 70, z = 1.78
.4932
z =
-.4625
Expected prob: .0307
For < 10:
Probability between 10 and the mean, 44.3 = (.0378 + .1145 + .2210
+ .1179) = .4913. Probability < 10 = .5000 - .4913 = .0087
For > 80:
Probability between 80 and the mean, 44.3 = (.0307 + .1004 + .2067 + .1554) =
.4932. Probability > 80 = .5000 - .4932 = .0068
Category
< 10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
> 80
Prob
.0087
.0378
.1146
.2210
.2733
.2067
.1004
.0307
.0068
expected frequency
.0087(129) = 0.99
.0378(129) = 4.88
14.78
28.51
35.26
26.66
12.95
3.96
0.88
Due to the small sizes of expected frequencies, category < 10 is folded into 10-20
and >80 into 70-80.
Chapter 12: Analysis of Categorical Data
8
( f0  fe )2
f0
.003
.041
.008
.213
.103
.672
.964
2.004
Category
fo
fe
10-20
20-30
30-40
40-50
50-60
60-70
70-80
6
14
29
38
25
10
7
5.87
14.78
28.51
32.26
26.66
12.95
4.84
Calculated  2  
( f0  fe )2
= 2.004
fe
df = k - 3 = 7 - 3 = 4,  = .05
2.05,1 = 9.48773
Since the observed 2 = 2.004 > 2.05,4 = 9.48773, the decision is to fail to reject
the null hypothesis. There is not enough evidence to declare that the observed
frequencies are not normally distributed.
12.5
Definition
fo
Exp.Prop.
fe
Happiness
Sales/Profit
Helping Others
Achievement/
Challenge
42
95
27
.39
.12
.18
227(.39)= 88.53
227(.12)= 27.24
40.86
63
227
.31
70.34
( f0  fe )2
f0
24.46
168.55
4.70
0.77
198.48
Ho: The observed frequencies are distributed the same as the expected
frequencies.
Ha: The observed frequencies are not distributed the same as the expected
frequencies.
Observed 2 = 198.48
df = k – 1 = 4 – 1 = 3,
2.05,3 = 7.81473
 = .05
Chapter 12: Analysis of Categorical Data
9
Since the observed 2 = 198.48 > 2.05,3 = 7.81473, the decision is to reject the
null hypothesis.
The observed frequencies for men are not distributed the same as the
expected frequencies which are based on the responses of women.
12.6
Age
fo
10-14
15-19
20-24
25-29
30-34
> 35
22
50
43
29
19
49
212
Prop. from survey
.09
.23
.22
.14
.10
.22
fe
(.09)(212)=19.08
(.23)(212)=48.76
46.64
29.68
21.20
46.64
( f0  fe )2
f0
0.45
0.03
0.28
0.02
0.23
0.12
1.13
Ho: The distribution of observed frequencies is the same as the distribution of
expected frequencies.
Ha: The distribution of observed frequencies is not the same as the distribution of
expected frequencies.
 = .01, df = k - 1 = 6 - 1 = 5
2.01,5 = 15.0863
The observed 2 = 1.13
Since the observed 2 = 1.13 < 2.01,5 = 15.0863, the decision is to fail to reject
the null hypothesis.
There is not enough evidence to declare that the distribution of observed
frequencies is different from the distribution of expected frequencies.
Chapter 12: Analysis of Categorical Data
12.7
Age
10-20
20-30
30-40
40-50
50-60
60-70
x
fo
16
44
61
56
35
19
231
 fM
n

fm
240
1,100
2,135
2,520
1,925
1,235
fm = 9,155
fm2
3,600
27,500
74,725
113,400
105,875
80,275
2
fm = 405,375
9,155
= 39.63
231
 fM 
2
s =
m
15
25
35
45
55
65
10
( fM ) 2
n 1
n

(9,155) 2
231 = 13.6
230
405,375 
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.
For Category 10-20
Prob
10  39.63
= -2.18
.4854
13.6
20  39.63
z =
= -1.44
-.4251
13.6
Expected prob. .0603
z =
For Category 20-30
for x = 20,
z =
z = -1.44
.4251
30  39.63
= -0.71
-.2611
13.6
Expected prob. .1640
For Category 30-40
for x = 30,
z =
Prob
z = -0.71
Prob
.2611
40  39.63
= 0.03
+.0120
13.6
Expected prob. .2731
Chapter 12: Analysis of Categorical Data
For Category 40-50
z =
Prob
50  39.63
= 0.76
13.6
for x = 40,
11
z = 0.03
.2764
-.0120
Expected prob. .2644
For Category 50-60
z =
Prob
60  39.63
= 1.50
13.6
for x = 50,
z = 0.76
.4332
-.2764
Expected prob. .1568
For Category 60-70
z =
Prob
70  39.63
= 2.23
13.6
for x = 60, z = 1.50
.4871
-.4332
Expected prob. .0539
For < 10:
Probability between 10 and the mean = .0603 + .1640 + .2611 = .4854
Probability < 10 = .5000 - .4854 = .0146
For > 70:
Probability between 70 and the mean = .0120 + .2644 + .1568 + .0539 = .4871
Probability > 70 = .5000 - .4871 = .0129
Age
< 10
10-20
20-30
30-40
40-50
Probability
fe
.0146
(.0146)(231) = 3.37
.0603
(.0603)(231) = 13.93
.1640
37.88
.2731
63.09
.2644
61.08
Chapter 12: Analysis of Categorical Data
50-60
60-70
> 70
.1568
.0539
.0129
12
36.22
12.45
2.98
Categories < 10 and > 70 are less than 5.
Collapse the < 10 into 10-20 and > 70 into 60-70.
Age
fo
fe
10-20
20-30
30-40
40-50
50-60
60-70
16
44
61
56
35
19
17.30
37.88
63.09
61.08
36.22
15.43
( f0  fe )2
f0
0.10
0.99
0.07
0.42
0.04
0.83
2.45
df = k - 3 = 6 - 3 = 3,  = .05
2.05,3 = 7.81473
Observed 2 = 2.45
Since the observed 2 < 2.05,3 = 7.81473, the decision is to fail to reject the null
hypothesis.
There is no reason to reject that the observed frequencies are normally
distributed.
12.8
Number
0
1
2
3
4
5
6 or more
=
f
18
28
47
21
16
11
9
f = 150
 f  number  358
150
f
(f) (number)
0
28
94
63
64
55
54
f(number) = 358
= 2.4
Ho: The observed frequencies are Poisson distributed.
Ha: The observed frequencies are not Poisson distributed.
Chapter 12: Analysis of Categorical Data
Number
0
1
2
3
4
5
6 or more
Probability
.0907
.2177
.2613
.2090
.1254
.0602
.0358
fo
fe
18
28
47
21
16
11
9
13.61
32.66
39.66
31.35
18.81
9.03
5.36
13
fe
(.0907)(150 = 13.61
(.2177)(150) = 32.66
39.20
31.35
18.81
9.03
5.36
( f0  fe )2
f0
1.42
0.66
1.55
3.42
0.42
0.43
2.47
10.37
The observed 2 = 10.27
 = .01, df = k – 2 = 7 – 2 = 5, 2.01,5 = 15.0863
Since the observed 2 = 10.27 < 2.01,5 = 15.0863, the decision is to fail to reject
the null hypothesis.
There is not enough evidence to reject the claim that the observed
frequencies are Poisson distributed.
12.9
H0: p = .28
Ha: p  .28
n = 270
x = 62
fo
Spend More
Don't Spend More
Total
fe
( f0  fe )2
f0
62
270(.28) = 75.6
2.44656
208
270(.72) = 194.4
0.95144
270
270.0
3.39800
Chapter 12: Analysis of Categorical Data
14
The observed value of 2 is 3.398
 = .05 and /2 = .025
df = k - 1 = 2 - 1 = 1
2.025,1 = 5.02389
Since the observed 2 = 3.398 < 2.025,1 = 5.02389, the decision is to fail to
reject the null hypothesis.
12.10
H0: p = .30
Ha: p  .30
n = 180 x= 42
42
180(.30) = 54
( f0  fe )2
f0
2.6666
138
180(.70) = 126
1.1429
180
180
3.8095
f0
Provide
Don't Provide
Total
fe
The observed value of 2 is 3.8095
 = .05 and /2 = .025
df = k - 1 = 2 - 1 = 1
2.025,1 = 5.02389
Since the observed 2 = 3.8095 < 2.025,1 = 5.02389, the decision is to fail to
reject the null hypothesis.
Chapter 12: Analysis of Categorical Data
15
12.11
Variable
One
Variable Two
203
326
68
110
271
436
529
178
707
Ho: Variable One is independent of Variable Two.
Ha: Variable One is not independent of Variable Two.
e11 =
(529)( 271)
= 202.77
707
e12 =
(529)( 436)
= 326.23
707
e21 =
(271)(178)
= 68.23
707
e22 =
(436)(178)
= 109.77
707
2 =
Variable Two
Variable (202.77) (326.23)
One
203
326
(68.23) (109.77)
68
110
529
271
707
436
178
(203  202.77) 2
(326  326.23) 2
(68  6.23) 2
(110  109.77)2
+
+
+
=
202.77
326.23
68.23
109.77
.00 + .00 + .00 + .00 = 0.00
 = .05, df = (c-1)(r-1) = (2-1)(2-1) = 1
2.05,1 = 3.84146
Since the observed 2 = 0.00 < 2.05,1 = 3.84146, the decision is to fail to reject
the null hypothesis.
Variable One is independent of Variable Two.
Chapter 12: Analysis of Categorical Data
16
12.12
Variable
One
24
93
117
Variable
Two
13
47
59
187
58
244
142
583
72
302
725
234
Ho: Variable One is independent of Variable Two.
Ha: Variable One is not independent of Variable Two.
e11 =
(142)(117)
= 22.92
725
e12 =
(142)(72)
= 14.10
725
e13 =
(142)( 234)
= 45.83
725
e14 =
(142)(302)
= 59.15
725
e21 =
(583)(117)
= 94.08
725
e22 =
(583)(72)
= 57.90
725
e23 =
(583)( 234)
= 188.17
725
e24 =
(583)(302)
= 242.85
725
Variable
Two
Variable (22.92) (14.10) (45.83) (59.15)
One
24
13
47
58
(94.08) (57.90) (188.17) (242.85)
93
59
187
244
117
72
234
302
142
583
725
(24  22.92) 2
(13  14.10) 2
(47  45.83) 2
(58  59.15)2
 =
+
+
+
+
22.92
14.10
45.83
59.15
2
(59  57.90) 2
(244  242.85) 2
(93  94.08) 2
(188  188.17)2
+
+
+
=
57.90
242.85
94.08
188.17
.05 + .09 + .03 + .02 + .01 + .02 + .01 + .01 = 0.24
Chapter 12: Analysis of Categorical Data
17
 = .01, df = (c-1)(r-1) = (4-1)(2-1) = 3
2.01,3 = 11.3449
Since the observed 2 = 0.24 < 2.01,3 = 11.3449, the decision is to fail to
reject the null hypothesis.
Variable One is independent of Variable Two.
12.13
Number
of
Children
0
1
2 or 3
>3
Social Class
Lower Middle Upper
7
18
6
9
38
23
34
97
58
47
31
30
97
184
117
31
70
189
108
398
Ho: Social Class is independent of Number of Children.
Ha: Social Class is not independent of Number of Children.
e11 =
(31)(97)
= 7.56
398
e31 =
(189)(97)
= 46.06
398
e12 =
(31)(184)
= 14.3
398
e32 =
(189)(184)
= 87.38
398
e13 =
(31)(117 )
= 9.11
398
e33 =
(189)(117)
= 55.56
398
e21 =
(70)(97)
= 17.06
398
e41 =
(108)(97)
= 26.32
398
e22 =
(70)(184)
= 32.36
398
e42 =
(108)(184)
= 49.93
398
e23 =
(70)(117)
= 20.58
398
e43 =
(108)(117)
= 31.75
398
Chapter 12: Analysis of Categorical Data
0
Number
of
Children
1
2 or 3
>3
18
Social Class
Lower Middle Upper
(7.56) (14.33) (9.11)
7
18
6
(17.06) (32.36) (20.58)
9
38
23
(46.06) (87.38) (55.56)
34
97
58
(26.32) (49.93) (31.75)
47
31
30
97
184
117
31
70
189
108
398
(18  14.33) 2
(9  17.06) 2
(7  7.56) 2
(6  9.11) 2
 =
+
+
+
+
14.33
17.06
7.56
9.11
2
(23  20.58) 2
(34  46.06) 2
(38  32.36)2
(97  87.38)2
+
+
+
+
46.06
32.36
87.38
20.58
(47  26.32) 2
(58  55.56)2
(31  49.93)2
(30  31.75)2
+
+
+
=
26.32
55.56
49.93
31.75
.04 + .94 + 1.06 + 3.81 + .98 + .28 + 3.16 + 1.06 + .11 + 16.25 +
7.18 + .10 = 34.97
 = .05, df = (c-1)(r-1) = (3-1)(4-1) = 6
2.05,6 = 12.5916
Since the observed 2 = 34.97 > 2.05,6 = 12.5916, the decision is to reject the
null hypothesis.
Number of children is not independent of social class.
Chapter 12: Analysis of Categorical Data
19
12.14
Region
NE
S
W
Type of Music Preferred
Rock
R&B Coun Clssic
140
32
5
18
134
41
52
8
154
27
8
13
428
100
65
39
195
235
202
632
Ho: Type of music preferred is independent of region.
Ha: Type of music preferred is not independent of region.
e11 =
(195)( 428)
= 132.6
632
e23 =
(235)(65)
= 24.17
632
e12 =
(195)(100)
= 30.85
632
e24 =
(235)(39)
= 14.50
632
e13 =
(195)( 65)
= 20.06
632
e31 =
(202)( 428)
= 136.80
632
e14 =
(195)(39)
= 12.03
632
e32 =
(202)(100)
= 31.96
632
e21 =
(235)( 428)
= 159.15
632
e33 =
(202)(65)
= 20.78
632
e22 =
(235)(100)
= 37.18
632
e34 =
(202)(39)
= 12.47
632
NE
Region
S
W
Type of Music Preferred
Rock
R&B Coun
Clssic
(132.06) (30.85) (20.06) (12.03)
140
32
5
18
(159.15) (37.18) (24.17) (14.50)
134
41
52
8
(136.80) (31.96) (20.78) (12.47)
154
27
8
13
428
100
65
39
195
235
202
632
Chapter 12: Analysis of Categorical Data
2 =
20
(141  132.06) 2
(5  20.06) 2
(18  12.03) 2
(32  30.85)2
+
+
+
+
132.06
20.06
12.03
30.85
(134  159.15) 2
(52  24.17) 2
(41  37.18)2
(8  14.50) 2
+
+
+
+
159.15
24.17
37.18
14.50
(27  31.96) 2
(8  20.78) 2
(13  12.47) 2
(154  136.80)2
+
+
+
=
31.96
20.78
12.47
136.80
.48 + .04 + 11.31 + 2.96 + 3.97 + .39 + 32.04 + 2.91 + 2.16 + .77 +
7.86 + .02 = 64.91
 = .01, df = (c-1)(r-1) = (4-1)(3-1) = 6
2.01,6 = 16.8119
Since the observed 2 = 64.91 > 2.01,6 = 16.8119, the decision is to
reject the null hypothesis.
Type of music preferred is not independent of region of the country.
12.15
Transportation Mode
Air
Train Truck
Industry Publishing
32
12
41
Comp.Hard.
5
6
24
37
18
65
85
35
120
H0: Transportation Mode is independent of Industry.
Ha: Transportation Mode is not independent of Industry.
e11 =
(85)(37)
= 26.21
120
e21 =
(35)(37)
= 10.79
120
e12 =
(85)(18)
= 12.75
120
e22 =
(35)(18)
= 5.25
120
e13 =
(85)(65)
= 46.04
120
e23 =
(35)(65)
= 18.96
120
Chapter 12: Analysis of Categorical Data
21
Transportation Mode
Air
Train Truck
Industry Publishing
(26.21) (12.75) (46.04)
32
12
41
Comp.Hard. (10.79) (5.25) (18.96)
5
6
24
37
18
65
85
35
120
(12  12.75) 2
(32  26.21) 2
(41  46.04)2
 =
+
+
+
12.75
26.21
46.04
2
(5  10.79) 2
(6  5.25) 2
(24  18.96) 2
+
+
=
10.79
5.25
18.96
1.28 + .04 + .55 + 3.11 + .11 + 1.34 = 6.43
 = .05, df = (c-1)(r-1) = (3-1)(2-1) = 2
2.05,2 = 5.99147
Since the observed 2 = 6.431 > 2.05,2 = 5.99147, the decision is to
reject the null hypothesis.
Transportation mode is not independent of industry.
12.16
Number of
Stories
1
2
Number of Bedrooms
<2
3
>4
116
101
57
90
325
160
206
426
217
274
575
849
H0: Number of Stories is independent of number of bedrooms.
Ha: Number of Stories is not independent of number of bedrooms.
e11 =
(274)( 206)
= 66.48
849
e21 =
(575)( 206)
= 139.52
849
e12 =
(274)( 426)
= 137.48
849
e22 =
(575)( 426)
= 288.52
849
Chapter 12: Analysis of Categorical Data
22
e13 =
(274)( 217)
= 70.03
849
2 =
(90  139.52) 2
(101  137.48) 2
(57  70.03) 2
(90  139.52) 2
+
+
+
+
139.52
137.48
70.03
139.52
e23 =
(575)( 217)
= 146.97
849
(325  288.52) 2
(160  146.97)2
+
=
288.52
146.97
2 = 36.89 + 9.68 + 2.42 + 17.58 + 4.61 + 1.16 = 72.34
 = .10
df = (c-1)(r-1) = (3-1)(2-1) = 2
2.10,2 = 4.60517
Since the observed 2 = 72.34 > 2.10,2 = 4.60517, the decision is to
reject the null hypothesis.
Number of stories is not independent of number of bedrooms.
12.17
Type
of
Store
Mexican Citizens
Yes
No
Dept.
24
17
Disc.
20
15
Hard.
11
19
Shoe
32
28
87
79
41
35
30
60
166
Ho: Citizenship is independent of store type
Ha: Citizenship is not independent of store type
e11 =
(41)(87)
= 21.49
166
e31 =
(30)(87)
= 15.72
166
e12 =
(41)(79)
= 19.51
166
e32 =
(30)(79)
= 14.28
166
e21 =
(35)(87)
= 18.34
166
e41 =
(60)(87)
= 31.45
166
Chapter 12: Analysis of Categorical Data
e22 =
(35)(79)
= 16.66
166
Type
of
Store
e42 =
23
(60)(79)
= 28.55
166
Mexican Citizens
Yes
No
Dept. (21.49) (19.51)
24
17
Disc.
(18.34) (16.66)
20
15
Hard. (15.72) (14.28)
11
19
Shoe
(31.45) (28.55)
32
28
87
79
41
35
30
60
166
(24  21.49) 2
(17  19.51) 2
(15  16.66) 2
(20  18.34) 2
 =
+
+
+
+
21.49
19.51
16.66
18.34
2
(11  15.72) 2
(19  14.28) 2
(28  28.55) 2
(32  31.45)2
+
+
+
=
15.72
14.28
28.55
31.45
.29 + .32 + .15 + .17 + 1.42 + 1.56 + .01 + .01 = 3.93
 = .05,
df = (c-1)(r-1) = (2-1)(4-1) = 3
2.05,3 = 7.81473
Since the observed 2 = 3.93 < 2.05,3 = 7.81473, the decision is to fail to
reject the null hypothesis.
Citizenship is independent of type of store.
Chapter 12: Analysis of Categorical Data
24
12.18  = .01, k = 7, df = 6
H0: The observed distribution is the same as the expected distribution
Ha: The observed distribution is not the same as the expected distribution
Use:
( f0  fe )2
 
fe
2
critical 2.01,7 = 18.4753
fo
fe
(f0-fe)2
214
235
279
281
264
254
211
206
232
268
284
268
232
206
64
9
121
9
16
484
25
2  
( f0  fe )2
f0
0.311
0.039
0.451
0.032
0.060
2.086
0.121
3.100
( f0  fe )2
= 3.100
fe
Since the observed value of 2 = 3.1 < 2.01,7 = 18.4753, the decision is to fail to
reject the null hypothesis. The observed distribution is not different from the
expected distribution.
12.19
Variable 1
12
8
7
27
Variable 2
23
17
11
51
e11 = 11.00
e12 = 20.85
e13 = 24.12
e21 = 8.87
e22 = 16.75
e23 = 19.38
21
20
18
59
56
45
36
137
Chapter 12: Analysis of Categorical Data
e31 = 7.09
2 =
e32 = 13.40
25
e33 = 15.50
(23  20.85) 2
(12  11.04)2
(21  24.12)2
(8  8.87) 2
+
+
+
+
20.85
11.04
24.12
8.87
(11  13.40) 2
(17  16.75) 2
(20  19.38)2
(7  7.09) 2
+
+
+
+
13.40
16.75
19.38
7.09
(18  15.50) 2
=
15.50
.084 + .222 + .403 + .085 + .004 + .020 + .001 + .430 + .402 = 1.652
df = (c-1)(r-1) = (2)(2) = 4
 = .05
2.05,4 = 9.48773
Since the observed value of 2 = 1.652 < 2.05,4 = 9.48773, the decision is to fail
to reject the null hypothesis.
12.20
Customer Industrial
Retail
NE
230
185
415
Location
W
S
115
68
143
89
258
157
413
417
830
e11 =
(413)( 415)
= 206.5
830
e21 =
(417)( 415)
= 208.5
830
e12 =
(413)( 258)
= 128.38
830
e22 =
(417)( 258)
= 129.62
830
e13 =
(413)(157)
= 78.12
830
e23 =
(417)(157)
= 78.88
830
Chapter 12: Analysis of Categorical Data
26
Location
NE
W
S
Customer Industrial (206.5) (128.38) (78.12)
230
115
68
Retail
(208.5) (129.62) (78.88)
185
143
89
415
258
157
2 =
413
417
830
(115  128.38) 2
(68  78.12) 2
(230  206.5) 2
+
+
+
128.38
78.12
206.5
(185  208.5) 2
(143  129.62) 2
(89  78.88)2
+
+
=
208.5
129.62
78.88
2.67 + 1.39 + 1.31 + 2.65 + 1.38 + 1.30 = 10.70
 = .10 and df = (c - 1)(r - 1) = (3 - 1)(2 - 1) = 2
2.10,2 = 4.60517
Since the observed 2 = 10.70 > 2.10,2 = 4.60517, the decision is to reject the
null hypothesis.
Type of customer is not independent of geographic region.
12.21 Cookie Type
fo
Chocolate Chip
189
Peanut Butter
168
Cheese Cracker
155
Lemon Flavored
161
Chocolate Mint
216
Vanilla Filled
165
fo = 1,054
Ho:
Ha:
Cookie Sales is uniformly distributed across kind of cookie.
Cookie Sales is not uniformly distributed across kind of cookie.
If cookie sales are uniformly distributed, then fe =
f
0
no.kinds

1,054
= 175.67
6
Chapter 12: Analysis of Categorical Data
fo
fe
189
168
155
161
216
165
175.67
175.67
175.67
175.67
175.67
175.67
27
( f0  fe )2
f0
1.01
0.33
2.43
1.23
9.26
0.65
14.91
The observed 2 = 14.91
 = .05
df = k - 1 = 6 - 1 = 5
2.05,5 = 11.0705
Since the observed 2 = 14.91 > 2.05,5 = 11.0705, the decision is to reject the
null hypothesis.
Cookie Sales is not uniformly distributed by kind of cookie.
12.22
Bought
Car
Y
N
Gender
M
F
207
65
811
984
1,018
1,049
272
1,795
2,067
Ho: Purchasing a car or not is independent of gender.
Ha: Purchasing a car or not is not independent of gender.
(272)(1,018)
= 133.96
2,067
(1,795)(1,018)
e21 =
= 884.04
2,067
e11 =
(27)(1,049)
= 138.04
2,067
(1,795)(1,049)
e22 =
= 910.96
2,067
e12 =
Chapter 12: Analysis of Categorical Data
Bought
Car
Gender
M
F
(133.96) (138.04)
207
65
272
(884.04) (910.96)
811
984
1,795
1,018
1,049
2,067
Y
N
2 =
28
(207  133.96) 2
(65  138.04) 2
(811  884.04) 2
+
+
+
133.96
138.04
884.04
(984  910.96) 2
=
910.96
 = .05
39.82 + 38.65 + 6.03 + 5.86 = 90.36
df = (c-1)(r-1) = (2-1)(2-1) = 1
2.05,1 = 3.841
Since the observed 2 = 90.36 > 2.05,1 = 3.841, the decision is to reject the
null hypothesis.
Purchasing a car is not independent of gender.
12.23
Arrivals
0
1
2
3
4
5
6
 =
fo
26
40
57
32
17
12
8
fo = 192
 ( f )(arrivals)  426
192
f
0
(fo)(Arrivals)
0
40
114
96
68
60
48
(fo)(arrivals) = 426
= 2.2
0
Ho: The observed frequencies are Poisson distributed.
Ha: The observed frequencies are not Poisson distributed.
Chapter 12: Analysis of Categorical Data
Arrivals
0
1
2
3
4
5
6
Probability
.1108
.2438
.2681
.1966
.1082
.0476
.0249
29
fe
(.1108)(192) = 21.27
(.2438)(192) = 46.81
51.48
37.75
20.77
9.14
4.78
fo
fe
26
40
57
32
17
12
8
21.27
46.81
51.48
37.75
20.77
9.14
4.78
( f0  fe )2
f0
1.05
2.18
0.59
0.88
0.68
0.89
2.17
8.44
Observed 2 = 8.44
 = .05
df = k - 2 = 7 - 2 = 5
2.05,5 = 11.0705
Since the observed 2 = 8.44 < 2.05,5 = 11.0705, the decision is to fail to reject
the null hypothesis. There is not enough evidence to reject the claim that the
observed frequency of arrivals is Poisson distributed.
Chapter 12: Analysis of Categorical Data
30
12.24 Ho: The distribution of observed frequencies is the same as the
distribution of expected frequencies.
Ha: The distribution of observed frequencies is not the same as the distribution of
expected frequencies.
Soft Drink
fo
Classic Coke
361
Pepsi
272
Diet Coke
192
Mt. Dew
121
Dr. Pepper
94
Sprite
102
Others
584
fo = 1,726
proportions
fe
.206 (.206)(1726) = 355.56
.145 (.145)(1726) = 250.27
.085
146.71
.063
108.74
.059
101.83
.062
107.01
.380
655.86
( f0  fe )2
f0
0.08
1.89
13.98
1.38
0.60
0.23
7.87
26.03
Calculated 2 = 26.03
 = .05
df = k - 1 = 7 - 1 = 6
2.05,6 = 12.5916
Since the observed 2 = 26.03 > 2.05,6 = 12.5916, the decision is to reject the
null hypothesis.
The observed frequencies are not distributed the same as the expected frequencies
from the national poll.
12.25
Position
Years
0-3
4-8
>8
Systems
Manager Programmer Operator Analyst
6
37
11
13
28
16
23
24
47
10
12
19
81
63
46
56
67
91
88
246
Chapter 12: Analysis of Categorical Data
31
e11 =
(67)(81)
= 22.06
246
e23 =
(91)( 46)
= 17.02
246
e12 =
(67)(63)
= 17.16
246
e24 =
(91)(56)
= 20.72
246
e13 =
(67)( 46)
= 12.53
246
e31 =
(88)(81)
= 28.98
246
e14 =
(67)(56)
= 15.25
246
e32 =
(88)(63)
= 22.54
246
e21 =
(91)(81)
= 29.96
246
e33 =
(88)( 46)
= 16.46
246
e22 =
(91)( 63)
= 23.30
246
e34 =
(88)(56)
= 20.03
246
Position
0-3
Years
4-8
>8
Systems
Manager Programmer Operator Analyst
(22.06)
(17.16)
(12.53)
(15.25)
6
37
11
13
(29.96)
(23.30)
(17.02)
(20.72)
28
16
23
24
(28.98)
(22.54)
(16.46)
(20.03)
47
10
12
19
81
63
46
56
67
91
88
246
(6  22.06) 2
(11  12.53) 2
(13  15.25) 2
(37  17.16)2
 =
+
+
+
+
22.06
12.53
15.25
17.16
2
(28  29.96) 2
(24  20.72) 2
(16  23.30)2
(23  17.02)2
+
+
+
+
29.96
20.72
23.30
17.02
(47  28.98) 2
(10  22.54)2
(12  16.46)2
(19  20.03)2
+
+
+
=
28.98
22.54
16.46
20.03
11.69 + 22.94 + .19 + .33 + .13 + 2.29 + 2.1 + .52 + 11.2 + 6.98 +
1.21 + .05 = 59.63
Chapter 12: Analysis of Categorical Data
 = .01
32
df = (c-1)(r-1) = (4-1)(3-1) = 6
2.01,6 = 16.8119
Since the observed 2 = 59.63 > 2.01,6 = 16.8119, the decision is to reject the
null hypothesis. Position is not independent of number of years of experience.
12.26 H0: p = .43
Ha: p  .43
n = 315
x = 120
 =.05
/2 = .025
fo
fe
( f0  fe )2
f0
More Work,
More Business
120
(.43)(315) = 135.45
1.76
Others
195
(.57)(315) = 179.55
1.33
Total
315
315.00
3.09
The calculated value of 2 is 3.09
 = .05 and /2 = .025
df = k - 1 = 2 - 1 = 1
2.025,1 = 5.02389
Since 2 = 3.09 < 2.025,1 = 5.02389, the decision is to fail to reject the null
hypothesis.
12.27
Number
of
Children
0
1
2
>3
Type of College or University
Community
Large
Small
College
University College
25
178
31
49
141
12
31
54
8
22
14
6
127
387
57
234
202
93
42
571
Ho: Number of Children is independent of Type of College or University.
Ha: Number of Children is not independent of Type of College or University.
Chapter 12: Analysis of Categorical Data
33
e11 =
(234)(127)
= 52.05
571
e31 =
(93)(127 )
= 20.68
571
e12 =
(234)(387)
= 158.60
571
e32 =
(193)(387)
= 63.03
571
e13 =
(234)(57)
= 23.36
571
e33 =
(93)(57)
= 9.28
571
e21 =
(202)(127)
= 44.93
571
e41 =
(42)(127)
= 9.34
571
e22 =
(202)(387)
= 136.91
571
e42 =
(42)(387)
= 28.47
571
e23 =
(202)(57)
= 20.16
571
e43 =
(42)(57)
= 4.19
571
Number
of
Children
0
1
2
>3
2 =
Type of College or University
Community
Large
Small
College
University College
(52.05)
(158.60) (23.36)
25
178
31
(44.93)
(136.91) (20.16)
49
141
12
(20.68)
(63.03)
(9.28)
31
54
8
(9.34)
(28.47)
(4.19)
22
14
6
127
387
57
234
202
93
42
571
(25  52.05) 2
(178  158.6) 2
(49  44.93) 2
(31  23.36) 2
+
+
+
+
52.05
158.6
44.93
23.36
(141  136.91) 2
(12  20.16)2
(31  20.68)2
(54  63.03)2
+
+
+
+
136.91
20.16
20.68
63.03
(22  9.34) 2
(8  9.28) 2
(14  28.47)2
(6  4.19) 2
+
+
+
=
9.34
9.28
28.47
4.19
Chapter 12: Analysis of Categorical Data
34
14.06 + 2.37 + 2.50 + 0.37 + 0.12 + 3.30 + 5.15 + 1.29 + 0.18 +
17.16 + 7.35 + 0.78 = 54.63
 = .05,
df= (c - 1)(r - 1) = (3 - 1)(4 - 1) = 6
2.05,6 = 12.5916
Since the observed 2 = 54.63 > 2.05,6 = 12.5916, the decision is to reject the
null hypothesis.
Number of children is not independent of type of College or University.
12.28 The observed chi-square is 30.18 with a p-value of .0000043. The chi-square
goodness-of-fit test indicates that there is a significant difference between the
observed frequencies and the expected frequencies. The distribution of responses
to the question are not the same for adults between 21 and 30 years of age as they
are to others. Marketing and sales people might reorient their 21 to 30 year old
efforts away from home improvement and pay more attention to leisure
travel/vacation, clothing, and home entertainment.
12.29 The observed chi-square value for this test of independence is 5.366. The
associated p-value of .252 indicates failure to reject the null hypothesis. There is
not enough evidence here to say that color choice is dependent upon gender.
Automobile marketing people do not have to worry about which colors especially
appeal to men or to women because car color is independent of gender. In
addition, design and production people can determine car color quotas based on
other variables.