12 ch ken black solution

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: chi-
square 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

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 Test of Independence
Chi-Square Distribution Contingency Analysis
Chi-Square Goodness-of-Fit Test Contingency Table

SOLUTIONS TO CHAPTER 16
12.1 f0
0
2
0 )(
f
ff e−
fe
53 68 3.309
37 42 0.595
32 33 0.030
28 22 1.636
18 10 6.400
15 8 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 ∑
−
=
e
e
f
ff 2
02 )(
χ = 18.095
df = k - 1 = 6 - 1 = 5, α = .05
χ2
.05,5 = 11.07
Since the observed χ2
= 18.095 > χ2
.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 fe
0
2
0 )(
f
ff e−
19 18 0.056
17 18 0.056
14 18 0.889
18 18 0.000
19 18 0.056
21 18 0.500
18 18 0.000
18 18 0.000
Σfo = 144 Σfe = 144 1.557

Ho: The observed frequencies are uniformly distributed.
Ha: The observed frequencies are not uniformly distributed.
8
1440
==
∑
k
f
x = 18
In this uniform distribution, each fe = 18
df = k – 1 = 8 – 1 = 7, α = .01
χ2
.01,7 = 18.4753
Observed ∑
−
=
e
e
f
ff 2
02 )(
χ = 1.557
= 1.557 < χ2
.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 (Number)(f0)
0 28 0
1 17 17
2 11 22
3 5 15
54
Ho: The frequency distribution is Poisson.
Ha: The frequency distribution is not Poisson.
λ =
61
54
=0.9
Expected Expected
Number Probability Frequency
0 .4066 24.803
1 .3659 22.312
2 .1647 10.047
> 3 .0628 3.831
Since fe for > 3 is less than 5, collapse categories 2 and >3:

Number fo fe
0
2
0 )(
f
ff e−
0 28 24.803 0.412
1 17 22.312 1.265
>2 16 13.878 0.324
61 60.993 2.001
df = k - 2 = 3 - 2 = 1, = .05
χ2
.05,1 = 3.84146
Calculated ∑
−
=
e
e
f
ff 2
02 )(
χ = 2.001
= 2.001 < χ2
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) Midpt. fm fm2
10-20 6 15 90 1,350
20-30 14 25 350 8,750
30-40 29 35 1,015 35,525
40-50 38 45 1,710 76,950
50-60 25 55 1,375 75,625
60-70 10 65 650 42,250
70-80 7 75 525 39,375
n = Σf = 129 Σfm = 5,715 Σfm2
= 279,825
129
715,5
==
∑
∑
f
fm
x = 44.3
s =
128
129
)715,5(
825,279
1
)( 22
2
−
=
−
−∑ ∑
n
n
fM
fM
= 14.43
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.

For Category 10 - 20 Prob
z =
43.14
3.4410 −
= -2.38 .4913
z =
43.14
3.4420 −
= -1.68 - .4535
Expected prob.: .0378
For Category 20-30 Prob
for x = 20, z = -1.68 .4535
z =
43.14
3.4430 −
= -0.99 -.3389
Expected prob: .1146
for x = 30, z = -0.99 .3389
z =
43.14
3.4440 −
= -0.30 -.1179
for x = 40, z = -0.30 .1179
z =
43.14
3.4450 −
= 0.40 +.1554
z =
43.14
3.4460 −
= 1.09 .3621
for x = 50, z = 0.40 -.1554

z =
43.14
3.4470 −
= 1.78 .4625
for x = 60, z = 1.09 -.3621
z =
43.14
3.4480 −
= 2.47 .4932
for x = 70, z = 1.78 -.4625
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 Prob expected frequency
< 10 .0087 .0087(129) = 0.99
10-20 .0378 .0378(129) = 4.88
20-30 .1146 14.78
30-40 .2210 28.51
40-50 .2733 35.26
50-60 .2067 26.66
60-70 .1004 12.95
70-80 .0307 3.96
> 80 .0068 0.88
Due to the small sizes of expected frequencies, category < 10 is folded into 10-20
and >80 into 70-80.

Category fo fe
0
2
0 )(
f
ff e−
10-20 6 5.87 .003
20-30 14 14.78 .041
30-40 29 28.51 .008
40-50 38 32.26 .213
50-60 25 26.66 .103
60-70 10 12.95 .672
70-80 7 4.84 .964
2.004
Calculated ∑
−
=
e
e
f
ff 2
02 )(
χ = 2.004
df = k - 3 = 7 - 3 = 4, α = .05
χ2
.05,1 = 9.48773
= 2.004 > χ2
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
0
2
0 )(
f
ff e−
Happiness 42 .39 227(.39)= 88.53 24.46
Sales/Profit 95 .12 227(.12)= 27.24 168.55
Helping Others 27 .18 40.86 4.70
Achievement/
Challenge 63 .31 70.34 0.77
227 198.48
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 χ2
= 198.48
df = k – 1 = 4 – 1 = 3, α = .05
χ2
.05,3 = 7.81473

= 198.48 > χ2
.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 Prop. from survey fe
0
2
0 )(
f
ff e−
10-14 22 .09 (.09)(212)=19.08 0.45
15-19 50 .23 (.23)(212)=48.76 0.03
20-24 43 .22 46.64 0.28
25-29 29 .14 29.68 0.02
30-34 19 .10 21.20 0.23
> 35 49 .22 46.64 0.12
212 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
α = .01, df = k - 1 = 6 - 1 = 5
χ2
.01,5 = 15.0863
The observed χ2
= 1.13
= 1.13 < χ2
There is not enough evidence to declare that the distribution of observed
frequencies is different from the distribution of expected frequencies.

12.7 Age fo m fm fm2
10-20 16 15 240 3,600
20-30 44 25 1,100 27,500
30-40 61 35 2,135 74,725
40-50 56 45 2,520 113,400
50-60 35 55 1,925 105,875
60-70 19 65 1,235 80,275
231 Σfm = 9,155 Σfm2
= 405,375
231
155,9
==
∑
n
fM
x = 39.63
s =
230
231
)155,9(
375,405
1
)( 22
2
−
=
−
−∑ ∑
n
n
fM
fM
= 13.6
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.
z =
6.13
63.3910 −
= -2.18 .4854
z =
6.13
63.3920 −
= -1.44 -.4251
Expected prob. .0603
for x = 20, z = -1.44 .4251
z =
6.13
63.3930 −
= -0.71 -.2611
for x = 30, z = -0.71 .2611
z =
6.13
63.3940 −
= 0.03 +.0120

z =
6.13
63.3950 −
= 0.76 .2764
for x = 40, z = 0.03 -.0120
z =
6.13
63.3960 −
= 1.50 .4332
for x = 50, z = 0.76 -.2764
z =
6.13
63.3970 −
= 2.23 .4871
for x = 60, z = 1.50 -.4332
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 Probability fe
< 10 .0146 (.0146)(231) = 3.37
10-20 .0603 (.0603)(231) = 13.93
20-30 .1640 37.88
30-40 .2731 63.09
40-50 .2644 61.08

50-60 .1568 36.22
60-70 .0539 12.45
> 70 .0129 2.98
Categories < 10 and > 70 are less than 5.
Collapse the < 10 into 10-20 and > 70 into 60-70.
Age fo fe
0
2
0 )(
f
ff e−
10-20 16 17.30 0.10
20-30 44 37.88 0.99
30-40 61 63.09 0.07
40-50 56 61.08 0.42
50-60 35 36.22 0.04
60-70 19 15.43 0.83
2.45
df = k - 3 = 6 - 3 = 3, α = .05
χ2
.05,3 = 7.81473
Observed χ2
= 2.45
< χ2
.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 f (f)⋅ (number)
0 18 0
1 28 28
2 47 94
3 21 63
4 16 64
5 11 55
6 or more 9 54
Σf = 150 Σf⋅(number) = 358
λ =
150
358
=
⋅
∑
∑
f
numberf
= 2.4
Ho: The observed frequencies are Poisson distributed.
Ha: The observed frequencies are not Poisson distributed.

Number Probability fe
0 .0907 (.0907)(150 = 13.61
1 .2177 (.2177)(150) = 32.66
2 .2613 39.20
3 .2090 31.35
4 .1254 18.81
5 .0602 9.03
6 or more .0358 5.36
fo fe
0
2
0 )(
f
ff e−
18 13.61 1.42
28 32.66 0.66
47 39.66 1.55
21 31.35 3.42
16 18.81 0.42
11 9.03 0.43
9 5.36 2.47
10.37
The observed χ2
= 10.27
α = .01, df = k – 2 = 7 – 2 = 5, χ2
.01,5 = 15.0863
= 10.27 < χ2
There is not enough evidence to reject the claim that the observed
frequencies are Poisson distributed.
12.9 H0: p = .28 n = 270 x = 62
Ha: p ≠ .28
fo fe
0
2
0 )(
f
ff e−
Spend More 62 270(.28) = 75.6 2.44656
Don't Spend More 208 270(.72) = 194.4 0.95144
Total 270 270.0 3.39800

The observed value of χ2
is 3.398
α = .05 and α/2 = .025 df = k - 1 = 2 - 1 = 1
χ2
.025,1 = 5.02389
= 3.398 < χ2
.025,1 = 5.02389, the decision is to fail to
reject the null hypothesis.
12.10 H0: p = .30 n = 180 x= 42
Ha: p ≠ .30
f0 fe
0
2
0 )(
f
ff e−
Provide 42 180(.30) = 54 2.6666
Don't Provide 138 180(.70) = 126 1.1429
Total 180 180 3.8095
The observed value of χ2
is 3.8095
α = .05 and α/2 = .025 df = k - 1 = 2 - 1 = 1
χ2
.025,1 = 5.02389
= 3.8095 < χ2

12.11
Variable Two
Variable
One
203 326 529
17868 110
271 436 707
Ho: Variable One is independent of Variable Two.
Ha: Variable One is not independent of Variable Two.
e11 =
707
)271)(529(
= 202.77 e12 =
707
)436)(529(
= 326.23
e21 =
707
)178)(271(
= 68.23 e22 =
707
)178)(436(
= 109.77
Variable Two
Variable
One
(202.77)
203
(326.23)
326 529
178
(68.23)
68
(109.77)
110
271 436 707
χ2
=
77.202
)77.202203( 2
−
+
23.326
)23.326326( 2
−
+
23.68
)23.668( 2
−
+
77.109
)77.109110( 2
−
=
.00 + .00 + .00 + .00 = 0.00
α = .05, df = (c-1)(r-1) = (2-1)(2-1) = 1
χ2
.05,1 = 3.84146
= 0.00 < χ2
Variable One is independent of Variable Two.

12.12
Variable
Two
Variable
One
24 13 47 58 142
58393 59 187 244
117 72 234 302 725
Ho: Variable One is independent of Variable Two.
Ha: Variable One is not independent of Variable Two.
e11 =
725
)117)(142(
= 22.92 e12 =
725
)72)(142(
= 14.10
e13 =
725
)234)(142(
= 45.83 e14 =
725
)302)(142(
= 59.15
e21 =
725
)117)(583(
= 94.08 e22 =
725
)72)(583(
= 57.90
e23 =
725
)234)(583(
= 188.17 e24 =
725
)302)(583(
= 242.85
Variable
Two
Variable
One
(22.92)
24
(14.10)
13
(45.83)
47
(59.15)
58 142
583
(94.08)
93
(57.90)
59
(188.17)
187
(242.85)
244
117 72 234 302 725
χ2
=
92.22
)92.2224( 2
−
+
10.14
)10.1413( 2
−
+
83.45
)83.4547( 2
−
+
15.59
)15.5958( 2
−
+
08.94
)08.9493( 2
−
+
90.57
)90.5759( 2
−
+
17.188
)17.188188( 2
−
+
85.242
)85.242244( 2
−
=
.05 + .09 + .03 + .02 + .01 + .02 + .01 + .01 = 0.24

α = .01, df = (c-1)(r-1) = (4-1)(2-1) = 3
χ2
.01,3 = 11.3449
= 0.24 < χ2
Variable One is independent of Variable Two.
12.13
Social Class
Number
of
Children
Lower Middle Upper
0
1
2 or 3
>3
7 18 6 31
70
189
108
9 38 23
34 97 58
47 31 30
97 184 117 398
Ho: Social Class is independent of Number of Children.
Ha: Social Class is not independent of Number of Children.
e11 =
398
)97)(31(
= 7.56 e31 =
398
)97)(189(
= 46.06
e12 =
398
)184)(31(
= 14.3 e32 =
398
)184)(189(
= 87.38
e13 =
398
)117)(31(
= 9.11 e33 =
398
)117)(189(
= 55.56
e21 =
398
)97)(70(
= 17.06 e41 =
398
)97)(108(
= 26.32
e22 =
398
)184)(70(
= 32.36 e42 =
398
)184)(108(
= 49.93
e23 =
398
)117)(70(
= 20.58 e43 =
398
)117)(108(
= 31.75

Social Class
Number
of
Children
Lower Middle Upper
0
1
2 or 3
>3
(7.56)
7
(14.33)
18
(9.11)
6 31
70
189
108
(17.06)
9
(32.36)
38
(20.58)
23
(46.06)
34
(87.38)
97
(55.56)
58
(26.32)
47
(49.93)
31
(31.75)
30
97 184 117 398
χ2
=
56.7
)56.77( 2
−
+
33.14
)33.1418( 2
−
+
11.9
)11.96( 2
−
+
06.17
)06.179( 2
−
+
36.32
)36.3238( 2
−
+
58.20
)58.2023( 2
−
+
06.46
)06.4634( 2
−
+
38.87
)38.8797( 2
−
+
56.55
)56.5558( 2
−
+
32.26
)32.2647( 2
−
+
93.49
)93.4931( 2
−
+
75.31
)75.3130( 2
−
=
.04 + .94 + 1.06 + 3.81 + .98 + .28 + 3.16 + 1.06 + .11 + 16.25 +
7.18 + .10 = 34.97
α = .05, df = (c-1)(r-1) = (3-1)(4-1) = 6
χ2
.05,6 = 12.5916
= 34.97 > χ2
null hypothesis.
Number of children is not independent of social class.

12.14
Type of Music Preferred
Region
Rock R&B Coun Clssic
195
235
202
632
NE 140 32 5 18
S 134 41 52 8
W 154 27 8 13
428 100 65 39
Ho: Type of music preferred is independent of region.
Ha: Type of music preferred is not independent of region.
e11 =
632
)428)(195(
= 132.6 e23 =
632
)65)(235(
= 24.17
e12 =
632
)100)(195(
= 30.85 e24 =
632
)39)(235(
= 14.50
e13 =
632
)65)(195(
= 20.06 e31 =
632
)428)(202(
= 136.80
e14 =
632
)39)(195(
= 12.03 e32 =
632
)100)(202(
= 31.96
e21 =
632
)428)(235(
= 159.15 e33 =
632
)65)(202(
= 20.78
e22 =
632
)100)(235(
= 37.18 e34 =
632
)39)(202(
= 12.47
Type of Music Preferred
Region
Rock R&B Coun Clssic
195
235
202
632
NE (132.06)
140
(30.85)
32
(20.06)
5
(12.03)
18
S (159.15)
134
(37.18)
41
(24.17)
52
(14.50)
8
W (136.80)
154
(31.96)
27
(20.78)
8
(12.47)
13
428 100 65 39

χ2
=
06.132
)06.132141( 2
−
+
85.30
)85.3032( 2
−
+
06.20
)06.205( 2
−
+
03.12
)03.1218( 2
−
+
15.159
)15.159134( 2
−
+
18.37
)18.3741( 2
−
+
17.24
)17.2452( 2
−
+
50.14
)50.148( 2
−
+
80.136
)80.136154( 2
−
+
96.31
)96.3127( 2
−
+
78.20
)78.208( 2
−
+
47.12
)47.1213( 2
−
=
.48 + .04 + 11.31 + 2.96 + 3.97 + .39 + 32.04 + 2.91 + 2.16 + .77 +
7.86 + .02 = 64.91
α = .01, df = (c-1)(r-1) = (4-1)(3-1) = 6
χ2
.01,6 = 16.8119
= 64.91 > χ2
.01,6 = 16.8119, the decision is to
Type of music preferred is not independent of region of the country.
12.15
Transportation Mode
Industry
Air Train Truck
85
35
120
Publishing 32 12 41
Comp.Hard. 5 6 24
37 18 65
H0: Transportation Mode is independent of Industry.
Ha: Transportation Mode is not independent of Industry.
e11 =
120
)37)(85(
= 26.21 e21 =
120
)37)(35(
= 10.79
e12 =
120
)18)(85(
= 12.75 e22 =
120
)18)(35(
= 5.25
e13 =
120
)65)(85(
= 46.04 e23 =
120
)65)(35(
= 18.96

Transportation Mode
Industry
Air Train Truck
85
35
120
Publishing (26.21)
32
(12.75)
12
(46.04)
41
Comp.Hard. (10.79)
5
(5.25)
6
(18.96)
24
37 18 65
χ2
=
21.26
)21.2632( 2
−
+
75.12
)75.1212( 2
−
+
04.46
)04.4641( 2
−
+
79.10
)79.105( 2
−
+
25.5
)25.56( 2
−
+
96.18
)96.1824( 2
−
=
1.28 + .04 + .55 + 3.11 + .11 + 1.34 = 6.43
α = .05, df = (c-1)(r-1) = (3-1)(2-1) = 2
χ2
.05,2 = 5.99147
= 6.431 > χ2
Transportation mode is not independent of industry.
12.16
Number of Bedrooms
Number of
Stories
< 2 3 > 4
274
575
1 116 101 57
2 90 325 160
206 426 217 849
H0: Number of Stories is independent of number of bedrooms.
Ha: Number of Stories is not independent of number of bedrooms.
e11 =
849
)206)(274(
= 66.48 e21 =
849
)206)(575(
= 139.52
e12 =
849
)426)(274(
= 137.48 e22 =
849
)426)(575(
= 288.52

e13 =
849
)217)(274(
= 70.03 e23 =
849
)217)(575(
= 146.97
χ2
=
52.139
)52.13990( 2
−
+
48.137
)48.137101( 2
−
+
03.70
)03.7057( 2
−
+
52.139
)52.13990( 2
−
+
52.288
)52.288325( 2
−
+
97.146
)97.146160( 2
−
=
χ2
= 36.89 + 9.68 + 2.42 + 17.58 + 4.61 + 1.16 = 72.34
α = .10 df = (c-1)(r-1) = (3-1)(2-1) = 2
χ2
.10,2 = 4.60517
= 72.34 > χ2
Number of stories is not independent of number of bedrooms.
12.17
Mexican Citizens
Type
of
Store
Yes No
41
35
30
60
Dept. 24 17
Disc. 20 15
Hard. 11 19
Shoe 32 28
87 79 166
Ho: Citizenship is independent of store type
Ha: Citizenship is not independent of store type
e11 =
166
)87)(41(
= 21.49 e31 =
166
)87)(30(
= 15.72
e12 =
166
)79)(41(
= 19.51 e32 =
166
)79)(30(
= 14.28
e21 =
166
)87)(35(
= 18.34 e41 =
166
)87)(60(
= 31.45

e22 =
166
)79)(35(
= 16.66 e42 =
166
)79)(60(
= 28.55
Mexican Citizens
Type
of
Store
Yes No
41
35
30
60
Dept. (21.49)
24
(19.51)
17
Disc. (18.34)
20
(16.66)
15
Hard. (15.72)
11
(14.28)
19
Shoe (31.45)
32
(28.55)
28
87 79 166
χ2
=
49.21
)49.2124( 2
−
+
51.19
)51.1917( 2
−
+
34.18
)34.1820( 2
−
+
66.16
)66.1615( 2
−
+
72.15
)72.1511( 2
−
+
28.14
)28.1419( 2
−
+
45.31
)45.3132( 2
−
+
55.28
)55.2828( 2
−
=
.29 + .32 + .15 + .17 + 1.42 + 1.56 + .01 + .01 = 3.93
α = .05, df = (c-1)(r-1) = (2-1)(4-1) = 3
χ2
.05,3 = 7.81473
= 3.93 < χ2
Citizenship is independent of type of store.

12.18 α = .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:
∑
−
=
e
e
f
ff 2
02 )(
χ
critical χ2
.01,7 = 18.4753
fo fe (f0-fe)2
0
2
0 )(
f
ff e−
214 206 64 0.311
235 232 9 0.039
279 268 121 0.451
281 284 9 0.032
264 268 16 0.060
254 232 484 2.086
211 206 25 0.121
3.100
∑
−
=
e
e
f
ff 2
02 )(
χ = 3.100
Since the observed value of χ2
= 3.1 < χ2
reject the null hypothesis. The observed distribution is not different from the
expected distribution.
12.19
Variable 2
Variable 1
12 23 21 56
8 17 20 45
7 11 18 36
27 51 59 137
e11 = 11.00 e12 = 20.85 e13 = 24.12
e21 = 8.87 e22 = 16.75 e23 = 19.38

e31 = 7.09 e32 = 13.40 e33 = 15.50
χ2
=
04.11
)04.1112( 2
−
+
85.20
)85.2023( 2
−
+
12.24
)12.2421( 2
−
+
87.8
)87.88( 2
−
+
75.16
)75.1617( 2
−
+
38.19
)38.1920( 2
−
+
09.7
)09.77( 2
−
+
40.13
)40.1311( 2
−
+
50.15
)50.1518( 2
−
=
.084 + .222 + .403 + .085 + .004 + .020 + .001 + .430 + .402 = 1.652
df = (c-1)(r-1) = (2)(2) = 4 α = .05
χ2
.05,4 = 9.48773
Since the observed value of χ2
= 1.652 < χ2
.05,4 = 9.48773, the decision is to fail
to reject the null hypothesis.
12.20
Location
NE W S
Customer Industrial 230 115 68 413
Retail 185 143 89 417
415 258 157 830
e11 =
830
)415)(413(
= 206.5 e21 =
830
)415)(417(
= 208.5
e12 =
830
)258)(413(
= 128.38 e22 =
830
)258)(417(
= 129.62
e13 =
830
)157)(413(
= 78.12 e23 =
830
)157)(417(
= 78.88

Location
NE W S
Customer Industrial (206.5)
230
(128.38)
115
(78.12)
68 413
Retail (208.5)
185
(129.62)
143
(78.88)
89 417
415 258 157 830
χ2
=
5.206
)5.206230( 2
−
+
38.128
)38.128115( 2
−
+
12.78
)12.7868( 2
−
+
5.208
)5.208185( 2
−
+
62.129
)62.129143( 2
−
+
88.78
)88.7889( 2
−
=
2.67 + 1.39 + 1.31 + 2.65 + 1.38 + 1.30 = 10.70
α = .10 and df = (c - 1)(r - 1) = (3 - 1)(2 - 1) = 2
χ2
.10,2 = 4.60517
= 10.70 > χ2
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
Σfo = 1,054
Ho: Cookie Sales is uniformly distributed across kind of cookie.
Ha: Cookie Sales is not uniformly distributed across kind of cookie.
If cookie sales are uniformly distributed, then fe =
6
054,1
.
0
=
∑
kindsno
f
= 175.67

fo fe
0
2
0 )(
f
ff e−
189 175.67 1.01
168 175.67 0.33
155 175.67 2.43
161 175.67 1.23
216 175.67 9.26
165 175.67 0.65
14.91
The observed χ2
= 14.91
α = .05 df = k - 1 = 6 - 1 = 5
χ2
.05,5 = 11.0705
= 14.91 > χ2
null hypothesis.
Cookie Sales is not uniformly distributed by kind of cookie.
12.22
Gender
M F
Bought
Car
Y 207 65 272
N 811 984 1,795
1,018 1,049 2,067
Ho: Purchasing a car or not is independent of gender.
Ha: Purchasing a car or not is not independent of gender.
e11 =
067,2
)018,1)(272(
= 133.96 e12 =
067,2
)049,1)(27(
= 138.04
e21 =
067,2
)018,1)(795,1(
= 884.04 e22 =
067,2
)049,1)(795,1(
= 910.96

Gender
M F
Bought
Car
Y (133.96)
207
(138.04)
65 272
N (884.04)
811
(910.96)
984 1,795
1,018 1,049 2,067
χ2
=
96.133
)96.133207( 2
−
+
04.138
)04.13865( 2
−
+
04.884
)04.884811( 2
−
+
96.910
)96.910984( 2
−
= 39.82 + 38.65 + 6.03 + 5.86 = 90.36
α = .05 df = (c-1)(r-1) = (2-1)(2-1) = 1
χ2
.05,1 = 3.841
= 90.36 > χ2
null hypothesis.
Purchasing a car is not independent of gender.
12.23 Arrivals fo (fo)(Arrivals)
0 26 0
1 40 40
2 57 114
3 32 96
4 17 68
5 12 60
6 8 48
Σfo = 192 Σ(fo)(arrivals) = 426
λ =
192
426))((
0
0
=
∑
∑
f
arrivalsf
= 2.2
Ho: The observed frequencies are Poisson distributed.
Ha: The observed frequencies are not Poisson distributed.

Arrivals Probability fe
0 .1108 (.1108)(192) = 21.27
1 .2438 (.2438)(192) = 46.81
2 .2681 51.48
3 .1966 37.75
4 .1082 20.77
5 .0476 9.14
6 .0249 4.78
fo fe
0
2
0 )(
f
ff e−
26 21.27 1.05
40 46.81 2.18
57 51.48 0.59
32 37.75 0.88
17 20.77 0.68
12 9.14 0.89
8 4.78 2.17
8.44
Observed χ2
= 8.44
α = .05 df = k - 2 = 7 - 2 = 5
χ2
.05,5 = 11.0705
= 8.44 < χ2
the null hypothesis. There is not enough evidence to reject the claim that the
observed frequency of arrivals is Poisson distributed.

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
Soft Drink fo proportions fe
0
2
0 )(
f
ff e−
Classic Coke 361 .206 (.206)(1726) = 355.56 0.08
Pepsi 272 .145 (.145)(1726) = 250.27 1.89
Diet Coke 192 .085 146.71 13.98
Mt. Dew 121 .063 108.74 1.38
Dr. Pepper 94 .059 101.83 0.60
Sprite 102 .062 107.01 0.23
Others 584 .380 655.86 7.87
∑fo = 1,726 26.03
Calculated χ2
= 26.03
α = .05 df = k - 1 = 7 - 1 = 6
χ2
.05,6 = 12.5916
= 26.03 > χ2
null hypothesis.
The observed frequencies are not distributed the same as the expected frequencies
from the national poll.
12.25
Position
Manager Programmer Operator
Systems
Analyst
Years
0-3 6 37 11 13 67
4-8 28 16 23 24 91
> 8 47 10 12 19 88
81 63 46 56 246

e11 =
246
)81)(67(
= 22.06 e23 =
246
)46)(91(
= 17.02
e12 =
246
)63)(67(
= 17.16 e24 =
246
)56)(91(
= 20.72
e13 =
246
)46)(67(
= 12.53 e31 =
246
)81)(88(
= 28.98
e14 =
246
)56)(67(
= 15.25 e32 =
246
)63)(88(
= 22.54
e21 =
246
)81)(91(
= 29.96 e33 =
246
)46)(88(
= 16.46
e22 =
246
)63)(91(
= 23.30 e34 =
246
)56)(88(
= 20.03
Position
Manager Programmer Operator
Systems
Analyst
Years
0-3 (22.06)
6
(17.16)
37
(12.53)
11
(15.25)
13
67
4-8 (29.96)
28
(23.30)
16
(17.02)
23
(20.72)
24
91
> 8 (28.98)
47
(22.54)
10
(16.46)
12
(20.03)
19
88
81 63 46 56 246
χ2
=
06.22
)06.226( 2
−
+
16.17
)16.1737( 2
−
+
53.12
)53.1211( 2
−
+
25.15
)25.1513( 2
−
+
96.29
)96.2928( 2
−
+
30.23
)30.2316( 2
−
+
02.17
)02.1723( 2
−
+
72.20
)72.2024( 2
−
+
98.28
)98.2847( 2
−
+
54.22
)54.2210( 2
−
+
46.16
)46.1612( 2
−
+
03.20
)03.2019( 2
−
=
11.69 + 22.94 + .19 + .33 + .13 + 2.29 + 2.1 + .52 + 11.2 + 6.98 +
1.21 + .05 = 59.63

α = .01 df = (c-1)(r-1) = (4-1)(3-1) = 6
χ2
.01,6 = 16.8119
= 59.63 > χ2
null hypothesis. Position is not independent of number of years of experience.
12.26 H0: p = .43 n = 315 α =.05
Ha: p ≠ .43 x = 120 α/2 = .025
fo fe
0
2
0 )(
f
ff e−
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 χ2
is 3.09
α = .05 and α/2 = .025 df = k - 1 = 2 - 1 = 1
χ2
.025,1 = 5.02389
Since χ2
= 3.09 < χ2
.025,1 = 5.02389, the decision is to fail to reject the null
hypothesis.
12.27
Type of College or University
Community
College
Large
University
Small
College
Number
of
Children
0 25 178 31 234
1 49 141 12 202
2 31 54 8 93
>3 22 14 6 42
127 387 57 571
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.

e11 =
571
)127)(234(
= 52.05 e31 =
571
)127)(93(
= 20.68
e12 =
571
)387)(234(
= 158.60 e32 =
571
)387)(193(
= 63.03
e13 =
571
)57)(234(
= 23.36 e33 =
571
)57)(93(
= 9.28
e21 =
571
)127)(202(
= 44.93 e41 =
571
)127)(42(
= 9.34
e22 =
571
)387)(202(
= 136.91 e42 =
571
)387)(42(
= 28.47
e23 =
571
)57)(202(
= 20.16 e43 =
571
)57)(42(
= 4.19
Type of College or University
Community
College
Large
University
Small
College
Number
of
Children
0 (52.05)
25
(158.60)
178
(23.36)
31
234
1 (44.93)
49
(136.91)
141
(20.16)
12
202
2 (20.68)
31
(63.03)
54
(9.28)
8
93
>3 (9.34)
22
(28.47)
14
(4.19)
6
42
127 387 57 571
χ2
=
05.52
)05.5225( 2
−
+
6.158
)6.158178( 2
−
+
36.23
)36.2331( 2
−
+
93.44
)93.4449( 2
−
+
91.136
)91.136141( 2
−
+
16.20
)16.2012( 2
−
+
68.20
)68.2031( 2
−
+
03.63
)03.6354( 2
−
+
28.9
)28.98( 2
−
+
34.9
)34.922( 2
−
+
47.28
)47.2814( 2
−
+
19.4
)19.46( 2
−
=

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
α = .05, df= (c - 1)(r - 1) = (3 - 1)(4 - 1) = 6
χ2
.05,6 = 12.5916
= 54.63 > χ2
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.

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