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.
2.
Understand the chi-square goodness-of-fit test and how to use it.
Analyze data using the chi-square test of independence.
CHAPTER OUTLINE
12.1
Chi-Square Goodness-of-Fit Test
Testing a Population Proportion by Using the Chi-square Goodness-of-Fit Test as an
Alternative Technique to the z Test
12.2
Contingency Analysis: Chi-Square Test of Independence
KEY WORDS
categorical data
chi-square distribution
chi-square goodness-of-fit test
chi-square test of independence
contingency analysis
contingency table
STUDY QUESTIONS
1. Statistical techniques based on assumptions about the population from which the sample data are
selected are called _______________ statistics.
2. Statistical techniques based on fewer assumptions about the population and the parameters are called
_______________ statistics.
3. A chi-square goodness-of-fit test is being used to determine if the observed frequencies from seven
categories are significantly different from the expected frequencies from the seven categories. The
degrees of freedom for this test are _______________.
4. A value of alpha = .05 is used to conduct the test described in question 3. The critical table chi-square
value is _______________.
5. A variable contains five categories. It is expected that data are uniformly distributed across these five
categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies
of 27, 30, 29, 21, 24. A value of .01 is specified for alpha. The degrees of freedom for this test are
_______________.
6. The critical table chi-square value of the problem presented in question 5 is _______________.
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7. The observed chi-square value for the problem presented in question five is _______________. Based
on this value and the critical chi-square value, a researcher would decide to _______________ the
null hypothesis.
8. A researcher believes that a variable is Poisson distributed across six categories. To test this, a
random sample of observations is made for the variable resulting in the following data:
Number of arrivals
0
1
2
3
4
5
Observed
47
56
38
23
15
12
Suppose alpha is .10, the critical table chi-square value used to conduct this chi-square goodness-offit test is _______________.
9. The value of the observed chi-square for the data presented in question 8 is _______________.
Based on this value and the critical value determined in question 8, the decision of
the researcher is to _______________ the null hypothesis.
10. The degrees of freedom used in conducting a chi-square goodness-of-fit test to determine if a
distribution is normally distributed are _______________.
11. In using the chi-square goodness-of-fit test, a statistician needs to make certain that none of the
expected values are less than _______________.
12. Suppose we want to test the following hypotheses using a chi-square goodness-of-fit test.
H0: p = .20 and Ha: p  .20
A sample of 150 data values is taken resulting in 37 items that possess the characteristic of interest.
Let  = .05. The degrees of freedom for this test are ____________. The critical chi-square value is
____________.
13. The calculated value of chi-square for question 12 is ________________. The
decision is to _____________________________________.
14. The chi-square ____________________ is used to analyze frequencies of two variables with multiple
categories.
15. A two-way frequency table is sometimes referred to as a _______________ table.
Chapter 12: Analysis of Categorical Data
223
16. Suppose a researcher wants to use the data below and the chi-square test of independence to
determine if variable one is independent of variable two.
Variable One
A
B
C
Variable
D
25
40
60
Two
E
10
15
20
The expected value for the cell of D and B is _______________.
17. The degrees of freedom for the problem presented in question 16 are _______________.
18. If alpha is .05, the critical chi-square value for the problem presented in question 16 is
_______________.
19. The observed value of chi-square for the problem presented in question 16 is _______________.
Based on this observed value of chi-square and the critical chi-square value determined in question
18, the researcher should decide to _______________ the null hypothesis that the two variables are
independent.
20. A researcher wants to statistically determine if variable three is independent of variable four using
the observed data given below:
Variable
Four
C
D
Variable Three
A
B
92
70
112
145
If alpha is .01, the critical chi-square table value for this problem is _______________.
21. The observed chi-square value for the problem presented in question 20 is _______________. Based
on this value and the critical value determined in question 20, the researcher should decide to
_______________ the null hypothesis.
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ANSWERS TO STUDY QUESTIONS
1.
Parametric Statistics
2.
Nonparametric Statistics
3.
6
4.
12.5916
5.
4
6.
13.2767
7.
2.091, Fail to Reject
8.
7.77944
9.
14.8, Reject
10. k – 3
11. 5
12. 1, 3.8416
13. 2.041, Fail to Reject
14. Test of Independence
15. Contingency
16. 40.44
17. 2
18. 5.99147
19. .19, Fail to Reject
20. 6.6349
21. 6.945, Reject
Chapter 12: Analysis of Categorical Data
SOLUTIONS TO ODD-NUMBERED PROBLEMS IN CHAPTER 12
12.1
f0
( f0  fe )2
f0
fe
53
37
32
28
18
15
68
42
33
22
10
8
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.
( f0  fe )2
Observed   
= 18.095
fe
2
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.
12.3
Number
f0
(Number)(f0)
0
1
2
3
28
17
11
5
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
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Since fe for > 3 is less than 5, collapse categories 2 and >3:
Number
fo
fe
( f0  fe )2
f0
0
1
>2
28
17
16
61
24.803
22.312
13.878
60.993
0.412
1.265
0.324
2.001
df = k – 2 = 3 – 2 = 1,
= .05
2.05,1 = 3.84146
( f0  fe )2
Observed   
= 2.001
fe
2
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.
12.5
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,
 = .05
2.05,3 = 7.81473
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.
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
s =
m
15
25
35
45
55
65
fm
240
1,100
2,135
2,520
1,925
1,235
fm = 9,155
2

( fM ) 2
n 1
n

(9,155) 2
405,375 
231 = 13.6
230
For Category 10-20
Prob
10  39.63
= –2.18
13.6
20  39.63
z =
= –1.44
13.6
z =
.4854
–.4251
Expected prob.
.0603
For Category 20-30
Prob
for x = 20, z = –1.44
.4251
30  39.63
= –0.71
13.6
Expected prob.
For Category 30-40
for x = 30, z = –0.71
z =
3,600
27,500
74,725
113,400
105,875
80,275
fm2 = 405,375
9,155
= 39.63
231
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.
z =
fm2
40  39.63
= 0.03
13.6
–.2611
.1640
Prob
.2611
+.0120
Expected prob. .2731
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For Category 40-50
z =
Prob
50  39.63
= 0.76
13.6
.2764
–.0120
for x = 40, z = 0.03
Expected prob.
For Category 50-60
z =
Prob
60  39.63
= 1.50
13.6
.4332
–.2764
for x = 50, z = 0.76
Expected prob.
For Category 60-70
z =
.2644
.1568
Prob
70  39.63
= 2.23
13.6
.4871
–.4332
for x = 60, z = 1.50
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
50-60
60-70
> 70
Probability
.0146
.0603
.1640
.2731
.2644
.1568
.0539
.0129
fe
(.0146)(231) = 3.37
(.0603)(231) = 13.93
37.88
63.09
61.08
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.
Chapter 12: Analysis of Categorical Data
( f0  fe )2
f0
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
 = .05
df = k – 3 = 6 – 3 = 3,
0.10
0.99
0.07
0.42
0.04
0.83
2.45
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.
12.9
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
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.
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12.11
Variable
One
Variable
Two
24
59
13
43
20
35
57 137
83
56
55
194
Ho: Variable one is independent of Variable Two.
Ha: Variable one is not independent of Variable Two.
e11 =
(83)(57)
= 24.39
194
e21 =
(56)(57)
= 16.45
194
e31 =
(55)(57)
= 16.16
194
e12 =
(83)(13)
= 58.61
194
e22 =
e32 =
(55)(137)
= 38.84
194
Variable
Two
(24.39)
Variable
24
One
(16.45)
13
(16.16)
20
57
2 =
(56)(137)
= 39.55
194
(58.61)
59
(39.55)
43
(38.84)
35
137
83
56
55
194
(24  24.39) 2
(59  58.61) 2
(13  16.45) 2
(43  39.55) 2
+
+
+
+
16.45
24.39
58.61
39.55
(20  16.16) 2
(35  38.84) 2
+
= .01 + .00 + .72 + .30 + .91 + .38 = 2.32
16.16
38.84
 = .05, df = (c – 1)(r – 1) = (2 – 1)(3 – 1) = 2
2.05,2 = 5.99147
Since the observed 2 = 2.32 < 2.05,2 = 5.99147, the decision is to fail to
reject the null hypothesis.
Variable One is independent of Variable Two.
Chapter 12: Analysis of Categorical Data
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
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
e31 =
(189)(97)
= 46.06
398
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0
Number
of
Children
1
2 or 3
>3
2 =
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
(7  7.56) 2
(18  14.33) 2
(6  9.11) 2
(9  17.06) 2
+
+
+
+
14.33
17.06
7.56
9.11
(38  32.36) 2
(23  20.58) 2
(34  46.06) 2
(97  87.38) 2
+
+
+
+
46.06
32.36
20.58
87.38
(58  55.56) 2
(47  26.32) 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
12.15
Industry
Transportation Mode
Air
Train
Publishing
32
12
Comp.Hard.
5
6
37
18
Truck
41
24
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
Industry
2 =
Transportation Mode
Air
Train
Publishing
(26.21) (12.75)
32
12
Comp.Hard. (10.79) (5.25)
5
6
37
18
Truck
(46.04)
41
(18.96)
24
65
85
35
120
(32  26.21) 2
(12  12.75) 2
(41  46.04) 2
+
+
+
12.75
26.21
46.04
(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.
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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
e22 =
(35)(79)
= 16.66
166
e42 =
(60)(79)
= 28.55
166
Chapter 12: Analysis of Categorical Data
Type
of
Store
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
(20  18.34) 2
(15  16.66) 2
 =
+
+
+
21.49
19.51
16.66
18.34
2
+
(11  15.72) 2
(19  14.28) 2
(32  31.45) 2
(28  28.55) 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.
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12.19
12
8
7
27
Variable 1
Variable 2
23
17
11
51
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
21
20
18
59
56
45
36
137
(12  11.04)2
(23  20.85) 2
(21  24.12) 2 (8  8.87) 2
 =
+
+
+
+
20.85
11.04
24.12
8.87
2
(17  16.75) 2
(20  19.38) 2
(7  7.09) 2
(11  13.40) 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.
12.21
Cookie Type
Chocolate Chip
Peanut Butter
Cheese Cracker
Lemon Flavored
Chocolate Mint
Vanilla Filled
Ho:
Ha:
fo
189
168
155
161
216
165
fo = 1,054
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
=
6
175.67
Chapter 12: Analysis of Categorical Data
fo
fe
189
168
155
161
216
165
237
( f0  fe )2
f0
175.67
175.67
175.67
175.67
175.67
175.67
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.
12.23
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.
Arrivals
0
1
2
3
4
5
6
Probability
.1108
.2438
.2681
.1966
.1082
.0476
.0249
fe
(.1108)(192) = 21.27
(.2438)(192) = 46.81
51.48
37.75
20.77
9.14
4.78
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Solutions Manual and Study Guide
fo
fe
26
40
57
32
17
12
8
( f0  fe )2
f0
21.27
46.81
51.48
37.75
20.77
9.14
4.78
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.
12.25
Position
Years
0-3
4-8
>8
Manager
6
28
47
81
Programmer
37
16
10
63
Operator
11
23
12
46
Systems
Analyst
13
24
19
56
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 =
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
(88)(81)
= 28.98
246
67
91
88
246
Chapter 12: Analysis of Categorical Data
239
Position
0-3
Years
4-8
>8
Manager
(22.06)
6
(29.96)
28
(28.98)
47
81
Programmer
(17.16)
37
(23.30)
16
(22.54)
10
63
Operator
(12.53)
11
(17.02)
23
(16.46)
12
46
Systems
Analyst
(15.25)
13
(20.72)
24
(20.03)
19
56
67
91
88
246
(6  22.06) 2
(37  17.16) 2
(11  12.53) 2 (13  15.25) 2
 =
+
+
+
+
22.06
12.53
15.25
17.16
2
(28  29.96) 2
(16  23.30) 2
(23  17.02) 2
(24  20.72) 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
 = .01
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.
240
Solutions Manual and Study Guide
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.
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
Chapter 12: Analysis of Categorical Data
Number
of
Children
0
1
2
>3
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
241
234
202
93
42
571
(25  52.05) 2
(178  158.6) 2
(31  23.36) 2
(49  44.93) 2
 =
+
+
+
+
52.05
158.6
44.93
23.36
2
(141  136.91) 2
(12  20.16) 2
(31  20.68) 2
(54  63.03) 2
+
+
+
+
136.91
20.16
20.68
63.03
(8  9.28) 2
(22  9.34) 2
(14  28.47) 2
(6  4.19) 2
+
+
+
=
9.34
9.28
28.47
4.19
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.
12.29 The observed chi-square value for this test of independence is 5.366. The associated pvalue 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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Solutions Manual and Study Guide