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Materi ini merupakan bahan ajar sebagai pelengkap e-materi mata kuliah statistika bisnis. Black, K. (2010). Business Statistics; for Contemporary Decision Making (6th Edition). wiley.
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Student’s Solutions Manual and Study Guide: Chapter 16 Page 1
Chapter 16
Analysis of Categorical Data
LEARNING OBJECTIVES
The overall objective of this chapter is to give you an understanding of two
statistical techniques used to analyze categorical data, thereby enabling
you to:
1. Use the chi-square goodness-of-fit test to analyze probabilities of
multinomial distribution trials along a single dimension and test a
population proportion
2. Use the chi-square test of independence to perform contingency
analysis
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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 2
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
_______________.
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 Observed
0 47
1 56
2 38
3 23
4 15
5 12
Suppose alpha is .10, the critical table chi-square value used to conduct this chi-square
goodness-of-fit 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 3
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 observed 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.
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
Two
D 25 40 60
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 _______________.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 4
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 Three
A B
Variable
Four
C 92 70
D 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 5
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. 16.2, 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. 5.9915
19. .19, Fail to Reject
20. 6.6349
21. 6.945, Reject
Student’s Solutions Manual and Study Guide: Chapter 16 Page 6
SOLUTIONS TO THE ODD-NUMBERED PROBLEMS IN CHAPTER 16
16.1 f0 fe 0
2
0 )(
f
ff e
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 )( = 18.095
df = k - 1 = 6 - 1 = 5, = .05
2
.05,5 = 11.0705
Since the observed 2 = 18.095 >
2.05,5 = 11.0705, the decision is to reject the
null hypothesis.
The observed frequencies are not distributed the same as the expected
frequencies.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 7
16.3 Number f0 (Number)(f0)
0 28 0
1 17 17
2 11 22
3 _5 15
61 54
Ho: The frequency distribution is Poisson.
Ha: The frequency distribution is not Poisson.
= 61
54=0.9
Expected Expected
Number Probability Frequency
0 .4066 24.803
1 .3659 22.320
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 22.320 1.268
>2 16 13.878 0.324
61 60.993 2.004
df = k - 2 = 3 - 2 = 1, = .05
2
.05,1 = 3.8415
Observed
e
e
f
ff 2
02 )( = 2.001
Since the observed 2 = 2.001 <
2.05,1 = 3.8415, 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 8
16.5 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 70.37 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 2 = 198.48
df = k – 1 = 4 – 1 = 3, = .05
2
.05,3 = 7.8147
Since the observed 2 = 198.48 >
2.05,3 = 7.8147, 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 9
16.7 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 fm = 9,155 fm2 = 405,375
231
155,9
n
fMx = 39.63
s = 230
231
)155,9(375,405
1
)( 22
2
n
n
fMfM
= 13.6
Ho: The observed frequencies are normally distributed.
Ha: The observed frequencies are not normally distributed.
For Category 10-20 Prob
z = 6.13
63.3910 = -2.18 .4854
z = 6.13
63.3920 = -1.44 -.4251
Expected prob. .0603
For Category 20-30 Prob
for x = 20, z = -1.44 .4251
z = 6.13
63.3930 = -0.71 -.2611
Expected prob. .1640
For Category 30-40 Prob
for x = 30, z = -0.71 .2611
z = 6.13
63.3940 = 0.03 +.0120
Expected prob. .2731
Student’s Solutions Manual and Study Guide: Chapter 16 Page 10
For Category 40-50 Prob
z = 6.13
63.3950 = 0.76 .2764
for x = 40, z = 0.03 -.0120
Expected prob. .2644
For Category 50-60 Prob
z = 6.13
63.3960 = 1.50 .4332
for x = 50, z = 0.76 -.2764
Expected prob. .1568
For Category 60-70 Prob
z = 6.13
63.3970 = 2.23 .4871
for x = 60, z = 1.50 -.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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 11
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, = .05
2
.05,3 = 7.8147
Observed 2 = 2.45
Since the observed 2 <
2.05,3 = 7.8147, the decision is to fail to reject the null
hypothesis.
There is no reason to reject that the observed frequencies are normally
distributed.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 12
16.9 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 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 13
16.11
Variable Two
Variable
One
203 326 529
178 68 110
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
2 =
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
= .01, df = (c-1)(r-1) = (2-1)(2-1) = 1
2
.01,1 = 6.6349
Since the observed 2 = 0.00 <
2.01,1 = 6.6349, the decision is to fail to reject
the null hypothesis.
Variable One is independent of Variable Two.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 14
16.13
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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 15
2 =
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
= .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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 16
16.15
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
2 =
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
= .05, df = (c-1)(r-1) = (3-1)(2-1) = 2
2
.05,2 = 5.9915
Since the observed 2 = 6.43 >
2.05,2 = 5.9915, the decision is to
reject the null hypothesis.
Transportation mode is not independent of industry.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 17
16.17
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
2 =
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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 18
= .05, df = (c-1)(r-1) = (2-1)(4-1) = 3
2
.05,3 = 7.8147
Since the observed 2 = 3.93 <
2.05,3 = 7.8147, the decision is to fail to
reject the null hypothesis.
Citizenship is independent of type of store.
16.19
Variable 2
Variable 1
12 23 21 56
8 17 20 45
7 11 18 36
27 51 59 137
e11 = 11.04 e12 = 20.85 e13 = 24.12
e21 = 8.87 e22 = 16.75 e23 = 19.38
e31 = 7.09 e32 = 13.40 e33 = 15.50
2 =
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 = .05
2
.05,4 = 9.4877
Since the observed value of 2 = 1.652 <
2.05,4 = 9.4877, the decision is to fail
to reject the null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 19
16.21 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: 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 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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 20
16.23 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
fo = 192 (fo)(arrivals) = 426
= 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 0.99
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
7.25
Observed 2 = 7.25
= .05 df = k - 2 = 7 - 2 = 5
2
.05,5 = 11.0705
Since the observed 2 = 7.25 <
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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 21
16.25
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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 22
2 =
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
= .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.
Student’s Solutions Manual and Study Guide: Chapter 16 Page 23
16.27
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
Student’s Solutions Manual and Study Guide: Chapter 16 Page 24
2 =
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
= .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.
16.29 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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