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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: 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
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 Test of Independence
Chi-Square Distribution Contingency Analysis Chi-Square Goodness-of-Fit Test Contingency Table
Chapter 12: Analysis of Categorical Data 3
SOLUTIONS TO CHAPTER 16
12.1 f0 0
20 )(
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 202 )(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
20 )(
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
Chapter 12: Analysis of Categorical Data 4
Ho: The observed frequencies are uniformly distributed. Ha: The observed frequencies are not uniformly distributed.
8
1440 == ∑k
fx = 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 202 )(χ = 1.557
Since the observed χ2 = 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:
Chapter 12: Analysis of Categorical Data 5
Number fo fe 0
20 )(
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 202 )(χ = 2.001
Since the observed χ2 = 2.001 < χ2
.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) 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
fmx = 44.3
s = 128
129
)715,5(825,279
1
)( 222 −
=−
−∑∑
nn
fMfM
= 14.43
Ho: The observed frequencies are normally distributed. Ha: The observed frequencies are not normally distributed.
Chapter 12: Analysis of Categorical Data 6
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 Category 30 - 40 Prob for x = 30, z = -0.99 .3389
z = 43.14
3.4440− = -0.30 -.1179
Expected prob: .2210 For Category 40 - 50 Prob for x = 40, z = -0.30 .1179
z = 43.14
3.4450− = 0.40 +.1554
Expected prob: .2733 For Category 50 - 60 Prob
z = 43.14
3.4460− = 1.09 .3621
for x = 50, z = 0.40 -.1554 Expected prob: .2067
Chapter 12: Analysis of Categorical Data 7
For Category 60 - 70 Prob
z = 43.14
3.4470− = 1.78 .4625
for x = 60, z = 1.09 -.3621 Expected prob: .1004 For Category 70 - 80 Prob
z = 43.14
3.4480− = 2.47 .4932
for x = 70, z = 1.78 -.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 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.
Chapter 12: Analysis of Categorical Data 8
Category fo fe 0
20 )(
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 202 )(χ = 2.004
df = k - 3 = 7 - 3 = 4, α = .05 χ2
.05,1 = 9.48773 Since the observed χ2 = 2.004 > χ2
.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 0
20 )(
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
Chapter 12: Analysis of Categorical Data 9
Since the observed χ2 = 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
20 )(
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 expected frequencies. α = .01, df = k - 1 = 6 - 1 = 5 χ2
.01,5 = 15.0863 The observed χ2 = 1.13 Since the observed χ2 = 1.13 < χ2
.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 10
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
fMx = 39.63
s = 230
231
)155,9(375,405
1
)( 222 −
=−
−∑∑
nn
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
Chapter 12: Analysis of Categorical Data 11
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 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
Chapter 12: Analysis of Categorical Data 12
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
20 )(
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 Since the observed χ2 < 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.
Chapter 12: Analysis of Categorical Data 13
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
20 )(
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 Since the observed χ2 = 10.27 < χ2
.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 n = 270 x = 62 Ha: p ≠ .28
fo fe 0
20 )(
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
Chapter 12: Analysis of Categorical Data 14
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 Since the observed χ2 = 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
20 )(
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 Since the observed χ2 = 3.8095 < χ2
.025,1 = 5.02389, the decision is to fail to reject the null hypothesis.
Chapter 12: Analysis of Categorical Data 15
12.11 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
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 Since the observed χ2 = 0.00 < χ2
.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 Two
Variable One
24 13 47 58 142 583 93 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
Chapter 12: Analysis of Categorical Data 17
α = .01, df = (c-1)(r-1) = (4-1)(2-1) = 3 χ2
.01,3 = 11.3449 Since the observed χ2 = 0.24 < χ2
.01,3 = 11.3449, the decision is to fail to reject the null hypothesis. 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
Chapter 12: Analysis of Categorical Data 18
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 Since the observed χ2 = 34.97 > χ2
.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
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
Chapter 12: Analysis of Categorical Data 20
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 Since the observed χ2 = 64.91 > χ2
.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 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
Chapter 12: Analysis of Categorical Data 21
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 Since the observed χ2 = 6.431 > χ2
.05,2 = 5.99147, the decision is to reject the null hypothesis. 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
Chapter 12: Analysis of Categorical Data 22
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 Since the observed χ2 = 72.34 > χ2
.10,2 = 4.60517, the decision is to reject the null hypothesis. 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
Chapter 12: Analysis of Categorical Data 23
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 Since the observed χ2 = 3.93 < χ2
.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 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 202 )(χ
critical χ2
.01,7 = 18.4753
fo fe (f0-fe)2
0
20 )(
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 202 )(χ = 3.100
Since the observed value of χ2 = 3.1 < χ2
.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 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
Chapter 12: Analysis of Categorical Data 25
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
Chapter 12: Analysis of Categorical Data 26
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 Since the observed χ2 = 10.70 > χ2
.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 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
Chapter 12: Analysis of Categorical Data 27
fo fe 0
20 )(
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 Since the observed χ2 = 14.91 > χ2
.05,5 = 11.0705, the decision is to reject the 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
Chapter 12: Analysis of Categorical Data 28
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
Since the observed χ2 = 90.36 > χ2.05,1 = 3.841, the decision is to reject the
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.
Chapter 12: Analysis of Categorical Data 29
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
20 )(
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 Since the observed χ2 = 8.44 < χ2
.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 proportions fe 0
20 )(
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 Since the observed χ2 = 26.03 > χ2
.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 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
Chapter 12: Analysis of Categorical Data 31
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
Chapter 12: Analysis of Categorical Data 32
α = .01 df = (c-1)(r-1) = (4-1)(3-1) = 6 χ2
.01,6 = 16.8119 Since the observed χ2 = 59.63 > χ2
.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 n = 315 α =.05 Ha: p ≠ .43 x = 120 α/2 = .025
fo fe 0
20 )(
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.
Chapter 12: Analysis of Categorical Data 33
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− =
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 α = .05, df= (c - 1)(r - 1) = (3 - 1)(4 - 1) = 6 χ2
.05,6 = 12.5916 Since the observed χ2 = 54.63 > χ2
.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.