Ken Black QA 5th chapter 12 Solution

8/3/2019 Ken Black QA 5th chapter 12 Solution

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Chapter 12: Analysis of Categorical Data 1

Chapter 12Analysis 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 areimportant 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 businessresearch 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. Onthese 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.


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The chi-square goodness-of-fit test examines the categories of one variable to

determine if the distribution of observed occurrences matches some expected ortheoretical 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 ofphenomena such as random arrivals that are often assumed to be Poisson distributed.

You will note that the degrees of freedom, 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- 3for 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 areestimated 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 frequenciesalong 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 inexploratory 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 possibilityof 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, wehave settled on the fairly widespread accepted criterion of five or more.


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CHAPTER OUTLINE

12.1 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

12.2 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


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SOLUTIONS TO THE ODD-NUMBERED PROBLEMS IN CHAPTER 12

12.1 f0 fee

eo

fff

2

)(

53 68 3.30937 42 0.595

32 33 0.030

28 22 1.63618 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 sameas the expected distribution.

Observed

=e

e

f

ff 202 )( = 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 thenull hypothesis.

The observed frequencies are not distributed the same as the expected

frequencies.


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12.2 f0 fee

eo

f

ff 2)(

19 18 0.056

17 18 0.05614 18 0.889

18 18 0.00019 18 0.056

21 18 0.50018 18 0.000

18 18 0.000

fo = 144 fe = 144 1.557

Ho: The observed frequencies are uniformly distributed.

Ha: The observed frequencies are not uniformly distributed.

8

1440==

k

fx = 18

In this uniform distribution, eachfe = 18

df = k 1 = 8 1 = 7, = .01

2.01,7 = 18.4753

Observed

=

e

e

f

ff 202 )( = 1.557

Since the observed 2 = 1.557 < 2.01,7 = 18.4753, the decision is to fail toreject

the null hypothesis

There is no reason to conclude that the frequencies are not uniformly

distributed.


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12.3 Number f0 (Number)(f0)

0 28 01 17 17

2 11 22

3 _5 1561 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 22.3202 .1647 10.047

> 3 .0628 3.831

Sincefe for > 3 is less than 5, collapse categories 2 and >3:

Number fo fee

eo

f

ff 2)(

0 28 24.803 0.4121 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 202 )( = 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.


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12.4

Category f(observed) Midpt. fm fm2

10-20 6 15 90 1,350

20-30 14 25 350 8,75030-40 29 35 1,015 35,525

40-50 38 45 1,710 76,950

50-60 25 55 1,375 75,62560-70 10 65 650 42,250

70-80 7 75 525 39,375

n = f= 129 fm = 5,715 fm2 = 279,825

129

715,5==

f

fmx = 44.3

s =

128

129

)715,5(825,279

1

)( 22

2

=

n

n

fMfM = 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

forx = 20, z= -1.68 .4535

z=43.14

3.4430 = -0.99 -.3389

Expected prob: .1146


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forx = 30, z= -0.99 .3389

z =

43.14

3.4440 = -0.30 -.1179



forx = 40, z= -0.30 .1179

z =43.14

3.4450 = 0.40 +.1554



z =43.14

3.4460 = 1.09 .3621

forx = 50, z= 0.40 -.1554Expected prob: .2067


z =43.14

3.4470

= 1.78 .4625

forx = 60, z= 1.09 -.3621Expected prob: .1004


z =43.14

3.4480 = 2.47 .4932

forx = 70, z= 1.78 -.4625


Forx < 10:

Probability between 10 and the mean, 44.3, = (.0378 + .1145 + .2210

+ .1179) = .4913. Probability < 10 = .5000 - .4912 = .0087


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Chapter 12: Analysis of Categorical Data 9Forx > 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) = 1.12

10-20 .0378 .0378(129) = 4.88

20-30 .1146 14.7830-40 .2210 28.51

40-50 .2733 35.26

50-60 .2067 26.66

60-70 .1004 12.9570-80 .0307 3.96

> 80 .0068 0.88

Due to the small sizes of expected frequencies, category < 10 is folded into 10-20and >80 into 70-80.

Category fo fee

eo

f

ff 2)(

10-20 6 6.00 .000

20-30 14 14.78 .04130-40 29 28.51 .008

40-50 38 35.26 .213

50-60 25 26.66 .103

60-70 10 12.95 .67270-80 7 4.84 .964

2.001

Calculated

=e

e

f

ff 202 )( = 2.001

df = k- 3 = 7 - 3 = 4, = .05

2.05,4 = 9.4877

Since the observed 2 = 2.004 < 2.05,4 = 9.4877, the decision is to fail to rejectthe null hypothesis. There is not enough evidence to declare that the observed

frequencies are not normally distributed.


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12.5 Definition fo Exp.Prop. fee

eo

f

ff 2)(

Happiness 42 .39 227(.39)= 88.53 24.46Sales/Profit 95 .12 227(.12)= 27.24 168.55

Helping Others 27 .18 40.86 4.70Achievement/

Challenge 63 .31 70.37 0.77227 198.48

Ho: The observed frequencies are distributed the same as the expectedfrequencies.

Ha: The observed frequencies are not distributed the same as the expectedfrequencies.

Observed 2 = 198.48

df = k 1 = 4 1 = 3, = .05

2.05,3 = 7.8147


The observed frequencies for men are not distributed the same as the

expected frequencies which are based on the responses of women.


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12.6 Age fo Prop. from survey fee

eo

f

ff 2)(

10-14 22 .09 (.09)(212)=19.08 0.4515-19 50 .23 (.23)(212)=48.76 0.03

20-24 43 .22 46.64 0.2825-29 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


= .01, df = k- 1 = 6 - 1 = 5

2.01,5 = 15.0863

The observed 2 = 1.13

Since the observed 2 = 1.13 < 2.01,5 = 15.0863, the decision is to fail to rejectthe null hypothesis.

There is not enough evidence to declare that the distribution of observedfrequencies is different from the distribution of expected frequencies.


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Chapter 12: Analysis of Categorical Data 1212.7 Age fo m fm fm

2

10-20 16 15 240 3,600

20-30 44 25 1,100 27,50030-40 61 35 2,135 74,725

40-50 56 45 2,520 113,400

50-60 35 55 1,925 105,87560-70 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

)( 222

=

n

n

fMfM

= 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


forx = 20, z= -1.44 .4251

z =6.13

63.3930 = -0.71 -.2611



forx = 30, z= -0.71 .2611

z =6.13

63.3940 = 0.03 +.0120



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z =6.13

63.3950 = 0.76 .2764

forx = 40, z= 0.03 -.0120



z =6.13

63.3960 = 1.50 .4332

forx = 50, z= 0.76 -.2764



z =6.13

63.3970 = 2.23 .4871

forx = 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


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Age Probability fe

< 10 .0146 (.0146)(231) = 3.3710-20 .0603 (.0603)(231) = 13.93

20-30 .1640 37.88

30-40 .2731 63.0940-50 .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 fee

eo

f

ff 2)(

10-20 16 17.30 0.1020-30 44 37.88 0.9930-40 61 63.09 0.07

40-50 56 61.08 0.42

50-60 35 36.22 0.0460-70 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 nullhypothesis.

There is no reason to reject that the observed frequencies are normally

distributed.


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12.8 Number f (f) (number)0 18 01 28 28

2 47 94

3 21 63

4 16 645 11 55

6 or more 9 54

f= 150 f (number) = 358

=150

358=

f

numberf= 2.4

Ho: The observed frequencies are Poisson distributed.

Ha: The observed frequencies are not Poisson distributed.

Number Probability fe0 .0907 (.0907)(150) = 13.611 .2177 (.2177)(150) = 32.66

2 .2613 39.20

3 .2090 31.354 .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 0.6647 39.20 1.55

21 31.35 3.42

16 18.81 0.42

11 9.03 0.439 5.36 2.47

10.37


= .01, df = k 2 = 7 2 = 5, 2.01,5 = 15.0863


There is not enough evidence to reject the claim that the observed

frequencies are Poisson distributed.


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12.9 H0: p = .28 n = 270 x = 62

Ha: p .28

fo fe e

eo

f

ff 2)(

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 toreject the null hypothesis.

12.10 H0: p = .30 n = 180 x= 42Ha: p < .30

f0 fee

eo

f

ff 2)(

Provide 42 180(.30) = 54 2.6666

Don't Provide 138 180(.70) = 126 1.1429

Total 180 180 3.8095


= .05 df = k- 1 = 2 - 1 = 1

2.05,1 = 3.8415



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Chapter 12: Analysis of Categorical Data 1712.11

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

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


Variable One is independent of Variable Two.


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VariableTwo

VariableOne

24 13 47 58 14258393 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

2 =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

= .01, df = (c-1)(r-1) = (4-1)(2-1) = 3, 2.01,3 = 11.3449

Since the observed 2 = 0.24 < 2.01,3 = 11.3449, the decision is to fail to


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Chapter 12: Analysis of Categorical Data 19reject 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

189108

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

(7.56)

7

(14.33

)

18

(9.11)

6

31

70

189

(17.06

)9

(32.36

)38

(20.58

)23


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>3

108

(46.06

)34

(87.38

)97

(55.56

)58

(26.32)

47

(49.93)

31

(31.75)

3097 184 117 398

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 thenull hypothesis.

Number of children is not independent of social class.

12.14

Type of Music Preferred

Region

Rock R&B Coun Clssic

195235

202632

NE 140 32 5 18S 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.


21/36


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

195235

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

2 =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 +


22/36

Chapter 12: Analysis of Categorical Data 227.86 + .02 = 64.91

= .01, df = (c-1)(r-1) = (4-1)(3-1) = 6

2.01,6 = 16.8119

Since the observed 2 = 64.91 > 2.01,6 = 16.8119, the decision is toreject the null hypothesis.

Type of music preferred is not independent of region of the country.

12.15

Transportation Mode

Industr

y

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

35120

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=


23/36

Chapter 12: Analysis of Categorical Data 231.28 + .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


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 160206 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

2 =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

=

2 = 36.89 + 9.68 + 2.42 + 17.58 + 4.61 + 1.16 = 72.34

= .10 df = (c-1)(r-1) = (3-1)(2-1) = 2

2.10,2 = 4.6052


24/36



Number of stories is not independent of number of bedrooms.


25/36


Mexican Citizens

Typeof

Store

Yes No

4135

3060

Dept. 24 17

Disc. 20 15

Hard. 11 19Shoe 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

ofStore

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

+


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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

= .05, df = (c-1)(r-1) = (2-1)(4-1) = 3

2.05,3 = 7.8147


Citizenship is independent of type of store.

12.18 = .01, k= 7, df = 6

H0: The observed distribution is the same as the expected distributionHa: The observed distribution is not the same as the expected distribution

Use:

=e

e

f

ff 202 )(

critical 2.01,6 = 16.8119

fo fe (f0-fe)2

e

eo

f

ff 2)(

214 206 64 0.311

235 232 9 0.039279 268 121 0.451

281 284 9 0.032

264 268 16 0.060254 232 484 2.086

211 206 25 0.121

3.100

=e

e

f

ff 202 )( = 3.100


27/36


Since the observed value of 2 = 3.1 < 2.01,6 = 16.8119, the decision is to failto

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 = 11.04 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 failto reject the null hypothesis.


28/36


12.20

Location

NE W S

Customer Industrial 230 115 68 413

Retail 185 143 89 417415 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

2 =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

= .10 and df = (c - 1)(r- 1) = (3 - 1)(2 - 1) = 2

2.10,2 = 4.6052



29/36

Chapter 12: Analysis of Categorical Data 29Type of customer is not independent of geographic region.

12.21 Cookie Type foChocolate Chip 189

Peanut Butter 168

Cheese Cracker 155Lemon 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, thenfe =6

054,1

.

0=

kindsno

f= 175.67

fo fee

eo

f

ff 2)(

189 175.67 1.01168 175.67 0.33

155 175.67 2.43

161 175.67 1.23216 175.67 9.26

165 175.67 0.65

14.91


= .05 df = k- 1 = 6 - 1 = 5

2.05,5 = 11.0705


Cookie Sales is not uniformly distributed by kind of cookie.


30/36


12.22

Gender

M F

Bought

Car

Y 207 65 272

N 811 984 1,7951,018 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

2 =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

= .05 df = (c-1)(r-1) = (2-1)(2-1) = 1

2.05,1 = 3.8415


Purchasing a car is not independent of gender.


31/36

Chapter 12: Analysis of Categorical Data 3112.23 Arrivals fo (fo)(Arrivals)

0 26 0

1 40 402 57 114

3 32 96

4 17 685 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) = 46.812 .2681 51.48

3 .1966 37.75

4 .1082 20.775 .0476 9.14

6 .0249 4.78

fo fe

e

eo

f

ff 2)(

26 21.27 1.05

40 46.81 0.99

57 51.48 0.5932 37.75 0.88

17 20.77 0.68

12 9.14 0.89

8 4.78 2.177.25

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 rejectthe null hypothesis. There is not enough evidence to reject the claim that the

observed frequency of arrivals is Poisson distributed.


32/36


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 fee

eo

f

ff 2)(

Classic Coke 314 .179 (.179)(1726) = 308.95 0.08

Pepsi 219 .115 (.115)(1726) = 198.49 2.12Diet Coke 212 .097 167.42 11.87

Mt. Dew 121 .063 108.74 1.38

Diet Pepsi 98 .061 105.29 0.50Sprite 93 .057 98.32 0.29

Dr. Pepper 88 .056 96.66 0.78

Others 581 .372 642.07 5.81fo = 1,726 22.83

Observed 2 = 22.83

= .05 df = k- 1 = 8 - 1 = 7

2.05,7 = 14.0671


The observed frequencies are not distributed the same as the expected frequenciesfrom the national poll.


33/36


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 Operato

r

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


34/36


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 thenull hypothesis. Position is not independent of number of years of experience.

12.26 H0: p = .43 n = 315 =.05

Ha: p .43 x = 120 /2 = .025

fo fee

eo

fff

2

)(

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


= .05 and /2 = .025 df = k- 1 = 2 - 1 = 1

2.025,1 = 5.0239

Since 2 = 3.09 < 2.025,1 = 5.0239, the decision is to fail to reject the nullhypothesis.


35/36


Type of College or University

Community

College

Large

University

Small

College

Number

ofChildren

0 25 178 31 234

1 49 141 12 2022 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


36/36


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


Number of children is not independent of type of College or University.

12.28 The observed chi-square is 30.18 with ap-value of .0000043. The chi-squaregoodness-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 is not the same for adults between 21 and 30 years of age as they

are for others. Marketing and sales people might reorient their 21 to 30 year oldefforts 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

associatedp-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.

Ken Black QA 5th chapter 12 Solution

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Transcript of Ken Black QA 5th chapter 12 Solution