Black Business Statistics Study Guide Ch16

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


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.


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 _______________.


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.


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


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.


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.


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



null hypothesis.

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

expected frequencies which are based on the responses of women.


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 x = 20, z = -1.44 .4251

z = 6.13

63.3930 = -0.71 -.2611



for x = 30, z = -0.71 .2611

z = 6.13

63.3940 = 0.03 +.0120




z = 6.13

63.3950 = 0.76 .2764

for x = 40, z = 0.03 -.0120



z = 6.13

63.3960 = 1.50 .4332

for x = 50, z = 0.76 -.2764



z = 6.13

63.3970 = 2.23 .4871

for x = 60, z = 1.50 -.4332


For < 10:

Probability between 10 and the mean = .0603 + .1640 + .2611 = .4854

Probability < 10 = .5000 - .4854 = .0146

For > 70:

Probability between 70 and the mean = .0120 + .2644 + .1568 + .0539 = .4871

Probability > 70 = .5000 - .4871 = .0129


Age Probability fe

< 10 .0146 (.0146)(231) = 3.37

10-20 .0603 (.0603)(231) = 13.93

20-30 .1640 37.88

30-40 .2731 63.09

40-50 .2644 61.08

50-60 .1568 36.22

60-70 .0539 12.45

> 70 .0129 2.98

Categories < 10 and > 70 are less than 5.

Collapse the < 10 into 10-20 and > 70 into 60-70.

Age fo fe 0

2

0 )(

f

ff e

10-20 16 17.30 0.10

20-30 44 37.88 0.99

30-40 61 63.09 0.07

40-50 56 61.08 0.42

50-60 35 36.22 0.04

60-70 19 15.43 0.83

2.45

df = k - 3 = 6 - 3 = 3, = .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.


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


2.025,1 = 5.02389, the decision is to fail to

reject the null hypothesis.


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



the null hypothesis.

Variable One is independent of Variable Two.


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


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



null hypothesis.

Number of children is not independent of social class.


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


2.05,2 = 5.9915, the decision is to


Transportation mode is not independent of industry.


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


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

2

.05,3 = 7.8147


2.05,3 = 7.8147, the decision is to fail to


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.


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.


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



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

observed frequency of arrivals is Poisson distributed.


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


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



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


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


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



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.

Documents

Black Business Statistics Study Guide Ch16