Statistical Inference About Means and Proportions with Two Populations

Chapter 11 Statistical Inference about Means and Proportions with Two Populations

Learning Objectives

1. Be able to develop interval estimates and conduct hypothesis tests about the difference between the proportions of two populations.

2. Know the properties of the sampling distribution of the difference between two proportions ( p1  p2 ) .

3. Be able to conduct a goodness of fit test when the population is hypothesized to have a multinomial probability distribution.

4. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent.

5. Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence.

11 - 1 Chapter 11

Solutions:

1. a. p1 p 2 = .48 - .36 = .12

p1(1 p 1 ) p 2 (1  p 2 ) b. p1 p 2  z .05  n1 n 2

.48(1 .48) .36(1  .36) .12 1.645  400 300

.12  .0614 (.0586 to .1814)

.48(1 .48) .36(1  .36) c. .12 1.96  400 300

.12  .0731 (.0469 to .1931)

n p n p 200(.22) 300(.16) 2. a. p 1 1 2 2   .1840 n1 n 2 200  300

p p .22 .16 z 1 2   1.70 1 1  1 1  p1 p  .1840 1 .1840     200 300  n1 n 2   

p - value = .5000 - .4554 = .0446

b. p-value  .05; reject H0.

3. p1 = 220/400 = .55 p2 = 192/400 = .48

p1(1 p 1 ) p 2 (1  p 2 ) b. p1 p 2  z .025  n1 n 2

.55(1 .55) .48(1  .48) .55 .48  1.96  400 400

.07  .0691 (.0009 to .1391)

7% more executives are predicting an increase in full-time jobs. The confidence interval shows the difference may be from 0% to 14%.

4. a. p1 = 150/250 = .46 Republicans

p2 = 98/350 = .28 Democrats

b. p1 p 2 = .46 - .28 = .18

Republicans have a .18, 18%, higher participation rate than Democrats.

11 - 2 Statistical Inference about Means and Proportions with Two Populations

p1(1 p 2 ) p 2 (1  p 2 ) c. z.025  n1 n 2

.46(1 .46) .28(1  .28) 1.96  .0777 250 350

d. Yes, .18  .0777 (.1023 to .2577)

Republicans have a 10% to 26% higher participation rate in online surveys than Democrats. Biased survey results of online political surveys are very likely.

5. a. p1 = 256/320 = .80

b. p2 = 165/250 = .66

c. p1 p 2 = .80 - .66 = .14

p1(1 p 1 ) p 2 (1  p 2 ) .14 z.025  n1 n 2

.80(1 .80) .66(1  .66) .14 1.96  320 250

.14  .0733 (.0667 to .2133)

6. a. p1 = 742/924 = .803

b. p2 = 714/841 = .849

c. H0: p1 - p2  0

Ha: p1 - p2 < 0

Support for Ha will show p2 > p1 which indicates an improvement in on-time performance.

n p n p 742 714 d. p 1 1 2 2   .8249 n1 n 2 924  841

p p .803 .849 z 1 2   2.54 1 1  1 1  p1 p  .8249 1 .8249     924 841  n1 n 2   

p-value = .5000 - .4945 = .0055

Reject H0. Conclude on-time performance has improved.

11 - 3 Chapter 11

7. a. H0: p1 - p2 = 0

Ha: p1 - p2  0

p1 = 63/150 = .42

p2 = 60/200 = .30

n p n p 63 60 p 1 1 2 2   .3514 n1 n 2 150  200

p p .42 .30 z 1 2   2.33 1 1  1 1  p1 p  .3514 1 .3514     150 200  n1 n 2   

p-value = 2(.5000 - .4901) = .0198

p-value  .05, reject H0. There is a difference between the recall rates for the two commercials.

p1(1 p 1 ) p 2 (1  p 2 ) b. p1 p 2  z .025  n1 n 2

.42(1 .42) .30(1  .30) .42 .30  1.96  150 200

.12  .1014 (.0186 to .2214)

Commercial A has the better recall rate.

8. a. p1 = proportion of under 30 liking the ad a lot

p2 = proportion of 30 to 49 liking the ad a lot

H0: p1 - p2 = 0

Ha: p1 - p2  0

b. p1 = 49/100 = .49

p2 = 54/150 = .36

p1 p 2 = .49 - .36 = .13

n p n p 100(.49) 150(.36) c. p 1 1 2 2   .412 n1 n 2 100  150

p p .49 .36 z 1 2   2.05 1 1  1 1  p1 p  .412 1 .412     100 150  n1 n 2   

p-value = 2(.5000 - .4798) = .0404

p-value  .05, reject H0. There is a difference between the response to the ad by the younger under 30 and the older 30 to 49 age groups.

d. There is a statistically significant difference between the population proportions for the two age groups. The stronger appeal is with the younger, under 30, age group. Miller Lite is most likely pleased and encouraged by the results of the poll. "The Miller Lite Girls" ad ranked among the top three Super Bowl ads in advertising effectiveness. In addition, 49% of the younger, under 30, group liked the ad a lot. While a response of 36% for the older age group was not bad, Miller Lite probably liked, and probably expected, the higher rating among the younger audience. Since a younger audience contains the newer beer drinkers, appealing to the younger audience could bring new customers to the Miller Lite product. The older age group may be less likely to change from their established personal favorite beer because of the commercial.

9. a. H0: p1 - p2 = 0

Ha: p1 - p2  0

b. p1 = 141/523 = .2696 (27%)

p2 = 81/477 = .1698 (17%)

n p n p 141 81 c. p 1 1 2 2   .2220 n1 n 2 523  477

p p .2696 .1698 z 1 2   3.79 1 1  1 1  p1 p  .222 1 .222     523 477  n1 n 2   

p-value  0

Reject H0. There is a significant difference in the population proportions. A higher flying rate in 2003 is observed.

d. It may be that the general population is more acceptable to flying on vacation in 2003. Also, frequent flyer awards and special discount air fares in 2003 may have made 2003 flying more economical.

Note: In 1993, a round trip Newark to San Francisco was $388. In 2003, a special fare for the same trip was $238.

10. H0: p1 - p2  0

11 - 5 Chapter 11

Ha: p1 - p2  0

n p n p 240(.40) 250(.32) p 1 1 2 2   .3592 n1 n 2 240  250

1 1   1 1  s p(1  p )   (.3592)(1  .3592)   .0434 p1 p 2     n1 n 2  240 250 

 p1 p 2   0 .40 .32 z    1.85 s .0434 p1 p 2

p-value = .5000 - .4678 = .0322

p-value < .05, reject H0. The proportion of users at work is greater in Washington D.C.

11. a. Expected frequencies: e1 = 200 (.40) = 80, e2 = 200 (.40) = 80

e3 = 200 (.20) = 40

Actual frequencies: f1 = 60, f2 = 120, f3 = 20

(60 80)2 (120  80) 2 (20  40) 2  2    80 80 40 400 1600 400    80 80 40 5  20  10  35

k - 1 = 2 degrees of freedom

Using chi-square distribution with df = 2,  2 = 35 shows p-value  0

p-value  .01, reject H0

b. .01 = 9.210

2 Reject H0 if   9.210

2  = 35, reject H0

12. Expected frequencies: e1 = 300 (.25) = 75, e2 = 300 (.25) = 75

e3 = 300 (.25) = 75, e4 = 300 (.25) = 75

Actual frequencies: f1 = 85, f2 = 95, f3 = 50, f4 = 70

(85 75)2 (95  75) 2 (50  75) 2 (70  75) 2  2     75 75 75 75 100 400 625 25     75 75 75 75 1150  75  15.33

Chi-square table shows p-value less than .005. Actual p-value = .0016

p-value  .05, reject H0

The populations proportions are not the same.

13. H0 = pABC = .29, pCBS = .28, pNBC = .25, pIND = .18

Ha = The proportions are not pABC = .29, pCBS = .28, pNBC = .25, pIND = .18

Expected frequencies: 300 (.29) = 87, 300 (.28) = 84

300 (.25) = 75, 300 (.18) = 54

e1 = 87, e2 = 84, e3 = 75, e4 = 54

Actual frequencies: f1 = 95, f2 = 70, f3 = 89, f4 = 46

(95 87)2 (70  84) 2 (89  75) 2 (46  54) 2  2     87 84 75 54  6.87

Using  2 table, p-value is between .05 and .10

Actual p-value = .0762

p-value > .05, do not reject H0. There has not been a significant change in the viewing audience proportions.

14. Observed Expected

11 - 7 Chapter 11

Hypothesized Frequency Frequency 2 Category Proportion (fi) (ei) (fi - ei) / ei Brown 0.30 177 151.8 4.18 Yellow 0.20 135 101.2 11.29 Red 0.20 79 101.2 4.87 Orange 0.10 41 50.6 1.82 Green 0.10 36 50.6 4.21 Blue 0.10 38 50.6 3.14 Totals: 506 29.51

Using  2 table,  2 = 29.51 shows p-value  0

p-value < .05, reject H0. The percentages reported by the company have changed.

15. Observed Expected Hypothesized Frequency Frequency 2 Outlet Proportion (fi) (ei) (fi - ei) / ei Wal-Mart .24 42 33.6 2.10 Dept Stores .11 20 15.4 1.37 J.C. Penney .08 8 11.2 0.91 Kohl's .08 10 11.2 0.13 Mail Order .12 21 16.8 1.05 Other .37 39 51.8 3.16 Totals: 140 140 8.73

Degrees of freedom = 5

Using  2 table,  2 = 8.73 shows p-value greater than .10

p-value > .05, cannot reject H0. We cannot conclude that women shoppers in Atlanta differ from the outlet preferences expressed in the U.S. Shopper Database.

16. a. Observed Expected Hypothesized Frequency Frequency 2 Method Proportion (fi) (ei) (fi - ei) / ei Credit Card .22 46 48.4 .12 Debit Card .21 67 46.2 9.36 Personal Check .18 33 39.6 1.10 Cash .39 74 85.8 1.62 Totals: 220 220 12.21

Using  2 table,  2 = 12.21 shows the p-value is between .005 and .01

p-value  .01, reject H0. Conclude that the percentages for the methods of in-store payments have changed over the four year period.

b. 2003 1999 % Change Credit Card 46/220 = 21% 22% -1% Debit Card 67/220 = 30% 21% +9% Personal Check 33/220 = 15% 18% -3% Cash 74/220 = 34% 39% -5%

The primary change is that the debit card usage shows the biggest increase in method of payment (up 9%). Cash and personal check have seen the biggest decline in usage, 5% and 3% respectively.

c. 21% + 30% = 51%. Over half of in-store purchases are made using plastic.

17. Expected frequencies: 20% each n = 60

e1 = 12, e2 = 12, e3 = 12, e4 = 12, e5 = 12

Actual frequencies: f1 = 5, f2 = 8, f3 = 15, f4 = 20, f5 = 12

(5 12)2 (8  12) 2 (15  12) 2 (20  12) 2 (12  12) 2  2      12 12 12 12 12  11.50

k -1 = 4 degrees of freedom

Using  2 table, p-value is between .025 and .01

Reject H0. Yes, the largest companies differ in performance from the 1000 companies. In general, the largest companies did not do as well as others. 15 of 60 companies (25%) are in the middle group and 20 of 60 companies (33%) are in the next lower group. These both are greater than the 20% expected. Relative few large companies are in the top A and B categories.

Note that this result is for the year 2002. This should not be generalized to other years without additional data.

18. H0: p1 = .03, p2 = .28, p3 = .45, p4 = .24

2 Rating Observed Expected (fi - ei) / ei Excellent 24 .03(400) = 12 12.00 Good 124 .28(400) = 112 1.29 Fair 172 .45(400) = 180 .36 Poor 80 .24(400) = 96 2.67 400 400 2 = 16.31

Degrees of freedom = k - 1 = 3

Using  2 table,  2 = 16.31 shows p-value < .005

11 - 9 Chapter 11

p-value  .01, reject H0. Conclude that the ratings differ. A comparison of observed and expected frequencies show telephone service is slightly better with more excellent and good ratings.

19. H0 = The column variable is independent of the row variable

Ha = The column variable is not independent of the row variable

Expected Frequencies:

A B C P 28.5 39.9 45.6 Q 21.5 30.1 34.4

(20 28.5)2 (44  39.9) 2 (50  45.6) 2 (30  21.5) 2 (26  30.1) 2(30  34.4) 2  2       28.5 39.9 45.6 21.5 30.1 34.4  7.86

Degrees of freedom = (2-1)(3-1) = 2

Using  2 table,  2 = 7.86 provides a p-value between .01 and .025

p-value  .05, reject H0. Conclude that the column variable is not independent of the row variable.

20. H0 = The column variable is independent of the row variable

Ha = The column variable is not independent of the row variable

A B C P 17.5000 30.6250 21.8750 Q 28.7500 50.3125 35.9375 R 13.7500 24.0625 17.1875

(20 17.5000)2 (30  30.6250) 2 (30  17.1875) 2  2     17.5000 30.6250 17.1875  19.77

Using  2 table,  2 = 19.77 shows a p-value less than .005

p-value  .05, reject H0. Conclude that the column variable is not independent of the row variable.

21. H0 : Type of ticket purchased is independent of the type of flight

Ha: Type of ticket purchased is not independent of the type of flight.

e11 = 35.59 e12 = 15.41 e21 = 150.73 e22 = 65.27 e31 = 455.68 e32 = 197.32

Observed Expected Frequency Frequency 2 Ticket Flight (fi) (ei) (fi - ei) / ei First Domestic 29 35.59 1.22 First International 22 15.41 2.82 Business Domestic 95 150.73 20.61 Business International 121 65.27 47.59 Full Fare Domestic 518 455.68 8.52 Full Fare International 135 197.32 19.68 Totals: 920 100.43

Using  2 table,  2 = 100.43 shows a p-value  0

p-value  .05, reject H0. Conclude that the type of ticket purchased is not independent of the type of flight.

22. a. Observed Frequency (fij)

Domestic European Asian Total Same 125 55 68 248 Different 140 105 107 352 Total 265 160 175 600

Expected Frequency (eij)

Domestic European Asian Total Same 109.53 66.13 72.33 248 Different 155.47 93.87 102.67 352 Total 265 160 175 600 2 Chi Square (fij - eij) / eij

Domestic European Asian Total Same 2.18 1.87 0.26 4.32 Different 1.54 1.32 0.18 3.04 2 = 7.36

Using  2 table,  2 = 7.36 shows p-value is between .025 and .05

p-value  .05, reject H0. Conclude brand loyalty is not independent of manufacturer.

b. Brand Loyalty

11 - 11 Chapter 11

Domestic 125/265 = .472 (47.2%)  Highest European 55/160 = .344 (34.4%) Asian 68/175 = .389 (38.9%)

23. a. Observed Frequencies Health Insurance Size of Company Yes No Total Small 36 14 50 Medium 65 10 75 Large 88 12 100 Total 189 36 225

Expected Frequencies Health Insurance Size of Company Yes No Total Small 42 8 50 Medium 63 12 75 Large 84 16 100 Total 189 36 225

Chi Square Health Insurance Size of Company Yes No Total Small .86 4.50 5.36 Medium .06 .33 .39 Large .19 1.00 1.19  2 = 6.94

Using  2 table,  2 = 6.94 shows p-value between .025 and .05

p-value  .05, reject H0. Health insurance coverage is not independent of the size of the company.

b. Percentage of no coverage by company size:

Small 14/50  28% Medium 10/75  13% Large 12/100  12%

Small companies have slightly more than twice the percentage of no coverage for medium and large companies.

Effect on Grades Hours Worked Positive None Negative Total 1-15 26 50 14 90

16-24 16 27 17 60 25-34 11 19 20 50 Total 53 96 51 200

Effect on Grades Hours Worked Positive None Negative Total 1-15 23.85 43.20 22.95 90 16-24 15.90 28.80 15.30 60 25-34 13.25 24.00 12.75 50 Total 53.00 96.00 51.00 200

2 Chi Square (fij - eij) / eij

Effect on Grades Hours Worked Positive None Negative Total 1-15 .19 1.07 3.49 4.75 16-24 .00 .11 .19 .30 25-34 .38 1.04 4.12 5.55 2 = 10.60

Using  2 table,  2 = 10.60 shows p-value between .025 and .05

p-value  .05, reject H0. The effect on grades is not independent of the hours worked per week.

b. Row percentages:

Effect on Grades Hours Worked Positive None Negative 1-15 29% 56% 16% 16-24 27% 45% 28% 25-34 22% 38% 40%

As hours worked increases, the negative effect increased from 16% to 40%. As hours worked increases, both positive and no effect percentages decline. Higher hours worked increases the negative effect on grades.

25. Expected Frequencies:

e11 = 17.16 e12 = 12.84 e21 = 14.88 e22 = 11.12 e31 = 28.03 e32 = 20.97 e41 = 22.31 e42 = 16.69

11 - 13 Chapter 11

e51 = 17.16 e52 = 12.84 e61 = 15.45 e62 = 11.55

Observed Expected Frequency Frequency 2 Magazine Appeal (fi) (ei) (fi - ei) / ei News Guilt 20 17.16 0.47 News Fear 10 12.84 0.63 General Guilt 15 14.88 0.00 General Fear 11 11.12 0.00 Family Guilt 30 28.03 0.14 Family Fear 19 20.97 0.18 Business Guilt 22 22.31 0.00 Business Fear 17 16.69 0.01 Female Guilt 16 17.16 0.08 Female Fear 14 12.84 0.11 African-American Guilt 12 15.45 0.77 African-American Fear 15 11.55 1.03 Totals: 201 3.41

Using  2 table,  2 = 3.41 shows p-value is greater than .10

p-value > .01, do not reject H0. The hypothesis of independence cannot be rejected.

Pharm Consumer Computer Telecom Total Correct 207 136 151 178 672 Incorrect 3 4 9 12 28 Total 210 140 160 190 700

Pharm Consumer Computer Telecom Total Correct 201.6 134.4 153.6 182.4 672 Incorrect 8.4 5.6 6.4 7.6 28 Total 210 140 160 190 700

Pharm Consumer Computer Telecom Total Correct .14 .02 .04 .11 .31 Incorrect 3.47 .46 1.06 2.55 7.53 2 = 7.85 Degrees of freedom = (2-1)(4-1) = 3

p-value  .05, reject H0. Conclude order fulfillment is not independent of industry.

b. The pharmaceutical industry is doing the best with 207 of 210 (98.6%) correctly filled orders.

27. a. Observed Frequencies Hours of Sleep Age Less than 6 6 to 6.9 7 to 7.9 8 or more Total 49 or younger 38 60 77 65 240 50 or older 36 57 75 92 260 Total 74 117 152 157 500

Expected Frequencies

Hours of Sleep Age Less than 6 6 to 6.9 7 to 7.9 8 or more Total 49 or younger 36 56 73 75 240 50 or older 38 61 79 82 260 Total 74 117 152 157 500

Chi Square

Hours of Sleep Age Less than 6 6 to 6.9 7 to 7.9 8 or more Total 49 or younger .17 .26 .22 1.42 2.08 50 or older .16 .24 .21 1.31 1.92 2 = 4.01

Using  2 table,  2 = 4.01 shows a p-value greater than .10

p-value > .05, do not reject H0. Cannot reject the assumption that age and hours of sleep are independent.

b. Since age does not appear to have an effect on sleep on weeknights, use the overall percentages.

Less than 6 74/500  14.8% 6 to 6.9 117/500  23.4% 7 to 7.9 152/500  30.4% 8 or more 157/500  31.4%

28. a. p1 = population proportion for men

p2 = population proportion for women

H0: p1 - p2 = 0

Ha: p1 - p2  0

b. p1 = 248/800 = .31

11 - 15 Chapter 11

p2 = 156/600 = .26

n p n p 800(.31) 600(.26) c. p 1 1 2 2   .2886 n1 n 2 800  600

p p (.31 .26) z 1 2   2.04 1 1  1 1  p1 p  .2886 1 .2886     800 600  n1 n 2   

p-value = 2(.5000 - .4793) = .0414

p-value  .05, reject H0. Conclude the population proportions are not equal. The proportion is higher for men.

p1(1 p 1 ) p 2 (1  p 2 ) d. p1 p 2  z .025  n1 n 2

.31(1 .31) .26(1  .26) (.31 .26)  1.96  800 600

.05  .0475

Margin of Error = .0475

95% Confidence Interval (.0025 to .0975)

29. a. p1 = 76/400 = .19

p2 = 90/900 = .10

n p n p 76 90 p 1 1 2 2   .1277 n1 n 2 400  900

p p .19 .10 z 1 2   4.49 1 1  1 1  p1 p  .1277 1 .1277     400 900  n1 n 2   

p-value  0

Reject H0; there is a difference between claim rates.

p1(1 p 1 ) p 2 (1  p 2 ) b. p1 p 2  z .025  n1 n 2

.19(1 .19) .10(1  .10) .19 .10  1.96  400 900

.09  .0432 (.0468 to .1332)

Claim rates are higher for single males.

30. p1 = 9/142 = .0634

p2 = 5/268 = .0187

n p n p 9 5 p 1 1 2 2   .0341 n1 n 2 142  268

p p .0634 .0187 z 1 2   2.37 1 1  1 1  p1 p  .0341 1 .0341     142 268  n1 n 2   

p-value = 2(.5000 - .4911) = .0178

p-value  .02, reject H0. There is a significant difference in drug resistance between the two states. New Jersey has the higher drug resistance rate.

31. a. .38(430) = 163.4 Estimate: 163

.23(285) = 65.55 Estimate: 66

b. p1 p 2 .38  .23  .15

.38(1 .38) .23(1  .23) s    .064 p1 p 2 163 66

Confidence interval: .15  1.96(.064) or .15  .125(.025 to .275)

c. Yes, since the confidence interval in part (b) does not include 0, I would conclude that the Kodak campaign is more effective than most.

32. a. p1  .38 p2  .22

Point estimate = p1 p 2 .38  .22  .16

b. H0: p1 - p2  0

Ha: p1 - p2  0

n p n p (200)(.38) (200)(.22) c. p 1 1 2 2   .30 n1 n 2 200  200

1 1   2  s p(1  p )   (.3)(.7)  .0458 p1 p 2     200 200   200 

11 - 17 Chapter 11

.38 .22 z   3.49 .0458

z.01 = 2.33

With z = 3.49 > 2.33 we reject H0 and conclude that expectations for future inflation have diminished.

33. Observed 48 323 79 16 63 Expected 37.03 306.82 126.96 21.16 37.03

(48 37.03)2 (323  306.82) 2 (63  37.03) 2  2      41.69 37.03 306.82 37.03

Degrees of freedom = 5 - 1 = 4

p-value  .01, reject H0. Mutual fund investors' attitudes toward corporate bonds differ from their attitudes toward corporate stock.

34. Hypothesized Observed Expected 2 Passenger Car Proportion Frequency Frequency (fi - ei) / ei Toyota Camry .37 480 444 2.92 Honda Accord .34 390 408 .79 Ford Taurus .29 330 348 .93 Totals: 1200 1200 4.64

Degrees of Freedom: 2

Using  2 table,  2 = 4.64 shows the p-value is between .05 and .10

p-value > .05, cannot reject H0. Toyota Camry's market share appears to have increased to 480/1200 = 40%. However, the sample does not justify the conclusion that the market shares have changed from their historical 37%, 34%, 29% levels.

All three manufacturers will want to watch for additional sales reports before drawing a conclusion. 35. Observed 13 16 28 17 16 Expected 18 18 18 18 18

 2 = 7.44

p-value > .05, do not reject H0. The assumption that the number of riders is uniformly distributed cannot be rejected.

36. Observed Expected Hypothesized Frequency Frequency 2 Category Proportion (fi) (ei) (fi - ei) / ei Very Satisfied 0.28 105 140 8.75 Somewhat Satisfied 0.46 235 230 0.11 Neither 0.12 55 60 0.42 Somewhat Dissatisfied 0.10 90 50 32.00 Very Dissatisfied 0.04 15 20 1.25 Totals: 500 42.53

p-value  .05, reject H0. Conclude that the job satisfaction for computer programmers is different than the job satisfaction for IS managers.

Quality Shift Good Defective 1st 368.44 31.56 2nd 276.33 23.67 3rd 184.22 15.78

 2 = 8.10

Degrees of freedom = (3 - 1)(2 - 1) = 2

p-value  .05, reject H0. Conclude that shift and quality are not independent.

e11 = 1046.19 e12 = 632.81 e21 = 28.66 e22 = 17.34 e31 = 258.59 e32 = 156.41 e41 = 516.55 e42 = 312.45

Observed Expected Frequency Frequency 2 Employment Region (fi) (ei) (fi - ei) / ei Full-Time Eastern 1105 1046.19 3.31 Full-Time Western 574 632.81 5.46 Part-Time Eastern 31 28.66 0.19

11 - 19 Chapter 11

Part-Time Western 15 17.34 0.32 Self-Employed Eastern 229 258.59 3.39 Self-Employed Western 186 156.41 5.60 Not Employed Eastern 485 516.55 1.93 Not Employed Western 344 312.45 3.19 Totals: 2969 23.37

p-value  .05, reject H0. Conclude that employment status is not independent of region.

39. Expected frequencies: Loan Approval Decision Loan Offices Approved Rejected Miller 24.86 15.14 McMahon 18.64 11.36 Games 31.07 18.93 Runk 12.43 7.57

 2 = 2.21

p-value > .05, do not reject H0. The loan decision does not appear to be dependent on the officer.

Never Married Married Divorced Total Men 234 106 10 350 Women 216 168 16 400 Total 450 274 26 750

Never Married Married Divorced Total Men 210 127.87 12.13 350 Women 240 146.13 13.87 400 Total 450 274 26 750

Never Married Married Divorced Total Men 2.74 3.74 .38 6.86 Women 2.40 3.27 .33 6.00 2 = 12.86

Using  2 table,  2 = 12.86 shows p-value is less than .005

p-value  .01, reject H0. Conclude martial status is not independent of gender.

b. Martial Status

Never Married Married Divorced Men 66.9% 30.3% 2.9% Women 54.0% 42.0% 4.0%

Men 100 - 66.9 = 33.1% have been married Women 100 - 54.0 = 46.0% have been married

41. Observed Frequencies

Church Attendance Age Yes No Total 20 to 29 31 69 100 30 to 39 63 87 150 40 to 49 94 106 200 50 to 59 72 78 150 Total 260 340 600

Expected Frequencies

Church Attendance Age Yes No Total 20 to 29 43 57 100 30 to 39 65 85 150 40 to 49 87 113 200 50 to 59 65 85 150 Total 260 340 600

Chi Square

Church Attendance Age Yes No 20 to 29 3.51 2.68 6.19 30 to 39 .06 .05 .11 40 to 49 .62 .47 1.10 50 to 59 .75 .58 1.33 2 = 8.73

Using  2 table,  2 = 8.73 shows the p-value is between .025 and .05.

11 - 21 Chapter 11

p-value  .05, reject H0. Conclude church attendance is not independent of age.

Attendance by age group:

20 - 29 31/100  31% 30 - 39 63/150  42% 40 - 49 94/200  47% 50 - 59 72/150  48%

Church attendance increases as individuals grow older.

42. a. Observed Binomial Prob. Expected x Frequencies n = 4, p = .30 Frequencies 0 30 .2401 24.01 1 32 .4116 41.16 2 25 .2646 26.46 3 10 .0756 7.56 4 3 .0081 .81 100 100.00

The expected frequency of x = 4 is .81. Combine x = 3 and x = 4 into one category so that all expected frequencies are 5 or more.

Observed Expected x Frequencies Frequencies 0 30 24.01 1 32 41.16 2 25 26.46 3 or 4 13 8.37 100 100.00

b.  2 = 6.17

Using  2 table,  2 = 6.17 shows p-value is greater than .10 Actual p-value = .1034

p-value > .05, do not reject H0. Conclude that the assumption of a binomial distribution cannot be rejected.

11 - 22