Statistical Inference About Means and Proportions with Two Populations

Chapter 10

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 two population means whenandare known.

2.Know the properties of the sampling distribution of .

3.Be able to use the t distribution to conduct statistical inferences about the difference between two population means whenandare unknown.

4.Learn how to analyze the difference between two population means when the samples are independent and when the samples are matched.

5.Be able to develop interval estimates and conduct hypothesis tests about the difference between two population proportions.

6.Know the properties of the sampling distribution of .

Solutions:

1.a. = 13.6 - 11.6 = 2

b.1.645

2  .98(1.02 to 2.98)

c.1.96

2  1.17(.83 to 3.17)

2.a.

b.p-value = 1.0000 - .9788 = .0212

c.p-value .05, reject H0.

3.a.

b.p-value = 2(.0630) = .1260

c.p-value > .05, do not reject H0.

4.a.= 2.04 - 1.72 = .32

c..32  .04(.28 to .36)

5.a.= 135.67 – 68.64 = 67.03

c.67.03  17.08(49.95 to 84.11) We estimate that men spend $67.03 more

than women on Valentine’s Day with a margin of error of $17.08.

6.= Mean loan amount for 2002

= Mean loan amount for 2001

H0:

Ha:

p-value = 1.0000 - .9850 = .0150

p-value .05; reject H0. The mean loan amount has increased between 2001 and 2002.

7.a.= Population mean 2002

= Population mean 2003

H0:

Ha:

b.With time in minutes,= 172 - 166 = 6 minutes

p-value = 1.0000 - .9955 = .0045

p-value .05; reject H0. The population mean duration of games in 2003 is less than the population mean in 2002.

6  4.5(1.5 to 10.5)

e.Percentage reduction: 6/172 = 3.5%. Management should be encouraged by the fact that steps taken in 2003 reduced the population mean duration of baseball games. However, the statistical analysis shows that the reduction in the mean duration is only 3.5%. The interval estimate shows the reduction in the population mean is 1.5 minutes (.9%) to 10.5 minutes (6.1%). Additional data collected by the end of the 2003 season would provide a more precise estimate. In any case, most likely the issue will continue in future years. It is expected that major league baseball would prefer that additional steps be taken to further reduce the mean duration of games.

8.a.

b.p-value = 2(1.0000 - .8599) = .2802

c.p-value > .05; do not reject H0. Cannot conclude that there is a difference between the population mean scores for the two golfers.

9.a.= 22.5 - 20.1 = 2.4

Use df = 45.

c.t.025 = 2.014

d.2.4 2.1(.3 to 4.5)

10.a.

Use df = 65

c.Using t table, area in tail is between .01 and .025

two-tail p-value is between .02 and .05.

Exactp-value corresponding to t = 2.18 is .0329

d.p-value .05, reject H0.

11.a.

c.= 9 - 7 = 2

Use df = 9, t.05 = 1.833

2 2.17(-.17 to 4.17)

12.a.= 22.5 - 18.6 = 3.9

Use df = 87, t.025 = 1.988

3.9 (.6 to 7.2)

13.a.

b.= 9.3 - 4.2 = 5.1 tons

Memphis is the higher volume airport and handled an average of 5.1 tons per day more than Louisville. Memphis handles more than twice the volume of Louisville.

Use df = 17, t.025 = 2.110

5.1  1.82(3.28 to 6.92)

14.a.H0:

Ha:

c.With n1 = 150 and n2 = 175, degrees of freedom will be well over 100. Using last row of the t table, the area in the tail at t= 2.18 is between .01 and .025. Thus, the two-tail p-value is between .02 and .05

Exactp-value = .03

d.p-value .05, reject H0. The population mean ages differ for the coastal and noncoastal areas.

15.1 for 2001 season

2 for 1992 season

H0:

Ha:

b.= 60 - 51 = 9 days

9/51(100) = 17.6% increase in number of days.

Using t table, p-value is between .005 and .01.

Exact p-value corresponding to t = 2.48 is .0076

p-value .01, reject H0. There is a greater mean number of days on the disabled list in 2001.

d.Management should be concerned. Players on the disabled list have increased 32% and time on the list has increased by 17.6%. Both the increase in inquiries to players and the cost of lost playing time need to be addressed.

16.a.1 = population mean verbal score parents college grads

2 = population mean verbal score parents high school grads

H0:

Ha:

= 525 - 487 = 38 points higher if parents are college grads

Use df = 25

Using t table, p-value is between .025 and .05

Exact p-value corresponding to t = 1.80 is .0420

d.p-value .05, reject H0. Conclude higher population mean verbal scores for students whose parents are college grads.

17.a.H0:

Ha:

Use df = 16

Using t table, p-value is between .025 and .05

Exact p-value corresponding to t = 1.99 is .0320

d.p-value .05, reject H0. The consultant with more experience has a higher population mean rating.

18.a.H0:

Ha:

Use df = 78

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

Exact p-value corresponding to t = -2.10 is .0195

p-value .05, reject H0. The improvement is less than the stated average of 120 points.

df = 78

7543(32 to 118)

d.This is a wide interval. A larger sample should be used to reduce the margin of error.

19.a.1, 2, 0, 0, 2

df = n - 1 = 4

Using t table, p-value is between .025 and .05

Exact p-value corresponding to t = 2.24 is .0443

Reject H0; conclude d > 0.

20.a.3, -1, 3, 5, 3, 0, 1

d.= 2

e.With 6 degrees of freedom t.025 = 2.447

2  1.93(.07 to 3.93)

21.Difference = rating after - rating before

H0: d 0

Ha: d > 0

= .625 and = 1.30

df = n - 1 = 7

Using t table, p-value is between .10 and .20

Exact p-value corresponding to t = 1.36 is .1080

Do not reject H0; we cannot conclude that seeing the commercial improves the mean potential to purchase.

22.Let di = current qtr. per share earnings – previous quarter per share earnings

With df = 24, t.025 = 2.064

.2064  2.064

Confidence interval: $.21  $.11 ($.10 to $.32)

Earnings have increased. The point estimate of the increase in earnings per

share is $.21 with a margin of error of $.11.

23.a.1 = population mean grocery expenditures

2 = population mean dining-out expenditures

H0:

Ha:

df = n - 1 = 41

p-value0

Conclude that there is a difference between the annual population mean expenditures for groceries and for dining-out.

c.Groceries has the higher mean annual expenditure by an estimated $850.

850  350 (500 to 1200)

24.H0:d≤ 0

Ha:d> 0

Differences 177, -21, 186, -131, 22, 212, -5, 14

df = n - 1 = 7

Using t table, p-value is greater than .10

Exact p-value corresponding to t = 1.32 is .1142

Since p-value > .10, do not reject H0. We cannot conclude that airfares fromDayton are higher thanthose from Louisville at a α = .05 level of significance.

25. a.H0:d= 0

Ha:d 0

Use difference data: -3, -2, -4, 3, -1, -2, -1, -2, 0, 0, -1, -4, -3, 1, 1

df = n - 1 = 14

Using t table, area is between .01 and .025.

Two-tail p-value is between .02 and .05.

Exact p-value corresponding to t = -2.36 is .0333

p-value .05, reject H0. Conclude that there is a difference between the population mean weekly usage for the two media.

b.hours per week for cable television.

hours per week for radio.

Radio has greater usage.

26.a.Differences .91, -.30, -.84, -.18, -.34, .53, -.20, .10, -.78, .07

df = n – 1 = 9

Using t table, area in tail is greater than .20

Two-tail p-value must be greater than .40

Exact p-value corresponding to t = -.60 is .5633

Cannot reject H0. No significant difference observed.

b.= -.103; Actual was less than predicted. Analysts overestimated earnings.

c.df = 9t = 2.262

.39

Given the point estimate of -.103, the margin of error is large. A larger sample is desired.

27.Difference = Price deluxe - Price Standard

H0: d = 10

Ha: d 10

= 8.86 and = 2.61

df = n - 1 = 6

Using t table, area is between .10 and .20

Two-tail p-value is between .20 and .40

Exact p-value corresponding to t = -1.16 is .2901

Do not reject H0; we cannot reject the hypothesis that a $10 price differential exists.

28.a.= .48 - .36 = .12

.12  .0614(.0586 to .1814)

.12  .0731(.0469 to .1931)

29.a.

p - value = 1.0000 - .9554 = .0446

b.p-value .05; reject H0.

30. = 220/400 = .55 = 192/400 = .48

.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%.

31.a.= 150/250 = .46 Republicans

= 98/350 = .28 Democrats

b.= .46 - .28 = .18

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

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.

32.a.

b. = 300/811 = .369937% of women would ask directions

c. = 255/750 = .340034% of men would ask directions

Upper tail p-value is the area to the right of the test statistic

Using normal table with z = 1.23: p-value = 1 - .8907 = .1093

p-value > α ; do not reject

We cannot conclude that women are more likely to ask directions.

33.a. = 256/320 = .80

b. = 165/250 = .66

c.= .80 - .66 = .14

.14  .0733(.0667 to .2133)

34.a. = 742/924 = .803

b. = 714/841 = .849

c.H0: p1 - p2 0

Ha: p1 - p2 < 0

Support for Ha will show p2p1 which indicates an improvement in on-time performance.

p-value = .0055

Reject H0. Conclude on-time performance has improved.

35.a.H0: p1 - p2 = 0

Ha: p1 - p2  0

= 63/150 = .42

= 60/200 = .30

p-value = 2(1.0000 - .9901) = .0198

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

.12  .1014(.0186 to .2214)

Commercial A has the better recall rate.

36.a. = proportion of under 30 liking the ad a lot

= proportion of 30 to 49 liking the ad a lot

H0: p1 - p2 = 0

Ha: p1 - p2  0

b.= 49/100 = .49

= 54/150 = .36

= .49 - .36 = .13

p-value = 2(1.0000 - .9798) = .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.

37.a.H0: p1 - p2 = 0

Ha: p1 - p2  0

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

= 81/477 = .1698(17%)

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.

38.H0: 1 - 2 = 0

Ha: 1 - 2 0

p-value = 2(1.0000 - .9974) = .0052

p-value .05, reject H0. A difference exists with system B having the lower mean checkout time.

39.a.

Difference = 660 - 330 = 330

Using sample mean prices, the 5-megapixel is twice as expensive as the 3-megapixel camera.

Use df = 10, t.025 = 2.228

330  160(170 to 490)

40.a.H0: 1 - 2 0

Ha: 1 - 2 > 0

b.n1 = 30n2 = 30

= 16.23= 15.70

s1 = 3.52s2 = 3.31

Use df = 57

Using t table, p-value is greater than .20

Exactp-value corresponding to t = .60 is .2754

p-value > .05, do not reject H0. Cannot conclude that the mutual funds with a load have a greater mean rate of return.

41. a.n1 = 10n2 = 8

= 21.2= 22.8

s1 = 2.70s2 = 3.55

= 21.2 - 22.8 = -1.6

Kitchens are less expensive by $1600.

Use df = 12, t.05 = 1.782

-1.6  2.7(-4.3 to 1.1)

42.a.Use difference data: 21, 38, 14, 5, 12, 9, 6, 28, 30, -4, 13, 6, 27, 10, 10

Note: Ocala at -4 is the only area showing a decrease in price over the one-year period.

= 225/15 = 15 ($15,000)

df = n - 1 = 14, t.05 = 1.761

15  5.19(9.81 to 20.19)

c.For 2002,

Percentage increase: 15/131 = 11.5%

43.a.p1 = population proportion for men

p2 = population proportion for women

H0: p1 - p2 = 0

Ha: p1 - p2  0

b. = 248/800 = .31

= 156/600 = .26

p-value = 2(1.0000 - .9793) = .0414

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

.05  .0475

Margin of Error = .0475

95% Confidence Interval(.0025 to .0975)

44.a.= 76/400 = .19

= 90/900 = .10

p-value  0

Reject H0; there is a difference between claim rates.

.09  .0432(.0468 to .1332)

Claim rates are higher for single males.

45.= 9/142 = .0634

= 5/268 = .0187

p-value = 2(1.0000 - .9911) = .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.

46. a.= 163/430 = .38

= 66/285 = .23

Confidence interval: .15  1.96(.0355) or .15  .0696(.0804 to .2196)

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.

47. a.

Point estimate =

b.H0: p1 - p2  0

Ha: p1 - p2  0

p-value ≈ .00

With p-value ≤ .01, we reject H0 and conclude that expectations for future inflation have diminished.

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