The Comparison of Two Populations

Chapter 8

The comparison of two populations

8-1.n = 25 = 19.08 = 30.67

H0: D = 0H1: D 0

t (24) = = 3.11

Reject H0 at = 0.01.

Paired Difference Test
Evidence
Size / 25 / n / Assumption
Average Difference / 19.08 / D / Populations Normal
Stdev. of Difference / 30.67 / sD
Note: Difference has been defined as
Test Statistic / 3.1105 / t
df / 24
Hypothesis Testing / At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0048 / Reject

8-2.n = 40 = 5 = 2.3

H0: D = 0H1: D 0

t(39) = = 13.75

Strongly reject H0. 95% C.I. for D 2.023(2.3/) = [4.26, 5.74].

8-3.n = 12 = 1.29 = 2.2(D = International – Domestic)

H0: D = 0H1: D 0

t (11) = = 2.034

At = 0.05, we cannot reject H0 of no difference in the average returns.

8-4.n = 60 = 0.2 = 1

H0: D 0H1: D > 0

t(24) = = 1.549. At = 0.05, we cannot reject H0.

Paired Difference Test
Evidence
Size / 60 / n / Assumption
Average Difference / 0.2 / D / Populations Normal
Stdev. of Difference / 1 / sD
Note: Difference has been defined as
Test Statistic / 1.5492 / t
df / 59
Hypothesis Testing / At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.1267
H0: 1 2 >= / 0 / 0.9367
H0: 1 2 <= / 0 / 0.0633

8-5.n = 15 = 3.2 = 8.436(D = After – Before)

H0: D 0H1: D > 0

t (14) = = 1.469

There is no evidence that the shelf facings are effective.

8-6.n = 25 = 4 = 8.436

H0: D 0H1: D > 0

t (24) = = 10

Reject H0. There is strong evidence that the average investment proportion did change in

mid-October in favor of Hong Kong.

8-7.Power at D = 0.1n = 60 = 1.0 = 0.01

H0: D 0H1: D > 0

C = + 2.326() = 0.30029We need:

P( > C | D = 0.1)

= P( > 0.30029 | D = 0.1)

= P

= P(Z > 1.55) = 0.0606

8-8.n = 20 = 1.25 = 42.896

H0: D = 0H1: D 0

t (19) = = 0.13

Do not reject H0; no evidence of a difference.

Paired Difference Test
Evidence
Size / 20 / n / Assumption
Average Difference / 1.25 / D / Populations Normal
Stdev. of Difference / 42.89 / sD
Note: Difference has been defined as
Test Statistic / 0.1303 / t
df / 19
Hypothesis Testing / At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.8977

8-9. = 45 = 32 = 26 = 21 = 8 = 6

H0: 0H1: < 0

z = = 3.13

Reject H0. There is evidence that LINC reduces average programming time.

8-10. = = 30

H0: = 0H1: 0

Nikon (1): = 8.5s1 = 2.1Minolta (2): = 7.8 = 1.8

z = = 1.386

Do not reject H0. There is no evidence of a difference in the average ratings of the two cameras.

8-11.Bel Air (1): = 32 = 345,650 = 48,500

Marin (2): = 35 = 289,440 = 87,090

H0: = 0H1: 0

z = = 3.3

Reject H0. There is evidence that the average Bel Air price is higher.

Evidence
Sample1 / Sample2
Size / 32 / 35 / n
Mean / 345650 / 289440 / x-bar
Popn. 1 / Popn. 2
Popn. Std. Devn. / 48500 / 87090 / 

Hypothesis Testing
Test Statistic / 3.2996 / z
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0010 / Reject

8-12.(Use template: “testing difference in means.xls”)

(need to use the t-test since the population std. dev. is unknown)

H0: μJ – μSP = 0 H1: μJ – μSP ≠ 0

t-Test for Difference in Population Means

Do not reject the null hypothesis. The funds perform equally.

8-13.Music: = 128 = 23.5 = 12.2

Verbal: = 212 = 18.0 = 10.5

H0: = 0H1: 0

z = = 4.24

Reject H0. Music is probably more effective.

Evidence
Sample1 / Sample2
Size / 128 / 212 / n
Mean / 23.5 / 18 / x-bar
Popn. 1 / Popn. 2
Popn. Std. Devn. / 12.2 / 10.5 / 

Hypothesis Testing
Test Statistic / 4.2397 / z
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0000 / Reject

8-14. = 13 = 13 = 20.385 = 10.385

= 7.622 = 4.292 = .05

H0: u1 = u2H1: u1 u2

Use a critical value of 2.064 for a two-tailed test. Reject H0. The two methods do differ.

8-15.Liz (1): = 32 = 4,238 = 1,002.5

Calvin (2): = 37 = 3,888.72 = 876.05

a.one-tailed:H0: 0H1: > 0

1. z = = 1.53
2. At = 0.5, the critical point is 1.645. Do not reject H0 that Liz Claiborne models do not get more money, on the average.
3. p-value = .5  .437 = .063(It is the probability of committing a Type I error if we choose to reject and H0 happens to be true.)

8-16.American patents (1): = 100 = 1,838.69 = 461

Japanese patents (2): = 80 = 1,050.22 = 560

H0: 0H1: > 0

z = = 10.14

Strong evidence to reject H0; the Japanese probably pay more in fees.

8-17.Non-research (1): = 255 = 0.64

Research (2): = 300 = 0.85

= 2.54

95% C.I. for is:()

= 2.54 1.96 = [2.416, 2.664] percent.

8-18.Audio (1): = 25 = 87 = 12

Video (2): = 20 = 64 = 23

H0: = 0H1: 0

t(43) = = 4.326

Reject H0. Audio is probably better (higher average purchase intent). Waldenbooks should concentrate in audio.

Evidence
Sample1 / Sample2
Size / 25 / 20 / n
Mean / 87 / 64 / x-bar
Std. Deviation / 12 / 23 / s
Assuming Population Variances are Equal
Pooled Variance / 314.116 / s2p
Test Statistic / 4.3257 / t
df / 43
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0001 / Reject

8-19.With training (1): = 13 = 55 = 8

Without training (2): = 15 = 48 = 6

H0: 4,000H1: > 4,000

t (26) = = 1.132

The critical value at = .05 for t (26) in a right-hand tailed test is 1.706. Since 1.132 < 1.706, there is no evidence at = .05 that the program executives get an average of $4,000 per year more than other executives of comparable levels.

8-20.(Use template: “testing difference in means.xls”)

(need to use the t-test since the population std. dev. is unknown)

H0: μC – μR = 0 H1: μC – μR ≠ 0

t-Test for Difference in Population Means

Reject the null hypothesis: the average cost of adoptions is not equal.

8-21.(Use template: “testing difference in means.xls”)

(need to use the t-test since the population std. dev. is unknown)

t-Test for Difference in Population Means

Data
Obese / Non-Ob
Sample1 / Sample2
45 / 54
54 / 69
39 / 36
67 / 77
33 / 56
62 / 30
39 / 72
50 / 80
48 / 44
51 / 73
37 / 55
61 / 60
70 / 49
35
42
40

H0: μo – μno ≤ 0 H1: μo – μno > 0

Accept the null hypothesis (p-value = 0.9673), obese people get paid less than non-obese people.

8-22.Old (1): = 19 = 8.26 = 1.43

New (2): = 23 = 9.11 = 1.56

H0: 0H1: > 0

t (40) = = 1.82

Some evidence to reject H0 (p-value = 0.038) for the t-distribution with df = 40, in a one-tailed test.

8-23.Take proposed route as population 1 and alternate route as 2. Assume equal variance for both populations.

H0: 0

H1: > 0

p-value from the template = 0.8674

cannot reject H0

8-24.nb = 18 = 7.4sb = 1.3

na = 23 = 8.2sa = 2.4

H0: 0H1: > 0

t(39) = = 1.273

There is no evidence that management decisions increase average customer satisfaction.

8-25.“Yes” (1): = 25 = 12 = 2.5

“No” (2): = 25 = 13.5 = 1

Assume independent random sampling from normal populations with equal population variances.

H0: 0H1: > 0

t(48) = = 2.785

At = 0.05, reject H0. Also reject at = 0.01. p-value = 0.0038.

Evidence
Sample1 / Sample2
Size / 25 / 25 / n
Mean / 12 / 13.5 / x-bar
Std. Deviation / 2.5 / 1 / s
Assuming Population Variances are Equal
Pooled Variance / 3.625 / s2p
Test Statistic / -2.7854 / t
df / 48
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0076 / Reject
H0: 1 2 >= / 0 / 0.0038 / Reject

8-26.H0: = 0H1: 0

z = = 0.8887

Do not reject H0. There is no evidence of a difference in average stock returns for the two periods.

8-27.Public sources (1): = 12 = 12,500 = 3,400

Private sources (2): = 18 = 21,000 = 5,000

t (28) = = 5.136

There is strong evidence that private sources lend more, on the average.

8-28.From Problem 8-25:

= = 25 = 12 = 13.5 = 2.5 = 1

We want a 95% C.I. for :

() 2.011

= (13.5 – 12) 2.001

= [0.4170, 2.5830] percent.

8-29.Before (1): = 85 = 100

After (2): = 68 = 100

H0: p1 – p2 0H1: p1 – p2 > 0

z = = = 2.835

Reject H0. On-time departure percentage has probably declined after NW’s merger with Republic. p-value = 0.0023.

Evidence / Sample 1 / Sample 2
Size / 100 / 100 / n
#Successes / 85 / 68 / x
Proportion / 0.8500 / 0.6800 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.7650
Test Statistic / 2.8351 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.0046 / Reject
H0: p1 - p2 >= 0 / 0.9977
H0: p1 - p2 <= 0 / 0.0023 / Reject

8-30.Small towns (1): = 1,000x1 = 850

Big cities (2): = 2,500x2= 1,950

H0: p1 – p2 0H1: p1 – p2 > 0

z = = 4.677

Reject H0. There is strong evidence that the percentage of word-of-mouth recommendations in small towns is greater than it is in large metropolitan areas.

8.31. = 31x1 = 11 = 50x2= 19

H0: p1 – p2 = 0H1: p1 – p2 0

z = = 0.228

Do not reject H0. There is no evidence that one corporate raider is more successful than the other.

8-32.Before campaign (1): = 2,060 = 0.13

After campaign (2): = 5,000 = 0.19

H0: p2p1 .05H1: p2 – p1 > .05

z = = = 1.08

No evidence to reject H0; cannot conclude that the campaign has increased the proportion of people who prefer California wines by over 0.05.

8-33.95% C.I. for p2p1:() 1.96

= .06 1.96 = [0.0419, 0.0781]

We are 95% confident that the increase in the proportion of the population preferring California wines is anywhere from 4.19% to 7.81%.

Confidence Interval
 / Confidence Interval
95% / 0.0600 / ± / 0.0181 / = / [ / 0.0419 / , / 0.0782 / ]

8-34.The statement to be tested must be hypothesized before looking at the data:

Chase Man. (1): = 650x1 = 48

Manuf. Han. (2): = 480x 2 = 20

H0: p 1 – p 2 0H1: p 1 – p 2 > 0

z = = 2.248

Reject H0. p-value = 0.0122.

8-35.American execs (1): = 120x1 = 34

European execs (2): = 200x 2 = 41

H0: p 1 – p 2 0H1: p 1 – p 2 > 0

z = = 1.601

At = 0.05, there is no evidence to conclude that the proportion of American executives who prefer the A320 is greater than that of European executives. (p-value = 0.0547.)

Evidence / Sample 1 / Sample 2
Size / 120 / 200 / n
#Successes / 34 / 41 / x
Proportion / 0.2833 / 0.2050 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.2344
Test Statistic / 1.6015 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.1093
H0: p1 - p2 >= 0 / 0.9454
H0: p1 - p2 <= 0 / 0.0546

8-36.Cleveland (1): = 1,000x1 = 75 = .075

Chicago (2): = 1,000x 2 = 72 = .072

H0: p 1 – p 2 = 0H1: p 1 – p 2 0 = (72 +75)/2,000 = .0735

z = = 0.257

We cannot reject H0. p. value = 0.7971

8-37.(Use template: “testing difference in proportions.xls”)

H0: pGQ – pE = 0 H1: pGQ – pE ≠ 0

Comparing Two Population Proportions

Evidence / Sample 1 / Sample 2
Size / 200 / 200 / n
#Successes / 48 / 61 / x
Proportion / 0.2400 / 0.3050 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.2725
Test Statistic / -1.4599 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.1443
H0: p1 - p2 >= 0 / 0.0722
H0: p1 - p2 <= 0 / 0.9278

Do not reject the null hypothesis, the number of subscribers is equal.

8-38.(Use template: “testing difference in proportions.xls”)

H0: pE – pW = 0 H1: pE – pW ≠ 0

Comparing Two Population Proportions

Evidence / Sample 1 / Sample 2
Size / 1000 / 1000 / n
#Successes / 483 / 551 / x
Proportion / 0.4830 / 0.5510 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.5170
Test Statistic / -3.0428 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.0023 / Reject
H0: p1 - p2 >= 0 / 0.0012 / Reject
H0:p1 - p2 <= 0 / 0.9988

Reject the null hypothesis: the proportion of returns filed by paid preparers is not equal.

8-39.Motorola (1): = 120x1 = 101p1 = .842

Blaupunkt (2): = 200x 2 = 110p2 = .550

H0: p 1p 2H1: p 1p 2 = (101 +110)/320 = .659

z = = 5.33

Strongly reject H0; Motorola’s system is superior (p-value is very small).

8-40.Old method (1): = 40 = 1,288

New method (2): = 15 = 1,112

H0:1222H1: 1222use = .05

F (39,14) = s 12/s 22 = 1,288/1,112 = 1.158

The critical point at = .05 is F (39,14) = 2.27 (using approximate df in the table). Do not reject H0. There is no evidence that the variance of the new production method is smaller.

F-Test for Equality of Variances
Sample 1 / Sample 2
Size / 40 / 15
Variance / 1288 / 1112
Test Statistic / 1.158273 / F
df1 / 39
df2 / 14
At an  of
Null Hypothesis / p-value / 5%
H0: 21 - 22 = 0 / 0.7977
H0: 21 - 22 >= 0 / 0.6012
H0: 21 - 22 <= 0 / 0.3988

8-41.Test the equal-variance assumption of Problem 8-27:

H0: 12 = 22H1: 1222

F (17,11) = (5,000)2/(3,400)2 = 2.16

Do not reject H0 at = 0.10. (The approximate critical point is 2.7.)

8-42.“Yes” (1): = 25s1= 2.5

“No” (2): = 25s2= 1

H0: 12 = 22H1: 1222

Put the larger s 2 in the numerator and use 2:

F (24,24) = s12/ s22 = (2.5) 2/(1) 2 = 6.25

From the F table using = .01, the critical point is F (24,24) = 2.66. Therefore, reject H0. The population variances are not equal at = 2(.01) = 0.02.

F-Test for Equality of Variances
Sample 1 / Sample 2
Size / 25 / 25
Variance / 6.25 / 1
Test Statistic / 6.25 / F
df1 / 24
df2 / 24
At an  of
Null Hypothesis / p-value / 5%
H0: 21 - 22 = 0 / 0.0000 / Reject
H0: 21 - 22 >= 0 / 1.0000
H0: 21 - 22 <= 0 / 0.0000 / Reject

8-43.n1 = 21s1 = .09n2 = 28s2 = .122

F (27,20) = (.122)2/(.09)2 = 1.838

At = .10, we cannot reject H0 because the critical point for = .05 from the table with df’s = 30, 20 is 2.04 and for df’s 24, 20 it is 2.08. We did not reject H0 at = .10 so we would also not reject it at = .02. Hence this particular C.I. contains the value 1.00.

8-44.Before (1): = 12 = 16,390.545

After (2): = 11 = 86,845.764

H0:12 = 22H1: 1222

F (10,11) = 5.298

The critical point from the table, using = .01, is F (10,11) = 4.54. Therefore, reject H0. The population variances are probably not equal. p-value < .02 (double the ).

F-Test for Equality of Variances
Sample 1 / Sample 2
Size / 11 / 12
Variance / 86845.76 / 16390.55
Test Statistic / 5.298528 / F
df1 / 10
df2 / 11
At an  of
Null Hypothesis / p-value / 1%
H0: 21 - 22 = 0 / 0.0109
H0: 21 - 22 >= 0 / 0.9945
H0: 21 - 22 <= 0 / 0.0055 / Reject

8-45.n1 = 25s1 = 2.5n2 = 25s2 = 3.1

H0: 12 = 22H1: 1222 = .02

F (24,24) = (3.1)2/(2.5)2 = 1.538

From the table: F .01(24,24) = 2.66. Do not reject H0. There is no evidence that the variances in the two waiting lines are unequal.

8-46.nA = 25sA2 = 6.52nB = 22sB2 = 3.47

H0: A2 = B2H1: A2B2 = .01

F (24,21) = 6.52/3.47 = 1.879

The critical point for = .01 is F (24,21) = 2.80. Do not reject H0. There is no evidence that stock A is riskier than stock B.

F-Test for Equality of Variances
Sample 1 / Sample 2
Size / 25 / 22
Variance / 6.52 / 3.47
Test Statistic / 1.878963 / F
df1 / 24
df2 / 21
At an  of
Null Hypothesis / p-value / 1%
H0: 21 - 22 = 0 / 0.1485
H0: 21 - 22 >= 0 / 0.9258
H0: 21 - 22 <= 0 / 0.0742

8-47.The assumptions we need are: independent random sampling from the populations in question, and normal population distributions. The normality assumption is not terribly crucial as long as no serious violations of this assumption exist. In time series data, the assumption of random sampling is often violated when the observations are dependent on each other through time. We must be careful.

8-48.Savannah (1):n1 = 17 = 6.8235294s1 = 1.4677915

Kingston (2):n2 = 9 = 8.2222222s2 = 2.3333333

H0: 12 = 0H1: 12 0

t (24) =

= = –1.88

Critical point: t .025(24) = 2.064. At = 0.05 there is no evidence that one port demands, on the

average, more containers than the other. (If we wanted to prove 21 , the answer would have been yes, because t .05(24) = 1.711 < 1.88.)

8-49.99% C.I. for sk:

() t .005(24)

= (6.8235294 – 8.2222222) 2.797(.743285) = [3.4776612, 0.6802753]

The C.I. contains zero as expected from the results of Problem 8-48.

Confidence Interval for difference in Population Means
 / Confidence Interval
99% / -1.3987 / ± / 2.07893 / = / [ / -3.4776 / , / 0.68024 / ]

8-50. = 51 = 4.636s d = 7.593

H0: u d 0H1: u d> 0

t (10) = = 2.025

Reject H0.Performance did improve after the sessions.

8-51.For Problem 8-50:

95% C.I.: t (10) s d/

= 4.636 2.228 = 4.636 5.101 = [0.465, 9.737]

Confidence Intervals for the Difference in Means
(1 - ) / Confidence Interval
95% / 4.636 / ± / 5.10105 / = / [ / -0.465 / , / 9.73705 / ]

8-52.Chrysler (1):n1 = 25 = 9,500s1 = 1,500

GM (2):n2 = 25 = 9,780s2 = 1,500

H0: 21 0H1: 21 > 0

t = = 0.66

Do not reject H0. There is evidence that a GM car costs more on the average.

8-53.99% C.I. for the difference between the average cost for GM and the average cost for Chrysler:

() 2.682

= 280 2.682(424.26) = [857.9, 1417.9]

The C.I. contains zero, as expected.

8-54.Using the method of Section 8-3 is easier and does not require the assumption of equal population variances. Redo using n1 = n2 = 100:

z = = = 1.32

We still cannot reject H0 (p-value = 0.0934).

8-55.x1 = 60n1 = 80x2= 65n2 = 100= 125/180 = .6944

H0: p1p2 = 0H1: p1p2 0

z = = = 1.447

Do not reject H0. (There is no evidence that one movie will be more successful than the other

(p-value = 0.1478).

Evidence / Sample 1 / Sample 2
Size / 80 / 100 / n
#Successes / 60 / 65 / x
Proportion / 0.7500 / 0.6500 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.6944
Test Statistic / 1.4473 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.1478

8-56.95% C.I. for the difference between the two population proportions:

() 1.96

= 0.10 1.96 = [0.0332, 0.2332]

Yes, 0 is in the C.I., as expected from the results of Problem 8-55.

8-57.K:nK = 12K = 12.55sK = .7342281

L:nL = 12L = 11.925sL = .3078517

H0: KL = 0H1: KL 0

t (22) = = 2.719

Reject H0. The critical points for t (22) at = .02 are 2.508. Critical points for t (22) at = .01 are 2.819. So .01 < p-value < .02. The L-boat is probably faster.

Evidence
Sample1 / Sample2
Size / 12 / 12 / n
Mean / 12.55 / 11.925 / x-bar
Std. Deviation / 0.73423 / 0.30785 / s
Assuming Population Variances are Equal
Pooled Variance / 0.31693 / s2p
Test Statistic / 2.7194 / t
df / 22
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0125 / Reject

8-58.Do Problem 8-57 with the data being paired. The differences KL are:

0.21.00.21.02.20.20.80.91.00.20.61.2

n = 12 = .625sD = .7723929

t (11)= = 2.803

2.718 < 2.803 < 3.106 (between the critical points of t (11) for = .01 and .02).

Hence, .01 < p-value < .02, which is as before, in Problem 8-57 (the pairing did not help much

herewe reach the same conclusion).

Paired Difference Test
Evidence
Size / 12 / n / Assumption
Average Difference / 0.625 / D / Populations Normal
Stdev. of Difference / 0.77239 / sD
Note: Difference has been defined as
Test Statistic / 2.8031 / t
df / 11
Hypothesis Testing / At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0172 / Reject
H0: 1 2 >= / 0 / 0.9914
H0: 1 2 <= / 0 / 0.0086 / Reject

8-59.(Use template: “testing difference in means.xls”; sheet:”t-test from stats”)

H0: μAIG – μFB = 0 H1: μAIG – μFB ≠ 0

t-Test for Difference in Population Means

(Use t-test since population std. dev. unknown.)

Reject the null hypothesis: the premiums are not the same.

8-60.IIT (1):n1 = 100 = 0.94

Competitor (2):n2 = 125 = 0.92

H0: p1p2 = 0H1: p1p2 0 = .92888

z = = 0.58

There is no evidence that one program is more successful than the other.

8-61.Design (1):n1 = 15 = 2.17333s1 = .3750555

Design (2):n2 = 13 = 2.5153846s2 = .3508232

H0: 21 = 0H1: 21 0

t (26) = = 2.479

p-value = .02. Reject H0. Design 1 is probably faster.

8-62.H0: 12 = 22H1: 1222

F (14,12) = s12/ s22 = (.3750555)2/(.3508232)2 = 1.143

Do not reject H0 at = 0.10. (Since 1.143 < 2.62. Also < 2.10, so the p-value > 0.20.) The solution of Problem 8-61 is valid from the equal-variance requirement.

8-63.A = After:nA = 16 = 91.75sA = 5.0265959

B = Before:nB = 15 = 84.7333sB = 5.3514573

H0: 5H1: > 5

t (29) = = 1.08

Do not reject H0. There is no evidence that advertising is effective.

8-64.H0: 12 = 22H1: 1222

F (14,15) = (5.3514573)2/(5.0265959)2 = 1.133

Do not reject H0 at = 0.10. There is no evidence that the population variances are not equal.

F-Test for Equality of Variances
Sample 1 / Sample 2
Size / 15 / 16
Variance / 28.6381 / 25.26667
Test Statistic / 1.133434 / F
df1 / 14
df2 / 15
At an  of
Null Hypothesis / p-value / 10%
H0: 21 - 22 = 0 / 0.8100
H0: 21 - 22 >= 0 / 0.5950
H0: 21 - 22 <= 0 / 0.4050

8-65.Savannah (1): n1 = 17s1 = 1.4677915

Kingston (2):n2 = 9s2 = 2.333333

H0: 1222H1: 1222

Since s12s22, do not reject H0.

8-66.H0: = H1:

F (11,11) = (.7342281)2/(.3078517)2 = 5.688

Critical point for = 0.02 is about 4.5. Therefore, reject H0. Thus the analysis in Problem 8-57 is not valid. We need to use the other test. The other test also gives t = 2.719 but the df are obtained using Equation (8-6):

df = = approximately 14 (rounded downward).

t.02(14) = 2.624 < 2.719 < 2.977 = t .01(14), hence 0.01 < p-value < 0.02. Reject H0.

8-67.Differences A – B:

11331481057221265102212

= 2.375sD = 9.7425185n = 16

t (15) = = 0.9751

Do not reject H0. There is no evidence that one package is better liked than the other.

Paired Difference Test
Evidence
Size / 16 / n / Assumption
Average Difference / -2.375 / D / Populations Normal
Stdev. of Difference / 9.74252 / sD
Note: Difference has been defined as
Test Statistic / -0.9751 / t
df / 15
Hypothesis Testing / At an  of
Null Hypothesis / p-value / 5%
H0: 1 2= / 0 / 0.3450
H0: 1 2 >= / 0 / 0.1725
H0: 1 2 <= / 0 / 0.8275

8-68.Supplier A:nA = 200xA = 12

Supplier B:nB = 250xB = 38

H0: pA – pB = 0H1: pA – pB 0 = (12 +38)/450 = .1111

z = = = 3.086

Reject H0. p-value = .002. Supplier A is probably more reliable as the proportion of defective components is lower.

8-69.95% C.I. for the difference in the proportion of defective items for the two suppliers:

() 1.96

=.092 1.96(.0282415) = [0.0366, 0.1474].

Confidence Interval
 / Confidence Interval
95% / 0.0920 / ± / 0.0554 / = / [ / 0.0366 / , / 0.1474 / ]

8-70.90% C.I. for the difference in average occupancy rate at the Westin Plaza Hotel before and after the advertising:

() 1.699

= 7.016667 3.1666375 = [3.85, 10.18] percent occupancy.

8-71.BellAtlantic: n = 24 = $13.1s = 1.2

Nynex: n = 24 = $13.4s = 1.2

H0: u1 = u2H1: u1u2sp2 = 1.44

t (46) = = –.866

Accept H0. Annual profits are equal.

8-72.(Use template: “testing difference in proportions.xls”)

H0: pE – pW = 0 H1: pE – pW ≠ 0

Comparing Two Population Proportions

Evidence / Sample 1 / Sample 2
Size / 3000 / 3000 / n
#Successes / 140 / 351 / x
Proportion / 0.0467 / 0.1170 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.0818
Test Statistic / -9.9376 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.0000 / Reject
H0: p1 - p2 >= 0 / 0.0000 / Reject
H0: p1 - p2 <= 0 / 1.0000

Reject the null hypothesis: the proportion of men and women having plastic surgery is not equal.

8-73.(Use template: “testing difference in means.xls”; sheet:”t-test from stats”)

H0: μ2 – μ1 = 0 H1: μ2 – μ1 ≠ 0

t-Test for Difference in Population Means

Do not reject the null hypothesis: the ratings are the same.

8-74.a. = 2500 = 39 = 2500 = 35

s1 = s2 = 2 = .05

H0: u1 = u2H1: u1u2

z = = 70.711

Reject H0. The average workweek has shortened.

1. 95% C.I.: (39 –35) 1.96

= 4.1109 = [3.8891, 4.1109]

8-75.Percentage making purchases did not increase. (p-value = 0.0687)

Comparing Two Population Proportions
Evidence / Sample 1 / Sample 2
Size / 684 / 663 / n
#Successes / 143.65 / 118 / x
Proportion / 0.2100 / 0.1780 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.1942
Test Statistic / 1.4858 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.1373
H0: p1 - p2 >= 0 / 0.9313
H0: p1 - p2 <= 0 / 0.0687

8-76.Yes. Lower income households are less likely to have internet access. (p-value = 0.0038)

Comparing Two Population Proportions
Evidence / Sample 1 / Sample 2
Size / 500 / 500 / n
#Successes / 350 / 310 / x
Proportion / 0.7000 / 0.6200 / p-hat
Hypothesis Testing
Hypothesized Difference Zero
Pooled p-hat / 0.6600
Test Statistic / 2.6702 / z
At an  of
Null Hypothesis / p-value / 5%
H0: p1 - p2 = 0 / 0.0076 / Reject
H0: p1 - p2 >= 0 / 0.9962
H0: p1 - p2 <= 0 / 0.0038 / Reject

8-77.Sale price has increased

t-Test for Difference in Population Means
Evidence / Assumptions
Sample1 / Sample2 / Populations Normal
Size / 400 / 350 / n / H0: Population Variances Equal
Mean / 128000 / 125000 / x-bar / F ratio / 1.08
Std. Deviation / 2500 / 2700 / s / p-value / 0.4563
Assuming Population Variances are Equal
Pooled Variance / 6735241 / s2p
Test Statistic / 15.7935 / t
df / 748
At an  of / Confidence Interval for difference in Population Means
Null Hypothesis / p-value / 5% /  / Confidence Interval
H0: 1 2 = / 0 / 0.0000 / Reject / 95% / 3000 / ± / 372.902

8-78The ration of the variances is 3.18. The degrees of freedom for both samples is 10 – 1 = 9. Using the F-table for 9 degrees of freedom in both the numerator and the denominator, we find a value of 3.18 when α = 0.05. Therefore, there is a 5% chance.

8-79(Use template: “testing difference in means.xls”; sheet:”t-test from data”)

1. Assuming equal variances:

H0: μ2 – μ1 = 0 H1: μ2 – μ1 ≠ 0

t-Test for Difference in Population Means

Data
Co.1 / Co.2
Sample1 / Sample2
2570 / 2055
2480 / 2940
2870 / 2850
2975 / 2475
2660 / 1940
2380 / 2100
2590 / 2655
2550 / 1950
2485 / 2115
2585
2710

Evidence:

Sample1 / Sample2
Size / 11 / 9 / n
Mean / 2623.18 / 2342.22 / x-bar
Std. Deviation / 174.087 / 393.55 / s
Assuming Population Variances are Equal
Pooled Variance / 85673.3 / s2p
Test Statistic / 2.1356 / t
df / 18
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0467 / Reject
H0: 1 2 >= / 0 / 0.9766
H0: 1 2 <= / 0 / 0.0234 / Reject

At 0.05 level of significance, reject the null hypothesis that the charges are the same.

2. Test the assumption of equal variances.

Assumptions
Populations Normal
H0: Population Variances Equal
F ratio / 5.11054
p-value / 0.0193

Reject null hypothesis: the variances are not equal.

3.Assuming unequal variances,

H0: μ2 – μ1 = 0 H1: μ2 – μ1 ≠ 0

Assuming Population Variances are Unequal
Test Statistic / 1.98846 / t
df / 10
At an  of
Null Hypothesis / p-value / 5%
H0: 1 2 = / 0 / 0.0748
H0: 1 2 >= / 0 / 0.9626
H0: 1 2 <= / 0 / 0.0374 / Reject

Accept the null hypothesis: the charges are not different.

1