Reserve Problems Chapter 10 Section 2 Problem 10

The following is a random sample of 15 measurements from high-flow rivers and 13 from low-

flow rivers of a total algae content (units are mg/L).

High

Low

High

Low

24.3

21.4

32

46.1

24.8

62.6

24.4

29

34.6

38.8

40.5

44.8

42.5

50.3

66

41.7

57

37.1

76.8

14.8

79.8

36.3

44.9

19.4

18.8

58

49.9

57.4

(a) Test the null hypothesis at

0.05

=

, that the amount of algae content is the same in both

high- and low-flow rivers. Assume that the variances are equal. Suppose that the alternative

hypothesis is

1

H

:

12

0

−

. Determine the value of test statistic. Determine the condition to

reject the null hypothesis. Select the correct conclusion. Is the alternative one- or two-sided?

(b) Find a 95% confidence interval for the difference in the mean algae content for the two flow

rates.

(c) Is the value zero contained in the 95% confidence interval? Explain the connection with the

conclusion you reached in part (a).

SOLUTION

(a)

The parameter of interest is the difference in mean algae content

12

−

, with

00=

.

(b)

Use a 95% two-sided confidence interval

Reserve Problems Chapter 10 Section 2 Problem 11

Olympic swimmers are seeded according to their previous 12-month performances with faster

swimmers going into the later heats. The last 24 swimmers, however, are distributed among the

last three heats more evenly. So we should see large differences in times of heats one–five but

not among the last three heats. The data of times from heats five–seven are in seconds for the

100m swim. NA indicates that the swimmer did not swim.

Time

Heat

Time

Heat

49.02

5

48.67

6

49.49

5

49.18

6

49.6

5

49.2

6

49.78

5

NA

6

49.95

5

48.54

7

50.08

5

48.67

7

NA

5

48.93

7

NA

5

48.93

7

48.19

6

48.97

7

48.29

6

49.03

7

48.54

6

49.29

7

48.6

6

Is there a statistically significant difference in the mean time of swimmers in heats 6 and 7 and

the mean time of swimmers in heat 5? Assume that the variances are equal. Use

0.05

=

.

Suppose that the hypotheses are

0

H

:

12

0

−=

versus

1

H

:

12

0

−

, where heats 6 and 7

are group 1 and heat 5 is group 2. Determine the value of test statistic. Select the correct

conclusion.

Is there evidence to suggest that the means of the heats differ for slower swimmers in heat five

and the faster swimmers in heat seven? Assume that the variances are equal. Use

0.05

=

.

Suppose that the hypotheses are

0

H

:

12

0

−=

versus

1

H

:

12

0

−

, where heat 5 is group

1 and heat 7 is group 2. Determine the value of test statistic. Select the correct conclusion.

What about the means of the two sets of elite swimmers in heats six and seven? Assume that the

variances are equal. Use

0.05

=

. Suppose that the hypotheses are

0

H

:

12

0

−=

versus

1

H

:

12

0

−

, where heat 6 is group 1 and heat 7 is group 2. Determine the value of test statistic.

Determine the condition to reject the null hypothesis. Select the correct conclusion.

SOLUTION

Heats 6 and 7 are group 1 and heat 5 is group 2. The parameter of interest is the difference in

mean time

12

−

, with

00=

.

0 1 2

: 0H

−=

Heat 6 is group 1 and heat 7 is group 2. The parameter of interest is the difference in mean time

12

−

, with

00=

.

Reserve Problems Chapter 10 Section 2 Problem 12

An article in Polymer Degradation and Stability (2006, Vol. 91) presented data from a nine-year

aging study on S537 foam. Foam samples were compressed to 50% of their original thickness

and stored at different temperatures for nine years. At the start of the experiment as well as

during each year, sample thickness was measured, and the thicknesses of the eight samples at

each storage condition were recorded. The data for two storage conditions follow.

50ºC

0.047

0.060

0.061

0.064

0.080

0.090

0.118

0.165

0.183

60ºC

0.062

0.105

0.118

0.137

0.153

0.197

0.210

0.250

0.375

(a) Is there evidence to support the claim that mean compression increases with the temperature

at the storage condition? Assume that the variances are equal. Use

0.05

=

. Suppose that the

hypotheses are

0

H

:

12

0

−=

versus

1

H

:

12

0

−

, where 1 indicates the storage at 50°C

and 2 indicates the storage at 60°C. Determine the value of test statistic. Determine the condition

to reject the null hypothesis. Select the correct conclusion.

(b) Find a 95% confidence interval for the difference in the mean compression for the two

temperatures.

(c) Is the value zero contained in the 95% confidence interval? Explain the connection with the

conclusion you reached in part (a).

SOLUTION

(a)

The parameter of interest is the difference in mean compression between storage conditions

12

−

, with

00=

.

Reserve Problems Chapter 10 Section 3 Problem 1

Consider the days, when a promo action of type A fired price tags with a discount) on one day

and a promo action of type B (every fifth pack is free) on another day were held in the

supermarket.

The values in the following table are indicators of the promo effectiveness (sales volume in

thousand dollars).

promo action of type A

348

81

60

238

150

promo campaign of type B

359

69

40

206

166

Use the Wilcoxon rank-sum test to test

0 1 2

:H

=

against the alternative

1 1 2

:H

, where 1

indicates type A and 2 indicates type B. Use

0.01

=

.

The parameters of interest are the means of the effectiveness indicator for the two types of

promotion.

0 1 2

:H

=

where 1 indicates type A and 2 indicates type B.

1 1 2

:H

. The alternative is ____-sided.

Because

0.01

=

and

12

nn==

___, the critical value of the rank sums is

0.01

ww

==

___. If

1

w

is ____

0.01

w=

____, we will reject

0 1 2

:H

=

.

The data are arranged in the ascending order and ranked as follows:

(Enter the missing values.)

Type

Volume of sales

Rank

__

___

1

A

60

2

B

69

3

A

81

4

__

___

5

__

___

6

B

206

7

A

238

8

__

___

9

B

359

10

The test statistic is

1

w=

__

Conclusion: Because

1

w

is ____

0.01

w

, we ____ reject the null hypothesis that both promo actions

exhibit the same mean effectiveness.

SOLUTION

6) The data are arranged in the ascending order and ranked as follows:

Type

Volume of sales

Rank

B

40

1

Reserve Problems Chapter 10 Section 3 Problem 2

The manufacturer of bicycles is interested in testing two different tires for bicycle wheels. The

tire that is the most wear-resistant would be preferable. The manufacturer obtains 10 samples of

each tire types and tests them. The data from the following table show how many kilometers the

bicycle will pass before the tire bursts.

Unit 1

954

999

1098

1090

985

1005

1009

1063

1086

998

Unit 2

1003

1053

900

997

1008

1011

972

1091

1073

1025

Use the Wilcoxon rank-sum test to test

0 1 2

:H

=

against the alternative

1 1 2

:H

. Use

0.01

=

.

The parameters of interest are the means of the effectiveness indicator for the two types of

promotion.

0 1 2

:H

=

1 1 2

:H

Because

0.01

=

and

12

nn==

___, the critical value of the rank sums is

0.01

ww

==

___. If

either

1

w

or

2

w

is ________

0.01

w=

___, we will reject

0 1 2

:H

=

.

The data are arranged in the ascending order and ranked as follows:

(Enter the missing values.)

Unit

Kilometers

Rank

2

900

1

1

954

2

_

___

3

_

___

4

2

997

5

1

998

6

_

___

7

_

___

8

1

1005

9

2

1008

10

1

1009

11

2

1011

12

2

1025

13

2

1053

14

1

1063

15

2

1073

16

1

1086

17

1

1090

18

2

1091

19

1

1098

20

The test statistics are

1

w=

___

and

2

w=

___

Conclusion: Because neither

1

w

or

2

w

is _________

0.01

w

, we _____ reject the null hypothesis

that both types of tires exhibit the same mean distance passed before the tires burst.

SOLUTION

6) The data are arranged in the ascending order and ranked as follows:

Unit

Kilometers

Rank

2

900

1

2

1003

8

1

1005

9

2

1008

10

1

1009

11

2

1011

12

The sum of the ranks for Unit 1 is

Reserve Problems Chapter 10 Section 4 Problem 1

To estimate efficiency of a drug for weight loss, the clinical trial was performed. The results are

presented in the table below.

Patient number

Weight before trial, kg

Weight after trial, kg

1

85.2

83.5

2

79.6

78.1

3

75.8

73.2

4

76.2

74.0

5

91

90.2

6

89.8

87.0

7

82.0

79.9

8

81.7

78.5

9

67.3

64.0

10

68.4

65.1

11

70.0

67.8

12

74

70.0

13

66.8

64.6

14

60

58.6

15

94

92.9

16

88.2

88.0

(a) Use the paired t-test to investigate the claim that the drug affects the weight. Test the null

hypothesis

0:0

d before after

H

= − =

against

1:0

d

H

using

0.01

=

.

The parameter of interest is the difference in mean weight,

d

, where

i

d=

weight before - weight

after.

0:0

d

H

=

1:0

d

H

Reject the null hypothesis if

0

t

________ for

0.01

=

.

Find the test statistic.

(b) Construct a 99% two-sided confidence interval for the difference in weight.

SOLUTION

(a)

1) The parameter of interest is the difference in mean weight,

d

, where

i

d=

weight before -

7) Conclusion: Because

8.45 2.947

, we reject the null hypothesis. There is evidence to support

Reserve Problems Chapter 10 Section 4 Problem 2

The table below contains the results of a students’ test before the course and after its completion.

Student number

Score before

Score after

1

64

72

0:0

d before after

1:0

d

0.1

=

The parameter of interest is the difference in mean weight,

d

, where

i

d=

weight before -

weight after.

0:0

d

H

=

1:0

d

H

Reject the null hypothesis if

0

t

________.

Find the test statictic.

(b) Calculate a one-sided confidence limit that can be used to answer the question in part (a).

SOLUTION

(a)

1) The parameter of interest is the difference in mean weight,

d

, where

i

d=

score before -

Reserve Problems Chapter 10 Section 4 Problem 3

Fifteen adult males between the ages of 35 and 50 participated in a study to evaluate the effect of

diet and exercise on blood cholesterol levels. The total cholesterol was measured in each subject

initially and then three months after participating in an aerobic exercise program and switching

to a low-fat diet. The data are shown in the following table.

Blood Cholesterol Level

Subject

Before

After

1

265

229

2

240

231

3

256

228

4

295

240

5

251

238

6

245

241

7

287

234

8

313

259

9

254

244

10

279

239

11

283

246

12

240

218

13

238

219

14

225

226

15

247

233

(a) Do the data support the claim that low-fat diet and aerobic exercise are of value in producing

a mean reduction in blood cholesterol levels? Use

0.05

=

. Suppose that the hypotheses are

0

H

:

0

d

=

versus

1

H

:

0

d

, where

i i i

d Before After=−

. What is the test statistic?

(b) Calculate a one-sided confidence limit that can be used to answer the question in part (a).

SOLUTION

(a)

1) The parameter of interest is the difference in blood cholesterol level,

d

, where

Reserve Problems Chapter 10 Section 4 Problem 4

An article in Neurology (1998, Vol. 50, pp. 1246-1252) discussed that monozygotic twins share

numerous physical, psychological, and pathological traits. The investigators measured an

intelligence score of 10 pairs of twins, and the data follow:

Pair

Birth order: 1 (first)

Birth order: 2 (second)

1

6.08

5.73

2

6.22

5.80

3

7.99

8.42

4

7.44

6.84

5

6.48

6.43

6

7.99

8.76

7

6.32

6.32

8

7.60

7.62

9

6.03

6.59

10

7.52

7.67

(a) Find 98% confidence interval on the difference in mean score.

(b) It is important to detect a mean difference in score of one point with a probability of at least

0.90. How many pairs should have been used?

SOLUTION

(a)

10n=

(b)

Reserve Problems Chapter 10 Section 4 Problem 5

Use the sign test on the blood cholesterol data in table.

Blood Cholesterol Level

Subject

Before

After

1

265

229

2

240

231

3

251

245

4

295

240

5

251

238

6

245

241

7

287

234

8

308

257

9

250

240

10

279

239

11

283

246

12

240

218

13

238

219

14

225

226

15

247

233

Is there evidence that diet and exercise reduce the median cholesterol level (

0.05

=

)?

What is the number of signs r which should be compared with the critical value?

What is the critical value?

There ___ enough evidence that diet and exercise reduce the median cholesterol level at

0.05

=

.

SOLUTION

1) Parameters of interest are the median cholesterol levels for two activities

˜

d

, where

6) Construct the table of differences and find

1r−=

.

Blood Cholesterol Level

Subject

Before

After

delta

sign

1

265

229

36

+

2

240

231

9

+

7) Conclusion: Because

13

, reject the null hypothesis. There is a significant difference in the

Reserve Problems Chapter 10 Section 4 Problem 6

Use the Wilcoxon signed-rank test on the blood cholesterol data in table.

Blood Cholesterol Level

Subject

Before

After

1

265

229

2

240

231

3

257

226

4

295

240

5

251

238

6

245

241

7

287

234

8

317

250

9

258

249

10

279

239

11

283

246

12

240

218

13

238

219

14

225

226

15

247

233

Is there evidence that diet and exercise reduce the mean cholesterol level at

0.05

=

?

What is the sum of signed-rank W which should be compared with the critical value?

What is the critical value

*

W

=

There ____ enough evidence that diet and exercise reduce the median cholesterol level at

0.05

=

.

SOLUTION

1) Parameters of interest are the median cholesterol levels for two activities

˜

d

, where

6) Construct the table of differences and find

1w−=

.

Blood Cholesterol Level

Subject

Before

After

delta

1

265

229

36

2

240

231

9

Reserve Problems Chapter 10 Section 5 Problem 1

Consider the hypothesis test

22

0 1 2

:H

=

against

22

1 1 2

:H

. Suppose the sample sizes are

116n=

and

221n=

and the sample standard deviations are

11.9s=

and

21.5s=

. Use

0.05

=

.

a) Test the hypothesis. Find the P-value.

b) Construct a 95% two-sided confidence interval of

2

relations.

SOLUTION

a) 1) The parameters of interest are the standard deviations,

1

and

2

.

6) Reject the null hypothesis if

12

0 1 /2, 1, 1 0,975,15,20

0.025,20,15

11

0.389

2.76

nn

f f f f

− − −

= = = =

or

b) 95% confidence interval:

Reserve Problems Chapter 10 Section 5 Problem 2

Hardness of water from two different water treatment facilities is investigated. Observed water

hardness (in ppm) for a random sample of faucets is as follows:

Facility 1:

62

56

57

61

65

57

60

59

54

61

58

60

58

Facility 2:

69

65

59

62

61

57

59

60

60

62

61

66

68

66

a) Is there evidence that the variance of water hardness is different for two facilities? Use

0.05

=

.

b) Construct a 95% two-sided confidence interval of

2

relations.

SOLUTION

a) 1) The parameters of interest are the standard deviations,

1

and

2

.

4) The test statistic is

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