(b)

(d)

00.638 0.635 0.23

−−

(e)

95% one-sided confidence interval

Reserve Problems Chapter 9 Section 3 Problem 6

An article in the ASCE Journal of Energy Engineering [“Overview of Reservoir Release

Improvements at 20 TVA Dams” (Vol. 125, April 1999, pp. 1–17)] presents data on dissolved

oxygen concentrations in streams below 20 dams in the Tennessee Valley Authority system. The

observations are (in milligrams per liter): 5.0, 3.4, 3.9, 1.3, 0.2, 0.9, 2.7, 3.7, 3.8, 4.1, 1.0, 1.0,

0.8, 0.4, 3.8, 4.5, 5.3, 6.1, 6.9, and 6.5.

(a) Test the hypothesis

0:4H

=

versus

1:4H

. Use

0.01

=

. Is it possible to reject

0

H

hypothesis at the 0.01 level of significance? Find the P-value.

(b) Check the normality assumption, using the normal probability plot.

(c) Compute the power of the test if the true mean dissolved oxygen concentration is as low as 3.

(d) What sample size would be required to detect a true mean dissolved oxygen concentration as

low as 2.5 if you wanted the power of the test to be at least 0.9?

(e) Which confidence interval should be used to consider

0

H

hypothesis? Find upper confidence

bound for this interval.

SOLUTION

(a)

In order to use t statistics in hypothesis testing, we need to assume that the underlying

distribution is normal.

(b)

(c)

034 0.47

2.127

d

−−

= = = =

(e)

The 99% two-sided confidence interval.

Reserve Problems Chapter 9 Section 3 Problem 7

A research engineer for a tire manufacturer is investigating tire life for a new rubber compound

and has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard

deviation are 60489.7 and 3645.94 kilometers.

(a) The engineer would like to demonstrate that the mean life of this new tire is in excess of

60,000 kilometers. Formulate the appropriate hypotheses. Test the appropriate hypotheses using

0.05

=

. Find

0

t

. Is it possible to reject

0

H

hypothesis at the 0.05 level of significance?

(b) Suppose that if the mean life is as long as 61,000 kilometers, the engineer would like to

detect this difference with probability at least 0.90. Was the sample size

16n=

used in part (a)

adequate?

SOLUTION

(a)

The parameter of interest is the true mean tire life, m. roughness.

0: 60000H

=

and

1: 60000H

Reserve Problems Chapter 9 Section 3 Problem 8

The United States Golf Association tests golf balls to ensure that they conform to the rules of

golf. Balls are tested for weight, diameter, roundness, and overall distance. The overall distance

test is conducted by hitting balls with a driver swung by a mechanical device nicknamed “Iron

Byron” after the legendary great Byron Nelson, who’s swing the machine is said to emulate.

Following are 100 distances (in yards) achieved by a particular brand of golf ball in the overall

distance test.

261.3 259.4 265.7 270.6 274.2 261.4 254.5 283.7

258.1 270.5 255.1 268.9 267.4 253.6 234.3 263.2

254.2 270.7 233.7 263.5 244.5 251.8 259.5 257.5

257.7 272.6 253.7 262.2 252.0 280.3 274.9 233.7

237.9 274.0 264.5 244.8 264.0 268.3 272.1 260.2

255.8 260.7 245.5 279.6 237.8 278.5 273.3 263.7

241.4 260.6 280.3 272.7 261.0 260.0 279.3 252.1

244.3 272.2 248.3 278.7 236.0 271.2 279.8 245.6

241.2 251.1 267.0 273.4 247.7 254.8 272.8 270.5

254.4 232.1 271.5 242.9 273.6 256.1 251.6

256.8 273.0 240.8 276.6 264.5 264.5 226.8

255.3 266.6 250.2 255.8 285.3 255.4 240.5

255.0 273.2 251.4 276.1 277.8 266.8 268.5

(a) Can you support a claim that mean distance achieved by this particular golf ball exceeds 280

yards? Test the hypothesis

0: 280H

=

versus

1: 280H

Use

0.05

=

. Is it possible to

reject

0

H

hypothesis at the 0.05 level of significance? Find the P-value.

(b) Check the normality assumption.

(c) Compute the power of the test if the true mean distance is 290 yards. What sample size would

be required to detect a true mean distance of 290 yards if you wanted the power of the test to be

at least 0.8?

SOLUTION

(a)

In order to use t statistics in hypothesis testing, we need to assume that the underlying

distribution is normal.

(b)

From the normal probability plot, the normality assumption seems reasonable.

(c)

0290 280 0.75

13.41

d

−−

= = = =

Reserve Problems Chapter 9 Section 3 Problem 9

Human oral normal body temperature is believed to be 98.6° F, but there is evidence that it

actually should be 98.2° F [Mackowiak, Wasserman, Steven and Levine, JAMA (1992, Vol.

268(12), pp. 1578–1580)]. From a sample of 52 healthy adults, the mean oral temperature was

98.285 with a standard deviation of 0.625 degrees.

(a) What are the null and alternative hypotheses?

(b) Test the null hypothesis at

0.05

=

. Find

0

t

. Is it possible to reject

0

H

hypothesis at the

0.05 level of significance?

(c) Construct a 95% confidence interval to answer the same question.

SOLUTION

(a)

(b)

0: 98.2H

=

(c)

95% two-sided confidence interval

Reserve Problems Chapter 9 Section 4 Problem 1

A group of 15 students has performed an experiment, they measured the coefficient of thermal

expansion for aluminum. The results are as follows:

22.0

25.9

25.6

23.1

22.7

25.6

24.9

21.9

26.1

24.3

23.5

20.7

21.4

23.5

20.4

a) Is there strong evidence to conclude that the standard deviation in this experiment exceeds 3?

Use

0.05

=

.

b) Find the P-value for this test.

c) Suppose that the true standard deviation is 1. How many measurements would be required to

detect this difference with the probability of at least 0.9? Use

0.05

=

.

SOLUTION

a) The parameter of interest is the true standard deviation

. However, the solution can be

found by performing a hypothesis test on

2

.

22

0: 3 9H

==

Reserve Problems Chapter 9 Section 4 Problem 2

The quantity of heat that is required to boil 2 liters of water was measured. The experiment was

performed 10 times. The results are as follows (kJ):

656

631

627

644

661

670

623

649

651

658

(a) Is there strong evidence to conclude that the variance in this experiment exceeds 100? Use

0.01

=

.

(b) Find the P-value for this test.

(c) Suppose that the true value of the standard deviation exceeds the hypothesized one by 50%.

What is the probability that this will be detected by performed measurements? Use

0.01

=

.

(d) How many measurements would be required to detect the difference described in (c) with the

probability of at least 0.7? Use

0.01

=

.

SOLUTION

(a) The parameter of interest is the true variance

2

.

2

0: 100H

=

(c) Using the chart below, with

1.5

=

and

10n=

we find

0.7

=

. Therefore, the probability is

only 30%.

Reserve Problems Chapter 9 Section 4 Problem 3

Consider the test of

2

0:5H

=

against

2

1:5H

. What are the critical values for the test

statistic

2

0

for the following significance levels and sample sizes?

(a)

0.01

=

and

20n=

.

(b)

0.05

=

and

15n=

.

(c)

0.10

=

and

16n=

.

SOLUTION

a)

0.01

=

,

20n=

, from Table V we find

2

1 , 1 7.63

n

−−

=

.

Reserve Problems Chapter 9 Section 4 Problem 4

Consider the hypothesis test of

2

0:7H

=

against

2

1:7H

.

Approximate the P-value for each of the following test statistics.

(a)

2

025.2

=

and

20n=

.

(b)

2

015.2

=

and

12n=

.

(c)

2

023.0

=

and

15n=

.

SOLUTION

(a)

2

025.2

=

and

20n=

.

2(0.1) < P-value < 2(0.5), then 0.2 < P-value < 1

Reserve Problems Chapter 9 Section 4 Problem 5

An article in Technometrics (1999, Vol. 41, pp. 202–211) studied the capability of a gauge by

measuring the weight of paper. The data for repeated measurements of one sheet of paper are in

the following table.

Observations

3.481

3.448

3.485

3.475

3.472

3.477

3.472

3.464

3.472

3.470

3.470

3.470

3.477

3.473

3.474

In summary, the sample standard deviation from 15 measurements was 0.0083 grams.

(a) Does the measurement standard deviation differ from 0.01 grams at

0.05

=

?

Find

2

0

if

2 2 2 2

01

: 0.01 , : 0.01 .HH

=

Is it possible to reject

0

H

hypothesis at the 0.05 level

of significance? Find the P-value for this test.

Which confidence interval for

should be used to consider

0

H

hypothesis?

Construct a confidence interval for

to answer the question in part (a).

(b) Find the P-value for this test.

(c) Construct a 99% two-sided confidence interval for

to consider

0

H

hypothesis.

SOLUTION

Does the measurement standard deviation differ from 0.01 grams at

0.05

=

?

Find

2

0

if

2 2 2 2

01

: 0.01 , : 0.01 .HH

=

In order to use the

2

statistic in hypothesis testing and confidence interval construction, we need

to assume that the underlying distribution is normal.

Reserve Problems Chapter 9 Section 4 Problem 6

An Izod impact test was performed on 19 specimens of PVC pipe, the sample standard deviation

was 0.25.

(a) Test the hypothesis that

0.10

=

against an alternative specifying that

0.10

, using

0.01

=

. State any necessary assumptions about the underlying distribution of the data.

Find

2

0

. Is it possible to reject

0

H

hypothesis at the 0.01 level of significance?

SOLUTION

(a)

1) The parameter of interest is the true standard deviation of Izod impact strength,

. However,

Reserve Problems Chapter 9 Section 4 Problem 7

The sugar content of the syrup in canned peaches is normally distributed. Suppose that the

variance is thought to be

22

18 ( )milligrams

=

. A random sample of

10n=

cans yields a

sample standard deviation of

4.8s=

milligrams.

(a) Test the hypothesis

2

0: 18H

=

versus

2

1: 18H

using

0.05

=

Find X2

0 .

Is it possible to reject

0

H

hypothesis at the 0.05 level of significance?

Find the P-value for this test.

(b) Suppose that the actual standard deviation is twice as large as the hypothesized value. What

is the probability that this difference will be detected by the test described in part (a)?

(c) Suppose that the true variance is

240

=

. How large a sample would be required to detect

this difference with probability at least 0.90?

SOLUTION

(a)

In order to use the

2

statistic in hypothesis testing and confidence interval construction, we need

to assume that the underlying distribution is normal.

1) The parameter of interest is the true variance of sugar content,

2

. The answer can be found

Reserve Problems Chapter 9 Section 5 Problem 1

A manufacturer of soap bubble liquid will test a new solution formula. The solution will be

approved, if the percent of produced parisons, in which the content does not allow the bubbles to

inflate, does not exceed 7%. A random sample of 700 parisons contains 55 defective parisons.

Use the z-values rounded to three decimal places to obtain the answers.

(a) Formulate and test an appropriate set of hypotheses to determine whether the solution can be

approved. Use

0.05

=

. Find the P-value.

(b) Find a 95% upper confidence bound on P.

SOLUTION

(a)

(b) The 95% upper confidence limit is:

Reserve Problems Chapter 9 Section 5 Problem 2

A new technology for the production of integrated circuits is being mastered. The technology

will be rejected if the percentage of the defective integrated circuits exceeds 1%.

How many integrated circuits have to be checked to make sure that the technology gives a

sufficiently small amounts of defects with a probability of 10% if the actual probability is

0.05p=

? Use

0.005

=

.

SOLUTION

( ) ( )

2

00

0

11z p p z p p

npp

− + −

=−

Reserve Problems Chapter 9 Section 5 Problem 3

A new technology for the production of integrated circuits is being mastered. The technology

will be rejected if the percentage of the defective integrated circuits exceeds 1%. 1,000,000

transistors are located in the integrated circuit. The integrated circuit is defective if there are

more than 10 malfunctioning transistors. Let X denote the number of defective transistors

observed in the integrated circuit. Of 50,000 integrated circuits inspected, the following data

were observed for the values of X:

Values

4 or less

5

6

7

8

9

10

11

12

13 or more

Observed

frequency

5021

4565

6074

6937

6938

6190

4962

3617

2376

3320

Does the assumption of the Poisson distribution seem appropriate as a probability model for

these data? Use

0.01

=

.

SOLUTION

The expected frequency is found from the Poisson distribution

Reserve Problems Chapter 9 Section 5 Problem 4

A new technology for the production of liquid-crystal displays (LCD) is being mastered.

8,000,000 pixels are located on the LCD. The LCD is defective if there are more than 5

malfunctioning pixels. Let X denote the number of defective pixels observed on the LCD. Of

3000 LCD inspected, the following data were observed for the values of X:

Values

1 or less

2

3

4

5

6

7

8 or more

Observed

frequency

125

266

437

523

517

420

327

385

Does the assumption of the Poisson distribution seem appropriate as a probability model for

these data? Use

0.95

=

.

SOLUTION

The expected frequency is found from the Poisson distribution

( )

!

x

e

P X x x

−

==

,

Values

1 or less

2

3

4

5

6

7

8 or more

The degrees of freedom are

16kp− − =

.

Reserve Problems Chapter 9 Section 5 Problem 5

Consider the following computer output

Test and CI for One Proportion

Test of

0.6p=

vs

0.6p

X

N

Sample p

95% Upper

Bound

Z-value

P-Value

287

494

?

?

?

?

(a) Is this a one-sided or a two-sided test?

(b) Is this a test based on the normal approximation? Is that appropriate?

(c) Complete the missing items.

Round your answers to two decimal places (e.g. 98.76).

Sample p =

95%upper bound =

Z Value−=

P Value−=

(d) Suppose that the alternative hypothesis was two-sided. What is the P-value for this situation?

SOLUTION

(b)

(c)

Reserve Problems Chapter 9 Section 5 Problem 6

In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish

roughness that exceeds the specifications. Do these data present strong evidence that the

proportion of crankshaft bearings exhibiting excess surface roughness exceeds 0.10?

(a) State the appropriate hypotheses. Test the appropriate hypotheses using

0.05

=

. Find

0

z

. Is

it possible to reject

0

H

hypothesis at the 0.05 level of significance?

(b) If it is really the situation that

0.16p=

, how likely is it that the test procedure in part (a) will

not reject the null hypothesis?

(c) If

0.16p=

, how large would the sample size have to be for us to have a probability of

correctly rejecting the null hypothesis of 0.9?

SOLUTION

(a)

The parameter of interest is the true proportion of engine crankshaft bearings exhibiting surface

roughness.

0: 0.10Hp=

and

1: 0.10Hp

(b)

0

0.16, 0.10, 85p p n= = =

and

/2 1.96z

=

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