For Exercises 1 through 20, perform each of the following steps . a . State the hypotheses and identify the claim. b . Find the critical value(s). c . Compute the test value. d . Make the decision. e . Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified . 1. Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime is not 18 months. He researched the lifetime of 50 $1 bills and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months. At α = 0.02, can it be concluded that the average lifetime of a circulated $1 bill differs from 18 months?

Question

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Answer 1

Textbook Question

Chapter 8, Problem 8.2.1RE

For Exercises 1 through 20, perform each of the following steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime is not 18 months. He researched the lifetime of 50 $1 bills and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months. At α = 0.02, can it be concluded that the average lifetime of a circulated $1 bill differs from 18 months?

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

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Answer 2

Textbook Question

Chapter 8, Problem 8.2.1RE

For Exercises 1 through 20, perform each of the following steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime is not 18 months. He researched the lifetime of 50 $1 bills and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months. At α = 0.02, can it be concluded that the average lifetime of a circulated $1 bill differs from 18 months?

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

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Answer 3

Textbook Question

Chapter 8, Problem 8.2.1RE

For Exercises 1 through 20, perform each of the following steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime is not 18 months. He researched the lifetime of 50 $1 bills and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months. At α = 0.02, can it be concluded that the average lifetime of a circulated $1 bill differs from 18 months?

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 8, Problem 8.2.1RE

For Exercises 1 through 20, perform each of the following steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime is not 18 months. He researched the lifetime of 50 $1 bills and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months. At α = 0.02, can it be concluded that the average lifetime of a circulated $1 bill differs from 18 months?

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Answer 8

a.

Expert Solution

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To state: The null and alternative hypotheses and identify the claim.

Answer 13

Answer to Problem 8.2.1RE

Null hypothesis: H0:μ=18

Alternative hypothesis: H1:μ≠18 (claim)

The claim is “the average lifetime of circulated $1 bills is not 18 months”.

Answer 14

Explanation of Solution

Given info:

A sample of 50 $1 bills selected and found the average lifetime was 18.8 months. The population standard deviation is 2.8 months.

Justification:

Here, the claim is that the average lifetime of circulated $1 bills is not 18 months. This can be written as μ≠18 . The complement of the claim is, μ=18 . In the given experiment, the alternative hypothesis indicates the claim.

The test hypotheses are given below:

Null hypothesis: The average lifetime of circulated $1 bills is 18 months

Alternative hypothesis (claim): The average lifetime of circulated $1 bills is not 18 months.

Answer 15

b.

Expert Solution

To determine

To find: The critical values.

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To find: The critical values.

Answer 20

Answer to Problem 8.2.1RE

The critical value is ±2.326.

Answer 21

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 1

From the output, the critical value is ±2.326.

Answer 22

c.

Expert Solution

To determine

To find: The test value.

Answer to Problem 8.2.1RE

The test value is 2.02.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To find: The test value.

Answer 27

Answer to Problem 8.2.1RE

The test value is 2.02.

Answer 28

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the test value using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data, enter the sample size as 50 and mean as 18.8.
In Standard deviation, enter 2.8.
In Perform hypothesis test, enter the test mean as 18.
Check Options; enter Confidence level as 98%.
Choose not equal in alternative.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 2

From the output, the test value is 2.02.

Answer 29

d.

Expert Solution

To determine

To make: The decision.

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

To make: The decision.

Answer 34

Answer to Problem 8.2.1RE

The decision is “fail to reject the null hypothesis”.

Answer 35

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability Value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.02.
Enter 2.02 under show reference lines at X values.
Click OK.

Output using the MINITAB software is given below:

ELEMENTARY STATISTICS: STEP BY STEP- ALE, Chapter 8, Problem 8.2.1RE , additional homework tip 3

From the output, it can be observed that the test statistic value do not falls in the critical region. Therefore, the null hypothesis is not rejected.

Answer 36

e.

Expert Solution

To determine

To summarize: The result.

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

To summarize: The result.

Answer 41

Answer to Problem 8.2.1RE

The conclusion is that, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.

Answer 42

Explanation of Solution

Justification:

From part d, the null hypothesis is not rejected. Therefore, there is no enough evidence to support the claim that the average lifetime of circulated $1 bills is not 18 months.