In Exercises 1−4. (a) identify the claim and state H 0 and H a , (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. 1. A researcher claims that the distribution of the lengths of visits at physician offices is different from the distribution shown in the pie chart. You randomly select 400 people and ask them how long their office visits with a physician were. The table shows the results. At a = 0.01, test the researcher’s claim, ( Adapted from Medscape ) Survey results Minutes Frequence, f less than 9 20 10−12 80 13−16 113 17−20 91 21−24 40 25 or more 56

Question

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Answer 1

Textbook Question

Chapter 10, Problem 10.1.1RE

In Exercises 1−4. (a) identify the claim and state H₀ and H_a, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.

1. A researcher claims that the distribution of the lengths of visits at physician offices is different from the distribution shown in the pie chart. You randomly select 400 people and ask them how long their office visits with a physician were. The table shows the results. At a = 0.01, test the researcher’s claim, (Adapted from Medscape)

Chapter 10, Problem 10.1.1RE, In Exercises 14. (a) identify the claim and state H0 and Ha, (b) find the critical value and

Survey results
Minutes	Frequence, f
less than 9	20
10−12	80
13−16	113
17−20	91
21−24	40
25 or more	56

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

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Students have asked these similar questions

(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…

= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.

Answer 2

Textbook Question

Chapter 10, Problem 10.1.1RE

In Exercises 1−4. (a) identify the claim and state H₀ and H_a, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.

1. A researcher claims that the distribution of the lengths of visits at physician offices is different from the distribution shown in the pie chart. You randomly select 400 people and ask them how long their office visits with a physician were. The table shows the results. At a = 0.01, test the researcher’s claim, (Adapted from Medscape)

Chapter 10, Problem 10.1.1RE, In Exercises 14. (a) identify the claim and state H0 and Ha, (b) find the critical value and

Survey results
Minutes	Frequence, f
less than 9	20
10−12	80
13−16	113
17−20	91
21−24	40
25 or more	56

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

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Answer 3

Textbook Question

Chapter 10, Problem 10.1.1RE

In Exercises 1−4. (a) identify the claim and state H₀ and H_a, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.

1. A researcher claims that the distribution of the lengths of visits at physician offices is different from the distribution shown in the pie chart. You randomly select 400 people and ask them how long their office visits with a physician were. The table shows the results. At a = 0.01, test the researcher’s claim, (Adapted from Medscape)

Chapter 10, Problem 10.1.1RE, In Exercises 14. (a) identify the claim and state H0 and Ha, (b) find the critical value and

Survey results
Minutes	Frequence, f
less than 9	20
10−12	80
13−16	113
17−20	91
21−24	40
25 or more	56

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

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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 10, Problem 10.1.1RE

In Exercises 1−4. (a) identify the claim and state H₀ and H_a, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.

1. A researcher claims that the distribution of the lengths of visits at physician offices is different from the distribution shown in the pie chart. You randomly select 400 people and ask them how long their office visits with a physician were. The table shows the results. At a = 0.01, test the researcher’s claim, (Adapted from Medscape)

Chapter 10, Problem 10.1.1RE, In Exercises 14. (a) identify the claim and state H0 and Ha, (b) find the critical value and

Survey results
Minutes	Frequence, f
less than 9	20
10−12	80
13−16	113
17−20	91
21−24	40
25 or more	56

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

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

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Answer 8

a.

Expert Solution

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To identify: The claim.

To state: The hypothesis H0 and Ha .

Answer 13

Answer to Problem 10.1.1RE

The claim is that, the distribution of the lengths differs from the expected distribution.

The hypothesis H0 and Ha is,

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Ha: The distribution of the lengths differs from the expected distribution.

Answer 14

Explanation of Solution

Given info:

The data shows the results of the distribution of the lengths of the visits at physician offices.

Calculation:

Here, the distribution of the lengths differs from the expected distribution is tested. Hence, the claim is that the distribution of the lengths differs from the expected distribution.

The hypotheses are given below:

Null hypothesis:

H0: The distribution of the lengths of visits at physician offices is not differs from the expected distribution.

Alternative hypothesis:

Ha: The distribution of the lengths differs from the expected distribution.

Answer 15

b.

Expert Solution

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To obtain: The critical value.

To identify: The rejection region.

Answer 20

Answer to Problem 10.1.1RE

The critical value is 15.086.

The rejection region is χ2>15.086 .

Answer 21

Explanation of Solution

Given info:

The level of significance is 0.01.

Calculation:

Critical value:

The critical value is calculated by using the (k−1) degrees of freedom.

Substitute k as 6 in degrees of freedom.

6−1=5

From the Table 6-Chi-Square Distribution, the critical value for 5 degrees of freedom for α=0.01 level of significance is 15.086.

Rejection region:

The null hypothesis would be rejected if χ2>15.086 .

Thus, the rejection region is χ2>15.086 .

Answer 22

c.

Expert Solution

To determine

To obtain: The chi-square test statistic.

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To obtain: The chi-square test statistic.

Answer 27

Answer to Problem 10.1.1RE

The chi-square test statistic is 18.770.

Answer 28

Explanation of Solution

Calculation:

Step by step procedure to obtain chi-square test statistic using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Frequency.
In Category names, enter the column of Minutes.
Under Test, select the column of Proportions in Proportions specified by historical counts.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics 7th.ed. Instructor's Review Copy, Chapter 10, Problem 10.1.1RE

Thus, the chi-square test statistic value is approximately 18.770.

Answer 29

d.

Expert Solution

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

To check: Whether the null hypothesis is rejected or fails to reject.

Answer 34

Answer to Problem 10.1.1RE

The null hypothesis is rejected.

Answer 35

Explanation of Solution

Conclusion:

From the result of (c), the test-statistic value is 18.770.

Here, the chi-square test statistic value is greater than the critical value.

That is, 18.770>15.086 .

Thus, it can be conclude that the null hypothesis is rejected.

Answer 36

e.

Expert Solution

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

To interpret: The decision in the context of the original claim.

Answer 41

Answer to Problem 10.1.1RE

The conclusion is that, there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.

Answer 42

Explanation of Solution

Interpretation:

From the results of part (d), it can be conclude that there is evidence to support the claim that the distribution of the lengths differs from the expected distribution.