A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last 5 years. Use the .01 significance level. (a) Can we use z as the test statistic? Tell why or why not. (b) State the null hypothesis and the alternate hypothesis. (c) Show the decision rule graphically. (d) Compute the value of z and state your decision regarding the null hypothesis. (e) Determine and interpret the p -value.

Question

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Answer 1

Textbook Question

Chapter 15, Problem 1SR

A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last 5 years. Use the .01 significance level.

(a) Can we use z as the test statistic? Tell why or why not.
(b) State the null hypothesis and the alternate hypothesis.
(c) Show the decision rule graphically.
(d) Compute the value of z and state your decision regarding the null hypothesis.
(e) Determine and interpret the p-value.

a.

Expert Solution

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

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Answer 2

Textbook Question

Chapter 15, Problem 1SR

A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last 5 years. Use the .01 significance level.

(a) Can we use z as the test statistic? Tell why or why not.
(b) State the null hypothesis and the alternate hypothesis.
(c) Show the decision rule graphically.
(d) Compute the value of z and state your decision regarding the null hypothesis.
(e) Determine and interpret the p-value.

a.

Expert Solution

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

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Answer 3

Textbook Question

Chapter 15, Problem 1SR

A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last 5 years. Use the .01 significance level.

(a) Can we use z as the test statistic? Tell why or why not.
(b) State the null hypothesis and the alternate hypothesis.
(c) Show the decision rule graphically.
(d) Compute the value of z and state your decision regarding the null hypothesis.
(e) Determine and interpret the p-value.

a.

Expert Solution

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

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Answer 4

Textbook Question

Chapter 15, Problem 1SR

A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last 5 years. Use the .01 significance level.

(a) Can we use z as the test statistic? Tell why or why not.
(b) State the null hypothesis and the alternate hypothesis.
(c) Show the decision rule graphically.
(d) Compute the value of z and state your decision regarding the null hypothesis.
(e) Determine and interpret the p-value.

a.

Expert Solution

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

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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

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Answer 8

a.

Expert Solution

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Check whether people can use z as the test statistic and explain the reason.

Answer 13

Answer to Problem 1SR

Yes, people can use z as the test statistic because both nπ and n (1–π) exceed 5.

Answer 14

Explanation of Solution

Requirements to check:

nπ>5n(1−π)>5

It is given that the sample size n is 100.

For nπ>5:

nπ=0.70×100=70>5

For n(1−π)>5:

nπ=100×(1−0.70)=30>5

Hence, the requirements are satisfied for using the z-statistic as the binomial distribution.

Answer 15

b.

Expert Solution

To determine

State the null and alternate hypotheses.

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

State the null and alternate hypotheses.

Answer 20

Explanation of Solution

In this case, the test is to check whether less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Let π represents population proportion of persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Therefore, the null and alternate hypotheses are shown below:

H0:π≥0.40H1:π<0.40

Answer 21

c.

Expert Solution

To determine

Show the decision rule graphically.

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

Answer 22

c.

Expert Solution

Answer 23

c.

Expert Solution

Answer 24

Expert Solution

Answer 25

To determine

Show the decision rule graphically.

Answer 26

Explanation of Solution

Step-by-step procedure to show the decision rule graphically using MINITAB software:

Choose Graph > Probability Distribution Plot > View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose Probability and Left Tail for the region of the curve to shade.
Enter the Probability as 0.01.
Click OK.

Output using MINITAB software is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 1

From the output, the critical value is –2.326.

Therefore, the decision rule is rejecting the null hypothesis if test statistic is less than –2.326.

Answer 27

d.

Expert Solution

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer to Problem 1SR

The value of chi-square is –0.87.

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

Answer 28

d.

Expert Solution

Answer 29

d.

Expert Solution

Answer 30

Expert Solution

Answer 31

To determine

Find the value of z-statistic and the write the decision regarding the null hypothesis.

Answer 32

Answer to Problem 1SR

The value of chi-square is –0.87.

Answer 33

Explanation of Solution

Calculation:

The sample size n is 200 and x is 74.

Step-by-step procedure to find the test statistic using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 74. In Number of trials, enter 200.
Enter Hypothesized proportion as 0.40.
Check Options, enter Confidence level as 99.0.
Choose less than in alternative.
Select Method as Normal approximation.
Click OK in all dialogue boxes.

Output is obtained as follows:

EBK STATISTICAL TECHNIQUES IN BUSINESS, Chapter 15, Problem 1SR , additional homework tip 2

From the output, the value of the test statistic is –0.87.

In this case, the critical values is –2.326 and the test statistic is –0.87.

Here, the test statistic value is less than the critical value.

That is, –0.87 > –2.326.

Therefore, do not reject the null hypothesis.

Answer 34

e.

Expert Solution

To determine

Find and interpret the p-value.

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Answer 35

e.

Expert Solution

Answer 36

e.

Expert Solution

Answer 37

Expert Solution

Answer 38

To determine

Find and interpret the p-value.

Answer 39

Explanation of Solution

From the output of Part (d), it can be observed that the p-value is 0.193 and it is more than the level of significance. Therefore, there is no sufficient evidence to conclude that less than 40% of the persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 years.

Answer 40

Ch. 15 - A recent insurance industry report indicated that...Ch. 15 - The following hypotheses are given. H0:.70H1:.70 A...Ch. 15 - The following hypotheses are given. H0:=.40H1:.40...Ch. 15 - Prob. 3E Ch. 15 - Prob. 4E Ch. 15 - Prob. 5E Ch. 15 - Prob. 6E Ch. 15 - Prob. 2SR Ch. 15 - Prob. 7E Ch. 15 - The null and alternate hypotheses are: H0:1=2H1:12...

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Answer to Problem 1SR

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