The claim and state H 0 and H 1 .

Question

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Answer 1

Question

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

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Answer 2

Question

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

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Answer 3

Question

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

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

Answer 4

Question

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

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Answer 5

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Answer 6

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

Answer 7

Chapter 9.5, Problem 13E

a.

To determine

To identify: The claim and state H0 and H1 .

Answer 8

a.

Expert Solution

Answer to Problem 13E

The claim is that “the variance in the number of calories differs between the two brands”.

Null hypothesis: H0:σ12=σ22

Alternative hypothesis: H1:σ12>σ22 (claim)

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info:

The data shows the areas (in square miles).

Justification:

Here, the claim is that “the variance in area is greater for eastern cities than for western cities”. This can be written as σ12>σ22 . The complement of the claim is σ12=σ22 . In the given experiment, the alternative hypothesis indicates the claim.

Answer 13

b.

To determine

To find: The critical value for 5% level and 1% level.

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

Answer 14

b.

To determine

To find: The critical value for 5% level and 1% level.

Answer 15

b.

Expert Solution

Answer to Problem 13E

The critical value at 5% level is 4.950 and the critical value at 1% level is 10.67.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The degrees of freedom for numerator is,

d.f.N=n1−1=7−1=6

The degrees of freedom for denominator is,

d.f.D=n2−1=6−1=5

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 1

From the output, the critical value is 4.950.

Level of significance, α=0.01 :

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 F.
Enter Numerator df as 6 and Denominator df as 5.
Click the Shaded Area tab.
Choose Probability value and Right Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 2

From the output, the critical value is 10.67.

Answer 20

c.

To determine

To find: The test value.

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

Answer 21

c.

To determine

To find: The test value.

Answer 22

c.

Expert Solution

Answer to Problem 13E

The test statistic value is 9.80.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Software Procedure:

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

Choose Stat > Basic Statistics >2 Variance.
Choose Each sample is in its own column.
In Sample 1, enter the column of Eastern.
In Sample 2, enter the column of Western.
Check Options; enter Confidence level as 95%.
Choose greater than in alternative.
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 3

From the output, the test value is 9.80.

Answer 27

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, 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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

Answer 28

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.05 and α=0.01 .

Answer 29

d.

Expert Solution

Answer to Problem 13E

For 5% level, the decision is “reject the null hypothesis”.

For 1% level, the decision is “fail to reject the null hypothesis”.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.05.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 4

From the output, it can be observed that the test statistic value falls in the rejection region. Therefore, the null hypothesis is rejected.

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 F.
Enter Numerator df as 8 and Denominator df as 8.
Click the Shaded Area tab.
Choose Probability value and Both Tail for the region of the curve to shade.
Enter the Probability value as 0.01.
Enter 9.80 under show reference lines at X values
Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9.5, Problem 13E , additional homework tip 5

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

Answer 34

e.

To determine

To summarize: The result.

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Answer 35

e.

To determine

To summarize: The result.

Answer 36

e.

Expert Solution

Answer to Problem 13E

The conclusion is that, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 5% level of significance.

The conclusion is that, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Justification:

For 5% level:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the variance in area is greater for eastern cities than for western citiesat 5% level of significance.

For 1% level:

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that the variance in area is greater for eastern cities than for western cities at 1% level of significance.