Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

Question

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Answer 1

Question

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

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Answer 2

Question

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

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

Answer 3

Question

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

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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 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

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

Answer 5

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

Answer 6

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Answer 7

Chapter 11, Problem 47DA

a.

To determine

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

Answer 8

a.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

In this context, let μ1 denotes the mean selling price of homes sold last year without a pool and μ2 denotes the mean selling price of homes sold last year with a pool.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=38+67−2=103

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 103.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 1

From the MINITAB output, the critical values are ±1.983

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without pool.
In Sample 2, enter the column of With pool.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.69.

Decision:

The critical values are ±1.983.

The value of test statistic is –0.69.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=−0.69)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean selling prices of homes with a pool and without a pool.

Answer 13

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

Answer 14

b.

To determine

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

Answer 15

b.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is evidence of difference in the mean selling prices of homes with an attached garage and homes without an attached garage.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year without an attached garage and μ2 denotes the mean selling price of homes sold last year with an attached garage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=27+78−2=103

From the Part a, the critical values are ±1.983.

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of Without garage.
In Sample 2, enter the column of With garage.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 3

From the given MINITAB output, the value of the test statistic is –4.28.

Decision:

The critical values are ±1.983.

The value of test statistic is –4.28.

The value of test statistic is less than the critical value –1.983.

That is, Test statistic(−4.28)< Critical value(±1.983).

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of difference in the mean selling prices of homes with an attached garage and without an attached garage.

Answer 20

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

Answer 21

c.

To determine

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

Answer 22

c.

Expert Solution

Answer to Problem 47DA

The conclusion is that there is no evidence of difference in the mean selling prices of homes that are in default on the mortgage and not in default on the mortgage.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

In this context, μ1 denotes the mean selling price of homes sold last year that are not default on the mortgage and μ2 denotes the mean selling price of homes sold last year that are default on the mortgage.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=70+35−2=103

From Part a, the critical values are ±1.983

Test statistic:

Software procedure:

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

Choose Stat > Basic Statistics > 2 sample t.
Choose Samples in different columns.
In Sample 1, enter the column of not default.
In Sample 2, enter the column of default.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 47DA , additional homework tip 4

From the given MINITAB output, the value of the test statistic is 0.39.

Decision:

The critical values are ±1.983.

The value of test statistic is 0.39.

The value of test statistic lies between the critical values.

That is, Critical value(=−1.983)<Test statistic(=0.39)<Critical value(=1.983).

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the difference in the mean selling price of homes that is in default on the mortgage and homes not default on the mortgage.

Answer 27

Ch. 11 - Tom Sevits is the owner of the Appliance Patch....Ch. 11 - Prob. 1E Ch. 11 - Prob. 2E Ch. 11 - Prob. 3E Ch. 11 - As part of a study of corporate employees, the...Ch. 11 - Prob. 5E Ch. 11 - Mary Jo Fitzpatrick is the vice president for...Ch. 11 - Prob. 2SR Ch. 11 - Prob. 7E Ch. 11 - For Exercises 7 and 8: (a) state the decision...

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

Videos

Check whether there is evidence that there is a difference in the mean selling price of homes with a pool and without a pool.

Answer to Problem 47DA

Explanation of Solution

Check whether there is evidence of a difference in the mean selling prices of homes with an attached garage and without an attached garage.

Answer to Problem 47DA

Explanation of Solution

Check whether there is evidence of a difference in the mean selling price of homes that are in default on the mortgage and not in default on the mortgage.

Answer to Problem 47DA

Explanation of Solution

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