Chapter 15, Problem 62DA

a.

To determine

Find the proportion of homes that have an attached garage.

Check whether people can conclude that more than 60% of the homes have an attached garage.

Find the p-value.

Answer to Problem 62DA

The proportion of homes that have an attached garage is 0.7429.

Yes, people can conclude that more than 60% of the homes have an attached garage.

The p-value is 0.001.

Explanation of Solution

Calculation:

In this case, the test is to check whether more than 60% of the homes have an attached garage.

Let $\pi$ represents hypothesized population proportion of homes have an attached garage.

$\begin{array}{l}{H}_0:\pi \le 0.60\\ H_1:\pi >0.60\end{array}$

The proportion of homes that have an attached garage is obtained as follows:

$\begin{array}{c}\text{Required proportion}=\frac{\text{Number of homes have an attached garage}}{\text{Total number of homes}}\\ =\frac{78}{105}\\ =0.7429\end{array}$

Thus, the proportion of homes that have an attached garage is 0.7429.

The level of significance is 0.05.

Therefore, the value of z score at 0.4500 $\left(=0.5000-0.0500\right)$ using the table B.3: Areas under the normal curve is 1.645.

Decision rule:

Reject the null hypothesis if z > 1.645.

The sample size n is 100 and p is 0.60. Therefore, the number of events x is 78.

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 78. In Number of trials, enter 105.
• Enter Hypothesized proportion as 0.60.
• Check Options, enter Confidence level as 95.0.
• Choose greater than in alternative.
• Select Method as Normal approximation.
• Click OK in all dialogue boxes.

Output is obtained as follows:

Thus, the value of the test statistic is 2.99 and the p-value is 0.001.

In this case, the critical value is 1.645 and the test statistic is 2.99.

Here, the test statistic 2.99 is greater than the critical value 1.645.

That is, 2.99 > 1.645.

Therefore, reject the null hypothesis.

Therefore, people can conclude that more than 60% of the homes have an attached garage.

b.

To determine

Find the proportion of homes that have a pool.

Check whether people can conclude that more than 60% of the homes have a pool.

Find the p-value.

Answer to Problem 62DA

The proportion of homes that have an attached garage is 0.6381.

Yes, people can conclude that more than 60% of the homes have a pool.

The p-value is 0.2119.

Explanation of Solution

Calculation:

In this case, the test is to check whether more than 60% of the homes have a pool.

Let $\pi$ represents hypothesized population proportion of homes have a pool.

$\begin{array}{l}{H}_0:\pi \le 0.60\\ H_1:\pi >0.60\end{array}$

The proportion of homes that have an attached garage is obtained as follows:

$\begin{array}{c}\text{Required proportion}=\frac{\text{Number of homes have a pool}}{\text{Total number of homes}}\\ =\frac{67}{105}\\ =0.6381\end{array}$

Thus, the proportion of homes that have an attached garage is 0.6381.

The level of significance is 0.05.

Therefore, the value of z score at 0.4500 $\left(=0.5000-0.0500\right)$ using the table B.3: Areas under the normal curve is 1.645.

Decision rule:

Reject the null hypothesis if z > 1.645.

The sample size n is 100 and p is 0.60. Therefore, the number of events x is 78.

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 67. In Number of trials, enter 105.
• Enter Hypothesized proportion as 0.60.
• Check Options, enter Confidence level as 95.0.
• Choose greater than in alternative.
• Select Method as Normal approximation.
• Click OK in all dialogue boxes.

Output is obtained as follows:

Thus, the value of the test statistic is 0.80 and the p-value is 0.213.

In this case, the critical value is 1.645 and the test statistic is 0.80.

Here, the test statistic 0.80 is less than the critical value 1.645.

That is, 0.80 < 1.645.

Therefore, do not reject the null hypothesis.

Therefore, people cannot conclude that more than 60% of the homes have a pool.

c.

To determine

Develop a contingency table that shows whether a home has a pool and the township in which the house is located.

Check whether there is an association between the variables pool and township.

Explanation of Solution

Calculation:

From the given information, the contingency table that shows whether a home has a pool and the township in which the house is located is shown below:

Pool	Township
	1	2	3	4	5	Total
No	9	8	7	11	3	38
Yes	6	12	18	18	13	67
Total	15	20	25	29	16	105

$H_0:$ There is no relationship between pool and township.

$H_1:$ There is a relationship between pool and township.

The number of degrees of freedom is obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=\left(\text{Rows}-1\right)\left(\text{Columns}-1\right)\\ =\left(2-1\right)\left(5-1\right)\\ =4\end{array}$

Therefore, the number of degrees of freedom is 4.

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

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Chi-Square’ distribution.
• Enter Degrees of freedom is 4.
• Click the Shaded Area tab.
• Choose Probability and Right Tail for the region of the curve to shade.
• Enter the data value as 0.05.
• Click OK.

Output using MINITAB software is obtained as follows:

From the output, the critical value of chi-square is 9.488.

The general decision rule is reject the null hypothesis if ${\chi }^2$> critical value.

Therefore, the decision rule is reject the null hypothesis if ${\chi }^2>9.488$.

Test statistic:

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

• Choose Stat > Tables > Chi-Square Test for Association.
• Choose Summarized data in a two-way table.
• In Columns containing the table, enter the columns of 1, 2, 3, 4 and 5.
• In Rows under Labels for the table, enter the column of Pool.
• Click OK.

Output is obtained as follows:

From the output, the test statistic is 6.68.

The critical value is 9.488.

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

That is, 6.68 < 9.488.

Thus, do not reject the null hypothesis.

Therefore, there is no sufficient evidence to conclude that there is an association between the variables pool and township.

d.

To determine

Develop a contingency table that shows whether a home has an attached garage and the township in which the house is located.

Check whether there is an association between the variables attached garage and township.

Explanation of Solution

Calculation:

From the given information, the contingency table that shows whether a home has an attached garage and the township in which the house is located is shown below:

Garage	Township
	1	2	3	4	5	Total
No	6	6	6	6	3	27
Yes	9	14	19	23	13	78
Total	15	20	25	29	16	105

$H_0:$ There is no relationship between attached garage and township.

$H_1:$ There is a relationship between attached garage and township.

The number of degrees of freedom is obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=\left(\text{Rows}-1\right)\left(\text{Columns}-1\right)\\ =\left(2-1\right)\left(5-1\right)\\ =4\end{array}$

Therefore, the number of degrees of freedom is 4.

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

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Chi-Square’ distribution.
• Enter Degrees of freedom is 4.
• Click the Shaded Area tab.
• Choose Probability and Right Tail for the region of the curve to shade.
• Enter the data value as 0.05.
• Click OK.

Output using MINITAB software is obtained as follows:

From the output, the critical value of chi-square is 9.488.

The general decision rule is reject the null hypothesis if ${\chi }^2$> critical value.

Therefore, the decision rule is reject the null hypothesis if ${\chi }^2>9.488$.

Test statistic:

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

• Choose Stat > Tables > Chi-Square Test for Association.
• Choose Summarized data in a two-way table.
• In Columns containing the table, enter the columns of 1, 2, 3, 4 and 5.
• In Rows under Labels for the table, enter the column of Garage.
• Click OK.

Output is obtained as follows:

From the output, the test statistic is 2.623.

The critical value is 9.488.

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

That is, 2.623 < 9.488.

Thus, do not reject the null hypothesis.

Therefore, there is no sufficient evidence to conclude that there is an association between the variables attached garage and township.