Chapter 15, Problem 62DE

Refer to the Real Estate data, which report information on homes sold in Goodyear, Arizona, and the surrounding area last year.

1. a. Determine the proportion of homes that have an attached garage. At the .05 significance level, can we conclude that more than 60% of the homes sold in the Goodyear, Arizona, area had an attached garage? What is the p-value?
2. b. Determine the proportion of homes that have a pool. At the .05 significance level, can we conclude that more than 60% of the homes sold in the Goodyear, Arizona, area had a pool? What is the p-value?
3. c. Develop a contingency table that shows whether a home has a pool and the township in which the house is located. Is there an association between the variables pool and township? Use the .05 significance level.
4. d. Develop a contingency table that shows whether a home has an attached garage and the township in which the home is located. Is there an association between the variables attached garage and township? Use the .05 significance level.

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 62DE

The proportion of homes that have an attached garage is 0.6762.

It cannot be concluded that more than 60% of the homes have an attached garage.

The p-value is 0.056.

Explanation of Solution

Let $\pi$ represent a hypothesized population proportion of homes that 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:

$\begin{array}{c}\text{ Proportion}=\frac{\text{Number of homes have an attached garage}}{\text{Total number of homes}}\\ =\frac{71}{105}\\ =0.6762\end{array}$

Thus, the proportion of homes that have an attached garage is 0.6762.

The sample size n is 100 and $\pi$ is 0.60. Therefore, the number of events x is 71.

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 71. 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 1.59 and the p-value is 0.056.

Decision rule:

If the p-value is less than the significance level, reject the null. Otherwise fail to reject the null hypothesis.

Conclusion:

The significance level is 0.05 and the p-value is 0.056.

Here, the p-value is greater than the significance level 0.05. By the rejection rule, fail to reject the null hypothesis.

Therefore, one cannot conclude that more than 60% of the homes have an attached garage.

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 62DE

The proportion of homes that have an attached garage is 0.6380.

It cannot be concluded that more than 60% of the homes have a pool.

The p-value is 0.213.

Explanation of Solution

Let $\pi$ represent a hypothesized population proportion of homes that 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 a pool is obtained as follows:

$\begin{array}{c}\text{ Proportion}=\frac{\text{Number of homes have a pool}}{\text{Total number of homes}}\\ =\frac{67}{105}\\ =0.6380\end{array}$

Thus, the proportion of homes that have an attached garage is 0.6380.

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.

Conclusion:

The significance level is 0.05 and the p-value is 0.213. Here, the p-value is greater than the significance level 0.05. By the rejection rule, fail to reject the null hypothesis.

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

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.

Answer to Problem 62DE

There is no association between pool and township.

Explanation of Solution

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

The null and alternative hypotheses are stated below:

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

$H_1:$ There is an association between pool and township.

The number of degrees of freedom for a contingency table 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 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 enter Pool, in the columns, enter Township.
• Click OK.

Output is obtained using MINITAB is as follows:

From the above output, the value of the test statistic is 6.68 and the p-value is 0.154.

Conclusion:

The significance level is 0.05 and the p-value is 0.154.

Here, the p-value is greater than the significance level 0.05. By the rejection rule, fail to reject the null hypothesis.

Thus, there is no association between pool and township.

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.

Answer to Problem 62DE

There is no association between garage and township.

Explanation of Solution

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	5	10	9	4	34
Yes	9	15	15	20	12	71
Total	15	20	25	29	16	105

The null and alternative hypotheses are stated below:

$H_0:$ 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 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 enter Garage, in the columns, enter Township.
• Click OK.

Output is obtained using MINITAB is as follows:

From the above output, the value of the test statistic is 1.98 and the p-value is 0.739.

Conclusion:

The significance level is 0.05 and the p-value is 0.739.

Here, the p-value is greater than the significance level 0.05. By the rejection rule, fail to reject the null hypothesis.

Thus, there is no association between garage and township.