Chapter 13, Problem 62DA

The North Valley Real Estate data reports information on homes on the market.

1. a. Let selling price be the dependent variable and size of the home the independent variable. Determine the regression equation. Estimate the selling price for a home with an area of 2,200 square feet. Determine the 95% confidence interval for all 2,200 square foot homes and the 95% prediction interval for the selling price of a home with 2,200 square feet.
2. b. Let days-on-the-market be the dependent variable and price be the independent variable. Determine the regression equation. Estimate the days-on-the-market of a home that is priced at $300,000. Determine the 95% confidence interval of days-on-the-market for homes with a mean price of $300,000, and the 95% prediction interval of days-on-the-market for a home priced at $300,000.
3. c. Can you conclude that the independent variables “days on the market” and “selling price” are positively correlated? Are the size of the home and the selling price positively correlated? Use the .05 significance level. Report the p-value of the test. Summarize your results in a brief report.

To determine

Find the regression equation.

Find the selling price of a home with an area of 2,200 square feet.

Construct a 95% confidence interval for all 2,200 square foot homes.

Construct a 95% prediction interval for the selling price of a home with 2,200 square feet.

Answer to Problem 62DA

The regression equation is $\text{Price}=-15,775.8836+108.3638\text{size}$.

The selling price of a home with an area of 2,200 square feet is 222,624.423.

The 95% confidence interval for all 2,200 square foot homes is $\left(209,802.242,235,446.605\right)$ .

The 95% prediction interval for the selling price of a home with 2,200 square feet is $\left(123,312.613,321,936.234\right)$.

Explanation of Solution

Here, the selling price is the dependent variable and size of the home is the independent variable.

Step-by-step procedure to obtain the ‘regression equation’ using MegaStat software:

• In an EXCEL sheet enter the data values of x and y.
• Go to Add-Ins > MegaStat > Correlation/Regression > Regression Analysis.
• Select input range as ‘Sheet1!$B$2:$B$106’ under Y/Dependent variable.
• Select input range ‘Sheet1!$A$2:$A$106’ under X/Independent variables.
• Select ‘Type in predictor values’.
• Enter 2,200 as ‘predictor values’ and 95% as ‘confidence level’.
• Click on OK.

Output obtained using MegaStat software is given below:

From the regression output, it is clear that

The regression equation is $\text{Price}=-15,775.8836+108.3638\text{size}$.

The selling price of a home with an area of 2,200 square feet is 222,624.423.

The 95% confidence interval for all 2,200 square foot homes is $\left(209,802.242,235,446.605\right)$

The 95% prediction interval for the selling price of a home with 2,200 square feet is $\left(123,312.613,321,936.234\right)$.

To determine

Find the regression equation.

Find day-on-the-market for homes with a mean price at $300,000.

Construct a 95% confidence interval of day-on-the-market for homes with a mean price at $300,000.

Construct a 95% prediction interval day-on-the-market for a home priced at $300,000.

Answer to Problem 62DA

The regression equation is $\text{days-on-the-market}=25.4531+0.00001159\text{price}$.

The day-on-the-market for homes with a mean price at $300,000 is 28.930.

The 95% confidence interval of day-on-the-market for homes with a mean price at $300,000 is $\left(26.885,30.974\right)$ .

The 95% prediction interval day-on-the-market for a home priced at $300,000 is $\left(9.091,48.768\right)$.

Explanation of Solution

Here, the selling price is the dependent variable and size of the home is the independent variable.

Step-by-step procedure to obtain the ‘regression equation’ using MegaStat software:

• In an EXCEL sheet enter the data values of x and y.
• Go to Add-Ins > MegaStat > Correlation/Regression > Regression Analysis.
• Select input range as ‘Sheet1!$C$2:$C$106’ under Y/Dependent variable.
• Select input range ‘Sheet1!$B$2:$B$106’ under X/Independent variables.
• Select ‘Type in predictor values’.
• Enter 300,000 as ‘predictor values’ and 95% as ‘confidence level’.
• Click on OK.

Output obtained using MegaStat software is given below:

From the regression output, it is clear that

The regression equation is $\text{days-on-the-market}=25.4531+0.00001159\text{price}$.

The day-on-the-market for homes with a mean price at $300,000 is 28.930.

The 95% confidence interval of day-on-the-market for homes with a mean price at $300,000 is $\left(26.885,30.974\right)$.

The 95% prediction interval day-on-the-market for a home priced at $300,000 is $\left(9.091,48.768\right)$.

To determine

Check whether the independent variables “day on the market” and “selling price” are positively correlated.

Check whether the independent variables “selling price” and “size of the home” are positively correlated.

Report the p-value of the test and summarize the result.

Answer to Problem 62DA

There is a positive association between “day on the market” and “selling price”.

There is a positive association between “selling price” and “size of the home”.

Explanation of Solution

Denote the population correlation as $\rho$.

Check the correlation between independent variables “day on the market” and “selling price” is positive

The hypotheses are given below:

Null hypothesis:

$H_0:\rho \le 0$

That is, the correlation between “day on the market” and “selling price” is less than or equal to zero.

Alternative hypothesis:

$H_1:\rho >0$

That is, the correlation between “day on the market” and “selling price” is positive.

Test statistic:

The test statistic is as follows:

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Here, the sample size is 105 and the correlation coefficient is 0.185.

The test statistic is as follows:

$\begin{array}{c}t=\frac{0.185\sqrt{105-2}}{\sqrt{1-{0.185}^2}}\\ =\frac{0.185\times \sqrt{103}}{\sqrt{0.965775}}\\ =1.91\end{array}$

The degrees of freedom is as follows:

$\begin{array}{c}df=n-2\\ =105-2\\ =103\end{array}$

The level of significance is 0.05. Therefore, $1-\alpha =0.95$.

Critical value:

Step-by-step software procedure to obtain the critical value using EXCEL software:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV (0.95, 103)”.

Output obtained using EXCEL is given as follows:

Decision rule:

Reject the null hypothesis H₀, if $t\text{-calculated}>t\text{-critical value}$. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is 1.91 and the critical value is 1.660.

Here, $t\text{-calculated}\left(=1.91\right)>t\text{-critical value}\left(=1.660\right)$.

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a positive association between “day on the market” and “selling price”.

The p-value of the test is 0.0591.

Check the correlation between independent variables “size of the home” and “selling price” is positive:

The hypotheses are given below:

Null hypothesis:

$H_0:\rho \le 0$

That is, the correlation between “size of the home” and “selling price” is less than or equal to zero.

Alternative hypothesis:

$H_1:\rho >0$

That is, the correlation between “size of the home” and “selling price” is positive.

Test statistic:

The test statistic is as follows:

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Here, the sample size is 105 and the correlation coefficient is 0.952.

The test statistic is as follows:

$\begin{array}{c}t=\frac{0.952\sqrt{105-2}}{\sqrt{1-{0.952}^2}}\\ =\frac{0.952\times \sqrt{103}}{\sqrt{0.093696}}\\ =31.56\end{array}$

Decision rule:

Reject the null hypothesis H₀, if $t\text{-calculated}>t\text{-critical value}$. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is 31.56 and the critical value is 1.660.

Here, $t\text{-calculated}\left(=31.56\right)>t\text{-critical value}\left(=1.660\right)$.

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a positive association between “size of the home” and “selling price”.

The p-value is approximately 0.