Chapter 4, Problem 9E

(a)

To determine

Estimated regression equation.

Explanation of Solution

The formula for regression equation is:

  $\begin{array}{l}Y^\text{=}\text{a+b<mo>X</mo><mo>^</mo>}\\ \text{Here,}\\ \text{a is estimated intercept coefficient}\\ \text{b is estimated slope coefficient}\end{array}$

Run the ordinary least squares method for the given data in excel. The results drawn are as follows:

  

Use the summary output to find the estimated regression equation as follows:

  $\begin{array}{c}\text{<mo>Y</mo><mo>^</mo>}\text{<mo> = 185</mo>.26 + 3}\text{.92<mo>X</mo><mo>^</mo><mo>1 + 3</mo>}\text{.59<mo>X</mo><mo>^</mo><mo>2 - 0</mo>}\text{.12<mo>X</mo><mo>^</mo><mo>3 - 2</mo>}\text{.83<mo>X</mo><mo>^</mo><mo>4</mo>}\\ \text{Here,}\\ \text{<mo>Y</mo><mo>^</mo>}\text{= estimated selling price in \$1,000}\\ \text{<mo>X</mo><mo>^</mo>}\text{1 = estimated size in 100 ft 2}\\ \text{<mo>X</mo><mo>^</mo>}\text{2 = estimated total number of rooms}\\ \text{<mo>X</mo><mo>^</mo>}\text{3 = estimated age}\\ \text{<mo>X</mo><mo>^</mo>}\text{4 = estimated garage}\end{array}$

(b)

To determine

Economic interpretation of each estimated regression coefficients.

Explanation of Solution

Interpretation of estimated slope (b) coefficient:

For a given level of total no. of rooms, age and attached garage, an additional 100 ft² will lead to rise in selling price by 3.92×$1000 = $3,920.

For a given level of size, age and attached garage, an additional room will lead to rise in selling price by 3.59×$1000 = $3,590.

For a given level of size, total no. of rooms and attached garage, an additional year of age will lead to fall in selling price by 0.12×$1000 = $120.

For a given level of size, total no. of rooms and age, an additional garage will lead to fall in selling price by 2.83×$1000 = $2,830.

(c)

To determine

Statistical significance of the independent variables at 0.05 level.

Explanation of Solution

Conduct the t-test to know the statistical significance of the independent variables X1, X2, X3 and X4. The test statistic can be calculated using following formula:

  $\begin{array}{l}\text{T = }\frac{{\text{<mover accent="true"><mo>b</mo><mo>^</mo> </mover>}}_{\text{1}}}{{\text{s}}_{{\text{b}}_{\text{1}}}}\\ \text{Here,}\\ {\text{<mover accent="true"><mo>b</mo><mo>^</mo> </mover>}}_{\text{1}}\text{ is coefficient of estimated independent variable}\\ {\text{s}}_{{\text{b}}_{\text{1}}}\text{ is standard error of coefficient of estimated independent variable}\end{array}$

The t-statistic follows t-distribution with n-1 degrees of freedom.

For variable X1, t-test is conducted as follows:

According to the summary output, the t-statistic for X1 variable is equal to 5.19.

At 5% significance level and 15-1= 14 degrees of freedom, the critical value is equal to 2.145.

  

In figure (1), since the calculated t-statistic lies in the critical region. Therefore, we reject the null hypothesis. This means that the variable X1is statistically significant.

For variable X2, t-test is conducted as follows:

According to the summary output, the t-statistic for X2 variable is equal to 0.80.

At 5% significance level and 15-1= 14 degrees of freedom, the critical value is equal to 2.145.

  

In figure (2), since the calculated t-statistic lies in the acceptance region. Therefore, we accept the null hypothesis. This means that the variable X2is not statistically significant.

For variable X3, t-test is conducted as follows:

According to the summary output, the t-statistic for X3 variable is equal to -0.18.

At 5% significance level and 15-1= 14 degrees of freedom, the critical value is equal to 2.145.

  

In figure (3), since the calculated t-statistic lies in the acceptance region. Therefore, we accept the null hypothesis. This means that the variable X3is not statistically significant.

For variable X4, t-test is conducted as follows:

According to the summary output, the t-statistic for X4 variable is equal to -0.29.

At 5% significance level and 15-1= 14 degrees of freedom, the critical value is equal to 2.145.

  

In figure (4), since the calculated t-statistic lies in the acceptance region. Therefore, we accept the null hypothesis. This means that the variable X4is not statistically significant.

(d)

To determine

Proportion of total variation in selling price explained by regression model.

Explanation of Solution

The coefficient of determination measures the proportion of variance predicted by the independent variable in the dependent variable. It is denoted as R².

According to the summary output, the value of R² is equal to 0.89. This means that the regression equation predicts 89% of the variance in selling price.

(e)

To determine

Overall explanatory power of model by performing F-test at 5 percent level of significance.

Explanation of Solution

The value of F-statistic is given as 20.85. And the critical value at 0.05 significance level is equal to 0.00.

Since, F-statistic is greater than the critical value. Thus, the overall model is statistically significant.

(f)

To determine

95 percent prediction interval for selling price of a 15-year-old house having 1,800 sq. ft., 7 rooms, and an attached garage.

Explanation of Solution

The confidence interval of a multiple linear regression model can be calculated using following formula:

  $\text{<mo>y</mo><mo>^</mo><mo> ± </mo>}\left(\text{t × s}\text{.e}\right)$

Here,

y is estimated selling price based on the given values of independent variables

t is critical t value or t-statistic

s.e is multiple standard error of the estimate

The estimated selling price based on the given values of independent variables can be calculated using the estimated regression equation as follows:

  $\begin{array}{c}\text{<mo>Y</mo><mo>^</mo><mo> = 185</mo>}\text{.26 + 3}\text{.92}\left(1800\right)\text{+ 3}\text{.59}\left(7\right)\text{- 0}\text{.12}\left(15\right)\text{- 2}\text{.83}\left(1\right)\\ =\text{185}\text{.26 + 7056 + 25}\text{.13 - 1}\text{.8 + 2}\text{.83}\\ \text{= 7267}\text{.42}\end{array}$

According to the regression statistics in the summary output t-statistic and value of multiple standard error of the estimate is equal to 2.14 and 11.13.

Plug the values in the above confidence interval formula as follows:

  $7267.42\pm \left(2.14\times 11.3\right)$

Thus, an approximate 95% prediction interval for the selling price of a house having an area of a 15-year-old having 1,800 sq. ft., 7 rooms, and an attached garage range from 7291.24 to 7243.60.