Question

Custom Homes. Following are the size and price data for custom homes from Exercise $4.60$

a. Compute$SST,SSR,SSE$

b. Compute the coefficient of determination

c. Determine the percentage of variation in the observed values of the response variable explained by the regression, and interpret your answer.

d. State how useful the regression equation appears to be for making predictions.

Short answer

(a)$$\begin{gathered} SST=25681.6 \\ SSR=24057.8913 \\ SSE=1623.7087 \end{gathered}$$

(b)$0.6868$

(c)$68.68\%$

(d) Utilising the regression equation to generate predictions isn't very useful, as the regression can only explain around $69\%$of the variation.

Step-by-step solution

Part (a) Step 1: Given information

The given data is

Part (a) Step 2: Explanation

The prices of $9$randomly selected houses with sizes are listed in the table given. The size is expressed in hundreds of square feet, while the price is expressed in thousands of dollars.

The formulas to calculate the sum of squares is

$SST=\sum y_i^2-{\left(\sum y_i\right)}^2/n$

$SSR=\frac{{\left[\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n\right]}^2}{\sum x_i^2-{\left(\sum x_i\right)}^2/n}$

$SSE=SST-SSR$

As shown in the table below, the relevant sums can be determined.

$$\begin{gathered} SST=3504412-\frac{{5552}^2}{9} \\ SST=79444.8889 \end{gathered}$$

$$\begin{gathered} SSR=\frac{|169993-270\times 5552÷9{|}^2}{8316-{270}^2÷9} \\ SSR=54562.4491 \end{gathered}$$

$$\begin{gathered} SSE=79444.8889-54562.4491 \\ SSE=24882.4398 \end{gathered}$$

Part (b) Step 1: Given information

The given data is

Part (b) Step 2: Explanation

The coefficient of determination is

$r^2=\frac{SSR}{SST}$

$$\begin{gathered} =\frac{54562.4491}{79444.8889} \\ =0.6868 \end{gathered}$$

Part (c) Step 1: Given information

The given data is

Part (c) Step 2: Explanation

The coefficient of determination restated as a percentage is the percentage of variation:

$0.6868=68.68\%$

Part (d) Step 1: Given information

The given data is

Part (d) Step 2: Ex[planation

The regression equation can be used to generate predictions if the estimated $r^2$is near to 1.

The computed $r^2=0.6868$, which is a long way from 1.

As a result, utilising the regression equation to generate predictions isn't very useful, as the regression can only explain around $69\%$of the variation.