Question

a. compute the three sums of squares, $SST,SSR,SSE$, using the defining formulas

b. verify the regression identity,$SST=SSR+SSE$

c. compute the coefficient of determination.

d. determine the percentage of variation in the observed values of the response variable that is required by the regression

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

$\hat{y}=2.875-0.625x$

Short answer

(a) $$\begin{gathered} SST=20 \\ SSR=9.375 \\ SSE=10.625 \end{gathered}$$

(b) The regression identity is verified

(c) $0.4688$

(d) $46.88\%$

(e) Utilising the regression equation to generate predictions is useless, because the regression can only explain $47\%$of the variation.

Step-by-step solution

Part (a) Step 1: Given information

The given data is

$\hat{y}=2.875-0.625x$

Part (a) Step 2: Explanation

The given regression equation is

$\hat{y}=2.875-0.625x$

The formulas to calculate the sum of squares is

$$\begin{gathered} SST=\sum {\left(y_i-\overline{y}\right)}^2 \\ SSR=\sum {\left({\hat{y}}_i-\overline{y}\right)}^2 \\ SSE=\sum {\left(y_i-\hat{y}\right)}^2 \end{gathered}$$

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

$$\begin{gathered} SST=20 \\ SSR=9.375 \\ SSE=10.625 \end{gathered}$$

Part (b) Step 1: Given information

The given data is

$\hat{y}=2.875-0.625x$

Part (b) Step 2: Explanation

From the above values

$SSR+SSE=9.375+10.625$

$$\begin{gathered} =20 \\ =SST \end{gathered}$$

Part (c) Step 1: Given information

The given data is

$\hat{y}=2.875-0.625x$

Part (c) Step 2: Explanation

The coefficient of determination is

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

$$\begin{gathered} =\frac{9.375}{20} \\ =0.4688 \end{gathered}$$

Part (d) Step 1: Given information

The given data is

$\hat{y}=2.875-0.625x$

Part (d) Step 2: Explanation

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

$0.4688=46.88\%$

Part (e) Step 1: Given information

The given data is

$\hat{y}=2.875-0.625x$

Part (e) Step 2: Explanation

The regression equation is useful to generate predictions if the calculated $r^2$is closed to $1$. The calculated $r^2=0.4688$, which is not closed to 1.

As a result, utilising the regression equation to generate predictions is useless, because the regression can only explain $47\%$of the variation.