91Ó°ÊÓ

The data about the bounce number and height is provided in the question. As a result, the scatterplot of the model that best describes the link between bounce number and height must have a roughly linear pattern. Because the top graph corresponds to an exponential model and the bottom graph corresponds to a power model, the exponential model will provide a better representation of the link between bounce number and height.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \mathrm{P}\\ \text{Constant}& 0.45374& 0.01385& 32.76& 0.000\\ \text{Bounce}& -0.117160& 0.004176& -28.06& 0.000\end{array}$

$S=0.0132043\mathrm{R}-\mathrm{Sq}=99.6\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=99.5\%$

Now, the query specifies that variable $x$represents the bounce and variable $y$represents the height. As a result, the general equation will be:

localid="1652853437967" $\log \hat{y}=a+bx$

The slope and constant of the regression line are given in the question as:

$a=0.45374$

$b=-0.117160$

As a result, the regression line appears to be as follows:

localid="1652853637700" $$\begin{gathered} \log \hat{y}=a+bx \\ =0.45374-0.117160x \end{gathered}$$

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \mathrm{P}\\ \text{Constant}& 0.45374& 0.01385& 32.76& 0.000\\ \text{Bounce}& -0.117160& 0.004176& -28.06& 0.000\end{array}$

$S=0.0132043\mathrm{R}-\mathrm{Sq}=99.6\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=99.5\%$

The regression line is:

$$\begin{gathered} \log \hat{y}=a+bx \\ =0.45374-0.117160x \end{gathered}$$

calculation to find the highest point is

$$\begin{gathered} \log \hat{y}=a+bx \\ =0.45374-0.117160x \\ =0.45374-0.117160×7 \\ =-0.36638 \\ ={10}^{-0.36638} \\ =0.43015 \end{gathered}$$

The regression line is:

$$\begin{gathered} \log \hat{y}=a+bx \\ =0.45374-0.117160x \end{gathered}$$

The question also includes a residual plot of the linear regression in section (b). As a result, we can see a rising pattern in the residual plot for bounces $3$, $4$, and $5$, and the residual will most likely climb even more for bounce seven. It is then assumed that the residual will be positive. As a result, the residual is equal to the actual value minus the anticipated value. The predicted value is supposed to be lower than the actual value since the residual is expected to be positive. As a result, our forecast is too low.