Students in Mr. Handford’s class dropped a kickball beneath a motion detector. The detector recorded the height of the ball as it bounced up and down several times. Here are the heights of the ball at the highest point on the first five bounces:

$\begin{array}{cc}\text{Bounce number}& \text{Height (feet)}\\ 1& 2.240\\ 2& 1.620\\ 3& 1.235\\ 4& 0.958\\ 5& 0.756\\ & \end{array}$

Here is the scatter plot of the data

(a) The following graphs show the results of two different transformations of the data. Would an exponential model or a power model provide a better description of the relationship between bounce number and height? Justify your answer.

(b) Minitab output from a linear regression analysis on the transformed data of log(height) versus bounce number is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use.

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

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

(c) Use your model from part (b) to predict the highest point the ball reaches on its seventh bounce. Show your work.

(d) A residual plot for the linear regression in part (b) is shown below. Do you expect your prediction part (c) to be too high, too low, or just right? Justify your answer.

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

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

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.