Chapter 12.2, Problem 47E

(a)

To determine

To Explain: the better relationship between percent made and distance.

Answer to Problem 47E

Exponential model

Explanation of Solution

Given:

  

  

  

It is observed that the given scatter plot between ln(percent made) and distance is not having strong curvature and therefore it would be appropriate to utilize a linear model between the two variables of the scatter plot. Therefore it is appropriate to utilize a linear model between ln(percent made) and distance.

General linear model to expect ln(count) from time

  $\ln \left(\text{percent}\text{made}\right)=a+b\left(\text{distance}\right)$

Taking the exponential

  $\begin{array}{c}\text{percent}\text{made}=e^{\ln \left(percentmade\right)}=e^{a+b\left(\text{distance}\right)}\\ =e^a{e}^{b\left(\text{distance}\right)}\end{array}$

Therefore the model associate with the model $\text{percent}\text{made}=e^a{e}^{b\left(\text{distance}\right)}$ it is exponential model between power made and distance. Although the linear model between $\ln \left(\text{percent}\text{made}\right)$ is appropriate, the exponential model between percent made and distance is also appropriate.

(b)

To determine

To find: the equation of the least-squares regression line on the basis of given computer output.

Answer to Problem 47E

  $\ln y=4.6649-0.1091x$

Explanation of Solution

Given:

  

Calculation:

General equation for the least square regression line

  $\widehat{y}=b_0+b_{1}x$

The calculated of the constant $b_0$ is mention in the row “constant” and in the column “Coef” of the output of the computer

  $b_0$ =4.6649

The calculated of the slope $b_1$ is mention in the row “Distance” and in the column “Coef” of the output of the computer

  $b_1$ =-0.1091

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=4.6649-0.1091x\end{array}$

Where x shows the distance and y shows the ln(percent made)

  $\ln y=4.6649-0.1091x$

Where $x$ is representing the distance and y is representing the percent made.

(c)

To determine

To Explain: the prediction of the percent made for putts of 21 feet on the basis of part (b).

Answer to Problem 47E

10.7381%

Explanation of Solution

Given:

  

Calculation:

From the part (b)

  $\ln y=4.6649-0.1091x$

Where $x$ is representing the distance and y is representing the percent made.

Putting the value of $x$ into the equation

  $\begin{array}{l}\ln y=4.6649-0.1091x\\ \ln y=4.6649-0.1091\left(21\right)\\ \ln y=2.3738\end{array}$

Taking the exponential

  $\begin{array}{c}\widehat{y}=e^{\ln y}=e^{2.3738}\\ =10.7381\end{array}$

Therefore the expected percent made is 10.7381

(d)

To determine

To Calculate: here is given linear regression in part (b), is it expected that prediction in part (c) to be too large, too small or about right, justify the answer.

Answer to Problem 47E

Too small

Explanation of Solution

Given:

  

It is observed that, made a prediction for a distance of $x$ =21 feet and in the residual plot, it is noticed that the all dots approximately 20 around are positive. So it is predicted that the residual $x$ =21 to positive as well.

Then residual is the difference between the predicted y-value and actual y-value. If a residual is positive, then this means that the actual y-value is larger than the expected y-value and therefore our prediction looks to have been too small.