How well do professional golfers putt from various distances to the hole? The scatterplot shows various distances to the hole (in feet) and the per cent of putts made at each distance for a sample of golfers.

The graphs show the results of two different transformations of the data. The first graph plots the natural logarithm of per cent made against distance. The second graph plots the natural logarithm of per cent made against the natural logarithm of distance.

a. Based on the scatterplots, would an exponential model or a power model provide a better description of the relationship between distance and per cent made? Justify your answer.

b. Here is computer output from a linear regression analysis of ln(per cent made) and distance. Give the equation of the least-squares regression line. Be sure to define any variables you use.

c. Use your model from part (b) to predict the per cent made for putts of 21 feet.

d. Here is a residual plot for the linear regression in part (b). Do you expect your prediction in part (c) to be too large, too small, or about right? Justify your answer.

The given data is

From the above-given data

Because the scatter plot between $\ln$(per cent made) and the distance does not have considerable curvature, a linear model between the two variables of the scatter plot would be acceptable.

As a result, a linear model between In(per cent made) and distance is adequate. Expect $\ln$(count) from time using a general linear model.

$\ln (percentmade)=a+b(dis\tan ce)$

Per cent maderole="math" localid="1654176006287" $$\begin{gathered} =e^{\ln \left(\text{percent made}\right)} \\ =e^{a+b\left(\text{distance}\right)} \end{gathered}$$

$=e^a{e}^{b\left(\text{distance}\right)}$

As a result, the model is associated with the per cent made$=e^a{e}^{b\left(dis\tan ce\right)}$ model, which is an exponential relationship between power and distance. While the linear model between In (per cent created) and distance is correct, the exponential model between per cent made and distance is also correct.

The given data is

The equation for the least square regression line is

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer output, the calculated constant $b_0$is mentioned.

$b_0=4.6649$

In the row "Distance" and the column "Coef" of the computer output, the calculated slope $b_1$is mentioned.

$b_1=-0.1091$

Then the value is

$\hat{y}=b_0+b_{1}x$

$\hat{y}=4.6649-0.1091x$

Therefore

$\ln y=4.6649-0.1091x$

Where x is the distance and y denotes the percentage achieved.

The given data is

From part b

$\ln y=4.6649-0.1091x$

Where x is the distance and y denotes the percentage achieved.
$$\begin{gathered} \ln y=4.6649-0.1090\left(21\right) \\ \ln y=2.3738 \end{gathered}$$

Take the exponential

$$\begin{gathered} \hat{y}=e^{\ln y} \\ \hat{y}=e^{2.3738} \end{gathered}$$

$=10.7381$

Then the per cent expected is$10.7381.$

The given data is

It is observed that when a forecast for a distance of $x=21$feet is made, all dots roughly $20$around are positive in the residual plot. As a result, the residual $x=21$is expected to be positive as well.

The difference between the projected $y-$value and the actual $y-$value is the residual. If a residual is positive, it signifies that the real $y-$value is higher than the projected $y-$value, implying that our prediction was too low.