Question

Exercises T12.4–T12.8 refer to the following setting. An old saying in golf is “You drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data from a random sample of 69 of the nearly 1000 players on the PGA Tour’s world money list are examined. The average number of putts per hole (fewer is better) and the player’s total winnings for the previous season are recorded and a least-squares regression line was fitted to the data. Assume the conditions for inference about the slope are met. Here is computer output from the regression analysis:

T12.7 Which of the following is the 95% confidence interval for the slope β1 of the population regression line?
a. $7,897,179\pm 3,023,782$
b. $7,897,179\pm 2.000(3,023,782$)
c. $-4,139,198\pm 1,698,371$
d. $-4,139,198\pm 1.960(1,698,371)$
e. $-4,139,198\pm 2.000(1,698,371)$

Short answer

The correct answer is option (e) $-4,139,198\pm 2.000(1,698,371)$.

Step-by-step solution

Given information

To determine the $95\%$ confidence interval for the slope ${\beta }_1$ of the population regression line.

Explanation

For shooting low scores and hence winning money, it is assumed that good putting is more significant than long driving.
A random sample of players is picked for analysis to see if this is the fact.
They assumed that the slope inference conditions were met, and that the data was calculated using the computer output provided in the question.
Assume the researcher is putting the theory to the test.
The test statistic has a P-value of $0.0087$. As a result, it is considered that,
$n=69$
$c=95\%$
The slope estimate $b_1$ is presented in the rows "Avg. putts" and "Coef" of the following computer output:
$b_1=-4139198$
In the row "Tapping time" and the column "Se coef" of the supplied output, the estimated standard deviation of the slope $SE{b}_1$ is given:
$S{E}_{b_1}=1698371$

Explanation

The critical $t$-value can be found in the appendix in the row of the Student's $T$ distribution table, and the degrees of freedom are:
$df=n-2$
$=69-2$
$=67$
Hence, $t^{*}=2$.
The boundaries of the confidence interval as follows:
$b_1\pm t^{*}\times S{E}_{b_1}$
$=-4139198\pm 2\times 1698371$
As a result, the option (e)$-4,139,198\pm 2.000(1,698,371)$ is the correct answer.