Question

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 and the player’s total winnings for the previous season are recorded. A least-squares regression line was fitted to the data. The following results were obtained from statistical software.

A $95\%$confidence interval for the slope $B$of the population regression line is

(a)$7,897,179\pm 3,023,782$

(b)$7,897,179\pm 6,047,564$

(c)$-4,139,198\pm 1,698,371$

(d)$-4,139,198\pm 3,328,807$

(e)$-4,139,198\pm 3,396,742$

Short answer

A $95\%$confidence interval for the slope $B$of the population regression line is (e) $-4,139,198\pm 3,396,742$.

Step-by-step solution

Given information

Given in the question that

$$\begin{gathered} b=-4139198 \\ S{E}_b=1698371 \\ n=69 \end{gathered}$$

Calculation

The confidence interval boundaries are

$b\pm t\times S{E}_b$

The degree of freedom is

$$\begin{gathered} df=n-2 \\ =69-2 \\ =67>60 \end{gathered}$$

The critical t-value is shown in table in the row of $df=60$and column of $c=95\mathrm{\%}$

$t^{\ast }=2.000$

The confidence interval boundaries become then :

$$\begin{gathered} b\pm t^{\ast }\times S{E}_b=-4139198\pm 2.000\times 1698371 \\ =-4139198\pm 3396742 \end{gathered}$$