Use the Following Information to Answer the Question $=65.84-0.15$

Use the following information to answer the question. A high school boys cross country coach performs a regression to
predict the finish times of runners in the 10k event from the number of minutes of training in the previous week. The output
is shown below. Assume that the conditions of the linear regression model hold. Regression Analysis: Training versus Finish Time
The regression equation is finish time $=65.84-0.15$ training minutes
$\text { Predicted Values: }$
$\begin{array}{|l|l|l|l|l|}\hline \text { X value } & \text { Pred. Y } & \text { s.e. (Pred. y) } & 95 \% \text { C.I. } & 95 \% \text { P.I. } \\\hline 180 & 38.84 & 1.44112 & (34.73,42.95) & (31.13,46.55) \end{array}$

-Suppose the coach's top runner trained for 180 minutes the previous week. If this runner participates in the 10k event, what is the coach's expected finish time for this runner? Can he be
Reasonably confident that this runner will beat the previous season's record of 43 minutes?
The previous season's record of 43 minutes because the interval contains the value of 43
Minutes.

A) Expected finish time is 38.84 minutes. The coach can be confident that this runner will beat
B) Expected finish time is 38.84 minutes. The coach cannot be confident that this runner will beat
The previous season's record because the interval contains the value of 43 minutes. To be
Confident the interval would have to lie completely below 43 minutes.
C) Expected finish time is 65.84 minutes. The coach can be confident that this runner will beat
The previous season's record of 43 minutes because the interval contains the value of 43
Minutes.
D) Expected finish time is 65.84 minutes. The coach cannot be confident that this runner will beat
The previous season's record because the interval contains the value of 43 minutes. To be
Confident the interval would have to lie completely below 43 minutes.