Use the Following Information to Answer the Question $= 69.71 - 0.13$

Use the following information to answer the question. A high school girls 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 $= 69.71 - 0.13$ training minutes
Predicted Values:
$\begin{array}{|l|l|l|l|l|}\hline \text { X value } & \text { Pred. Y } & \text { s.e. (Pred. y) } & \text { 95\% C.I. } & \text { 95\% P.I. } \\\hline 145 & 50.86 & 1.82543 & (46.09,55.63) & (44.85,56.87) \\\hline\end{array}$
-Suppose the coach's top runner trained for 145 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 time she had at the last meet of 51 minutes?

A) Expected finish time is 50.86 minutes. The coach can be confident that this runner will beat the previous meet's time of 51 minutes because the interval contains the value of 51 minutes.
B) Expected finish time is 88.56minutes. The coach cannot be confident that this runner will beat
The previous meet's time because the interval contains the value of 51 minutes. To be
Confident the interval would have to lie completely below 51 minutes.
C) Expected finish time is 88.56 minutes. The coach can be confident that this runner will beat
The previous meet's time of 51 minutes because the interval contains the value of 51 minutes.
D) Expected finish time is 50.86 minutes. The coach cannot be confident that this runner will beat
The previous meet's time because the interval contains the value of 51 minutes. To be
Confident the interval would have to lie completely below 51 minutes.