Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the
population of paired differences is normally distributed.
-A coach uses a new technique in training middle distance runners. The times for 9 different athletes to run 800 meters before and after this training are shown below.
$\begin{array}{r|ccccccccc}
\text { Athlete } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } \
\hline \text { Time before training (seconds) } & 115.2 & 120.9 & 108.0 & 112.4 & 107.5 & 119.1 & 121.3 & 110.8 & 122.3 \
\hline \text { Time after training (seconds) } & 116.0 & 119.1 & 105.1 & 111.9 & 109.1 & 115.2 & 118.5 & 110.7 & 120.9
\end{array}$

Construct a $99 \%$ confidence interval for the mean difference of the "before" minus "after" times.
A) $- 0.85

Question

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the
population of paired differences is normally distributed.
-A coach uses a new technique in training middle distance runners. The times for 9 different athletes to run 800 meters before and after this training are shown below.
$\begin{array}{r|ccccccccc}
\text { Athlete } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } \
\hline \text { Time before training (seconds) } & 115.2 & 120.9 & 108.0 & 112.4 & 107.5 & 119.1 & 121.3 & 110.8 & 122.3 \
\hline \text { Time after training (seconds) } & 116.0 & 119.1 & 105.1 & 111.9 & 109.1 & 115.2 & 118.5 & 110.7 & 120.9
\end{array}$

Construct a $99 \%$ confidence interval for the mean difference of the "before" minus "after" times.
A) $- 0.85 < \mu _ { \mathrm { d } } < 3.29$
B) $- 0.76 < \mu _ { \mathrm { d } } < 3.20$
C) $- 0.82 < \mu _ { \mathrm { d } } < 3.26$
D) $- 0.54 < \mu _ { \mathrm { d } } < 2.98$

Quizplus · Accepted Answer

To construct a confidence interval for the mean difference of the "before" minus "after" times, we first need to calculate the differences for each athlete:

$\begin{array}{r|ccccccccc}
\text { Athlete } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } \
\hline \text { Difference (after - before) } & 0.8 & -1.8 & -2.9 & -0.5 & 1.6 & -3.9 & -2.8 & -0.1 & -1.4 \
\end{array}$

Then we can calculate the sample mean and standard deviation of these differences:

$\overline { d } = - 1.33$
$s_{d}=2.0397$

Using a t-distribution with 8 degrees of freedom (since there are 9 pairs of data and we are estimating one parameter), we can find the t-value for a 99% confidence interval:

$t_{0.005,8} = 3.3554$

Using this t-value, we can construct the confidence interval:

$\overline { d } \pm t_{0.005,8} \cdot \frac { s_{d} } { \sqrt { n } }$

$- 1.33 \pm 3.3554 \cdot \frac { 2.0397 } { \sqrt { 9 } }$

$- 0.82 < \mu _ { \mathrm { d } } < 3.26$

Therefore, the best choice is C.

Construct a Confidence Interval For μd\mu _ { \mathrm { d } }μd​

Construct a Confidence Interval For $\mu _ { \mathrm { d } }$