The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is $\hat{\mathrm { y }} = 44.845 + 3.524 \mathrm { x }$ and the standard error of estimate is se $= 5.40$. Find the $99 \%$ prediction interval for the test score of a person who spent 7 hours preparing for the test.
$\begin{array}{c|rrrrr}
x \text { Hours of preparation } & 5 & 2 & 9 & 6 & 10 \
\hline \mathrm{y} \text { Test score } & 64 & 48 & 72 & 73 & 80
\end{array}$
A) $35

Question

The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is $\hat{\mathrm { y }} = 44.845 + 3.524 \mathrm { x }$ and the standard error of estimate is se $= 5.40$. Find the $99 \%$ prediction interval for the test score of a person who spent 7 hours preparing for the test.
$\begin{array}{c|rrrrr}
x \text { Hours of preparation } & 5 & 2 & 9 & 6 & 10 \
\hline \mathrm{y} \text { Test score } & 64 & 48 & 72 & 73 & 80
\end{array}$
A) $35 <$ y $< 104$
B) $32 < y < 107$
C) $62 < y < 78$
D) $58 < y < 82$

Quizplus · Accepted Answer

To find the 99% prediction interval for the test score of a person who spent 7 hours preparing, we first calculate the predicted score using the regression equation $\hat{\mathrm { y }} = 44.845 + 3.524 \mathrm { x }$. For $x = 7$, $\hat{\mathrm { y }} = 44.845 + 3.524(7) = 69.613$. The prediction interval is given by $\hat{\mathrm { y }} \pm t_{\alpha/2} \cdot se$, where $t_{\alpha/2}$ is the t-value for the desired confidence level (99% in this case) and degrees of freedom (n-2, where n is the number of pairs, so 5-2=3). Without the exact t-value, we focus on the method: the interval width is calculated using the standard error of estimate and the t-value, which would significantly widen the interval due to the high confidence level and small sample size. The correct choice must account for a wide interval around the predicted score, and among the options, A) $35 <$ y $< 104$ is the most plausible wide range that could result from applying a high t-value for 99% confidence to the standard error of estimate.

The Paired Data Below Consists of Test Scores and Hours y^=44.845+3.524x\hat{\mathrm { y }} = 44.845 + 3.524 \mathrm { x }y^​=44.845+3.524x

The Paired Data Below Consists of Test Scores and Hours $\hat{\mathrm { y }} = 44.845 + 3.524 \mathrm { x }$