Construct the indicated prediction interval for an individual y.
-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.
$\begin{array}{c|rrrrr}
\text { x Hours of preparation } & 5 & 2 & 9 & 6 & 10 \
\hline \mathrm{y} \text { Test score } & 64 & 48 & 72 & 73 & 80
\end{array}$

A) $32

Question

Construct the indicated prediction interval for an individual y.
-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.
$\begin{array}{c|rrrrr}
\text { x Hours of preparation } & 5 & 2 & 9 & 6 & 10 \
\hline \mathrm{y} \text { Test score } & 64 & 48 & 72 & 73 & 80
\end{array}$

A) $32 < y < 107$
B) $58 < y < 82$
C) $62 < y < 78$
D) $35 < y < 104$

Quizplus · Accepted Answer

The prediction interval is calculated using the formula $\hat{y} \pm t \cdot se \cdot \sqrt{1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{\sum (x_i - \bar{x})^2}}$. For 7 hours, $\hat{y} = 44.845 + 3.524 \times 7 = 69.513$. Using a t-value for 99% confidence with 3 degrees of freedom (approximately 4.541), the interval is approximately $35 < y < 104$.

Construct the Indicated Prediction Interval for an Individual Y y^=44.845+3.524x\hat { \mathrm { y } } = 44.845 + 3.524 \mathrm { x }y^​=44.845+3.524x

Construct the Indicated Prediction Interval for an Individual Y $\hat { \mathrm { y } } = 44.845 + 3.524 \mathrm { x }$