Construct the indicated prediction interval for an individual y.
-The regression equation for the given paired data is $\hat { y } = 73.012 + 0.8859 x$ and the standard error of estimate is $\mathrm { s_e } = 2.78807 .$ Find the $95 \%$ prediction interval of $\mathrm { y }$ for $\mathrm { x } = 37$.
$\begin{array}{r|rrrrrr}
\mathrm{x} & 25 & 26 & 36 & 36 & 40 & 48 \
\hline \mathrm{y} & 95 & 95 & 102 & 109 & 110 & 114
\end{array}$

A) $73.0

Question

Construct the indicated prediction interval for an individual y.
-The regression equation for the given paired data is $\hat { y } = 73.012 + 0.8859 x$ and the standard error of estimate is $\mathrm { s_e }  = 2.78807 .$ Find the $95 \%$ prediction interval of $\mathrm { y }$ for $\mathrm { x } = 37$.
$\begin{array}{r|rrrrrr}
\mathrm{x} & 25 & 26 & 36 & 36 & 40 & 48 \
\hline \mathrm{y} & 95 & 95 & 102 & 109 & 110 & 114
\end{array}$

A) $73.0 < \mathrm { y } < 114.2$
B) $97.4 < \mathrm { y } < 114.2$
C) $73.0 < y < 105.8$
D) $97.4 < \mathrm { y } < 10.8$

Quizplus · Accepted Answer

First we need to find the prediction interval formula:
$$ \hat { y } \pm t^* \cdot s_e \sqrt {1 + \frac { 1 } { n } + \frac { ( x - \bar { x } ) ^ { 2 } } { ( n - 1 ) s _ { x } ^ { 2 } } } $$
where $\hat{y} = 73.012 + 0.8859(37) = 105.64$ is the predicted value of y for x = 37, $t^*$ is the critical value from t-distribution with degree of freedom = n-2 = 4 and $\alpha$ = 0.05/2 = 0.025, which is 2.776, $s_e$ is given as 2.78807, $\bar{x}$ is the sample mean of x, 34.167, $s_x$ is the sample standard deviation of x, 8.752, and n is the sample size, 6.

Plugging in the values, we get:
$$105.64 \pm 2.776 \cdot 2.78807 \sqrt {1 + \frac { 1 } { 6 } + \frac { ( 37 - 34.167 ) ^ { 2 } } { 5 \cdot 8.752 ^ { 2 } } } $$
which simplifies to:
$$97.37 < y < 113.91$$
Therefore, the best choice is B, with the correct range of values for the prediction interval.

Construct the Indicated Prediction Interval for an Individual Y y^=73.012+0.8859x\hat { y } = 73.012 + 0.8859 xy^​=73.012+0.8859x

Construct the Indicated Prediction Interval for an Individual Y $\hat { y } = 73.012 + 0.8859 x$