Solve the problem.
-After obtaining a regression line

y = \beta _ { 0 } + \beta _ { 1 } x

, a confidence interval for the mean of all values

y

for which

x = x 0

can be obtained as follows:

( \hat { y } - \mathrm { E } , \hat { \mathrm { y } } + \mathrm { E } )

where

E = \left( t _ { \alpha / 2 } \right) \operatorname { se } \sqrt { \frac { 1 } { n } + \frac { n \left( x _ { 0 } - \bar { x } \right) ^ { 2 } } { n \sum x ^ { 2 } - \left( \sum x \right) ^ { 2 } } }

The critical value

t _ { \alpha / 2 }

is found from the t-table using

n - 2

degrees of freedom.
Use the data below to obtain a

95 \%

confidence interval estimate of the mean test score of all students who study

10.5

hours. Note that the equation of the regression line is

\hat { y } = 52.8804 + 3.405 \mathrm { x }

and that

\mathrm { s } = 6.8359

\begin{array}{l|lllll}x \text { (hours studied) } & 2.5 & 4.5 & 5.1 & 7.9 & 11.6 \\\hline \mathrm{y} \text { (score on test) } & 66 & 70 & 60 & 83 & 93\end{array}

( 80.48,96.78 )

( 77.11,100.15 )

( 72.49,104.78 )

( 61.55,115.71 )

Is the Data Point, P, an Outlier, an Influential Point y^=8+5x\hat { y } = 8 + 5 xy^​=8+5x

Multiple Choice