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After Obtaining a Regression Line y=β0+β1xy = \beta _ { 0 } + \beta _ { 1 } x

Question 86

Multiple Choice

After obtaining a regression line y=β0+β1xy = \beta _ { 0 } + \beta _ { 1 } x , a confidence interval for the mean of all values yy for which x=x = x0\mathrm { x } _ { 0 } can be obtained as follows:
(y^Ery^+E) ,\left( \hat { y } - E _ { r } \hat { y } + E \right) ,
where E=(tα/2) se1n+n(x0xˉ) 2nx2(x) 2E = \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 tα/2t _ { \alpha / 2 } is found from the tt -table using n2n - 2 degrees of freedom.
Use the data below to obtain a 95%95 \% confidence interval estimate of the mean test score of all students who study hours. Note that the equation of the regression line is y^=52.8804+3.405x\hat { y } = 52.8804 + 3.405 \mathrm { x } and that se=6.8359\mathrm { s } _ { \mathrm { e } } = 6.8359 .
x (hours studied)  2.54.55.17.911.6y (score on test)  6670608393\begin{array}{l|ccccc}\mathrm{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}


A) (77.11,100.15) ( 77.11,100.15 )
B) (72.49,104.78) ( 72.49,104.78 )
C) (80.48,96.78) ( 80.48,96.78 )
D) (61.55,115.71) ( 61.55,115.71 )

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