A confidence interval for the slope $\beta _ { 1 }$ for a regression line $\mathrm { y } = \beta _ { 0 } + \beta _ { 1 } \mathrm { x }$ can be found by evaluating the limits in $\mathrm { t }$ interval below:
$$\begin{array} { l }
\mathrm { b } _ { 1 } - \mathrm { E }

Question

A confidence interval for the slope $\beta _ { 1 }$ for a regression line $\mathrm { y } = \beta _ { 0 } + \beta _ { 1 } \mathrm { x }$ can be found by evaluating the limits in $\mathrm { t }$ interval below:
$$\begin{array} { l } 
\mathrm { b } _ { 1 } - \mathrm { E } < \beta _ { 1 } < \mathrm { b } _ { 1 } + \mathrm { E } _ { g } \
\text { where } \mathrm { E } = \frac { \left( \mathrm { t } _ { \alpha / 2 } \right) \mathrm { se } _ { \mathrm { e } } } { \sqrt { \sum \mathrm { x } ^ { 2 } - \left( \sum \mathrm { x } \right) ^ { 2 } / \mathrm { n } } } .
\end{array}$$
The critical value $t _ { \alpha / 2 }$ is found from the t-table using $\mathrm { n } - 2$ degrees of freedom and $\mathrm { b } _ { 1 }$ is calculated in the usual v from the sample data.
Use the data below to obtain a $95 \%$ confidence interval estimate of $\beta _ { 1 }$.
$\begin{array}{c|ccccc}
\mathrm{x} \text { (hours studied) } & 2.5 & 4.5 & 5.1 & 7.9 & 11.6 \
\hline \mathrm{y} \text { (score on test) } & 66 & 70 & 60 & 83 & 93
\end{array}$
A) $0.322 < \beta _ { 1 } < 6.488$
B) $1.936 < \beta _ { 1 } < 4.874$
C) $0.134 < \beta _ { 1 } < 6.676$
D) $0.686 < \beta _ { 1 } < 6.124$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as follows:$\mathrm { b } _ { 1 } = \frac { \sum \mathrm { x } \mathrm { y } - \left( \sum \mathrm { x } \right) \left( \sum \mathrm { y } \right) / \mathrm { n } } { \sum \mathrm { x } ^ { 2 } - \left( \sum \mathrm { x } \right) ^ { 2 } / \mathrm { n } } = 3.556$$\mathrm { se } _ { \mathrm { e } } = \sqrt { \frac { \sum \left( \mathrm { y } - \hat { \mathrm { y } } \right) ^ { 2 } } { \mathrm { n } - 2 } } = 11.481$$\mathrm { t } _ { \alpha / 2 } = 2.776$$\mathrm { E } = \frac { \left( \mathrm { t } _ { \alpha / 2 } \right) \mathrm { se } _ { \mathrm { e } } } { \sqrt { \sum \mathrm { x } ^ { 2 } - \left( \sum \mathrm { x } \right) ^ { 2 } / \mathrm { n } } } = 2.581$Therefore, the $95 \%$ confidence interval is:$\mathrm { b } _ { 1 } - \mathrm { E } < \beta _ { 1 } < \mathrm { b } _ { 1 } + \mathrm { E }$$3.556 - 2.581 < \beta _ { 1 } < 3.556 + 2.581$$0.975 < \beta _ { 1 } < 6.137$

A Confidence Interval for the Slope β1\beta _ { 1 }β1​

A Confidence Interval for the Slope $\beta _ { 1 }$