Question

The students in Mr. Shenk’s class measured the arm spans and heights (in inches) of a random sample of $18$students from their large high school. Here is computer output from a least-squares regression analysis of these data. Construct and interpret a $90\%$confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met.

$$\begin{gathered} \mathrm{Predictor}\mathrm{Coef}\mathrm{Stdev}\mathrm{t}-\mathrm{ratio}\mathrm{P} \\ \mathrm{Constant}11.5475.6002.060.056 \\ \mathrm{Armspan}0.840420.0809110.390.000 \\ \mathrm{S}=1.613\mathrm{R}-\mathrm{Sq}=87.1\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=86.3\% \end{gathered}$$

Short answer

We are $90\%$confident that the slope of the true regression line is between $0.69915114\mathrm{and}0.98168886$.

Step-by-step solution

Given Information

We need to construct and interpret a $90\%$confidence interval for the slope of the population regression line.

Simplify

Consider:

$$\begin{gathered} \mathrm{n}=18 \\ \mathrm{b}=0.84042 \\ {\mathrm{SE}}_{\mathrm{b}}=0.08091 \end{gathered}$$

The degrees of freedom in sample size decreased by $2$:

$\mathrm{df}=\mathrm{n}-2=18-2=16$

The critical $\mathrm{t}$-value can be found in table B in the row of $\mathrm{df}=16$and the column of $\mathrm{c}=90\%$

$\mathrm{t}\text{'}=1.746$

The boundaries of the confidence interval then become:

$b-t\text{'}\times S{E}_b=0.84042-1.746\times 0.08091=0.69915114$
role="math" localid="1654161040846" $b+t\text{'}\times S{E}_b=0.84042+1.746\times 0.08091=0.98168886$