Question

Do beavers benefit beetles? Researchers laid out $23$circular plots, each $4$meters in diameter, at random in an area where beavers were cutting down cottonwood trees. In each plot, they counted the number of stumps from trees cut by beavers and the number of clusters of beetle larvae. Ecologists think that the new sprouts from stumps are more tender than other cottonwood growth so beetles prefer them. If so, more stumps should produce more beetle larvae

Here is computer output for regression analysis of these data. Construct and interpret a$99\%$ confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met.

role="math" localid="1654159204144" $$\begin{gathered} \mathrm{Predictor}\mathrm{Coef}\mathrm{SE}\mathrm{Coef}\mathrm{T}\mathrm{P} \\ \mathrm{Constant}-1.2862.853-0.450.657 \\ \mathrm{Stumps}11.8941.13610.470.000 \\ \mathrm{S}=6.41939\mathrm{R}-\mathrm{Sq}=83.9\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=83.1\% \end{gathered}$$

Short answer

We are $99\%$confident that the slope of the true regression line is between .

Step-by-step solution

Given Information

We need to construct and interpret a $99\%$confidence interval for the slope of the population regression line.

Simplify

Consider:

$$\begin{gathered} \mathrm{n}=23 \\ \mathrm{b}=11.894 \\ {\mathrm{SE}}_{\mathrm{b}}=1.136 \end{gathered}$$

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

$\mathrm{df}=\mathrm{n}-2==23-2=21$

The critical $\mathrm{t}$-value can be found in table B in the row of $\mathrm{df}=21$and in the column of $\mathrm{c}=99\%$

$\mathrm{t}\text{'}=2.831$

The boundaries of the confidence interval then become:

$b-t\text{'}\times SEb=11.894-2.831\times 1.136=8.677984$

$b+t\text{'}\times SEb=11.894+2.831\times 1.136=115.110016$