Weeds among the corn Refer to Exercise 13.

(a) Construct and interpret a 90% confidence interval for the slope of the true regression line. Explain how your results are consistent with the significance test in Exercise 13.

(b) Interpret each of the following in context:

(i) 8

(ii) $r^2$

(iii) The standard error of the slope

Given:

$n=16$

$b=-1.0987$

$S{E}_b=0.5712$

The degrees of freedom is the sample size decreased by $2$:

$$\begin{gathered} df=n-2 \\ =16-2 \\ =14 \end{gathered}$$

The critical t-value can be found in table B in the row of $df=14$and in the column of $c=90\%$:

$t^{*}=1.761$

The boundaries of the confidence interval then become:

localid="1650694151582" $$\begin{gathered} b-t^{*}\times S{E}_b=-1.0987-1.761\times 0.5712 \\ =-2.1046 \end{gathered}$$

localid="1650694166451" $$\begin{gathered} b+t^{*}\times S{E}_b=-1.0987+1.761\times 0.5712 \\ =-0.0928 \end{gathered}$$

Given:

$s=7.97665$

$$\begin{gathered} r^2=20.9\% \\ =0.209 \end{gathered}$$

$S{E}_b=0.5823$

Given:

$s=7.97665$

$$\begin{gathered} r^2=20.9\% \\ =0.209 \end{gathered}$$

$S{E}_b=0.5823$

(i) $s=7.97665$indicates that the expected error when making predictions using the regression line is about $7.97665$bushels.

(ii) $r^2=20.9\%$indicates that $20.9\%$of the variation between the variables has been explained by the linear regression line.

(iii) $S{E}_b=0.5823$indicates that the estimation of the slope of the least-squares regression line is expected to vary on average by $0.5823$over all possible samples.