91影视

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

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

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.