Chapter 12.1, Problem 15E

(a)

To determine

To construct and interpret a $90\%$ confidence interval for the slope of the true regression line and explain how your results are consistent with the significance test.

Answer to Problem 15E

  $(-2.1046,-0.0928)$ .

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}n=16\\ b=-1.0987\\ {\text{SE}}_b=0.5712\end{array}$

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

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

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

  $t^{*}=1.761$

Thus, the confidence interval is as:

  $\begin{array}{l}b-t^{*}\times {\text{SE}}_b\\ =1.0987-1.761\times 0.5712\\ =-2.1046\\ b+t^{*}\times {\text{SE}}_b\\ =1.0987+1.761\times 0.5712\\ =-0.0928\end{array}$

We are $90\%$ confident that the slope of the true regression line is between $(-2.1046,-0.0928)$ .

(b)

To determine

To interpret s , $r^2$ and the standard error of the slope.

Explanation of Solution

Now, from the computer output of this data, we have,

  $\begin{array}{l}s=7.97665\\ r^2=20.9\%=0.209\\ {\text{SE}}_b=0.5823\end{array}$

Thus, we have to interpret the three as:

1. $s=7.97665$ indicates that the expected error when making predictions using the regression line is about $7.97665$ bushels.
2. $r^2=20.9\%$ indicates that $20.9\%$ of the variation between the variables has been explained by the linear regression line
3. ${\text{SE}}_b=0.5823$ 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.