Chapter 12.1, Problem 16E

(a)

To determine

To construct and interpret a $98\%$ confidence interval for the slope of the true regression line and explain how your results are consistent with the significance test.

Answer to Problem 16E

  $(-5.2458,-0.9084)$ .

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}n=20\\ b=-3.0771\\ {\text{SE}}_b=0.8498\end{array}$

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

  $\begin{array}{l}df=n-2\\ =20-2\\ =18\end{array}$

The critical t -value can be found in table B in the row of $df=18$ and in the column of $c=98\%$ as:

  $t^{*}=2.552$

Thus, the confidence interval is as:

  $\begin{array}{l}b-t^{*}\times {\text{SE}}_b\\ =-3.0771-2.552\times 0.8498\\ =-5.2458\\ b+t^{*}\times {\text{SE}}_b\\ =-3.0771+2.552\times 0.8498\\ =-0.9084\end{array}$

We are $98\%$ confident that the slope of the true regression line is between $(-5.2458,-0.9084)$ .

(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,

  $\begin{array}{l}s=23.3980\\ r^2=42.1\%=0.421\\ {\text{SE}}_b=0.8498\end{array}$

Thus, we have to interpret the three as:

1. $s=23.3980$ indicates that the expected error when making predictions using the regression line is about $23.3980$ calories.
2. $r^2=42.1\%$ indicates that $42.1\%$ of the variation between the variables has been explained by the linear regression line
3. ${\text{SE}}_b=0.8498$ indicates that the estimation of the slope of the least-squares regression line is expected to vary on average by $0.8498$
4. over all possible samples.