Chapter 24, Problem 11E

(a)

To determine

To find a $\text{95\%}$ confidence interval for the difference in average commuting time for the two routes.

Answer to Problem 11E

The confidence interval will be: $(1.31,4.69)$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=43\\ {\overline{x}}_2=40\\ n_1=20\\ n_2=20\\ s_1=2\\ s_2=3\\ \alpha =0.05\end{array}$

Now, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (20-1,20-1)\\ =19\end{array}$

Thus the t value will be:

  $t=2.093$

The confidence interval will be as:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(43-40)-2.093\times \sqrt{\frac{2^2}{20}+\frac{3^2}{20}}\\ =1.31\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(43-40)+2.093\times \sqrt{\frac{2^2}{20}+\frac{3^2}{20}}\\ =4.69\end{array}$

Thus, the confidence interval will be: $(1.31,4.69)$ .

(b)

To determine

To explain should the man believe the old-timer’s claim that he can save an average of five minutes a day by always driving Route A.

Answer to Problem 11E

No, the man does not believe the old timer’s claim.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=43\\ {\overline{x}}_2=40\\ n_1=20\\ n_2=20\\ s_1=2\\ s_2=3\\ \alpha =0.05\end{array}$

And the confidence interval will be: $(1.31,4.69)$ . Thus, the man should not believe the old-timer’s claim that he can save an average of five minutes a day by always driving Route A as according to the confidence interval that is, on average the minutes travelled by the man on both the routes is between $(1.31,4.69)$ and five minutes does not lie in the interval. Thus, the claim by the old man is false ad not appropriate as the five minutes is beyond the high end of the interval.