Chapter 27, Problem 21E

a)

To determine

To find the 95% confidence interval for average fuel efficiency among cars weighing 2500 pounds.

Answer to Problem 21E

We are 95% confidence that the average fuel efficient among cars weighing 2500 pounds is between 27.3396 mpg and 29.0709 mpg.

Explanation of Solution

Given:

  

Formula:

Confidence interval for average of response variable is,

  $(\widehat{y}\mp t_c\times S{E}_{\widehat{\mu }})$

Standard error is,

  $S{E}_{\widehat{\mu }}=\sqrt{S{E}_b{}^2{(x-\overline{x})}^2+\frac{s{e}^2}{n}}$

We know, the 2500 pounds means 2.5 thousand pounds.

Therefore, predicted value becomes,

  $\widehat{y}=48.7393-8.21362\times 2.5=28.20525$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{{0.6738}^2{(2.5-2.88780)}^2+\frac{{2.413}^2}{50}}\\ S{E}_{\widehat{\mu }}=0.4298\end{array}$

The confidence level = 0.95

So, level of significance = $\alpha$ = 0.05

First need to find critical t value.

tc = 2.014 …Using excel formula, =TINV(0.05,48)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(28.20525-2.014\times 0.4298,28.20525+2.014\times 0.4298)\\ (27.3396,29.0709)\end{array}$

Hence, we are 95% confidence that the average fuel efficient among cars weighing 2500 pounds is between 27.3396 mpg and 29.0709 mpg.

b)

To determine

To find 95% prediction interval for the gas mileage you might get driving your new 3450-pound SUV.

Answer to Problem 21E

We are 95% confidence that the average fuel efficient among cars weighing 3450 pounds is between 15.4352 mpg and 25.3694 mpg.

Explanation of Solution

Given:

  

Formula:

Confidence interval for average of response variable is,

  $(\widehat{y}\mp t_c\times S{E}_{\widehat{\mu }})$

Standard error is,

  $S{E}_{\widehat{\mu }}=\sqrt{S{E}_b{(x-\overline{x})}^2+\frac{s{e}^2}{n}+s{e}^2}$

We know, the 3450 pounds means 3.45 thousand pounds.

Therefore, predicted value becomes,

  $\widehat{y}=48.7393-8.21362\times 3.45=20.402311$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{{0.6738}^2\times {(3.45-2.88780)}^2+\frac{{2.413}^2}{50}+{2.413}^2}\\ S{E}_{\widehat{\mu }}=2.466276\end{array}$

The confidence level = 0.95

So, level of significance = $\alpha$ = 0.05

First need to find critical t value.

tc = 2.014 …Using excel formula, =TINV(0.05,48)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(20.402311-2.466276\times 0.4298,20.402311+2.014\times 2.466276)\\ (15.4352,25.3694)\end{array}$

Hence, we are 95% confidence that the average fuel efficient among cars weighing 3450 pounds is between 15.4352 mpg and 25.3694 mpg.