Chapter 27, Problem 36E

a)

To determine

To find a 90% confidence interval for the slope of the regression line of % Body Fact on Weight.

Answer to Problem 36E

(0.1447%, 0.3551%).

Explanation of Solution

Given:

  

Using excel,

  

Therefore, we are 90% confident that the percentage of body fat increases on average between 0.1447% and 0.3551%.

b)

To determine

To interpret a 90% confidence interval for the slope of the regression line of % Body Fact on Weight.

Answer to Problem 36E

We are 90% confident that the percentage of body fat increases on average between 0.1447% and 0.3551%.

Explanation of Solution

Given:

  

Therefore, we are 90% confident that the percentage of body fat increases on average between 0.1447% and 0.3551%.

c)

To determine

To find the 95% prediction interval for the

Answer to Problem 36E

We are 95% confidence that the percentage of body fat of an individual who weighs 165 pounds is between -5.6286% and 33.3346%.

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,

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

Therefore, predicted value becomes,

  $\widehat{y}=-27.3763+0.2499\times 165=13.8530$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{0.0607\times {(165-188.6)}^2+\frac{{7.049}^2}{20}+{7.049}^2}\\ S{E}_{\widehat{\mu }}=9.2726\end{array}$

The confidence level = 0.95

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

First need to find critical t value.

tc = 2.101 …Using excel formula, =TINV(0.05,18)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(13.853-2.101\times 9.2726,13.853+2.101\times 9.2726)\\ (-5.6286,33.3346)\end{array}$

Hence, we are 95% confidence that the percentage of body fat of an individual who weighs 165 pounds is between -5.6286% and 33.3346%.