Chapter 27, Problem 32E

a)

To determine

To find the 95% confidence interval for the slope of the regression line.

Answer to Problem 32E

We are 95% confident that the slope of the population of regression line is between 0.08 and 0.11.

Explanation of Solution

Given:

  

Formula:

Confidence interval for slope is,

  $(Coefficient(b)\mp t_c\times S{E}_b)$

The confidence level = 0.95

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

First need to find critical t value.

tc = 1.992 …Using excel formula, =TINV(0.05,75)

Using formula, 95% confidence interval for the slope of the regression line.

  $\begin{array}{l}(0.092498-1.992\times 0.0075,0.092498+1.992\times 0.0075)\\ (0.08,0.11)\end{array}$

Therefore, we are 95% confident that the slope of the population of regression line is between 0.08 and 0.11.

b)

To determine

To find the 95% confidence interval for the mean company’s profit.

Answer to Problem 32E

We are 95% confidence that the company’s profits are between 532.59 million dollars and 779.08 million dollars

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,

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

Therefore, predicted value becomes,

  $\widehat{y}=-176.644+0.092498\times 9000=655.838$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{{0.0075}^2{(9000-\mathrm{41.78.29})}^2+\frac{{446.2}^2}{79}}\\ S{E}_{\widehat{\mu }}=61.8703\end{array}$

The confidence level = 0.95

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

First need to find critical t value.

tc = 1.992 …Using excel formula, =TINV(0.05,77)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(655.838-1.992\times 61.8703,655.838+1.992\times 61.8703)\\ (532.59,779.08)\end{array}$

Hence, we are 95% confidence that the company’s profits are between 532.59 million dollars and 779.08 million dollars.