Chapter 27, Problem 35E

a)

To determine

To test an appropriate hypothesis.

Answer to Problem 35E

There is sufficient evidence to support a claim that there is an association between Waist size and percentage of body fat.

Explanation of Solution

Given:

  

The null and alternative hypotheses:

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_a:{\beta }_1\ne 0\end{array}$

Using excel,

  

Test statistic:

t = 8.144

Therefore, p-value would be,

P-value = 0

Decision: P-value < 0.05, reject H₀.

Conclusion: There is sufficient evidence to support a claim that there is an association between Waist size and percentage of body fat.

b)

To determine

To find the 95% confidence interval for the mean percentage of body fat.

Answer to Problem 35E

We are 95% confidence interval for the mean percentage of body fat found in people with a 40- inch Waist is between 23.5810% and 29.0233%.

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,

  $\widehat{y}=-62.5573+\mathrm{0.2.2215}\times 40=26.3027$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{{0.2728}^2{(40-37.05)}^2+\frac{{4.5397}^2}{20}}\\ S{E}_{\widehat{\mu }}=1.2954\end{array}$

The confidence level = 0.95

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

First need to find critical t value.

tc = 1.101 …Using excel formula, =TINV(0.05,18)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(26.3027-1.101\times 1.2954,26.3027+1.101\times 1.2954)\\ (23.5810,29.0244)\end{array}$

Hence, we are 95% confidence interval for the mean percentage of body fat found in people with a 40- inch Waist is between 23.5810% and 29.0233%.