Chapter 10.1, Problem 9E

(a)

To determine

To construct and interpret a $99\%$ confidence interval for the difference in the true proportions of men and women aged $19\text{ to }25$ who lives in their parents’ homes.

Answer to Problem 9E

There is a $99\%$ confidence that the proportion of men aged $19\text{ to }25$ who live in their parents’ house is between $0.051\text{ and 0}\text{.123}$ higher than the proportion of women aged $19\text{ to }25$ who live in their parents’ house.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=986\\ x_2=923\\ n_1=2253\\ n_2=2629\\ c=99\%=0.99\end{array}$

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the samples are independent random samples.

Independent: It is satisfied because the $2253$ U.S. men are less than $10\%$ of all U.S. men and the same is for the U.S. women also.

Normal: It is satisfied because there are $983$ successes and $2253-983=1270$ failures in the first sample and there are $923$ successes and $2629-923=1706$ failures in the second sample, which are all at least ten.

Thus, all the conditions are satisfied.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{986}{2253}\\ =0.438\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{923}{2629}\\ =0.351\end{array}$

Now, for the confidence interval $1-\alpha =0.99$ , we have to find out the value of $z_{\alpha /2}=z_{0.005}$ using the table. Thus the z score will be:

  $z_{\alpha /2}=2.575$

Thus, the confidence interval will be:

  $\begin{array}{l}({\widehat{p}}_1-{\widehat{p}}_2)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.438-0.351)-2.575\times \sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}\\ \approx 0.051\\ ({\widehat{p}}_1-{\widehat{p}}_2)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.438-0.351)+2.575\times \sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}\\ \approx 0.123\end{array}$

Thus, we conclude that, we are $99\%$ confident that the proportion of men aged $19\text{ to }25$ who live in their parents’ house is between $0.051\text{ and 0}\text{.123}$ higher than the proportion of women aged $19\text{ to }25$ who live in their parents’ house.

(b)

To determine

To explain does your interval from part (a) give a convincing evidence of a difference between the population proportions.

Answer to Problem 9E

Yes, it gives.

Explanation of Solution

The confidence interval in part (a) is calculated as:

  $\left(0.051,0.123\right)$

Thus, this confidence interval does not contain zero, then it is very unlikely that the population proportions are equal and thus there is a convincing evidence of a difference between the population proportions. Thus, confidence interval calculated in part (a) gives the convincing evidence.