Chapter 10.1, Problem 11E

To determine

To construct and interpret a $90\%$ confidence interval for the difference in the proportions of people like these who would say they support president Obama’s decision when asked by Hassan versus when asked by Ken.

Answer to Problem 11E

The confidence interval is $(-0.4142,-0.1004)$ .

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=11\\ x_2=21\\ n_1=50\\ n_2=44\\ c=90\%=0.90\end{array}$

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 assuming that $50$ people are less than $10\%$ of all people walking on main street of the town.

Normal: It is satisfied because there are $11$ successes and $50-11=39$ failures in the first sample and there are $21$ successes and $44-21=23$ 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,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{11}{50}\\ =0.22\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{21}{44}\\ =0.4773\end{array}$

Now, for the confidence interval $1-\alpha =0.90$ , we have to find out the value of $z_{\alpha /2}=z_{0.05}$ using the table. Thus the z score will be:

  $z_{\alpha /2}=1.645$

Thus, the confidence interval will be:

  $\begin{array}{l}({\widehat{p}}_1-{\widehat{p}}_2)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.22-0.4773)-1.645\times \sqrt{\frac{0.22(1-0.22)}{50}+\frac{0.4773(1-0.4773)}{44}}\\ \approx -0.4142\\ ({\widehat{p}}_1-{\widehat{p}}_2)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.22-0.4773)+1.645\times \sqrt{\frac{0.22(1-0.22)}{50}+\frac{0.4773(1-0.4773)}{44}}\\ \approx -0.1004\end{array}$

Thus, we conclude that, we are $90\%$ confident that the true proportion of people who say they support president Obama’s decision when asked by Hassan is between $0.4142\text{ and 0}\text{.1004}$ lower than the true proportion of people who say they support president Obama’s decision when asked by Ken.