Chapter 8.2, Problem 46E

(a)

To determine

To construct and interpret a $90\%$ confidence interval for the proportion of all U.S. adults who could name all three branches of government.

Answer to Problem 46E

We are $90\%$ confident that the true proportion of all U.S. adults who could name all three branches of government is between $(0.3350,0.3850)$ .

Explanation of Solution

It is given that,

  $\begin{array}{l}c=90\%=0.90\\ \widehat{p}=0.36\\ n=1416\end{array}$

Conditions to be met are:

The three conditions for calculating a confidence interval for the population proportion p are:

Random: It is satisfied because the U.S. adults are randomly selected.

Independent: It is satisfied because the sample of $1416$ U.S. adults is less than $10\%$ of the population of all U.S. adults.

Normal: It is satisfied because there are $1416(0.36)=509.76$ successes and $1416(1-0.36)=906.24$ failures, which are bout at least ten.

Thus, all the three conditions are satisfied.

The calculation is as:

For the confidence interval $1-\alpha =0.90$ determine $z_{\alpha /2}=z_{0.05}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=1.96$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =1.96\times \sqrt{\frac{0.36(1-0.36)}{1416}}\\ =0.0250\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.6-0.0250\\ =0.3350\\ \widehat{p}+E=0.6+0.0250\\ =0.3850\end{array}$

So, we are $90\%$ confident that the true proportion of all U.S. adults who could name all three branches of government is between $(0.3350,0.3850)$ .

(b)

To determine

To explain does the interval from part (a) provide convincing evidence that less than half of all U.S. adults could name all three branches of government.

Answer to Problem 46E

Yes, there is convincing evidence that less than half of all U.S. adults could name all three branches of government.

Explanation of Solution

It is given that,

  $\begin{array}{l}c=90\%=0.90\\ \widehat{p}=0.36\\ n=1416\end{array}$

For the confidence interval $1-\alpha =0.90$ determine $z_{\alpha /2}=z_{0.05}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=1.96$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =1.96\times \sqrt{\frac{0.36(1-0.36)}{1416}}\\ =0.0250\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.6-0.0250\\ =0.3350\\ \widehat{p}+E=0.6+0.0250\\ =0.3850\end{array}$

Thus, the confidence interval $(0.3350,0.3850)$ does not contain $0.5$ and contains only values below $0.5$ which indicates that it is likely that the population proportion is less than $0.5$ . This then implies that there is convincing evidence that less than half of all U.S. adults could name all three branches of government.