Question: Misleading Survey Responses: In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.

(a) Find a 99% confidence interval estimate of the proportion of people who say that they voted.

(b) Are the survey results consistant with the actual voter turnout of 61%? Why or why not?

Solution: (a) Using the data provided, the 99% confidence interval for the actual population proportion is computed as

\[CI=\left( \hat{p}-{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}},\,\,\hat{p}+{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}} \right)\]

In this case, we have that

\[\begin{aligned} & \hat{p}=\,\text{Sample Proportion}=\frac{x}{n}=\frac{701}{1002}={0.6996} \\ & {{z}_{\alpha /2}}=\text{ Critical z-value}={2.58} \\ \end{aligned}\]

where n corresponds to the sample size. The confidence interval corresponds to

\[CI=\left( {0.6996}-{2.58}\sqrt{\frac{{0.6996}\left( 1-{0.6996} \right)}{1002}},\,\,{0.6996}+{2.58}\sqrt{\frac{{0.6996}\left( 1-{0.6996} \right)}{1002}} \right)=({0.6623},\,\,{0.7369})\]

(b) No, because 0.61 is NOT contained by the confidence interval.

Finding Sample Size: in exercise 14, use the given information to find the minimum sample size required to estimate an unknown population mean u (with a long line).

14. Breaking Distance: How many cars must be randomly selected and tested in order to estimate the mean braking distance of registered cars in the United States. We want 99% confidence that the sample mean is within 2 ft of the population, mean, and the population standard deviation is known to be 7ft.