he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 who smoke. Step 2 of 2: Suppose a sample of 830 Americans over 45 is drawn. Of these people, 631 don't smoke. Using the data, construct the 85% confidence interval for the population proportion of Americans over 45 who smoke. Round your answers to three decimal places.

Answer:The 85% confidence interval is (0.2187, 0.2613).Step-by-step explanation:We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is [tex]\hat{p} = 199/830 = 0.24[/tex] and an estimate of the standard deviation of [tex]\hat{p}[/tex] is [tex]\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}[/tex] = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148[/tex] where [tex]z_{0.075}[/tex] is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is [tex]0.24\pm (1.4395)(0.0148)[/tex] or equivalently (0.2187, 0.2613).