The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .31. a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence.

Answer:A sample of 2017 people should be taken.Step-by-step explanation:In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]In whichz is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].The margin of error is:[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]Suppose a 95% confidence level:So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex]. Preliminary estimate of the proportion who smoke of .30.This means that [tex]\pi = 0.3[/tex]a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)This is n for which M = 0.02. So[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex][tex]0.02 = 1.96\sqrt{\frac{0.3*0.7}{n}}[/tex][tex]0.02\sqrt{n} = 1.96\sqrt{0.3*0.7}[/tex][tex]\sqrt{n} = \frac{1.96\sqrt{0.3*0.7}}{0.02}[/tex][tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.3*0.7}}{0.02})^2[/tex][tex]n = 2016.84[/tex]To the nearest whole number, 2017.A sample of 2017 people should be taken.