he 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .26.
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.

b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)?

c. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

The margin of error is the 1.96(sqrt(pq/n) and that has to equal .02. We know p =.26, so q = .74

So we have 1.96(sqrt(.26*.74/n) = .02

3.8416(.1924)/n = .0004

0.73912384/n = .0004

n = 1848

b) point estimate is 524/1848 = .2835

c) .26 +/- .02 = (.24, .28)