Question

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 .29.

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)?

Answer 1

a)
Estimate of proportion, pcap = 0.29
Margin of Error, ME = 0.02
For 0.95 Confidence level, the z-value = 1.96

Sample size, n = (z/ME)^2 * pcap * (1 - pcap)
n = (1.96 / 0.02 )^2 * 0.29 * (1 - 0.29)
n = 1977.46
Rounding to the nearest integer
n = 1977

b)
point estimate = 520/1977 = 0.263

c)
sample proportion, pcap = 0.263
sample size, n = 1977
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.263 * (1 - 0.263)/1977) = 0.0099

For 95% Confidence level, the z-value = 1.96
CI = (pcap - z*SE, pcap + z*SE)
CI = (0.263 - 1.96 * 0.0099 , 0.263 + 1.96 * 0.0099)
CI = (0.2436 , 0.2824)