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 .30.

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)
The following information is provided,
Significance Level, α = 0.05, Margin of Error, E = 0.02

The provided estimate of proportion p is, p = 0.3
The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.3*(1 - 0.3)*(1.96/0.02)^2
n = 2016.84

Ans : Sample size, n = 2017

b)
point estimate, pcap = 520/2017 = 0.2578

c)
sample proportion, pcap = 0.2578
sample size, n = 2017
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.2578 * (1 - 0.2578)/2017) = 0.0097

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.2578 - 1.96 * 0.0097 , 0.2578 + 1.96 * 0.0097)
CI = (0.2388 , 0.2768)