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 (rounded up to the next 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

Solution,

Given that,

a) = 0.29

1 - = 1 - 0.29 = 0.71

margin of error = E = 0.02

At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.02 )² * 0.29 * 0.71

= 1977.46

sample size = n = 1978

b) Point estimate = sample proportion = = x / n = 520 / 1978 = 0.2629

1 - = 1 - 0.2629 = 0.7371

c) Z/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.2629 * 0.7371) / 1978)

= 0.0194

A 95% confidence interval for population proportion p is ,

± E

= 0.2629 ± 0.0194

= ( 0.2435, 0.2823 )