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 .32. 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)

Sample size = Z²/2 * p( 1 -p ) / E² , Where E is margin of error.

= 1.96² * 0.32 * 0.68 / 0.02²

= 2089.83

Sample size = 2090 (Rounded up to nearest integer)

b)

Sample proportion = 520 / 2090 = 0.2488

is point estimate of population proportion.

So, point estimate of proportion = 0.2488

c)

95% confidence interval for proportion p is

- Z/2 * sqrt( ( 1 - ) / n) < p < + Z/2 * sqrt( ( 1 - ) / n)

0.2488 - 1.96 * sqrt(0.2488 * 0.7512 / 2090) < p < 0.2488 + 1.96 * sqrt(0.2488 * 0.7512 / 2090)

0.2303 < p < 0.2673

95% CI is ( 0.2303 , 0.2673)