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 .31. 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. 637 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)? .31 c. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

Answer 1

Solution,

Given that,

a) = 0.31

1 - = 1 - 0.31 = 0.69

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 * 0.31 * 0.69

= 2054.29

sample size = n = 2055

b) Point estimate = sample proportion = = x / n = 520 / 2055 = 0.2530

1 - = 1 - 0.2530 = 0.7470

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

= 1.96 (((0.2530 * 0.7470) / 2055 )

= 0.0188

A 95% confidence interval for population proportion p is ,

± E   

= 0.2530  ± 0.0188

= ( 0.2342, 0.2718 )