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 .34.

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

= 1.962 * 0.34 ( 1 - 0.34) / 0.022

= 2155.14

Sample size = 2156 (Rounded up to nearest integer)

b)

Point estimate = = 520 / 2156 = 0.2412

c)

margin of error E = Z/2 * sqrt [ ( 1 - ) / n ]

= sqrt [ 1.96 * 0.2412 ( 1 - 0.2412) / 2156 ]

= 0.0181

95% CI is

- E < p < + E

0.2412 - 0.0181 < p < 0.2412 + 0.0181

0.2231 < p < 0.2593

95% CI is ( 0.2231 , 0.2593 )