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 .28.

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

Solution,

Given that,

a) = 0.28

1 - = 1 - 0.28 = 0.72

margin of error = E = 0.02

At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= (1.96 / 0.02)2 * 0.28 * 0.72

= 1936.16

sample size = n = 1937

b) x = 520

Point estimate = sample proportion = = x / n = 520 / 1937 = 0.2685

1 - = 1 - 0.2685 = 0.7315

c) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.96 (((0.2685 * 0.7315) / 1937)

= 0.0197

A 95% confidence interval for population proportion p is ,

± E  

= 0.2685  ± 0.0197

= ( 0.2488, 0.2882 )