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 .27.

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) = 0.27

1 - = 1 - 0.27 = 0.73

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 * 0.27 * 0.73

= 1892.94

sample size = n = 1893

b) x = 520

Point estimate = sample proportion = = x / n = 520 / 1893 = 0.275

1 - = 1 - 0.275 = 0.725

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 (((0.275 *0.725) / 1893)

= 0.0201

A 95% confidence interval for population proportion p is ,

± E   

= 0.275  ± 0.0201

= ( 0.2549, 0.2951 )