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

Solution :

Given that,

= 0.32

1 - = 1 - 0.32 = 0.68

margin of error = E = 0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

a)

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

= (1.96 / 0.02)² *0.32*0.68

= 2089.83

sample size = 2090

b)

n = 2090

x = 520

Point estimate = sample proportion = = x / n = 520/2090 = 0.2488

1 - = 1 - 0.2488 = 0.7512

c)

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

= 1.96 * (((0.2488*0.7512) / 2090)

= 0.0185

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2488 - 0.0185 < p < 0.2488 + 0.0185

0.2303 < p < 0.2673

The 95% confidence interval for the population proportion p is : ( 0.2303 , 0.2673 )