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 .35.

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.

the answer for A is 2185

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)?

The answer for B is 0.2380

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

( , )

What is the answers to C because I can't figure it out? Please write legibly! and thank you!

Answer 1

Solution :

Given that,

c) n = 2185

Point estimate = sample proportion = = 0.2380

1 - = 1 - 0.2380 = 0.7620

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

= 1.96 (((0.2380 * 0.7620) / 2185 )

= 0.0179

A 95% confidence interval for population proportion p is ,

± E   

= 0.2380 ± 0.0179

= (0.2201, 0.2559)