The 2003 Statistical Abstract of the United States reported the percentage of pe

Question

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .26.

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 sample proportion of smokers (to 4 decimals)?

c. Based on the answer in part (b), what is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

Please show computations, thank you.

Explanation / Answer

a) The formula for finding the sample size needed to estimate a population proportion is

n = phat * (1-phat) * (z/E)^2, where

phat is the sample proportion, which according to the preliminary estimate is 0.26,

z is the multiplier based on the confidence level. Since this is a 95% confidence interval, then we would use 1.96, and

E is the desired margin of error, which is 0.02. So

n = 0.26 * 0.64 * (1.96/0.02)^2 = 1598.1056, which should be rounded up to 1599.

b) If there are 520 smokers out of these 1599, then the point estimate, which is your best guess, is

phat = 520/1599 = 0.32520

c) The 95% confidence interval for the population proportion is found by computing

phat +- z*sqrt{phat*(1-phat)/n}

=> 0.32520 +- 1.96 * sqrt{0.32520*(1-0.32520)/1599}
=> 0.32520 +- 0.02296

So I am 95% confident that the proportion of smokers in the population is between 0.32520-0.02296 = 0.30224 and 0.32520+0.02296 = 0.34816.

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