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 .34.

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)?

( , )

Explanation / Answer

a.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.34
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.34*0.66
= 2155.138 ~ 2156
b.
No. of success(x)=520
Sample Size(n)=2156
Sample proportion = x/n =0.241

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.241 ±Z a/2 ( Sqrt ( 0.241*0.759) /2156)]
= [ 0.241 - 1.96* Sqrt(0) , 0.241 + 1.96* Sqrt(0) ]
= [ 0.223,0.259]

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