The 2003 Statistical Abstract of the United States reported the percentage of pe
ID: 3183660 • Letter: T
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 .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.
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 = 0.35
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.35*0.65
= 2184.91 ~ 2185
b.
no.of observed in part a(x)=520
Sample Size(n)=2185
Sample proportion = x/n =0.238
c.
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 = [ 0.238 ±Z a/2 ( Sqrt ( 0.238*0.762) /2185)]
= [ 0.238 - 1.96* Sqrt(0) , 0.238 + 1.96* Sqrt(0) ]
= [ 0.22,0.256]
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