The 2003 statistical Abstract of the united States reported the percentage of pe
ID: 3390902 • 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 smokes and monsmokers uses a preliminary estimate of the proportion who smoke of .30 How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02? Use 95% confidence. Remove all cormas from your answer before submitting. Assume that the study uses sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the of smokers in the population?Explanation / Answer
a)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.3
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.3*0.7
= 2016.84 ~ 2017
b)
No. of people who smoke(x)=520
Sample Size(n)=2017
Point of estimate = Sample proportion = x/n =0.2578
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 = [ 0.2578 ±Z a/2 ( Sqrt ( 0.2578*0.7422) /2017)]
= [ 0.2578 - 2.576* Sqrt(0.0001) , 0.2578 + 2.58* Sqrt(0.0001) ]
= [ 0.2327,0.2829]
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