 # 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 designe pancha3 2021-01-15 Answered
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 to be .30.
a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 2%? Use 95% confidence.
b) Assume that the study uses your sample size recommendation above and finds 520 smokers. What is the point estimate of the proportion of smokers in the population?
c) What is the 95% confidence interval for the proportion of smokers in the population?
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Step 1
Given data
Error = 0.02
Confidence level = 0.95
Significance level $=\alpha =1-0.95=0.05$
${Z}_{\frac{0.05}{2}}={Z}_{0.025}=±1.96$
(From Excel = NORM.S.INV(0.025))
sample proportion(p)=0.30
a)
Sample size is given by(n)
Margin of error formula is given by
$E={Z}_{\frac{\alpha }{2}}×\sqrt{\frac{p-\left(1-p\right)}{n}}$
Simplifying the above formula
$n=p\left(1-p\right)×{\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}=0.3×\left(1-0.3\right)×{\left(\frac{1.96}{0.02}\right)}^{2}=2016.84\approx 2017$
Step 2
b)
n=2017
No of smokers = 520
Point estimate is given by
$\stackrel{^}{p}=\frac{520}{2017}=0.258$
Step 3
c)
95% confidence interval for the proportion of smokers in the population is given by
Confidence level = 0.95
Significance level $=\alpha =1-0.95=0.05$
${Z}_{\frac{0.05}{2}}={Z}_{0.025}=±1.96$
(From Excel = NORM.S.INV(0.025))
Confidence interval is given by
$\stackrel{^}{p}+{Z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}-\left(1-\stackrel{^}{p}\right)}{n}}=0.258±1.96\sqrt{\frac{0.258\left(1-0.258\right)}{2017}}=\left(0.239,0.277\right)$
Step 4
Result:
a) 2017
b) 0.258
c) (0.239,0.277)