 # In 2014, the Centers for Disearse reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new OlmekinjP 2020-11-20 Answered
In 2014, the Centers for Disearse 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 0.30.
a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? Use 95% confidence.(Round your answer up to the nearest integer.)
b)Assume that the study uses your sample size recommendation in part (a) and finds 470 smokers. What is the point estimate of the proportion of smokers in the population? (Round your answer to four decimal places.)
c) What is the 95% confidence interval for the proportion of smokers in the population? (Round your answer to four decimal places.)
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Step 1
Given information
Error = 3% = 0.02
Confidence level = 0.95
Significance level $=\alpha =1-0.95=0.05$
${Z}_{\frac{0.05}{2}}={Z}_{0.025}=±1.9600\left(FromExcel=NORM.S.INV\left(0.025\right)\right)$
Population proportion = 0.30
Margin of error formula is given by
$E={Z}_{\frac{\alpha }{2}}×\sqrt{\frac{p\left(1-p\right)}{n}}$
a)
$n=p\left(1-p\right)×{\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}=0.30×0.70×{\left(\frac{1.9600}{0.02}\right)}^{2}=2016.84\approx 2017$
Sample size = 2017
Step 2
b)
x=470
n=2017
$\stackrel{^}{p}=\frac{x}{n}=\frac{470}{2017}=0.2330$
c)
95% Confidence interval given by
$\stackrel{^}{p}±{Z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$0.2330±1.9600\sqrt{\frac{0.2330\left(1-0.2330\right)}{2017}}=\left(0.2146,0.2514\right)$
${Z}_{\frac{\alpha }{2}}={Z}_{\frac{0.05}{2}}={Z}_{0.025}=±1.9600\left(FromExcel=NORM.S.INV\left(0.025\right)\right)$