# The centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a

The centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and Questions Navigation Menu preliminary estimate of the proportion who smoke of .26.
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 point estimate of the proportion of smokers in the population (to 4 decimals)?
c) What is the 95% confidence interval for the proportion of smokers in the population?(to 4 decimals)?
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Step 1
Given:
Margin of error (E) = 0.02
Confidence level = 95%
Population proportion (p) = 0.26
Step 2
(a)
The sample size could be calculated using the following formula:
$n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}$
Step 3
The sample size at 95% confidence level can be calculated as:
The value of $\frac{{Z}_{\alpha }}{2}$ corresponding to 95% confidence level is 1.96
$n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}$
$=0.26\left(1-0.26\right){\left(\frac{1.96}{0.02}\right)}^{2}$
$=0.1924×9604$
=1847.810
$\approx 1848$
Step 4
(b)
Here, X = 520 and n = 1848.
The point estimate can be calculated as:
$\stackrel{^}{p}=\frac{X}{n}$
$=\frac{520}{1848}$
=0.2814
Step 5
(c)
The 95% confidence interval for the proportion of smokers in the population can be calculated as:
$CI=\stackrel{^}{p}±{Z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.2814±1.96×\sqrt{\frac{0.2814\left(1-0.2814\right)}{1848}}$
$=0.2814±0.0205$
=(0.2609,0.3019)