# 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

The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke $⟨CDCwebsite,December14,2014⟩$. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30.
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
a.
The given information is as follows:
Given that the preliminary estimate of proportion of smokers among the people of 18 years and older age is 0.30.
That is, p = 0.30.
The confidence level is $\left(1–\alpha \right)=95\mathrm{%}$.
From this, the level of significance is $\alpha =0.05$.
Furthermore, it is given that the margin of error is 0.02. That is, E = 0.02.
Step 2
Obtain the critical value for the given situation:
The confidence level is 95%.
The critical value corresponding to the given situation is obtained as ${Z}_{\frac{\alpha }{2}}=1.96$ from the calculation given below:
$1-\alpha =1-0.95$
$\alpha =0.05$
$\frac{\alpha }{2}=\frac{0.05}{2}$
=0.025
Hence, cumulative area to the left is, Area to the left = 1 - Area to the right
=1-0.025
=0.975
${Z}_{\frac{\alpha }{2}}=1.96\left[Excelf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mu la:NORMINV\left(0.975,0,1\right)\right]$
Step 3
Obtain the required sample size:
Here, ${z}_{\frac{\alpha }{2}}=1.96,p=0.3,E=0.02$.
The required sample size is obtained as 2,017 from the calculation given below:
$n=\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)×{\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}$
$=\left(0.3×\left(1-0.3\right)\right)×{\left(\frac{1.96}{0.02}\right)}^{2}$
$=0.21×9,604$
=2,016.84
$\approx 2,017$(Rounded to nearest whole number)
Step 4
b.
Obtain the point estimate of the proportion of smokers in the population using the sample size obtained in part (a):
It is given that among a sample of 2,017 people of age 18 years and older, 520 are identified as smokers.
The number of people sampled is n = 2,017.
The number of smokers among the sampled people is x = 520.
The point estimate of the proportion of smokers in the population is obtained as 0.2578 from the calculation given below:
$\stackrel{^}{p}=\frac{x}{n}$
$=\frac{520}{2},017$
=0.2578(Rounded to 4 decimal places)
Step 5
c.
Obtain the 95% confidence interval for the proportion of smokers in the population:
Here, $p-\stackrel{^}{=}0.2578,n=2,017\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{\frac{\alpha }{2}}=1.96$.
Denote the proportion of smokers in the population as p.
The 95% confidence interval for the proportion of smokers in the population is obtained as $0.2387\le p\le 0.2769$ from the calculation given below:
$CI\left(\stackrel{^}{p}-{Z}_{\frac{\alpha }{2}}×\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}\le p\le \stackrel{^}{p}+{Z}_{\frac{\alpha }{2}}×\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}}$
$=\left(0.2578-1.96×\sqrt{\frac{0.2578\left(1-0.2578\right)}{2},017}\le p\le 0.2578+1.96×\sqrt{\frac{0.2578\left(1-0.2578\right)}{2},017}\right)$
$=\left(0.2578-0.0191\le p\le 0.2578+0.0191\right)$
$=\left(0.2387\le p\le 0.2769\right)$
Step 6
Result:
a. The sample size satisfying the given requirement is n = 2,017.
b. The point estimate of the proportion of smokers in the population is 0.2578.
c. The 95% confidence interval for the proportion of smokers in the population is obtained as $0.2387\le p\le 0.2769$.