# 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 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.

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Step 1
Solution:
a.
From the given information, confidence level is $$0.95, p=0.30\ and\ E=0.02.$$
$$\displaystyle{1}-\alpha={0.95}$$
$$\displaystyle\alpha={1}-{0.95}$$
$$\displaystyle\alpha={0.05}$$
Step 2
Then,
$$\displaystyle{n}={\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}{p}{\left({1}-{p}\right)}$$
$$\displaystyle={\left(\frac{{Z}_{{\frac{{0.05}}{{2}}}}}{{0.02}}\right)}^{{2}}{0.30}{\left({1}-{0.30}\right)}$$
$$\displaystyle={\left(\frac{{Z}_{{0.025}}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}$$
$$\displaystyle={\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}{\left[{U}{\sin{{g}}}{e}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n}={N}{O}{R}{M}.{I}{N}{V}{\left({0.025},{0},{1}\right)}\right]}$$
$$=2016.84$$
$$\displaystyle\approx{2017}$$
Thus, the required sample size is 2017.