# 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)?

Question
Study design
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)?

2021-02-01
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:
$$\displaystyle{n}={p}{\left({1}-{p}\right)}{\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}$$
Step 3
The sample size at 95% confidence level can be calculated as:
The value of $$\displaystyle\frac{{Z}_{\alpha}}{{2}}$$ corresponding to 95% confidence level is 1.96
$$\displaystyle{n}={p}{\left({1}-{p}\right)}{\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}$$
$$\displaystyle={0.26}{\left({1}-{0.26}\right)}{\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}$$
$$\displaystyle={0.1924}\times{9604}$$
=1847.810
$$\displaystyle\approx{1848}$$
Step 4
(b)
Here, X = 520 and n = 1848.
The point estimate can be calculated as:
$$\displaystyle\hat{{{p}}}=\frac{{X}}{{n}}$$
$$\displaystyle=\frac{{520}}{{1848}}$$
=0.2814
Step 5
(c)
The 95% confidence interval for the proportion of smokers in the population can be calculated as:
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{Z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}$$
$$\displaystyle={0.2814}\pm{1.96}\times\sqrt{{\frac{{{0.2814}{\left({1}-{0.2814}\right)}}}{{1848}}}}$$
$$\displaystyle={0.2814}\pm{0.0205}$$
=(0.2609,0.3019)

### Relevant Questions

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