Question # 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

Study design
ANSWERED The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke $$\displaystyle{\left\langle{C}{D}{C}{w}{e}{b}{s}{i}{t}{e},{D}{e}{c}{e}{m}{b}{e}{r}{14},{2014}\right\rangle}$$. 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)? 2021-02-22
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 $$\displaystyle{\left({1}–\alpha\right)}={95}\%$$.
From this, the level of significance is $$\displaystyle\alpha={0.05}$$.
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 $$\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={1.96}$$ from the calculation given below:
$$\displaystyle{1}-\alpha={1}-{0.95}$$
$$\displaystyle\alpha={0.05}$$
$$\displaystyle\frac{\alpha}{{2}}=\frac{{0.05}}{{2}}$$
=0.025
Hence, cumulative area to the left is, Area to the left = 1 - Area to the right
=1-0.025
=0.975
$$\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={1.96}{\left[{E}{x}{c}{e}{l}{f}{\quad\text{or}\quad}\mu{l}{a}:{N}{O}{R}{M}{I}{N}{V}{\left({0.975},{0},{1}\right)}\right]}$$
Step 3
Obtain the required sample size:
Here, $$\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.96},{p}={0.3},{E}={0.02}$$.
The required sample size is obtained as 2,017 from the calculation given below:
$$\displaystyle{n}=\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}\times{\left(\frac{{{Z}_{{\frac{\alpha}{{2}}}}}}{{E}}\right)}^{{2}}$$
$$\displaystyle={\left({0.3}\times{\left({1}-{0.3}\right)}\right)}\times{\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}$$
$$\displaystyle={0.21}\times{9},{604}$$
=2,016.84
$$\displaystyle\approx{2},{017}$$(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:
$$\displaystyle\hat{{{p}}}=\frac{{x}}{{n}}$$
$$\displaystyle=\frac{{520}}{{2}},{017}$$
=0.2578(Rounded to 4 decimal places)
Step 5
c.
Obtain the 95% confidence interval for the proportion of smokers in the population:
Here, $$\displaystyle{p}-\hat{=}{0.2578},{n}={2},{017}{\quad\text{and}\quad}{z}_{{\frac{\alpha}{{2}}}}={1.96}$$.
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 $$\displaystyle{0.2387}≤{p}≤{0.2769}$$ from the calculation given below:
$$\displaystyle{C}{I}{\left(\hat{{{p}}}-{Z}_{{\frac{\alpha}{{2}}}}\times\sqrt{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{n}}\le{p}\le\hat{{{p}}}+{Z}_{{\frac{\alpha}{{2}}}}\times\sqrt{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}}}\right.}$$
$$\displaystyle={\left({0.2578}-{1.96}\times\sqrt{{\frac{{{0.2578}{\left({1}-{0.2578}\right)}}}{{2}},{017}}}\le{p}\le{0.2578}+{1.96}\times\sqrt{{\frac{{{0.2578}{\left({1}-{0.2578}\right)}}}{{2}},{017}}}\right)}$$
$$\displaystyle={\left({0.2578}-{0.0191}\le{p}\le{0.2578}+{0.0191}\right)}$$
$$\displaystyle={\left({0.2387}\le{p}\le{0.2769}\right)}$$
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 $$\displaystyle{0.2387}≤{p}≤{0.2769}$$.