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 $\left(1\u2013\alpha \right)=95\mathrm{\%}$.

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

${Z}_{\frac{\alpha}{2}}=1.96[Excelf{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\mu la:NORMINV(0.975,0,1)]$

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:

$n=\hat{p}(1-\hat{p})\times {\left(\frac{{Z}_{\frac{\alpha}{2}}}{E}\right)}^{2}$

$=(0.3\times (1-0.3))\times {\left(\frac{1.96}{0.02}\right)}^{2}$

$=0.21\times 9,604$

=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:

$\hat{p}=\frac{x}{n}$

$=\frac{520}{2},017$

=0.2578(Rounded to 4 decimal places)

Step 5

c.

Obtain the 95% confidence interval for the proportion of smokers in the population:

Here, $p-\hat{=}0.2578,n=2,017{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{z}_{\frac{\alpha}{2}}=1.96$.

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:

$CI(\hat{p}-{Z}_{\frac{\alpha}{2}}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}\le p\le \hat{p}+{Z}_{\frac{\alpha}{2}}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}$

$=(0.2578-1.96\times \sqrt{\frac{0.2578(1-0.2578)}{2},017}\le p\le 0.2578+1.96\times \sqrt{\frac{0.2578(1-0.2578)}{2},017})$

$=(0.2578-0.0191\le p\le 0.2578+0.0191)$

$=(0.2387\le p\le 0.2769)$

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 $0.2387\le p\le 0.2769$.

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