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 \(\displaystyle{0.2387}≤{p}≤{0.2769}\).

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 \(\displaystyle{0.2387}≤{p}≤{0.2769}\).