Step 1

a)

Here, \(\displaystyle{x}={3332}{\quad\text{and}\quad}{n}={5805}.\)

The point estimate of the population proportion is calculated as follows:

\(\displaystyle\hat{{p}}=\frac{x}{{n}}\)

\(\displaystyle=\frac{3332}{{5805}}\)

\(\displaystyle={0.573987}\)

\(\displaystyle\approx{0.5740}\)

Step 2

b)

Since the required confidence interval is \(99\%\), the two-tailed z-critical value is 2.5758 using the Excel function \(\displaystyle“={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.995}\right)}\text{"}\).

The \(99\%\) confidence interval for the population proportion is calculated as follows:

\(\displaystyle{C}{I}=\hat{{p}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{p}}{\left({1}-\hat{{p}}\right)}}}{{n}}}}\)

\(\displaystyle={0.5740}\pm{\left({2.5758}\right)}\sqrt{{{0.5740}\frac{{{1}-{0.5740}}}{{5805}}}}\)

\(\displaystyle={0.5740}\pm{0.0167}\)

\(\displaystyle={\left({0.5573},{0.5907}\right)}\)

\(\displaystyle\approx{\left({0.557},{0.591}\right)}\)

Lower limit \(\displaystyle={0.557}\)

Upper limit \(\displaystyle={0.591}\)

Step 3

c)

The normal approximation to binomial is possible only if,

\(\displaystyle{n}{p}>{5}\)

\(\displaystyle{n}{q}>{5}\)

Checking the conditions for normality:

\(\displaystyle{n}{p}={5805}\times\frac{3332}{{5805}}\)

\(\displaystyle={3332}\)

\(\displaystyle>{5}\)

\(\displaystyle{n}{q}={5805}\times{\left({1}-\frac{3332}{{5805}}\right)}\)

\(\displaystyle={2473}\)

\(\displaystyle>{5}\)

Thus, normal approximation to binomial is possible.

Yes, \(\displaystyle{n}{p}>{5}{\quad\text{and}\quad}{n}{q}>{5}\)

a)

Here, \(\displaystyle{x}={3332}{\quad\text{and}\quad}{n}={5805}.\)

The point estimate of the population proportion is calculated as follows:

\(\displaystyle\hat{{p}}=\frac{x}{{n}}\)

\(\displaystyle=\frac{3332}{{5805}}\)

\(\displaystyle={0.573987}\)

\(\displaystyle\approx{0.5740}\)

Step 2

b)

Since the required confidence interval is \(99\%\), the two-tailed z-critical value is 2.5758 using the Excel function \(\displaystyle“={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.995}\right)}\text{"}\).

The \(99\%\) confidence interval for the population proportion is calculated as follows:

\(\displaystyle{C}{I}=\hat{{p}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{p}}{\left({1}-\hat{{p}}\right)}}}{{n}}}}\)

\(\displaystyle={0.5740}\pm{\left({2.5758}\right)}\sqrt{{{0.5740}\frac{{{1}-{0.5740}}}{{5805}}}}\)

\(\displaystyle={0.5740}\pm{0.0167}\)

\(\displaystyle={\left({0.5573},{0.5907}\right)}\)

\(\displaystyle\approx{\left({0.557},{0.591}\right)}\)

Lower limit \(\displaystyle={0.557}\)

Upper limit \(\displaystyle={0.591}\)

Step 3

c)

The normal approximation to binomial is possible only if,

\(\displaystyle{n}{p}>{5}\)

\(\displaystyle{n}{q}>{5}\)

Checking the conditions for normality:

\(\displaystyle{n}{p}={5805}\times\frac{3332}{{5805}}\)

\(\displaystyle={3332}\)

\(\displaystyle>{5}\)

\(\displaystyle{n}{q}={5805}\times{\left({1}-\frac{3332}{{5805}}\right)}\)

\(\displaystyle={2473}\)

\(\displaystyle>{5}\)

Thus, normal approximation to binomial is possible.

Yes, \(\displaystyle{n}{p}>{5}{\quad\text{and}\quad}{n}{q}>{5}\)