Chapter 8.1, Problem 32E

(a)

To determine

To test: Whether an 99% confidence interval would be wider or narrower than the 95% confidence interval obtained in Exercise 8.31.

Answer to Problem 32E

Solution: The 99% confidence interval would be wider. An 99% confidence interval is obtained as $\left(0.3962,0.4638\right)$, which is wider than the 95% confidence interval $\underset{\_}{\left(0.4043,0.4557\right)}$ obtained in Exercise 8.31.

Explanation of Solution

Calculation: The 95% confidence interval is obtained as $\left(0.4043,0.4557\right)$ in the previous Exercise 8.31. Compute an 99% confidence interval for the same data.

The formula for 99% confidence interval for population proportion p is defined as:

$\text{Confidence interval}=\widehat{p}\pm m$

Where, $\widehat{p}$ the proportion of sample and m is the margin of error.

The sample proportion is provided as:

$\begin{array}{c}\widehat{p}=43\%\\ =\frac{43}{100}\\ =0.43\end{array}$

Therefore, the sample proportion $\widehat{p}$ is obtained as 0.43.

The formula for margin of error m is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

In the above formula, $z^{*}$ is the critical value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error.

The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion $\widehat{p}$ is obtained as 0.43 in the previous step. Substitute the obtained sample proportion of 0.43 and sample size of 1430 in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =\sqrt{\frac{0.43\times \left(1-0.43\right)}{1430}}\\ =\sqrt{\frac{0.2451}{1430}}\\ =0.0131\end{array}$

Therefore, the standard error is obtained as 0.0131. The value of $z^{*}$ for 99% confidence level is $z^{*}=2.58$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =2.58\times 0.0131\\ =0.033798\end{array}$

Therefore, the margin of error is obtained as 0.033798.

Substitute the obtained values of margin of error and sample proportion in the formula for confidence interval. Therefore, an 99% confidence interval is obtained as:

$\begin{array}{c}\widehat{p}\pm m=\left(0.43\pm 0.033798\right)\\ =\left(0.43-0.033798,0.43+0.033798\right)\\ =\left(0.3962,0.4638\right)\end{array}$

Therefore, an 99% confidence interval is obtained as $\left(0.3962,0.4638\right)$.

The width of the 99% confidence interval is obtained as:

$0.4638-0.3962=0.0676$

The width of the 95% confidence interval is obtained as:

$0.4557-0.4043=0.0514$

Conclusion: The obtained widths of the two confidence levels show that the 99% confidence interval is wider than the 95% confidence interval.

(b)

To determine

To test: Whether a 90% confidence interval would be wider or narrower than the 95% confidence interval obtained in Exercise 8.31.

Answer to Problem 32E

Solution: A 90% confidence interval would be wider. A 90% confidence interval is obtained as $\left(0.40845,0.45155\right)$ is narrower than the 95% confidence interval $\underset{\_}{\left(0.4043,0.4557\right)}$ obtained in Exercise 8.31.

Explanation of Solution

Calculation: The 95% confidence interval is obtained as $\left(0.4043,0.4557\right)$ in the previous exercise 8.31. Compute a 90% confidence interval for the same data.

The formula for 90% confidence interval for p is defined as:

$\text{Confidence interval}=\widehat{p}\pm m$

The sample proportion is provided as:

$\begin{array}{c}\widehat{p}=43\%\\ =\frac{43}{100}\\ =0.43\end{array}$

Therefore, the sample proportion $\widehat{p}$ is obtained as 0.43.

The formula for margin of error m is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

In the above formula, $z^{*}$ is standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error.

The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion $\widehat{p}$ is obtained as 0.43 in the previous step. Substitute the obtained sample proportion of 0.43 and sample size of 1430 in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =\sqrt{\frac{0.43\times \left(1-0.43\right)}{1430}}\\ =\sqrt{\frac{0.2451}{1430}}\\ =0.0131\end{array}$

Therefore, the standard error is obtained as 0.0131. The value of $z^{*}$ for 90% confidence level is $z^{*}=1.645$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.645\times 0.0131\\ =0.02155\end{array}$

Therefore, the margin of error is obtained as 0.02155.

Substitute the obtained values of margin of error and sample proportion in the formula for confidence interval. Therefore, a 90% confidence interval is obtained as:

$\begin{array}{c}\widehat{p}\pm m=\left(0.43\pm 0.02155\right)\\ =\left(0.43-0.02155,0.43+0.02155\right)\\ =\left(0.40845,0.45155\right)\end{array}$

Therefore, a 90% confidence interval is obtained as $\left(0.40845,0.45155\right)$.

The width of the 90% confidence interval is obtained as:

$0.40845-0.45155=0.043$

The width of the 95% confidence interval is obtained as:

$0.4557-0.4043=0.0514$

Conclusion: The obtained widths of the two confidence levels show that 90% confidence interval is narrower than the 95% confidence interval.