Chapter 8, Problem 59SE

a.

To determine

Find the 95% confidence interval for the proportion of workers in each of the industries who feel their industry is understaffed.

Answer to Problem 59SE

There is 95% confidence interval for the proportion of workers in the government sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.3032,0.4368\right)}$.

There is 95% confidence interval for the proportion of workers in the health care sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.2649,0.3951\right)}$.

There is 95% confidence interval for the proportion of workers in the education sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.2179,0.3421\right)}$.

Explanation of Solution

Calculation:

In the government sector, 37% of the respondents said that they were understaffed; in the health care sector, 33% said that they were understaffed; and in the education sector, 28% said that they were understaffed.

The sampling distribution of $\overline{p}$ is the population proportion p.

Confidence interval for government sector:

The formula for margin of error is as follows:

$\text{Margin of error}=z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}$

Here, the value of $\overline{p}$ is 0.37.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.37\left(1-0.37\right)}{200}}\\ =1.96\sqrt{\frac{0.2331}{200}}\\ =1.96\times 0.0341\\ =0.0668\end{array}$

Thus, the value of margin of error is 0.0668.

The 95% confidence interval for population proportion is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.37\pm 0.0668\\ =\left(0.37-0.0668,0.37+0.0668\right)\\ =\left(0.3032,0.4368\right)\end{array}$

Thus, the 95% confidence interval for the proportion of workers in the government sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.3032,0.4368\right)}$.

Confidence interval for the health care sector:

Here, the value of $\overline{p}$ is 0.33.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.33\left(1-0.33\right)}{200}}\\ =1.96\sqrt{\frac{0.2211}{200}}\\ =1.96\times 0.0332\\ =0.0651\end{array}$

Thus, the value of margin of error is 0.0651.

The 95% confidence interval for population proportion is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.33\pm 0.0651\\ =\left(0.33-0.0651,0.33+0.0651\right)\\ =\left(0.2649,0.3951\right)\end{array}$

Thus, the 95% confidence interval for the proportion of workers in health care sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.2649,0.3951\right)}$.

Confidence interval for the education sector:

Here, the value of $\overline{p}$ is 0.28.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.28\left(1-0.28\right)}{200}}\\ =1.96\sqrt{\frac{0.2016}{200}}\\ =1.96\times 0.0317\\ =0.0621\end{array}$

Thus, the value of margin of error is 0.0621.

The 95% confidence interval for population proportion is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.28\pm 0.0621\\ =\left(0.28-0.0621,0.28+0.0621\right)\\ =\left(0.2179,0.3421\right)\end{array}$

Thus, 95% confidence interval for the proportion of workers in education sector industries who feel their industry is understaffed is $\underset{\_}{\left(0.2179,0.3421\right)}$.

b.

To determine

Compute the sample that needs to be ensured that the margin of error is 0.05 or less for each of the three confidence intervals.

Answer to Problem 59SE

The sample that needs to be ensured that the margin of error is 0.05 or less for each of the three confidence intervals is 359.

Explanation of Solution

Calculation:

From the results, the margin of error for the government sector is greater.

The formula for sample size for an interval estimate of a population proportion is as follows:

$n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^{2}p\ast \left(1-p\ast \right)}{E^2}$

The value of $p\ast$ is 0.37.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The sample that needs to be ensured that the margin of error is 0.05 or less for each of the three confidence intervals is as follows:

$\begin{array}{c}n=\frac{{\left(1.96\right)}^2\left(0.37\right)\left(1-0.37\right)}{{\left(0.05\right)}^2}\\ =\frac{3.8416\times \text{0}\text{.2331}}{\text{0}\text{.0025}}\\ =\frac{\text{0}\text{.8955}}{\text{0}\text{.0025}}\\ =\text{358}\text{.2}\\ \approx \text{359}\end{array}$

Thus, the sample that needs to be ensured that the margin of error is 0.05 or less for each of the three confidence intervals is 359.