Chapter 8, Problem 63SE

The American Enterprise Institute (AEI) Public Opinion Study found that 86% of Americans are satisfied with their job (AEI website, December 4, 2015). Is this consistent across age groups? A follow-up survey revealed that 72% of Americans age 18–30 are satisfied with their job, 85% of Americans age 31–50 are satisfied with their job, and 93% of Americans age 51–70 are satisfied with their job. Suppose that 200 workers were surveyed in each age group.

1. a. Construct a 95% confidence interval for the proportion of workers in each of these age groups who is satisfied with her or his job. Do these results surprise you? Why or why not?
2. b. Assuming the same sample size will be used in each age group, how large would the sample need to be to ensure that the margin of error is .05 or less for each of the three confidence intervals?

To determine

Find the 95% confidence interval for the population proportion of people in each of the given age groups who are satisfied with the job.

Check whether the results surprised or not.

Answer to Problem 63SE

The 95% confidence interval for the population proportion of people in 18‑30 age group, who are satisfied with their job, is (0.6579, 0.7821).

The 95% confidence interval for the population proportion of people in 31‑50 age group, who are satisfied with their job, is (0.8005, 0.8995).

The 95% confidence interval for the population proportion of people in 51‑70 age group, who are satisfied with their job, is (0.8947, 0.9653).

No, the results are not surprised.

Explanation of Solution

Calculation:

It is given that 86% of Americans are satisfied with their job according to AEI. The sample size n is 200.

For 18‑30 age group, the percentage of people who are satisfied with their job is 72%.

For 31‑50 age group, the percentage of people who are satisfied with their job is 85%.

For 50‑70 age group, the percentage of people who are satisfied with their job is 93%.

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

Margin of error (ME) for18-30 age group:

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.72\left(1-0.72\right)}{200}}\\ =1.96\sqrt{\frac{0.2016}{200}}\\ =1.96\times 0.0317\\ =0.0621\end{array}$

Thus, the margin of error is 0.0621.

Confidence interval:

The 95% confidence interval for the population proportion of people in 18‑30 age group who are satisfied with their job is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.72\pm 0.0621\\ =\left(0.72-0.0621,0.72+0.0621\right)\\ =\left(0.6579,0.7821\right)\end{array}$

Thus, the 95% confidence interval for the population proportion of people in 18‑30 age group, who are satisfied with their job, is (0.6579, 0.7821).

Margin of error for31-50 age group:

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.85\left(1-0.85\right)}{200}}\\ =1.96\sqrt{\frac{0.1275}{200}}\\ =1.96\times 0.0252\\ =0.0494\end{array}$

Thus, the margin of error is 0.0494.

Confidence interval:

The 95% confidence interval for the population proportion of people in 31‑50 age group who are satisfied with their job is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.85\pm 0.0494\\ =\left(0.85-0.0494,0.85+0.0494\right)\\ =\left(0.8005,0.8995\right)\end{array}$

Thus, the 95% confidence interval for the population proportion of people in 31‑50 age group who are satisfied with their job is (0.8005, 0.8995).

Margin of error for 51-70 age group:

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.93\left(1-0.93\right)}{200}}\\ =1.96\sqrt{\frac{0.0651}{200}}\\ =1.96\times 0.0180\\ =0.0353\end{array}$

Confidence interval:

The 95% confidence interval for the population proportion of people in 51‑70 age group who are satisfied with their job is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.93\pm 0.0353\\ =\left(0.93-0.0353,0.93+0.0353\right)\\ =\left(0.8947,0.9653\right)\end{array}$

Thus, the 95% confidence interval for the population proportion of people in 51‑70 age group who are satisfied with their job is (0.8947, 0.9653).

From the results, it is observed that young workers satisfaction on their jobs is lesser than older workers satisfaction.

Also, it is observed that the results are not surprising because of the places of these workers are on the career cycle.

To determine

Find the number of sample needed if no planning value for the proportion could be specified.

Answer to Problem 63SE

The number of sample needed if no planning value for the proportion could be specified is 310.

Explanation of Solution

Calculation:

From the results of Part (a), it is observed that age group of 18‑30 has the largest margin of error. Therefore, the sample size should be large to reduce this margin of error to 0.05 or less.

In this case, consider the margin of error is 0.05.

From Part (a), the value of $p\ast$ for age group 18‑30 is 0.72.

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

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

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

The required sample size is as follows:

$\begin{array}{c}n=\frac{{\left(1.96\right)}^2\left(0.72\right)\left(1-0.72\right)}{{\left(0.05\right)}^2}\\ =\frac{3.8416\times 0.2016}{0.0025}\\ =\frac{0.7745}{0.0025}\\ =309.7\\ \approx 310\end{array}$

Hence, the number of sample needed if no planning value for the proportion could be specified is 310.