Questions: Number of Jobs A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 6.8. The population standard deviation is 1.8. Part 1 of 4 (a) Find the best point estimate of the mean. The best point estimate of the mean is 6.8. Part 2 of 4 (b) Find the 95% confidence interval of the mean number of jobs. Round intermediate and final answers to one decimal place. Part 3 of 4 (c) Find the 99% confidence interval of the mean number of jobs. Round intermediate and final answers to one decimal place.

Solution Steps

The best point estimate of the mean number of jobs is given by the sample mean, which is:

\[ \bar{x} = 6.8 \]

To calculate the 95% confidence interval for the mean number of jobs, we use the formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 6.8\) (sample mean)
• \(z \approx 2.0\) (z-value for 95% confidence level)
• \(\sigma = 1.8\) (population standard deviation)
• \(n = 50\) (sample size)

Calculating the margin of error:

\[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 2.0 \cdot \frac{1.8}{\sqrt{50}} \approx 0.5 \]

Thus, the 95% confidence interval is:

\[ (6.8 - 0.5, 6.8 + 0.5) = (6.3, 7.3) \]

For the 99% confidence interval, we again use the formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 6.8\) (sample mean)
• \(z \approx 2.6\) (z-value for 99% confidence level)
• \(\sigma = 1.8\) (population standard deviation)
• \(n = 50\) (sample size)

Calculating the margin of error:

\[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 2.6 \cdot \frac{1.8}{\sqrt{50}} \approx 0.7 \]

Thus, the 99% confidence interval is:

\[ (6.8 - 0.7, 6.8 + 0.7) = (6.1, 7.5) \]

Final Answer