Question

A company has offices in two different countries. Suppose that the average age (in years) of the employees at location A is \[40.0\] with a standard deviation of \[5.5\], and the average at location B is \[36.0\] with a standard deviation of \[6.3\]. Every month, the company takes separate random samples of \[50\] employees from location A and \[75\] from location B for a survey. Each time, they look at the difference in the mean age sampled from each location \[\left( \bar{x}_\text{A} - \bar{x}_\text{B} \right)\]. What are the mean and standard deviation (in years) of the sampling distribution of \[\bar{x}_\text{A}-\bar{x}_\text{B}\]?

Answer 1

The mean of the sampling distribution of \(\bar{x}_\text{A}-\bar{x}_\text{B}\) is 4.0 years, and the standard deviation is 1.065 years, calculated using the differences between the population means and combining the variances of each sample appropriately.

Step-by-step explanation:

To determine the mean and standard deviation of the sampling distribution of \(\bar{x}_\text{A}-\bar{x}_\text{B}\), we use the formulas for the mean and standard deviation of the difference between two independent sample means.

First, we find the mean (expected value) of the sampling distribution of the difference between two means:

\(\mu_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \mu_{\bar{x}_\text{A}} - \mu_{\bar{x}_\text{B}}\)

This is simply the difference between the two population means, which is:

\(40.0 - 36.0 = 4.0\) years.

Next, to find the standard deviation of the sampling distribution of the difference between two means, we use the following formula:

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \sqrt{\frac{\sigma^2_{\text{A}}}{n_{\text{A}}} + \frac{\sigma^2_{\text{B}}}{n_{\text{B}}}}\)

Plugging in the numbers:

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \sqrt{\frac{5.5^2}{50} + \frac{6.3^2}{75}}\)

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \sqrt{\frac{30.25}{50} + \frac{39.69}{75}}\)

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \sqrt{0.605 + 0.5292}\)

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = \sqrt{1.1342}\)

\(\sigma_{\bar{x}_\text{A}-\bar{x}_\text{B}} = 1.065\) years.

Therefore, the mean of the sampling distribution of \(\bar{x}_\text{A}-\bar{x}_\text{B}\) is 4.0 years and the standard deviation is 1.065 years.