To construct a confidence interval for the difference between two population means $\mu_{1}-\mu_{2}$, use the formula shown below when both population standard deviations are known, and either both populations are normally distributed or both $n_{1} \geq 30$ and $n_{2} \geq 30$. Also, the samples must be randomly selected and independent. \[ \left(\bar{x}_{1}-\bar{x}_{2}\right)-z_{c} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}<\mu_{1}-\mu_{2}<\left(\bar{x}_{1}-\bar{x}_{2}\right)+z_{c} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}} \] The descriptive statistics for the annual salaries from a random sample of microbiologists from two regions are shown below. Construct a $95 \%$ confidence interval for the difference between the mean annual salaries. \[ \bar{x}_{1}=\$ 101,860, n_{1}=41 \text {, and } \sigma_{1}=\$ 8820 ; \bar{x}_{2}=\$ 85,350, n_{2}=35 \text {, and } \sigma_{2}=\$ 8845 \] Complete the $95 \%$ confidence interval for $\mu_{1}-\mu_{2}$ below. \[ \$ \square<\mu_{1}-\mu_{2}<\$ \square \] (Round to the nearest dollar as needed.)
Solution
Step 1 ：Given the sample means, sample sizes, and standard deviations for both regions, we can substitute these values into the formula to calculate the confidence interval. The z-value for a 95% confidence interval is approximately 1.96.
Step 2 ：First, calculate the difference between the sample means: \(\bar{x}_{1} - \bar{x}_{2} = 101860 - 85350 = 16510\).
Step 3 ：Next, calculate the standard error (SE) using the formula: \(SE = z_{c} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} = 1.96 \sqrt{\frac{8820^{2}}{41} + \frac{8845^{2}}{35}} = 2032.89\).
Step 4 ：Then, calculate the lower and upper bounds of the confidence interval using the formula: \(\bar{x}_{1} - \bar{x}_{2} \pm z_{c} \times SE\). The lower bound is \(16510 - 1.96 \times 2032.89 = 12525.54\) and the upper bound is \(16510 + 1.96 \times 2032.89 = 20494.46\).
Step 5 ：Finally, round the lower and upper bounds to the nearest dollar to get \$12526 and \$20494 respectively.
Step 6 ：Final Answer: The 95% confidence interval for the difference between the mean annual salaries of microbiologists from the two regions is \(\boxed{\$12526} < \mu_{1} - \mu_{2} < \boxed{\$20494}\).
