Question

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\
ight) - 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\
ight) + 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 = \\$103,400,\\, n_1 = 40,\\, \text{and } \sigma_1 = \\$9150;\\, \bar{x}_2 = \\$85,100,\\, n_2 = 45,\\, \text{and } \sigma_2 = \\$9125\\\\\
ule{\textwidth}{0.5pt}\complete the 95% confidence interval for $\mu_1 - \mu_2$ below.\\\\$\square < \mu_1 - \mu_2 < \\$\square\\\\(\text{round to the nearest dollar as needed.})\\

Explanation:

Step1: Find \( \bar{x}_1 - \bar{x}_2 \)

\( \bar{x}_1 = 103400 \), \( \bar{x}_2 = 85100 \)
\( \bar{x}_1 - \bar{x}_2 = 103400 - 85100 = 18300 \)

Step2: Find \( z_c \) for 95% confidence

For 95% confidence, \( z_c = 1.96 \) (from standard normal table)

Step3: Calculate the standard error term \( \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \)

\( \sigma_1 = 9150 \), \( n_1 = 40 \), \( \sigma_2 = 9125 \), \( n_2 = 45 \)
\( \frac{\sigma_1^2}{n_1} = \frac{9150^2}{40} = \frac{83722500}{40} = 2093062.5 \)
\( \frac{\sigma_2^2}{n_2} = \frac{9125^2}{45} = \frac{83265625}{45} \approx 1850347.22 \)
Sum: \( 2093062.5 + 1850347.22 = 3943409.72 \)
Square root: \( \sqrt{3943409.72} \approx 1985.79 \)

Step4: Calculate the margin of error \( z_c \times \) standard error

\( 1.96 \times 1985.79 \approx 3892.15 \)

Step5: Find the lower and upper bounds

Lower bound: \( 18300 - 3892.15 = 14407.85 \approx 14408 \)
Upper bound: \( 18300 + 3892.15 = 22192.15 \approx 22192 \)

Answer:

\( \$14408 < \mu_1 - \mu_2 < \$22192 \) (rounded to nearest dollar)