Salary information regarding male and female employees of a large company is shown below. $$\begin{array} { l r r }
& \text { Male } & \text { Female } \
\text { Sample Size } & 64 & 36 \
\text { Sample Mean Salary } ( \text { in } \$ 1000 ) & 44 & 41 \
\text { Population Variance } \left( \sigma ^ { 2 } \right)& 128 & 72
\end{array}$$ At 95% confidence, the margin of error is

Question

Salary information regarding male and female employees of a large company is shown below. $$\begin{array} { l r r } 
& \text { Male } & \text { Female } \
\text { Sample Size } & 64 & 36 \
\text { Sample Mean Salary } ( \text { in } \$ 1000 )  & 44 & 41 \
\text { Population Variance } \left( \sigma ^ { 2 } \right)& 128 & 72
\end{array}$$   At 95% confidence, the margin of error is

Quizplus · Accepted Answer

The margin of error for the difference in means is calculated using the formula: $ ME = z \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} $. Using $ z = 1.96 $ for 95% confidence, $ \sigma_1^2 = 128 $, $ n_1 = 64 $, $ \sigma_2^2 = 72 $, and $ n_2 = 36 $, the margin of error is approximately 3.920.

Salary Information Regarding Male and Female Employees of a Large