Question

In a random sample of 74 women at a company, the mean salary is $39,902 with a standard deviation of $3270 in a random sample of 40 men at the company the mean salary is $36454 with a standard deviation of $4677.
which interval is the 95% confidence interval for the difference between the mean salaries of all women and men at the company?

Please help, I do not understand this at all.

Answer 1

The correct answer is the first choice, ($1818.30, $5077.70.)

To find this, we first find the z-score based on the confidence level:
Convert 95% to a decimal:  95%=95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

Next, we identify 
$\overline{x_1}=39902; \overline{x_2}=36454; n_1=74; n_2=40; \sigma_1=3270; \sigma_2=4677$

Next we find
$(\overline{x_1}-\overline{x_2})=(39902-36454) = 3448$

Next we find 
$\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}} \\ \\=\sqrt{\frac{3270^2}{74}+\frac{4677^2}{40}}=\sqrt{\frac{10692900}{74}+\frac{21874329}{40}} \\ \\=831.479$

Next, we multiply this value by z:
1.96(831.479) = 1629.70

The confidence interval is given by
$3448\pm1629.70 \\ \\=(3448-1629.70, 3448+1629.70) \\ \\=(1818.30, 5077.70)$