In a random sample of 63 women at a company, the mean salary is $48,902 with a standard deviation of $5270. In a random sample of 50 men at the company, the mean salary is $53,454 with a standard deviation of $4677. Construct a 95% confidence interval for the difference between the mean salaries of all women and men at the company.
(Points : 4)
($2134.05, $6969.95)

($2715.11, $6388.89)

($3614.81, $5489.19)

($4083.40, $5020.60)

Question

In a random sample of 63 women at a company, the mean salary is $48,902 with a standard deviation of $5270. In a random sample of 50 men at the company, the mean salary is $53,454 with a standard deviation of $4677. Construct a 95% confidence interval for the difference between the mean salaries of all women and men at the company.
(Points : 4)
       ($2134.05, $6969.95)

       ($2715.11, $6388.89)

       ($3614.81, $5489.19)

       ($4083.40, $5020.60)

kropot72 · Answer

The 95% confidence interval for the difference between the mean salaries is found from the following:
$$\large (\bar{x}_{1}-\bar{x}_{2})\pm1.96\sqrt{\frac{s _{1}^{2}}{n _{1}}+\frac{s _{2}^{2}}{n _{2}}}$$
where s1 is the standard deviation of the first sample and s2 is the standard deviation of the second sample.

kropot72 · Answer

@Kyliejones Just substitute the given values and you will find the result agrees with one of the answer choices.

perl · Answer

*