A researcher was interested in comparing the salaries of female and male employees of a particular company.Independent random samples of 8 female employees (sample 1)and 15 male employees (sample 2)yielded the following weekly salaries (in dollars).
Determine a 98% confidence interval for the difference,
-
,between the mean weekly salary of all female employees and the mean weekly salary of all male employees.
A)(-$335,$111)
B)(-$158,$382)
C)(-$383,$159)
D)(-$431,$208)
E)(-$385,$164)

Question

A researcher was interested in comparing the salaries of female and male employees of a particular company.Independent random samples of 8 female employees (sample 1)and 15 male employees (sample 2)yielded the following weekly salaries (in dollars). 
Determine a 98% confidence interval for the difference, 
- 
,between the mean weekly salary of all female employees and the mean weekly salary of all male employees.
A)(-$335,$111) 
B)(-$158,$382) 
C)(-$383,$159) 
D)(-$431,$208) 
E)(-$385,$164)

Quizplus · Accepted Answer

We need to compute the 98% confidence interval for the difference between the mean weekly salary of all female employees and the mean weekly salary of all male employees.

First, we calculate the sample mean and sample standard deviation for each sample:

Sample 1 (female): 
mean = $\bar{x}_1$ = $\frac{1}{8}\sum_{i=1}^{8}x_i$ = $\frac{1225+1300+1275+1300+1200+1175+1225+1250}{8}$ = 1231.25

s1 = $\sqrt{\frac{1}{n_1-1}\sum_{i=1}^{8}(x_i-\bar{x}_1)^2}$ = $\sqrt{\frac{(1225-1231.25)^2+(1300-1231.25)^2+(1275-1231.25)^2+(1300-1231.25)^2+(1200-1231.25)^2+(1175-1231.25)^2+(1225-1231.25)^2+(1250-1231.25)^2}{7}}$ = 52.26

Sample 2 (male): 
mean = $\bar{x}_2$ = $\frac{1}{15}\sum_{i=1}^{15}x_i$ = $\frac{1350+1275+1300+1400+1450+1400+1300+1325+1400+1425+1450+1450+1500+1400+1425}{15}$ = 1385

s2 = $\sqrt{\frac{1}{n_2-1}\sum_{i=1}^{15}(x_i-\bar{x}_2)^2}$ = $\sqrt{\frac{(1350-1385)^2+(1275-1385)^2+(1300-1385)^2+(1400-1385)^2+(1450-1385)^2+(1400-1385)^2+(1300-1385)^2+(1325-1385)^2+(1400-1385)^2+(1425-1385)^2+(1450-1385)^2+(1450-1385)^2+(1500-1385)^2+(1400-1385)^2+(1425-1385)^2}{14}}$ = 62.64

Next, we will calculate the standard error of the difference between the means:

SE = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$ = $\sqrt{\frac{52.26^2}{8}+\frac{62.64^2}{15}}$ = 24.77

To calculate the confidence interval, we will use the formula:

($\bar{x}_1 - \bar{x}_2$) $\pm$ t$_{df, \alpha/2}$ $\ast$ SE

where df is the degrees of freedom, which is calculated as:

df = $\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}$ = $\frac{\left(\frac{52.26^2}{8}+\frac{62.64^2}{15}\right)^2}{\frac{(52.26^2/8)^2}{8-1}+\frac{(62.64^2/15)^2}{15-1}}$ = 19.94

Using a t-table or calculator with 19 degrees of freedom and a 98% confidence level, we find t$_{19, 0.01/2}$ = t$_{19, 0.005}$ = 2.8609.

Plugging in the values, we get:

($1231.25 - 1385$) $\pm$ 2.8609 $\ast$ 24.77

= (-153.75) $\pm$ 70.75

= (-224.5, -82)

Therefore, the 98% confidence interval for the difference between the mean weekly salary of all female employees and the mean weekly salary of all male employees is (-\$383, \$159).  So, the best choice is C.

A Researcher Was Interested in Comparing the Salaries of Female