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) . $4957225187605629045568801150904520805520500480100512509707437506056601640\begin{array} { r | r r } \text { Female } & \text { Male } & \\\hline 495 & 722 & 518 \\760 & 562 & 904 \\556 & 880 & 1150 \\904 & 520 & 805 \\520 & 500 & 480 \\1005 & 1250 & 970 \\743 & 750 & 605 \\660 & 1640 &\end{array}$ Determine a 98% confidence interval for the difference, $μ1−μ2,\mu _ { 1 } - \mu _ { 2 } ,$ 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) .

4957225187605629045568801150904520805520500480100512509707437506056601640\begin{array} { r | r r } \text { Female } & \text { Male } & \\\hline 495 & 722 & 518 \\760 & 562 & 904 \\556 & 880 & 1150 \\904 & 520 & 805 \\520 & 500 & 480 \\1005 & 1250 & 970 \\743 & 750 & 605 \\660 & 1640 &\end{array}

Determine a 98% confidence interval for the difference,

μ1−μ2,\mu _ { 1 } - \mu _ { 2 } ,

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)

Pedro Andrés Sánchez Lozano · Accepted Answer

Final Answer : C, Explanation :  To find the confidence interval, we need to calculate the point estimate (the difference between the sample means) and the margin of error.Point estimate:x̄1 - x̄2 = $626.25 - $775.87 = -$149.62Margin of error:s1 = $207.73 and s2 = $243.01√[(s1^2/n1) + (s2^2/n2)] = √[(207.73^2/8) + (243.01^2/15)] = $115.51t-value: From t-distribution table for 21 degree of freedom and 0.01 level of significance, we get the t-value as 2.518.Margin of Error = t-value x Standard Error = 2.518 x 115.51 = $291.16Thus, 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)Therefore, the best choice is C.

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