Construct the 95% confidence interval for the mean starting salary of the IT employees with an under graduate degree using t-distribution.
Sample size, (n_1 ) = 4
Sample Mean (x ¯_1) = $44,250.00
Sample Standard Deviation (s_1) = $9,945.69
Level of significance, α = 0.05
The two-tailed t-critical value at 0.05 level with n - 1 = 4 - 1 = 3 degrees of freedom is
t_(0.025,3)= 3.182.
The margin of error for the 95% confidence interval,
E_1= (t_(0.025,3) s_1)/√(n_1 )= (3.182 × $9,945.69)/√4=$15,825.81
The 95% confidence interval for the mean starting salary of the IT employees with an under graduate degree is,
(x ¯_1-E_1,x ¯_1-E_1 )= ($44,250.00-$15,825.81,$44,250.00+$15,825.81)
= ($28,424.19,$60,075.81)
Construct the 95% confidence interval for the mean starting salary of the IT employees with a graduate degree using t-distribution.
Sample size, (n_2 ) = 6
Sample Mean (x ¯_2) = $72,833.33
Sample Standard Deviation (s_2) = $6,794.61
Level of significance, α = 0.05
The two-tailed t-critical value at 0.05 level with n - 1 = 6 - 1 = 5 degrees of freedom is
t_(0.025,5)=2.571.
The margin of error for the 95% confidence interval,
E_2= (t_(0.025,5) s_2)/√(n_2 )= (2.571 × $6,794.61)/√6=$7,130.50
The 95% confidence interval for the mean starting salary of the IT employees with a graduate degree is,
(x ¯_2-E_2,x ¯_2-E_3 )= ($72,833.33-$7,130.50,$72,833.33+$7,130.50)
= ($65,702.83,$79,963.83)
Construct the 95% confidence interval for the mean current salary of the IT employees with an under graduate degree using t-distribution.
Sample size, (n_3 ) = 4
Sample Mean (x ¯_3) = $54,250.00
Sample Standard Deviation (s_3) = $9,945.69
Level of significance, α = 0.05
The two-tailed t-critical value at 0.05 level with n - 1 = 4 - 1 = 3 degrees of freedom is
t_(0.025,3)= 3.182.
The margin of error for the 95% confidence interval,
E_3= (t_(0.025,3) s_3)/√(n_3 )= (3.182 × $9,945.69)/√4=$15,825.81
The 95% confidence interval for the mean current salary of the IT employees with an undergraduate degree is,
(x ¯_3-E_3,x ¯_3-E_3 )= ($54,250.00-$15,825.81,$54,250.00+$15,825.81)
= ($38,424.19,$70,075.81)
Construct the 95% confidence interval for the mean current salary of the IT employees with a graduate degree using t-distribution.
Sample size, (n_4 ) = 6
Sample Mean (x ¯_4) = $82,833.33
Sample Standard Deviation (s_4) = $6,794.61
Level of significance, α = 0.05
The two-tailed t-critical value at 0.05 level with n - 1 = 6 - 1 = 5 degrees of freedom is
t_(0.025,5)= 2.571.
The margin of error for the 95% confidence interval,
E_4= (t_(0.025,5) s_4)/√(n_4 )= (2.571 × $6,794.61)/√6=$7,130.50
The 95% confidence interval for the mean current salary of the IT employees with a graduate degree is,
(x ¯_4-E_4,x ¯_4-E_4 )= ($82,833.33-$7,130.50,$82,833.33+$7,130.50)
= ($75,702.83,$89,963.83)
The results indicated that the mean starting salary for graduate IT employees is much larger than that of undergraduate IT employees. But the width of the confidence interval for mean starting salary is larger for under graduate IT employees than that of for graduate IT employees due to larger variability for under graduate IT employees. It is same for the current salaries for graduate IT employees and for under graduate IT employees.
Attachment:- U3 Discussion.xlsx
Attachment:- U4 Discussion.docx