Question

The following data are the monthly salaries y and the grade point averages x for students who obtained a bachelor's degree in business administration.

GPA	Monthly Salary ($)
2.6	3,500
3.4	3,900
3.6	4,300
3.2	3,800
3.5	4,200
2.9	2,200

The estimated regression equation for these data is y=-674.3+1,351.4x and MSE=395,912.

Use Table 1 of Appendix B.

a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal).

b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals).

(  ,  )

c. Develop a  prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).

(  ,  )

d. Discuss the differences in your answers to parts (b) and (c).

Answer 1

	X	Y	X * Y	X2	Ŷ	Sxx =Σ (Xi - X̅ )2	Syy = Σ( Yi - Y̅ )2	Sxy = Σ (Xi - X̅ ) * (Yi - Y̅)
	2.6	3500	9100	6.76	2839.1892	0.36	22500	90
	3.4	3900	13260	11.56	3920.2703	0.04	62500	50
	3.6	4300	15480	12.96	4190.5405	0.16	422500	260
	3.2	3800	12160	10.24	3650.0000	0	22500	0
	3.5	4200	14700	12.25	4055.4054	0.09	302500	165
	2.9	2200	6380	8.41	3244.5946	0.09	2102500	435
Total	19.2	21900	71080	62.18	244.0800	0.74	2935000	1000

X̅ = Σ (Xi / n ) = 19.2/6 = 3.2
Y̅ = Σ (Yi / n ) = 21900/6 = 3650

Part a)

Point Estimate

Ŷ = -674.3243 + 1351.3514X
Ŷ = 3379.7

Part b)

Estimated Error Variance (σ̂2) =
S2 = ( 2935000 - 1351.3514 * 1000 ) / 6 - 2
S2 = 395912.15
S = 629.2155

Confidence Interval of
Ŷ = -674.3243 + 1351.3514X
Ŷ = 3379.73

t(α/2) = t(0.05/2) = 2.776
X̅ = (Xi / n ) = 19.2/6 = 3.2


95% confidence interval is ( 2559.12 < < 4200.34 )

Part c)

Predictive Confidence Interval of
Ŷ = -674.3243 + 1351.3514X
Ŷ = 3379.73

t(α/2) = t(0.05/2) = 2.776
X̅ = (Xi / n ) = 19.2/6 = 3.2
= 3379.73

95% Predictive confidence interval is ( 1449.85 < < 5309.61 )

Part d)

The difference in the confidence interval is, in prediction interval it covers future single observation than that found in part b)