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	3,600
3.4	3,800
3.6	4,300
3.2	3,700
3.4	4,100
3	2,400

The estimated regression equation for these data is y= 350 + 1031.3x and MSE = 383594.

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 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).

$ (  ,  )

Answer 1

	ΣX		Σ(x-x̅)²
total sum	19.20		0.64
mean	3.20		SSxx

a)

Predicted Y at X=   3   is          
Ŷ=   350.0000   +   1031.2500   *3=   3443.8

b)

X Value=   3              
Confidence Level=   95%              
                  
                  
Sample Size , n=   6              
Degrees of Freedom,df=n-2 =   4              
critical t Value=tα/2 =   2.776   [excel function: =t.inv.2t(α/2,df) ]          
                  
X̅ =    3.20              
Σ(x-x̅)² =Sxx   0.64              
Standard Error of the Estimate,Se=   619.3495              
                  
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    296.491              
margin of error,E=t*Std error=t* S(ŷ) =   2.7764   *   296.491   =   823.1909
                  
Confidence Lower Limit=Ŷ +E =    3443.750   -   823.191   =   2620.56
Confidence Upper Limit=Ŷ +E =   3443.750   +   823.191   =   4266.94

c)

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   686.6591              
margin of error,E=t*std error=t*S(ŷ)=    2.776   *   686.659   =   1906.4712
                  
Prediction Interval Lower Limit=Ŷ -E =   3443.750   -   1906.471   =   1537.28
Prediction Interval Upper Limit=Ŷ +E =   3443.750   +   1906.471   =   5350.22