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 ($)
2.7	3,600
3.5	3,800
3.6	4,200
3.2	3,700
3.4	4,200
2.8	2,400

The estimated regression equation for these data is y^ = -464.3 + 1285.7x and MSE = 259,464

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

b. Develop a  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).

(  ,   )

Answer 1

a)

Predicted Y at X=   3   is                  
Ŷ =   -464.28571   +   1285.714286   *   3   =   3392.9

...

b)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    240.978              
margin of error,E=t*Std error=t* S(ŷ) =   2.7764   *   240.9784   =   669.0632
                  
Confidence Lower Limit=Ŷ +E =    3392.857   -   669.063   =   2723.79
Confidence Upper Limit=Ŷ +E =   3392.857   +   669.063   =   4061.92
....

c)

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   563.5023              
margin of error,E=t*std error=t*S(ŷ)=    2.7764   *   563.50   =   1564.5333
                  
Prediction Interval Lower Limit=Ŷ -E =   3392.857   -   1564.53   =   1828.32
Prediction Interval Upper Limit=Ŷ +E =   3392.857   +   1564.53   =   4957.39

....................

Please revert back in case of any doubt.

Please upvote. Thanks in advance.