The following data are the monthly salaries y and the grade

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.

The estimated regression equation for these data is = -1,100 + 1,484.4x and MSE =451,211.

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).
$ ( , )

GPA	Monthly Salary ($)
2.7	3,600
3.4	3,900
3.6	4,300
3.1	3,800
3.5	4,200
2.9	2,100

Solution

The estimated regression equation for these data is = -1,100 + 1,484.4x and MSE =451,211.

Excel used

Regression Analysis

r²

0.439

n

6

r

0.662

k

1

Std. Error

671.722

Dep. Var.

Monthly Salary ($)

ANOVA table

Source

SS

df

MS

F

p-value

Regression

1,410,156.2500

1  

1,410,156.2500

3.13

.1518

Residual

1,804,843.7500

4  

451,210.9375

Total

3,215,000.0000

5  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=4)

p-value

95% lower

95% upper

Intercept

-1,100.0000

2,700.8474

-0.407

.7047

-8,598.7546

6,398.7546

GPA

1,484.3750

839.6530

1.768

.1518

-846.8753

3,815.6253

Predicted values for: Monthly Salary ($)

95% Confidence Interval

95% Prediction Interval

GPA

Predicted

lower

upper

lower

upper

Leverage

3

3,353.125

2,460.324

4,245.926

1,285.440

5,420.810

0.229

a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal).
$3353.1

b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals).
$ (2460.32 , 4245.93)

c. Develop a 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).
$ ( 1285.44, 5420.81)

Regression Analysis
	r²	0.439	n	6
	r	0.662	k	1
	Std. Error	671.722	Dep. Var.	Monthly Salary ($)
ANOVA table
Source	SS	df	MS	F	p-value
Regression	1,410,156.2500	1	1,410,156.2500	3.13	.1518
Residual	1,804,843.7500	4	451,210.9375
Total	3,215,000.0000	5
Regression output				confidence interval
variables	coefficients	std. error	t (df=4)	p-value	95% lower	95% upper
Intercept	-1,100.0000	2,700.8474	-0.407	.7047	-8,598.7546	6,398.7546
GPA	1,484.3750	839.6530	1.768	.1518	-846.8753	3,815.6253
Predicted values for: Monthly Salary ($)
		95% Confidence Interval	95% Prediction Interval
GPA	Predicted	lower	upper	lower	upper	Leverage
3	3,353.125	2,460.324	4,245.926	1,285.440	5,420.810	0.229