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The following data are the monthly salaries y and the grade point averages x for

ID: 3258047 • Letter: T

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.

The estimated regression equation for these data is  = -350 + 1,250x and MSE =236,875.


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 3.5 3,900 3.6 4,200 3.1 3,700 3.5 4,100 2.8 2,400

Explanation / Answer

Answer:

The estimated regression equation for these data is  y = -350 + 1,250x and MSE =236,875.


a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal).
$3400.0

b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals).
$ (2767.20  ,4032.80 )

c. Develop a 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).
$ ( 1907.88 , 4892.12 )

Regression Analysis

0.556

n

6

r

0.746

k

1

Std. Error

486.698

Dep. Var.

Monthly Salary ($)

ANOVA table

Source

SS

df

MS

F

p-value

Regression

1,187,500.0000

1  

1,187,500.0000

5.01

.0887

Residual

947,500.0000

4  

236,875.0000

Total

2,135,000.0000

5  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=4)

p-value

95% lower

95% upper

Intercept

-350.0000

1,797.5144

-0.195

.8551

-5,340.7001

4,640.7001

GPA

1,250.0000

558.2810

2.239

.0887

-300.0364

2,800.0364

Predicted values for: Monthly Salary ($)

95% Confidence Interval

95% Prediction Interval

GPA

Predicted

lower

upper

lower

upper

Leverage

3

3,400.000

2,767.200

4,032.800

1,907.880

4,892.120

0.219

Regression Analysis

0.556

n

6

r

0.746

k

1

Std. Error

486.698

Dep. Var.

Monthly Salary ($)

ANOVA table

Source

SS

df

MS

F

p-value

Regression

1,187,500.0000

1  

1,187,500.0000

5.01

.0887

Residual

947,500.0000

4  

236,875.0000

Total

2,135,000.0000

5  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=4)

p-value

95% lower

95% upper

Intercept

-350.0000

1,797.5144

-0.195

.8551

-5,340.7001

4,640.7001

GPA

1,250.0000

558.2810

2.239

.0887

-300.0364

2,800.0364

Predicted values for: Monthly Salary ($)

95% Confidence Interval

95% Prediction Interval

GPA

Predicted

lower

upper

lower

upper

Leverage

3

3,400.000

2,767.200

4,032.800

1,907.880

4,892.120

0.219

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