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

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 = -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

Explanation / Answer

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

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