Does it pay to stay in school? A report looked at the median hourly wage gain pe

Question

Does it pay to stay in school? A report looked at the median hourly wage gain per additional year of schooling in 2007. The report states that workers with a high school diploma had a median hourly wage that was 10% higher than those who had only completed 11 years of school, workers who had completed 1 year of college (13 years of education) had a median hourly wage that was 11 % higher than that of the workers who had completed only 12 years of school. The added gain in median hourly wage for each additional year of school is shown in the accompanying table. The entry for 15 years of schooling has been intentionally omitted from the table. The entry for 1s years oknswho had completed any t of school Workrs who ha 2007 Median Hourly Wage Gain for the Additional Year (percent) 10 Years of Schooling 12 13 14 16 17 18 13 16 18 19 (a) Use the given data to predict the median hourly wage gain for the 15th year of schooling b) The actual wage gain for 15th year of schooling was 14% How close as the actual value to the predicted age gain percent rom part ? se as pre te actual

Explanation / Answer

Result:

a).

the regression line is y = -9.0714+1.5714*x

when x=15, predicted y = -9.0714+1.5714*15 =14.4996

=14.5%

b). 14-14.5 = -0.5

0.5%

Regression Analysis

r²

0.995

n

6

r

0.997

k

1

Std. Error

0.299

Dep. Var.

wage

ANOVA table

Source

SS

df

MS

F

p-value

Regression

69.1429

1  

69.1429

774.40

9.92E-06

Residual

0.3571

4  

0.0893

Total

69.5000

5  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=4)

p-value

95% lower

95% upper

Intercept

-9.0714

0.8558

-10.600

.0004

-11.4474

-6.6954

year

1.5714

0.0565

27.828

9.92E-06

1.4146

1.7282

Predicted values for: wage

95% Confidence Interval

95% Prediction Interval

year

Predicted

lower

upper

lower

upper

Leverage

15

14.500

14.161

14.839

13.604

15.396

0.167

Regression Analysis

r²

0.995

n

6

r

0.997

k

1

Std. Error

0.299

Dep. Var.

wage

ANOVA table

Source

SS

df

MS

F

p-value

Regression

69.1429

1  

69.1429

774.40

9.92E-06

Residual

0.3571

4  

0.0893

Total

69.5000

5  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=4)

p-value

95% lower

95% upper

Intercept

-9.0714

0.8558

-10.600

.0004

-11.4474

-6.6954

year

1.5714

0.0565

27.828

9.92E-06

1.4146

1.7282

Predicted values for: wage

95% Confidence Interval

95% Prediction Interval

year

Predicted

lower

upper

lower

upper

Leverage

15

14.500

14.161

14.839

13.604

15.396

0.167

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.