Question

1. The table below gives information on GPAs and starting salaries (rounded to the nearest thousand dollars) of 7 recent college graduates.

GPA                       | 2.90   3.81   3.20   2.42   3.94   2.05   2.25

Starting salary        |    38      48     38      35      50      31      37

1. List the independent variable ___________ and the dependent variable_________.
2. Build a table, with column totals containing x, y, x², y², and xy.
3. Calculate average of y & average of x.
4. Show formula & work for a, b, SS_xx, SS_xy, SS_yy.
5. Write the regression line and predict the starting salary when GPA is 2.05
6. Solve for s_e, r, and r². Based on the value of r², explain if this is a good model.

Answer 1

a) Independent variable = GPA

Dependent variable = Starting salary

b)

X	Y	XY	X²	Y²
2.90	38	110.2	8.41	1444
3.81	48	182.88	14.5161	2304
3.20	38	121.6	10.24	1444
2.42	35	84.7	5.8564	1225
3.94	50	197	15.5236	2500
2.05	31	63.55	4.2025	961
2.25	37	83.25	5.0625	1369
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
20.57	277	843.18	63.8111	11247

c)

x̅ = Ʃx/n = 20.57/7 = 2.938571

y̅ = Ʃy/n = 277/7 = 39.571429

d)     

SSxx = Ʃx² - (Ʃx)²/n = 63.8111 - (20.57)²/7 = 3.364686

SSyy = Ʃy² - (Ʃy)²/n = 11247 - (277)²/7 = 285.714286

SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 843.18 - (20.57)(277)/7 = 29.195714

Slope, b = SSxy/SSxx = 29.19571/3.36469 = 8.6771

y-intercept, a = y̅ -b* x̅ = 39.57143 - (8.6771)*2.93857 = 14.07315

Regression equation :   

ŷ = 14.0732 + (8.6771) x  

e)

Predicted value of y at x = 2.05

ŷ = 14.0732 + (8.6771) * 2.05 = 31.8612  

f)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 285.71429 - (29.19571)²/3.36469 = 32.380154

Standard error, se = √(SSE/(n-2)) = √(32.38015/(7-2)) = 2.5448

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 29.19571/√(3.36469*285.71429) = 0.9416

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = (29.19571)²/(3.36469*285.71429) = 0.8867

88.67% variation in y is explained by the least squares model.

This is a good model.