The regression equation

 

Question 1

The regression equation is intended to be the “best fitting” straight line for a set of data. What is the criterion for “best fitting”?

 

The best fitting line is determined by the error between the predicted Y values on the line and the actual Y values in the data.  The regression equation is determined by the line with the smallest total squared error.

 

Question 2

A set of n=15 pairs of scores (X and Y values) has a SS_X = 15 and SP = -60. If the mean for the X values is M_X= 6 and the mean for the Y values is M_y= 12, find the regression equation for predicting Y from the X values.

b = SP/ SS_X

b = -60/15 = -4

a = M_Y – bM_X

a = 12 – (-4)(6) = 12 + 24 = 36

Ŷ = bX + a

Ŷ = -4X + 36

Question 3

A set of n =25 pairs of scores (X and Y values) produces a regression equation of Ŷ = 3X – 6. Find the predicted Y value for each of the following X scores:

1. 0

Ŷ = 3(0) – 6 = -6

1. 1

Ŷ = 3(1) – 6 = -3

1. 3

Ŷ = 3(3) – 6 = 3

1. -3

Ŷ = 3(-3) – 6 = -15

Question 4

 

We have to find a and b: First we find b

b = SP/ SS_X

SP = ƩXY – ƩXƩY

n

SS_X = Ʃ(X – M_X)²

 

 

 

 

 

 

1. We find the numerator of b: SP

X	Y	XY
1	2	2
4	7	28
3	5	15
2	1	2
5	8	40
3	7	21

ƩX = 18        ƩY = 30          ƩXY = 108

SP = ƩXY – ƩXƩY

n

SP = 108 – (18)(30) = 108-90 = 18

6

1. We find the denominator of b: SS_X

SS_X = Ʃ(X – M_X)²

M_X= ƩX/n = 18/6 =3

 

X	X – MX	(X – MX)2	Y
1	-2	4	2
4	1	1	7
3	0	0	5
2	-1	1	1
5	2	4	8
3	0	0	7

                                            SS_X = Ʃ(X – M_X)²= 10

SS_X = 10

 

 

1. Calculate b

b = SP/ SS_X

b = 18/10 = 1.8

Now we find a

a = M_Y – bM_X

M_X= ƩX/n = 18/6 =3

M_Y= ƩY/n = 30/6 =5

a = 5 – (1.8)(3) = 5- 5.4 =  -0.4

The linear regression equation for predicting Y from X is:

Ŷ = bX + a 

Ŷ = 1.8X – 0.4   

 

Question 5

We have to find a and b: First we find b

b = SP/ SS_X

SP = ƩXY – ƩXƩY

n

SS_X = Ʃ(X – M_X)²

 

1. We find the numerator of b: SP

X	Y	XY
3	8	24
6	4	24
3	5	15
3	5	15
5	3	15

ƩX = 20        ƩY = 25         ƩXY = 93

SP = ƩXY – ƩXƩY

n

SP = 93 – (20)(25) = 93 -100 = -7

5

1. We find the denominator of b: SS_X

SS_X = Ʃ(X – M_X)²

M_X= ƩX/n = 20/5 =4

 

X	X – MX	(X – MX)2	Y
3	-1	1	8
6	2	4	4
3	-1	1	5
3	-1	1	5
5	1	1	3

                                            SS_X = Ʃ(X – M_X)²= 8

 

SS_X = 8

1. Calculate b

b = SP/ SS_X

b = -7/8 = -.875

Now we find a

a = M_Y – bM_X

M_X= ƩX/n = 20/5 =4

M_Y= ƩY/n = 25/5 =5

a = 5 – (-.875)(4) = 5 + 3.5 =  8.5

The linear regression equation for predicting Y from X is:

Ŷ = bX + a 

Ŷ = -.875X + 8.5

Question 6

1. Use the data to find the regression equation for predicting memory scores from age

We have to find a and b: First we find b

b = SP/ SS_X

SP = ƩXY – ƩXƩY

n

SS_X = Ʃ(X – M_X)²

 

 

 

 

 

 

1. We find the numerator of b: SP

Age (X)	Memory Score (Y)	XY
25	10	250
32	10	320
39	9	351
48	9	432
56	7	392

ƩX = 200       ƩY = 45                       ƩXY = 1745

SP = ƩXY – ƩXƩY

n

SP = 1745 – (200)(45) = 1745- 1800 = -55

5

1. We find the denominator of b: SS_X

SS_X = Ʃ(X – M_X)²

M_X= ƩX/n = 200/5 =40

X	X – MX	(X – MX)2	Y
25	-15	225	10
32	-8	64	10
39	-1	1	9
48	8	64	9
56	16	256	7

                                            SS_X = Ʃ(X – M_X)²= 610

 

SS_X = 610

1. Calculate b

b = SP/ SS_X

b = –55/610 = -.09

 

Now we find a

a = M_Y – bM_X

M_X= ƩX/n = 200/5 =40

M_Y= ƩY/n = 45/5 =9

a = 9 – (-.09)(40) = 9 + 3.6 = 12.6

The linear regression equation for predicting Y from X is:

Ŷ = bX + a 

Ŷ = -.09X + 12.6

Questions 7 to 9

Use the regression equation you found in question 6 to find the predicted memory scores for the following ages: 28, 43, and 50

Ŷ = -0.09(28) + 12.6 = 10.08

Ŷ = -0.09(43) + 12.6 = 8.73

Ŷ = -0.09(50) + 12.6 = 8.1