Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

DEMONSTRATION 16.1 555
DEMONSTRATION 16.1
LINEAR REGRESSION
The following data will be used to demonstrate the process of linear regression. The scores and
summary statistics are as follows.
Person X Y
A 0 4
B 2 1
C 8 10
D 6 9
E 4 6
M X
= 4 with SS X
= 40
M Y
= 6 with SS Y
= 54
SP = 40
These data produce a Pearson correlation of r = 0.861.
STEP 1
Compute the values for the regression equation The general form of the regression
equation is
Ŷ = bX + a where b = SP and a = M
SS Y
– bM X
X
For these data, b = 40 = 1.00 and a = 6 − 1(4) = +2.00
40
Thus, the regression equation is Ŷ = (1)X + 2.00 or simply, Ŷ = X + 2.
STEP 2
Evaluate the significance of the regression equation The null hypothesis states that the
regression equation does not predict a significant portion of the variance for the Y scores. To
conduct the test, the total variability for the Y scores, SS Y
= 54, is partitioned into the portion
predicted by the regression equation and the residual portion.
SS regression
= r 2 (SS Y
) = 0.741(54) = 40.01 with df = 1
SS residual
= (1 – r 2 )(SS Y
) = 0.259(54) = 13.99 with df = n – 2 = 3
The two MS values (variances) for the F-ratio are
And the F-ratio is
MS regression
=
SS regression
df
MS residual
= SS residual
df
F =
5 40.01
1
5 13.99
3
= 40.01
= 4.66
MS regression
MS residual
5 40.01
4.66 = 8.59
With df = 1, 3 and α = .05, the critical value for the F-ratio is 10.13. Therefore, we fail
to reject the null hypothesis and conclude that the regression equation does not predict a
significant portion of the variance for the Y scores.