The ANOVA F is on all the variables in the model. Right?
But you have a "variable of interest" so what you want is the test on
that. It does not matter whether there were covariates before it that
accounted for 95% of the variance for a single score, or for 0.001% of
the variance with a dozen -- How should those affect the hypothesis?
(Well, they do shrink the error Sum of Squares, slightly. But they
don't dominate the test.) You need to have enough d.f. to have a
decent error term, but the test on your V. of I. has just the df
accounted for by the v. of i.
--
Rich Ulrich, wpi...@pitt.edu
http://www.pitt.edu/~wpilib/index.html
> On Wed, 27 Sep 2000 14:05:50 +1000, Kate <ka...@onthenet.com.au> wrote:
At step 1, you have R^2y.234, which is not significant because
(SSreg/3)/(SSerr/(n-3-1)) with 3, n-3-1 df is not
significant. Note that it is _possible_ that 1 or more of the
predictors does make a significant contribution, but that the
effect is nonsignificant when diluted across all 3 predictors.
The same reasoning applies at the next step. Now you have
R^2y.1234, which is not sig, F = (SSreg/4)/(SSerr/(n-4-1)) with
4, n-4-1 df. But the unique contribution of one of the
predictors (e.g., x1) could (and is in your case)) be
significant.
The unique contribution of x1 is given by R^2change = R^2y.1234 -
R^2y.234 (or corresponding operations with SSreg's) and is tested
for significance by F = (SSchange/1)/(SSerr/(n-4-1))with 1, n-4-1
df. Note there is no dilution of the SSchange because it has a
single df. Fchange = t^2 for the significance of the regression
coefficent for x1, when x2, x3, and x4 are also in the equation.
Best wishes
Jim
============================================================================
James M. Clark (204) 786-9757
Department of Psychology (204) 774-4134 Fax
University of Winnipeg 4L05D
Winnipeg, Manitoba R3B 2E9 cl...@uwinnipeg.ca
CANADA http://www.uwinnipeg.ca/~clark
============================================================================