On 10/11/2008 3:31 PM, Ted Harding wrote:
Hi Folks,
I'm comparing some output from R with output from SPSS.
The coefficients of the independent variables (which are
all factors, each at 2 levels) are identical.
However, R's Intercept (using default contr.treatment)
differs from SPSS's
Looks like the contrast matrix for indicator is contr.SAS(n),
for deviation is contr.sum(n) and for simple is:
(diag(n) - 1/n)[, -1]
That works at least for the n = 3 example in the link.
Perhaps the others could be checked against SPSS
for a variety of values of n to be sure.
On Sun, Oct 12,
The formula should be (diag(n) - 1/n)[, -n]
On Sun, Oct 12, 2008 at 1:36 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Looks like the contrast matrix for indicator is contr.SAS(n),
for deviation is contr.sum(n) and for simple is:
(diag(n) - 1/n)[, -1]
That works at least for the n = 3
I found this link:
http://webs.edinboro.edu/EDocs/SPSS/SPSS%20Regression%20Models%2013.0.pdf
which indicates that the contrast in SPSS that is used
depends not only on the contrast selected but also on the
reference category selected and the two can be chosen
independently. Thus one could have
Very many thanks, Chuck and Gabor, for the comments and the
references to on-line explanations. It is beginning to become
clear!
Most grateful.
Ted.
On 12-Oct-08 18:03:53, Gabor Grothendieck wrote:
I found this link:
http://webs.edinboro.edu/EDocs/SPSS/SPSS%20Regression%20Models%2013.0.pd
f
Don't know but perhaps you could just use each of:
contr.helmert, contr.poly, contr.sum, contr.treatment, contr.SAS
in turn on the R side until you get one that matches. Once you
find out adding a contr.SPSS to R might be nice.
On Sat, Oct 11, 2008 at 3:31 PM, Ted Harding
[EMAIL PROTECTED]
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