Either lm(y~ a-1 + a:x) or lm (y~ a + (a-1):x) or lm(y~ a+ a:(x-1))
-- Bert Gunter
Genentech
-----Original Message-----
From: bgunter
Sent: Friday, May 16, 2008 8:39 AM
To: Rolf Turner
Subject: RE: [R] Making slope coefficients ``relative to 0''.
lm(y ~ a + a:(x-1))
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On
Behalf Of Rolf Turner
Sent: Thursday, May 15, 2008 8:55 PM
To: R-help forum
Subject: [R] Making slope coefficients ``relative to 0''.
I am interested in whether the slopes in a linear model are different
from 0.
I.e. I would like to obtain the slope estimates, and their standard
errors,
``relative to 0'' for each group, rather than relative to some baseline.
Explicitly I would like to write/represent the model as
y = a_i + b_i*x + E
i = 1, ..., K, where x is a continuous variate and i indexes groups
(levels of a factor with K levels).
The ``usual'' structure (using ``treatment contrasts'') gives
y = a + a_i + b*x + b_i*x + E
i = 2, ..., K. (So that b is the slope for the baseline group, and
b_i measures
how much the slope for group i differs from that for the baseline group.
I can force the *intercepts* to be ``relative to 0'' by putting a
``-1'' into the formula:
lm(y ~ g*x-1)
But I don't really care about the intercepts; it's the slopes I'm
interested in.
And there doesn't seem to a way to do the thing equivalent to the
``-1'' trick
for slopes. Or is there?
There are of course several work-arounds. (E.g. calculate my b_i-
hats and their
standard errors from the information obtained from the usual model
structure.
Or set up my own dummy variable to regress upon. Easy enough, and I
could do that.)
I just wanted to know for sure that there wasn't a sexier way, using
some aspect
of the formula machinery with which I am not yet familiar.
Thanks for any insights.
cheers,
Rolf Turner
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