> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
>
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens Dustin Fife
> Verzonden: dinsdag 29 mei 2012 14:56
> Aan: Ru
ject.org] Namens
Dustin Fife
Verzonden: dinsdag 29 mei 2012 14:56
Aan: Rui Barradas
CC: r-help
Onderwerp: Re: [R] setting parameters equal in lm
That did it. Thanks! One more follow-up questions. How do I set a parameter to
a particular value? I tried I(.5*X2), but that didn't do what I expected.
Simple, just make y2 <- y - 0.5*X2
Rui Barradas
Em 29-05-2012 13:55, Dustin Fife escreveu:
That did it. Thanks! One more follow-up questions. How do I set a parameter
to a particular value? I tried I(.5*X2), but that didn't do what I
expected.
On Tue, May 29, 2012 at 6:39 AM, Rui Barradas wro
That did it. Thanks! One more follow-up questions. How do I set a parameter
to a particular value? I tried I(.5*X2), but that didn't do what I
expected.
On Tue, May 29, 2012 at 6:39 AM, Rui Barradas wrote:
> Hello,
>
> Your model is equivalent of
>
> y = b1(X1 + X3) + b2X2
>
> (plus error)
>
> S
Hello,
Your model is equivalent of
y = b1(X1 + X3) + b2X2
(plus error)
So, use I() to add X1 and X3. You don't need to create an extra variable
X13 <- X1 + X3. See the help page for it. The point on function formula.
?I
linMod2 = lm(Y ~ -1 + I(X1 + X3) + X2, data=data.set)
summary(linMod2
I don't know how it ties into the tools car gives you, but one (quick
and dirty) way to do this is to simply regress on
Y ~ aX2 + b(X1+X3)
or in R code something like:
lm(Y ~ X2 + I(X1+X3), data = data.set)
which gives a linear model you can play around with. Note the I()
function [that's the c
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