[R] [EXTERNAL] Re: Question About lm()

[Bromaghin, Jeffrey F]

Rolf,

Yes, that is what is happening. I thought that lm() recognized the presence of a categorical variable as a replacement, if you will, for the intercept, but either my memory is faulty or the behavior of the function changed somewhere along the way.

Thank you,
Jeff

-----Original Message-----
From: Rolf Turner <r.turner using auckland.ac.nz> 
Sent: Thursday, February 10, 2022 12:09 PM
To: David Winsemius <dwinsemius using comcast.net>; r-help using r-project.org
Cc: Bromaghin, Jeffrey F <jbromaghin using usgs.gov>
Subject: [EXTERNAL] Re: [R] Question About lm()



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On Wed, 9 Feb 2022 23:16:13 -0800
David Winsemius <dwinsemius using comcast.net> wrote:

> The models are NOT equivalent. Why would you’ll think they were?

'Scuse me, David, but they *are* "equivalent".  They are simply different parametrisations of the same model.  *However* the different parametrisations imply, according to the conventions of lm(), different
*null* models.

For yResp ~ xCat + xCont, the null model is yResp ~ 1.

For yResp ~ -1 + xCat + xCont, the null model is yResp ~ 0.

For the first null model, the residual sum of squares is
ssr1 = sum((yResp-mean(yResp))^2).

For the second null model, the residual sum of squares is
ssr2 = sum(yResp^2).

Thus for the first parametrisation one gets

RSquared = 1 - ssr1/ssr

and for the second parametrisation one gets

RSquared = 1 - ssr2/ssr

in both cases "ssr" is the sum of squares of the residuals from the full model (which is of course the same in both cases).

I hope that this clarifies things a bit for the OP.

One further comment:  RSquared is a bit of a dubious concept;  it is particularly dubious for models with no intercept.  Caveat utilitor.

cheers,

Rolf

--
Honorary Research Fellow
Department of Statistics
University of Auckland
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