[R] Question About lm()

Rolf Turner r@turner @end|ng |rom @uck|@nd@@c@nz
Thu Feb 10 22:08:30 CET 2022


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
Phone: +64-9-373-7599 ext. 88276



More information about the R-help mailing list