Thanks Duncan.
Yes i do have variation in the lidar metrics (be, ch, crr, and home) although i
have a quite high correlation between ch and home. But even if i eliminate one
metric (either ch or home) i end up with a deviation of 99.99. The species has
values of 0 and 1 since i try to predict presence / absence.
Do you think it is still a valid result?
Thanks again,
Monica> Date: Mon, 31 Mar 2008 08:47:48 -0400> From: [EMAIL PROTECTED]> To:
[EMAIL PROTECTED]> CC: r-help@r-project.org> Subject: Re: [R] unexpected GAM
result - at least for me!> > On 3/31/2008 8:34 AM, Monica Pisica wrote:> > > >
Hi> > > > > > I am afraid i am not understanding something very fundamental....
and does not matter how much i am looking into the book "Generalized Additive
Models" of S. Wood i still don't understand my result.> > > > I am trying to
model presence / absence (presence = 1, absence = 0) of a species using some
lidar metrics (i have 4 of these). I am using different models and such ....
and when i used gam i got this very weird (for me) result which i thought it is
not possible - or i have no idea how to interpret it.> > > >> can3.gam <-
gam(can>0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial')> >>
summary(can3.gam)> > Family: binomial> > Link function: logit> > Formula:> >
can> 0 ~ s(be) + s(crr) + s(ch) + s(home)> > Parametric coefficien!
ts:> > Estimate Std. Error z value Pr(>|z|)> > (Intercept) 85.39 162.88 0.524
0.6> > Approximate significance of smooth terms:> > edf Est.rank Chi.sq
p-value> > s(be) 1.000 1 0.100 0.751> > s(crr) 3.929 8 0.380 1.000> > s(ch)
6.820 9 0.396 1.000> > s(home) 1.000 1 0.314 0.575> > R-sq.(adj) = 1 Deviance
explained = 100%> > UBRE score = -0.81413 Scale est. = 1 n = 148> > > > Is this
a perfect fit with no statistical significance, an over-estimating or what????
It seems that the significance of the smooths terms is "null". Of course with
such a model i predict perfectly presence / absence of species.> > > > Again, i
hope you don't mind i'm asking you this. Any explanation will be very much
appreciated.> > Look at the data. You can get a perfect fit to a logistic
regression > model fairly easily, and it looks as though you've got one. (In
fact, > the huge intercept suggests that all predictions will be 1. Do you >
actually have any variation in the data?)> > Duncan Murdoch
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