[R] non-linear curve fitting

[Douglas Bates]

On 3/22/07, Hufkens Koen <koen.hufkens at ua.ac.be> wrote:
> Is there a means of getting an F-statistic (p-value) out of all of this.

> Because least-square criterion / r-square only tell me how good the fit is and not necessarily how solid this fit is. An F-statistic (p-value) would be nice...

What would the F-statistic be?  For a linear model with an intercept
the F-statistic represents a comparison of the model that you have fit
to the trivial model (intercept only).  It is important that the
models being compared are nested - otherwise the F statistic is of
questionable validity.

For the logistic growth model you described (which is not quite the
one fit by SSlogis - that model has one more parameter, a scale factor
on the response) the response always goes to zero as x -> -\Infty and
to one as x -> \Infty.  The trivial model is not nested within this
model for finite parameter values so I'm not sure what hypotheses
would be tested by an F-statistic.

> Regards,
> Koen
>
>
> > -----Original Message-----
> > From: Philippe Grosjean [mailto:phgrosjean at sciviews.org]
> > Sent: donderdag 22 maart 2007 14:02
> > To: Hufkens Koen; r-help at stat.math.ethz.ch
> > Subject: Re: [R] non-linear curve fitting
> >
> > Hello,
> >
> > If a least-square criterion is fine for you, you should use
> > nls(). For the logistic curve, you have a convenient
> > self-starting model available:
> > SSlogis(). Look at:
> >
> > ?nls
> > ?SSlogis
> >
> > Best,
> >
> > Philippe Grosjean
> >
> > ..............................................<°}))><........
> >   ) ) ) ) )
> > ( ( ( ( (    Prof. Philippe Grosjean
> >   ) ) ) ) )
> > ( ( ( ( (    Numerical Ecology of Aquatic Systems
> >   ) ) ) ) )   Mons-Hainaut University, Belgium
> > ( ( ( ( (
> > ..............................................................
> >
> > Hufkens Koen wrote:
> > > Hi list,
> > >
> > > I have a little curve fitting problem.
> > >
> > > I would like to fit a sigmoid curve to my data using the
> > following equation:
> > >
> > > f(x) = 1/(1 + exp(-(x-c)*b)) (or any other form for that matter)
> > >
> > > Where x is the distance/location within the dataframe, c is
> > the shift of the curve across the dataframe and b is the
> > steepness of the curve.
> > >
> > > I've been playing with glm() and glm.fit() but without any luck.
> > >
> > > for example the most simple example
> > >
> > > x = -10:10
> > > y = 1/(1 + exp(-x))
> > > glm(y ~ x, family=binomial(link="logit"))
> > >
> > > I get a warning:
> > > non-integer #successes in a binomial glm! in: eval(expr, envir,
> > > enclos)
> > >
> > > and some erratic results
> > >
> > > This is the most simple test to see if I could fit a curve
> > to this perfect data so since this didn't work out, bringing
> > in the extra parameters is a whole other ballgame so could
> > someone give me a clue?
> > >
> > > Kind regards,
> > > Koen
> > >
> > >
> >
> > --
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> >
>
> --
>
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