To add to Brian's points (which you should heed!) -- you **may** find it
also useful to look at (possibly smoothed) residuals to compare lack of fit
from your alternative models. If any shows up, some subject matter knowledge
might lead you to choose one or the other of your models -- or neither.
-- Bert Gunter
Genentech
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of Prof Brian Ripley
Sent: Monday, August 11, 2008 11:34 PM
To: Nazareno Andrade
Cc: r-help
Subject: Re: [R] No answer in anova.nls
The reason for no F test showing up is that the additional df is 0 and the
F value is Inf. But the underlying problem is that your models are not
nested and so ANOVA between them is invalid.
I suggest you seek help from a local statistician: your misunderstanding
and your question about model adequacy are subtle statistical issues and
not help on R.
On Mon, 11 Aug 2008, Nazareno Andrade wrote:
> Dear R-helpers,
>
> I am trying to check whether a model of the form y(t) = a/(1 +b*t) fits
the
> curve of downloads per day of a file in a specific online community better
> than a model of the form y(t) = a*exp(-b*t). For that, I used nls to fit
> both models and I am now trying to compare the fits with anova. The
problem
> I find is that anova does not report an F statistic or a p-value when I
> compare these two models.
>
> The data for a file is typically the following:
>> d
> V1 V2
> 1 1 293
> 2 2 101
> 3 3 63
> 4 4 53
> 5 5 42
> 6 6 19
> 7 7 28
> 8 8 23
> 9 9 18
> 10 10 17
> 11 11 14
> 12 12 18
> 13 13 5
> 14 14 9
> 15 15 10
> 16 15 0
>
> My code:
>
> d <-
read.table(url("http://ece.ubc.ca/~nazareno/85247.arrivalRates<http://ece.ub
c.ca/%7Enazareno/85247.arrivalRates>
> "))
> plot(d)
> f.exp.nw <- nls(V2 ~ a. * exp(-b. * V1), data = d, list( a. = d$V2[1], b.
=
> 0.05))
> f.exp5.nw <- nls(V2 ~ a. / (1+ b. *V1), data = d, list( a. = d$V2[1], b. =
> 2))
> lines(d$V1, predict(f.exp.nw), col = "royalblue")
> lines(d$V1, predict(f.exp5.nw), col = "orange")
>
> anova(f.exp.nw, f.exp5.nw)
>
> However, the output from anova.nls is:
>
> Analysis of Variance Table
>
> Model 1: V2 ~ a. * exp(-b. * V1)
> Model 2: V2 ~ a./(1 + b. * V1)
> Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
> 1 13 4994.9
> 2 13 314.7 0 0.0
>
> and I cannot interpretate the lack of an F value. Looking at the
> implementation of the anova.nls() function, this seems to be related to
the
> fact that the residuals' degrees of freedom are the same, but I could not
> find anywhere more information on whether they were required to be
> different. Thus, I'd greatly appreciate if you could spot any mistakes I
> might be doing or a (preferably online) reference for more on this issue.
>
>
> As a side question, it would be great also if someone with more experience
> on this matter could confirm with me that the proper direction for
checking
> whether "the y(t) = a/(1 +b*t) form models more precisely the behavior of
> downloads of files in this communtiy" by quantifying for how many files it
> outperforms the exponential model.
>
> thank you very much in advance,
> Nazareno
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.