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.ubc.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
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