[R] Interpreting Results from LOF.test() from qpcR package
Ben Bolker
bbo|ker @end|ng |rom gm@||@com
Mon Aug 21 03:34:01 CEST 2023
The p-values are non-significant by any standard cutoff (e.g.
p<=0.05, p<=0.1) but note that this is a *lack-of-fit* test -- i.e.,
"does my function fit the data well enough?", **not** a "significant
pattern" test (e.g., "does my function fit the data better than a
reasonable null model?"). In other words, this test tells you that you
*can't* reject the null hypothesis that the model is "good enough" in
some sense.
To test against a constant null model, you could do
nullmod <- nlsr(y ~ const,
data = mod14data2_random,
start = list(const = 0.45))
anova(nlregmod3, nullmod)
(This question seems to be verging on "general question about
statistics" rather than "question about R", so maybe better for a venue
like https://stats.stackexchange.com ??)
On 2023-08-20 9:01 p.m., Paul Bernal wrote:
> I am using LOF.test() function from the qpcR package and got the following
> result:
>
>> LOF.test(nlregmod3)
> $pF
> [1] 0.97686
>
> $pLR
> [1] 0.77025
>
> Can I conclude from the LOF.test() results that my nonlinear regression
> model is significant/statistically significant?
>
> Where my nonlinear model was fitted as follows:
> nlregmod3 <- nlsr(formula=y ~ theta1 - theta2*exp(-theta3*x), data =
> mod14data2_random,
> start = list(theta1 = 0.37,
> theta2 = -exp(-1.8),
> theta3 = 0.05538))
> And the data used to fit this model is the following:
> dput(mod14data2_random)
> structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L,
> 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4,
> 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4
> ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34,
> 28)), row.names = c(NA, -15L), class = "data.frame")
>
> Cheers,
> Paul
>
> [[alternative HTML version deleted]]
>
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