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
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