[R] Interpreting Results from LOF.test() from qpcR package
Bert Gunter
bgunter@4567 @end|ng |rom gm@||@com
Mon Aug 21 04:00:17 CEST 2023
I would suggest that a simple plot of residuals vs. fitted values and
perhaps plots of residuals vs. the independent variables are almost always
more useful than omnibus LOF tests. (many would disagree!) However,as Ben
noted, this is wandering outside R-Help's strict remit, and you would be
better served by statistics discussion/help sites rather than R-Help.
Though with this small a data set and this complex a model, I would be
surprised if there could be LOF unless it were glaringly obvious from
simple plots.
Cheers,
Bert
-- Bert
On Sun, Aug 20, 2023 at 6:02 PM Paul Bernal <paulbernal07 using gmail.com> 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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>
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