Thank you so much for your valuable feedback Berwin.
Have a great day.
Cheers,
Paul
El El dom, 20 de ago. de 2023 a la(s) 10:21 p. m., Berwin A Turlach <
berwin.turl...@gmail.com> escribió:
> G'day Paul,
>
> On Sun, 20 Aug 2023 12:15:08 -0500
> Paul Bernal wrote:
>
> > Any idea on how to procee
G'day Paul,
On Sun, 20 Aug 2023 12:15:08 -0500
Paul Bernal wrote:
> Any idea on how to proceed in this situation? What could I do?
You are fitting a simple asymptotic model for which nls() can find good
starting values if you use the self starting models (SSxyz()). Well,
Doug (et al.) choose t
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
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?").
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 a
Thanks a lot Bert, I appreciate your help.
Kind regards,
Paul
El El dom, 20 de ago. de 2023 a la(s) 2:39 p. m., Bert Gunter <
bgunter.4...@gmail.com> escribió:
> Basic algebra and exponentials/logs. I leave those details to you or
> another HelpeR.
>
> -- Bert
>
> On Sun, Aug 20, 2023 at 12:17
Basic algebra and exponentials/logs. I leave those details to you or
another HelpeR.
-- Bert
On Sun, Aug 20, 2023 at 12:17 PM Paul Bernal wrote:
> Dear Bert,
>
> Thank you for your extremely valuable feedback. Now, I just want to
> understand why the signs for those starting values, given the f
Dear Bert,
Thank you for your extremely valuable feedback. Now, I just want to
understand why the signs for those starting values, given the following:
> #Fiting intermediate model to get starting values
> intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random)
> summary(intermediatemod)
Oh, sorry; I changed signs in the model, fitting
theta0 + theta1*exp(theta2*x)
So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 =
+.055 as starting values.
-- Bert
On Sun, Aug 20, 2023 at 11:50 AM Paul Bernal wrote:
> Dear Bert,
>
> Thank you so much for your kind a
I haven't looked to see whether you or Bert made an algebraic mistake
in translating the parameters of the log-linear model to their
equivalents for the nonlinear model, but nls() gives me the same answer
as nls() in this case (I called my data 'dd2'):
n1 <- nlxb(y~theta1 - theta2*exp(
Dear Bert,
Thank you so much for your kind and valuable feedback. I tried finding the
starting values using the approach you mentioned, then did the following to
fit the nonlinear regression model:
nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
start =
l
My answer is WAY longer than Bert Gunter's but maybe useful nonetheless.
(Q for John Nash: why does the coef() method for nlmrt objects
return the coefficient vector **invisibly**? That seems confusing!)
Here's what I did:
* as a preliminary step, adjust the formula so that I don't ha
I got starting values as follows:
Noting that the minimum data value is .38, I fit the linear model log(y -
.37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37,
exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2
in the nonlinear model. This converged without pr
Dear friends,
This is the dataset I am currently working with:
>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, 3
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