Sorry, just got back into town.
I wonder if AIC, BIC, or cross-validation scoring couldn't also be used as
criteria for model selection - I've seen it mostly in the context of variable
selection rather than 'form' selection but in principle might apply here?
--- Dieter Menne [EMAIL PROTECTED]
On 24-Jul-07 01:09:06, Andrew Clegg wrote:
Hi folks,
I've looked through the list archives and online resources, but I
haven't really found an answer to this -- it's pretty basic, but I'm
(very much) not a statistician, and I just want to check that my
solution is statistically sound.
I think your way is probably the easiest (shockingly). For instance, here are
some alternatives - I think in both cases you have to calculate the
coefficient of determination (R^2) manually. My understanding is that
multiple R^2 in your case is the usual R^2 because you only have one
predictor
Well spoken. And since log transformations are nonlinear and 'compresses' the
data, it's not surprising to find that the fit doesn't look so nice while the
fit metrics tell you that a model does a good job.
--- [EMAIL PROTECTED] wrote:
On 24-Jul-07 01:09:06, Andrew Clegg wrote:
Hi folks,
Stephen, Ted -- thanks for your input. I'm glad to know I was barking
up the right-ish tree at least.
On 7/24/07, Ted Harding [EMAIL PROTECTED] wrote:
There are not enough data to properly identify the non-linearity,
but the overall appearance of the data plot suggests to me that
you should
On 7/24/07, Stephen Tucker [EMAIL PROTECTED] wrote:
Hope these help for alternatives to lm()? I show the use of a 2nd order
polynomial as an example to generalize a bit.
Great, thanks. If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for
Andrew Clegg:
Great, thanks. If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for that? Given
that neither of nls() or optim() provide R-squared.
You really need to *very* careful when trying to interprete R² (which can
be defined in
Andrew Clegg andrew.clegg at gmail.com writes:
... If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for that? Given
that neither of nls() or optim() provide R-squared.
To supplement Karl's comment, try Douglas Bates' (author of nls)
Hi folks,
I've looked through the list archives and online resources, but I
haven't really found an answer to this -- it's pretty basic, but I'm
(very much) not a statistician, and I just want to check that my
solution is statistically sound.
Basically, I have a data file containing two columns
