dear Julian,
Il 18/01/2012 14.36, crimsonengineer87 ha scritto:
Thanks for the comments. Yes, I also had segmented and then I went away from
that. I can't remember. I've tried using it but I get some sort of strange
error. Here's some code ...
it is difficult for me to help you without knowin
Thanks for the comments. Yes, I also had segmented and then I went away from
that. I can't remember. I've tried using it but I get some sort of strange
error. Here's some code ...
pavlu.glm <- glm(Na ~ yield, data=pavludata, family=gaussian)
pavlu.seg <- segmented(pavlu.glm, seg.Z=~yield, psi=100
On Tue, 17 Jan 2012, crimsonengineer87 wrote:
Thanks for the comments everyone. I was hoping to not have to find someone in
the stats department ... well, we'll see.
So in response to Z's comment ... I have tried breakpoints(Na ~ yield)
and I did expect to get something continuous.
You won't
In respect of fitting piecewise linear regressions, have you looked at
the "segmented" package?
cheers,
Rolf Turner
On 18/01/12 04:30, crimsonengineer87 wrote:
Dear Forum,
I have been wracking my head over this problem for the past few days. I have
a dataset of (x,y). I have been
Thanks for the comments everyone. I was hoping to not have to find someone in
the stats department ... well, we'll see.
So in response to Z's comment ... I have tried breakpoints(Na ~ yield) and I
did expect to get something continuous. The idea was to get two or three
linear functions making up t
On Tue, 17 Jan 2012, Bert Gunter wrote:
On Tue, Jan 17, 2012 at 8:06 AM, Kenneth Frost wrote:
Sorry, that wasn't to helpful...I see that the intervals and se.fit argument
are currently ignored.
Yes, because the fitted values are nonlinear in the parameters, which
makes finding exact confide
On Tue, 17 Jan 2012, crimsonengineer87 wrote:
Dear Forum,
I have been wracking my head over this problem for the past few days. I have
a dataset of (x,y). I have been able to obtain a nonlinear regression line
using nls. However, we would like to do some statistical analysis. I would
like to ob
On Tue, Jan 17, 2012 at 8:06 AM, Kenneth Frost wrote:
> Sorry, that wasn't to helpful...I see that the intervals and se.fit argument
> are currently ignored.
Yes, because the fitted values are nonlinear in the parameters, which
makes finding exact confidence regions impossible. I think the "usua
Hi Ken,
Thx for that advice. I took a brief look at it. I already have my curve by
just using the curve() function using the parameters a and b given by the
nls. Would se.fit and interval have computed the CI?
Maybe where I'm confused is at how I can break up my curve into pieces of
linear regres
Sorry, that wasn't to helpful...I see that the intervals and se.fit argument
are currently ignored.
On 01/17/12, crimsonengineer87 wrote:
> Dear Forum,
>
> I have been wracking my head over this problem for the past few days. I have
> a dataset of (x,y). I have been able to obtain a nonlinear
Hi, Julian-
I'm not sure if this will be what you want but you could start by taking a look
at:
?predict.nls
Ken
On 01/17/12, crimsonengineer87 wrote:
> Dear Forum,
>
> I have been wracking my head over this problem for the past few days. I have
> a dataset of (x,y). I have been able to obtain
Dear Forum,
I have been wracking my head over this problem for the past few days. I have
a dataset of (x,y). I have been able to obtain a nonlinear regression line
using nls. However, we would like to do some statistical analysis. I would
like to obtain a confidence interval for the curve. We thou
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