On Thu, 14 Aug 2003 09:08:26 -0700, Spencer Graves
[EMAIL PROTECTED] wrote :
This seems to identify a possible bug in R 1.7.1 under Windows 2000:
tstDf - data.frame(y = 1:11, x=1:11)
fit - nls(y~a/x, data=tstDf, start=list(a=1))
predict(fit, se.fit=TRUE)
[1] 7.0601879 3.5300939 2.3533960
On Thu, 14 Aug 2003, Spencer Graves wrote:
This seems to identify a possible bug in R 1.7.1 under Windows 2000:
tstDf - data.frame(y = 1:11, x=1:11)
fit - nls(y~a/x, data=tstDf, start=list(a=1))
predict(fit, se.fit=TRUE)
[1] 7.0601879 3.5300939 2.3533960 1.7650470 1.4120376
the approximation will work.
Hope this helps,
david paul
-Original Message-
From: Enrique Portilla [mailto:[EMAIL PROTECTED]
Sent: Thursday, August 14, 2003 9:28 AM
To: [EMAIL PROTECTED]
Subject: [R] nls confidence intervals
Hi,
Does anyone know how to compute the confidence prediction
This seems to identify a possible bug in R 1.7.1 under Windows 2000:
tstDf - data.frame(y = 1:11, x=1:11)
fit - nls(y~a/x, data=tstDf, start=list(a=1))
predict(fit, se.fit=TRUE)
[1] 7.0601879 3.5300939 2.3533960 1.7650470 1.4120376 1.1766980 1.0085983
[8] 0.8825235 0.7844653 0.7060188
[mailto:[EMAIL PROTECTED]
Sent: Thursday, August 14, 2003 9:28 AM
To: [EMAIL PROTECTED]
Subject: [R] nls confidence intervals
Hi,
Does anyone know how to compute the confidence prediction intervals for a
nonlinear least squares models (nls)?
I was trying to use the function 'predict' as I
On Thu, 14 Aug 2003 12:43:25 -0400, Duncan Murdoch [EMAIL PROTECTED]
wrote :
Perhaps the description below of what se.fit is supposed to do should
be modified.
I've done that now in the development version (to become 1.8.0).
Err, I mean in the patch version (but it should still end up in
On Thu, 14 Aug 2003 12:43:25 -0400, Duncan Murdoch [EMAIL PROTECTED]
wrote :
Perhaps the description below of what se.fit is supposed to do should
be modified.
I've done that now in the development version (to become 1.8.0).
Duncan Murdoch
__
[EMAIL
