David Winsemius comcast.net> writes:
>
> It's really not that difficult to get the variance covariance matrix.
> What is not so clear is why you think differential weighting of a set
> that has a perfect fit should give meaningfully different results than
> a fit that has no weights.
Aga
Peter Dalgaard biostat.ku.dk> writes:
>
> The point is that R (as well as almost all other mainstream statistical
> software) assumes that a "weight" means that the variance of the
> corresponding observation is the general variance divided by the weight
> factor. The general variance is still
Mauricio O Calvao wrote:
Unfortunately you eschewed answering objectively any of my questions; I insist
they do make sense. Don't mention the data are perfect; this does not help to
make any progress in understanding the choice of convenient summary info the lm
method provides, as compared to wh
On Nov 14, 2009, at 1:50 PM, Mauricio O Calvao wrote:
David Winsemius comcast.net> writes:
Which means those x, y, and "error" figures did not come from an
experiment, but rather from theory???
The fact is I am trying to compare the results of:
(1) lm under R and
(2) the Java applet at
David Winsemius comcast.net> writes:
>
> Which means those x, y, and "error" figures did not come from an
> experiment, but rather from theory???
>
The fact is I am trying to compare the results of:
(1) lm under R and
(2) the Java applet at http://omnis.if.ufrj.br/~carlos/applets/reta/reta
On Wed, 11 Nov 2009, David Winsemius wrote:
On Nov 11, 2009, at 7:45 PM, Mauricio Calvao wrote:
When I try:
fit_mod <- lm(y~x,weights=1/error^2)
I get
Warning message:
In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
extra arguments weigths are just disregarded.
(Ac
On Nov 11, 2009, at 7:45 PM, Mauricio Calvao wrote:
Hi there
Sorry for what may be a naive or dumb question.
I have the following data:
> x <- c(1,2,3,4) # predictor vector
> y <- c(2,4,6,8) # response vector. Notice that it is an exact,
perfect straight line through the origin and slope
Hi there
Sorry for what may be a naive or dumb question.
I have the following data:
> x <- c(1,2,3,4) # predictor vector
> y <- c(2,4,6,8) # response vector. Notice that it is an exact,
perfect straight line through the origin and slope equal to 2
> error <- c(0.3,0.3,0.3,0.3) # I have (equ
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