From: John Fox <jfox_at_mcmaster.ca>

Date: Wed, 09 May 2007 07:10:54 -0400

R-help_at_stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 09 May 2007 - 11:14:41 GMT

Date: Wed, 09 May 2007 07:10:54 -0400

Dear Adai,

> -----Original Message-----

*> From: Adaikalavan Ramasamy [mailto:ramasamy_at_cancer.org.uk]
**> Sent: Tuesday, May 08, 2007 8:38 PM
**> To: S Ellison
**> Cc: h.wickham_at_gmail.com; jfox_at_mcmaster.ca; R-help_at_stat.math.ethz.ch
**> Subject: Re: [R] Weighted least squares
**>
**> http://en.wikipedia.org/wiki/Weighted_least_squares gives a
**> formulaic description of what you have said.
**>
**> I believe the original poster has converted something like this
**>
**> y x
**> 0 1.1
**> 0 2.2
**> 0 2.2
**> 0 2.2
**> 1 3.3
**> 1 3.3
**> 2 4.4
**> ...
**>
**> into something like the following
**>
**> y x freq
**> 0 1.1 1
**> 0 2.2 3
**> 1 3.3 2
**> 2 4.4 1
**> ...
**>
**> Now, the variance of means of each row in table above is ZERO
**> because the individual elements that comprise each row are
**> identical. Therefore your method of using inverse-variance
**> will not work here.
**>
**> Then is it valid then to use lm( y ~ x, weights=freq ) ?
*

No, because the weights argument gives inverse-variance weights not case weights.

Regards,

John

*>
*

> Regards, Adai

*>
**>
**>
**> S Ellison wrote:
**> > Hadley,
**> >
**> > You asked
**> >> .. what is the usual way to do a linear regression when you have
**> >> aggregated data?
**> >
**> > Least squares generally uses inverse variance weighting.
**> For aggregated data fitted as mean values, you just need the
**> variances for the _means_.
**> >
**> > So if you have individual means x_i and sd's s_i that arise
**> from aggregated data with n_i observations in group i, the
**> natural weighting is by inverse squared standard error of the
**> mean. The appropriate weight for x_i would then be
**> n_i/(s_i^2). In R, that's n/(s^2), as n and s would be
**> vectors with the same length as x. If all the groups had the
**> same variance, or nearly so, s is a scalar; if they have the
**> same number of observations, n is a scalar.
**> >
**> > Of course, if they have the same variance and same number
**> of observations, they all have the same weight and you
**> needn't weight them at all: see previous posting!
**> >
**> > Steve E
**> >
**> >
**> >
**> > *******************************************************************
**> > This email and any attachments are confidential. Any use,
**> > co...{{dropped}}
**> >
**> > ______________________________________________
**> > R-help_at_stat.math.ethz.ch mailing list
**> > https://stat.ethz.ch/mailman/listinfo/r-help
**> > PLEASE do read the posting guide
**> > http://www.R-project.org/posting-guide.html
**> > and provide commented, minimal, self-contained, reproducible code.
**> >
**> >
**> >
**>
*

>

R-help_at_stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 09 May 2007 - 11:14:41 GMT

Archive maintained by Robert King, hosted by
the discipline of
statistics at the
University of Newcastle,
Australia.

Archive generated by hypermail 2.2.0, at Wed 09 May 2007 - 12:31:31 GMT.

*
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help.
Please read the posting
guide before posting to the list.
*