On Thu, Jun 24, 2010 at 9:20 AM, Viechtbauer Wolfgang (STAT) <wolfgang.viechtba...@stat.unimaas.nl> wrote: > The weights in 'aa' are the inverse standard deviations. But you want to use > the inverse variances as the weights: > > aa <- (attributes(summary(f1)$modelStruct$varStruct)$weights)^2 > > And then the results are essentially identical. >
We might now ask how we might have found Wolfgang's answer via calculation. Lets redo the gls calculation of variance from scratch by iterated re-weighted least squares (just one iteration here) and compare that to the gls aa calculated by the original poster: # estimate beta fm <- lm(Petal.Width ~ Species / Petal.Length, iris) # estimate variance v <- fitted(lm(resid(fm)^2 ~ Species, iris)) v <- v/v[1] # compare to aa from original poster lm(log(aa) ~ log(v)) The last line gives: (Intercept) log(v) -4.212e-07 -5.000e-01 which suggsts: aa = 1/sqrt(variance) ______________________________________________ R-help@r-project.org 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.
