Hi, Yes, that is basically the idea. It is defined as:
stddev <- (colSums(as.matrix(resids^2))/(n - p))^0.5 Where n is the number of rows in the residual matrix and p is the rank of the QR decomposition. I believe the reason they are slightly different is that the mean of the residuals is not necessarily exactly 0. Cheers, Josh On Sat, Jan 22, 2011 at 2:54 PM, MM <finjulh...@gmail.com> wrote: > Hello, > > Is the "std.dev" component of ls.diag( lsfit(x,y) ) the sample standard > deviation of the residuals of the fit? > > I have > ls.diag(lsfit(xx,yy))$std.dev > different from > sd(lsfit(xx,yy)$residuals) > > where xx and yy are vectors of 5 elements. > > Regards, > > ______________________________________________ > 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. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ ______________________________________________ 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.
