Hi Ken, Thx for that advice. I took a brief look at it. I already have my curve by just using the curve() function using the parameters a and b given by the nls. Would se.fit and interval have computed the CI?
Maybe where I'm confused is at how I can break up my curve into pieces of linear regressions. Then doing CI's from there? Thanks. Julian -- View this message in context: http://r.789695.n4.nabble.com/breakpoints-and-nonlinear-regression-tp4303629p4303763.html Sent from the R help mailing list archive at Nabble.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.
