> Perhaps a coding error on my part (or on your part). Perhaps different > methods (none of which you describe)?
> > I suspect that my method only used the first two points (I just > checked by plotting and -2.7 is closer to the paper and pen result I > get than is -3.28. Perhaps you made an extrapolation from a linear fit > of a dataset that is not co-linear? > > > lm(c(0,5) ~ c(16,45)) > > Call: > lm(formula = c(0, 5) ~ c(16, 45)) > > Coefficients: > (Intercept) c(16, 45) > -2.7586 0.1724 > > It not that "R is different", it is merely that I used it differently > than you used your other tools. > > Here's another method ( using all points and again reversing the roles > of x and y) : > > lm(c(0,5,10,15,20) ~ c(16,45,77,101,125)) > > Call: > lm(formula = c(0, 5, 10, 15, 20) ~ c(16, 45, 77, 101, 125)) > > Coefficients: > (Intercept) c(16, 45, 77, 101, 125) > -3.2332 0.1818 My understanding from gnuplot manual is that a marquart-levenberg algorithm is used, which I applied to the data to perform a least squares best fit linear curve. Gnuplot returns values for the intercept and gradient which I then apply to solve the linear equation y=mx+c. Similarly with scilab, where the regress(ion?) function was applied. Qtiplot performed non-weighted linear regression to output values similar to those from gnuplot. Why reverse the roles of x and y in your method? Although your revised value is closer to those from other programs, how do I understand and explain the discrepancy? ______________________________________________ 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.
