Question about curve fitting...
As you appeared to have received no reply ... 1) Use nls() on the original equation 2) Transforming first and using linear fitting is **NOT** the same. The error structures differ and therefore you get different results. The greatest effect is on inferences -- i.e confidence intervals for the parameters: the usual asymptotic intervals would be symmetric based on the original scale, asymmetric based on the transformed. For reasonably well-behaved data the point estimates shouldn't be too much different, however. In any case, the linearization is a good way to find starting values. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA "The business of the statistician is to catalyze the scientific learning process." - George E. P. Box
-----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of S.O. Nyangoma Sent: Wednesday, August 10, 2005 3:22 AM To: Dan Bolser Cc: R mailing list Subject: Re: [R] Question about curve fitting... I see that log(y)=log(k1)+k2*log(x) use lm? ----- Original Message ----- From: Dan Bolser <dmb at mrc-dunn.cam.ac.uk> Date: Wednesday, August 10, 2005 11:41 am Subject: [R] Question about curve fitting...
Meta: This question is somewhat long and has two parts, I would be very happyfor someone just to nudge me in the right direction with the manual / tutorial, as I am somewhat lost... 1) How do I fit a curve of the form "y = k1 * x^k2" ? I want to estimate values of k1 and k2 given the x/y data I have, and I can't work out how to get R to calculate and return their estimates. 2) Given the value of k1 and k2 for population A, how can I test if population B has significantly different values of k1 and k2? Sorry for the basic question. I think I just need to read
up on a few
functions. I have about 50 xy pairs in total if that makes a difference. Dan.
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