Why would you expect the confidence limits on the parameters of a non-linear model to be symmetrical? In general they are not. For example, if you fit the von Bertalanffy curve to data for young fish the confidence range for L-infinity can run from some finite value to infinity.
I have not been following this discussion so perhaps the following reference is off-topic, but my pragmatic approach to nonlinear parameter estimation can be seen in: Silvert, W. 1979. Practical curve fitting. Limnol. Oceanogr. 24:767-773. and the PDF can be downloaded from http://bill.silvert.org/pdf/Practical%20Curve%20Fitting.pdf. Bill Silvert ----- Original Message ----- From: "Ben Bond-Lamberty" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Monday, April 16, 2007 2:26 PM Subject: Re: calculating standard error of turnover times > Erika, > > If you use a nonlinear regression package to fit your curve to existing > data, it should compute standard errors for each parameter estimate. > Ideally, I would just report these; certainly don't take 1/k as the > "standard error of turnover time," because this is a nonlinear > transformation. Perhaps calculate > > 1/k (turnover time estimate) > 1/(k+se(k)) (turnover estimate plus error) > 1/(k-se(k)) (turnover estimate minus error) > > and report those. > > Regards, > Ben > > > On 4/13/07 5:50 PM, "Erika Marín-Spiotta" <[EMAIL PROTECTED]> wrote: > >> Hello, I have a quick question on how to convert the standard error of a >> slope (k) of an equation of the type: >> ln (y) = b + kx >> where y = fraction of mass remaining >> x = time in years, and >> where 1/k gives you a turnover time in years. >> In other words, how do you report standard error of turnover time? > >
