My opinions only below; consume at your own risk. On Thu, Aug 26, 2010 at 2:20 PM, Marlin Keith Cox <marlink...@gmail.com> wrote: > The background you requested are energetic level (joules) in a group of > starved fish over a time period of 45 days. Weekly, fish (n=5) were removed > killed and measured for energy. This was done at three temperatures. I am > comparing the rates at which the fish consume stored body energy at each of > the three temperatures. Initial data looks like the colder fish > have different rates (as would be expected) than do warmer fish. In all > cases the slope is greatest at the beginning of the curve and flattens after > several weeks. This is what is interesting - where in time the line > starts to flatten out. > > By calculating a non-linear equation of a line, I was hoping to use the > first and second derivatives of the function to compare and explain > differences between the three temperature.
Bad idea. Derivatives from fitted curves are generally pretty imprecisely determined. And you don't need them: If the curves are being (adequately/appropriately) parameterized as Weibull, then all the information is in the parameters anyway, which can be directly modeled, fitted, and compared as functions of temperature -- provided that the design permits this (i.e. provides sufficient precision for the characterizations/comparisons). If you don't know how to do this, seek further statistical help. -- Bert Gunter Genentech Nonclinical Statistics > > The data originally posted was an example of one of the curves experienced. > > kc > > On Thu, Aug 26, 2010 at 9:48 AM, David Winsemius > <dwinsem...@comcast.net>wrote: > >> >> On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote: >> >> I need the parameters estimated for a non-linear equation, an example of >>> the >>> data is below. >>> >>> >>> # rm(list=ls()) I really wish people would add comments to destructive >>> pieces of code. >>> >> >> Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, >>> 4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8) >>> Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40, 41, 50, >>> 47, >>> 44, 44, 42, 38, 40, 37, 37, 35, 40, 34, 32, 20, 22, 25, 27, >>> 29) >>> plot(Time,Level,pch=16) >>> >> >> You did not say what sort of "non-linear equation" would best suit, nor did >> you offer any background regarding the domain of study. There must be many >> ways to do this. After looking at the data, a first pass looks like this: >> >> > lm(log(Level) ~Time ) >> >> Call: >> lm(formula = log(Level) ~ Time) >> >> Coefficients: >> (Intercept) Time >> 4.4294 -0.1673 >> >> > exp(4.4294) >> [1] 83.88107 >> > points(unique(Time), exp(4.4294 -unique(Time)*0.1673), col="red", pch=4) >> >> Maybe a Weibull model would be more appropriate. >> >> >> -- >> >> David Winsemius, MD >> West Hartford, CT >> >> > > > [[alternative HTML version deleted]] > > ______________________________________________ > 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. > -- Bert Gunter Genentech Nonclinical Biostatistics 467-7374 http://devo.gene.com/groups/devo/depts/ncb/home.shtml ______________________________________________ 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.