On Aug 26, 2010, at 5:20 PM, Marlin Keith Cox 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.
>
> The data originally posted was an example of one of the curves
> experienced.
You might be interest in what happens when you expand the plot a bit
using the fm1 model:
plot(Level~Time, xlim=c(0,40), ylim=c(0,max(Level)) )
lines(0:40, predict(fm1, data.frame(Time=0:40 ) ) , col="red", pch=4)
abline(h=0)
And not forgetting that extrapolation beyond the limits of data is a
dangerous maneuver. In this instance any Weibull model would predict
some sort of decay to 0, which may or may not be scientifically
plausible.
--
David.
>
> 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
>
>
>
>
> --
> M. Keith Cox, Ph.D.
> Alaska NOAA Fisheries, National Marine Fisheries Service
> Auke Bay Laboratories
> 17109 Pt. Lena Loop Rd.
> Juneau, AK 99801
> keith....@noaa.gov
> marlink...@gmail.com
> U.S. (907) 789-6603
David Winsemius, MD
West Hartford, CT
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