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
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