[R] non-linear plot parameters

[Bert Gunter]

My opinions only below; consume at your own risk.

On Thu, Aug 26, 2010 at 2:20 PM, Marlin Keith Cox <marlinkcox at 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 <dwinsemius at 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
>>
>>
>
>

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-- 
Bert Gunter
Genentech Nonclinical Biostatistics
467-7374
http://devo.gene.com/groups/devo/depts/ncb/home.shtml