Date: Wed, 28 Dec 2011 07:15:19 -0800
Reply-To: David Marso <david.marso@gmail.com>
Sender: "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From: David Marso <david.marso@gmail.com>
Subject: Re: Fitting logistic curve to growth data
In-Reply-To: <1325070409338-5105041.post@n5.nabble.com>
Content-Type: text/plain; charset=UTF-8
See the NLR (Non Linear Regression) Procedure.
Following is an example of simulated data and the set up and results of NLR.
Note that the parameter estimates used to generate the data are recovered
quite nicely!
You will wish to consult the docs on how to set up and run NLR from the
dialog boxes.
HTH, David
---
INPUT PROGRAM.
COMPUTE #K=1.5 .
COMPUTE #A=4.0.
COMPUTE #i= 1.7 .
LOOP ID=1 TO 1000.
COMPUTE d= Uniform(1)+4.
COMPUTE Y=#A/(1+EXP(-#K * (d-#i)))+ UNIFORM(.01).
END CASE.
END LOOP.
END FILE.
END INPUT PROGRAM.
EXE.
GRAPH
/SCATTERPLOT(BIVAR)=d WITH y
/MISSING=LISTWISE .
* NonLinear Regression.
MODEL PROGRAM A=3 K=1 I=1 .
COMPUTE PRED_ = A/(1+EXP(-K * (d-I))).
NLR y
/PRED PRED_
/CRITERIA SSCONVERGENCE 1E-8 PCON 1E-8 .
All the derivatives will be calculated numerically.
Iteration Residual SS A K I
1 1064.692791 3.00000000 1.00000000 1.00000000
1.1 1.348304567 3.99063649 1.49442721 1.88384875
2 1.348304567 3.99063649 1.49442721 1.88384875
2.1 .0146986193 4.00583423 1.47908955 1.69332655
3 .0146986193 4.00583423 1.47908955 1.69332655
3.1 .0080481338 4.00572338 1.47960600 1.66912002
4 .0080481338 4.00572338 1.47960600 1.66912002
4.1 .0080459976 4.00572123 1.47971278 1.66884117
5 .0080459976 4.00572123 1.47971278 1.66884117
5.1 .0080459976 4.00572119 1.47971372 1.66884238
Run stopped after 10 model evaluations and 5 derivative evaluations.
Iterations have been stopped because the relative reduction between
successive
residual sums of squares is at most SSCON = 1.000E-08
Nonlinear Regression Summary Statistics Dependent Variable Y
Source DF Sum of Squares Mean Square
Regression 3 15524.41480 5174.80493
Residual 997 8.045998E-03 8.070208E-06
Uncorrected Total 1000 15524.42285
(Corrected Total) 999 .74860
R squared = 1 - Residual SS / Corrected SS = .98925
Asymptotic 95 %
Asymptotic Confidence Interval
Parameter Estimate Std. Error Lower Upper
A 4.005721193 .001171063 4.003423162 4.008019224
K 1.479713721 .026491205 1.427728805 1.531698638
I 1.668842378 .036748380 1.596729333 1.740955424
Asymptotic Correlation Matrix of the Parameter Estimates
A K I
A 1.0000 -.9791 -.9656
K -.9791 1.0000 .9981
I -.9656 .9981 1.0000
DP_Sydney wrote
>
> Hi all,
>
> I need to fit a curve to data on mass gain, which follows a sigmoidal
> pattern and is best described using a logistic equation (based on similar
> published data, with which I wish to compare my data) of the form:
>
> Y = A/(1 + e^-K(d-i))
> where Y = mass
> A = asymptotic mass
> K = growth rate
> d = age (time)
> i = inflection point
> (note ^ indicates to the power)
>
> The data I have are body mass recorded daily for several individuals
> (nestling birds), which will be analysed separately.
>
> I need to output the parameters (A, K, i) rather than just produce a
> curve. I've searched the archives and Google but haven't found
> instructions for how to do this (I have instructions for running this in
> GENSTAT but of course do not have GENSTAT).
>
> Any help would be great...but beware I am a drop-down menus and buttons
> person when it comes to software :-)
>
> Regards,
> Dean
>
--
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