On 2011-04-01 05:44, stephen sefick wrote:
Setting Z=Q-A would be the incorrect dimensions. I could Z=Q/A. Is
fitting a nls model the same as fitting an ols? These data are
hydraulic data from ~47 sites. To access predictive ability I am
removing one site fitting a new model and then accessing the fit with
a myriad of model assessment criteria. I should get the same answer
with ols vs nls? Thank you for all of your help.
No, ols and nls won't give the same result.
If you use ols on the logged data, you're assuming
additive errors on the log scale. With nls, you
assume additive errors on the original scale.
But your model looks simple enough - why not run
it through both functions and see what the difference is.
Ultimately, everything depends on what assumptions
you're comfortable with.
Peter Ehlers
Stephen
On Thu, Mar 31, 2011 at 8:34 PM, Steven McKinney<smckin...@bccrc.ca> wrote:
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of stephen sefick
Sent: March-31-11 3:38 PM
To: R help
Subject: [R] Linear Model with curve fitting parameter?
I have a model Q=K*A*(R^r)*(S^s)
A, R, and S are data I have and K is a curve fitting parameter. I
have linearized as
log(Q)=log(K)+log(A)+r*log(R)+s*log(S)
I have taken the log of the data that I have and this is the model
formula without the K part
lm(Q~offset(A)+R+S, data=x)
What is the formula that I should use?
Let Z = Q - A for your logged data.
Fitting lm(Z ~ R + S, data = x) should yield
intercept parameter estimate = estimate for log(K)
R coefficient parameter estimate = estimate for r
S coefficient parameter estimate = estimate for s
Steven McKinney
Statistician
Molecular Oncology and Breast Cancer Program
British Columbia Cancer Research Centre
Thanks for all of your help. I can provide a subset of data if necessary.
--
Stephen Sefick
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