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Re: st: FW: Model SS/R-square in nl
From
Steven Samuels <[email protected]>
To
[email protected]
Subject
Re: st: FW: Model SS/R-square in nl
Date
Thu, 30 Jun 2011 18:24:09 -0400
Some people agree with you, e.g.
H. A. Gordon. Errors in Computer Packages. Least Squares Regression Through the Origin
Journal of the Royal Statistical Society. Series D (The Statistician) Vol. 30, No. 1 (Mar., 1981), pp. 23-29
But others don't, and the "error" is well-established. If you take your point of view, you have to justify an ANOVA table with the following d.f., taking p = 1 regressor.
SS d.f.
Model 1
Error n - 1
Total n - 1 ?
This problem arises because the mean-only model is _not_ nested in the no-constant model as standard LS theory requires.
You can achieve a "nesting" by fitting no mean, getting:
Model 1
Error n - 1
Total n
The main benefit to the "SST must be the same for all models" approach, I think, is that one can compare R2 consistently for the same data set as R2 = 1 - SSE/SST.
Steve
[email protected]
On Jun 30, 2011, at 4:24 PM, CJ Lan wrote:
If you look at the residual SS, i.e., sum of (yi-yhat)^2, the 1st model renders 28315 and the 2nd model renders 28427, which sounds reasonable because one parameter is eliminated. My point is the Total SS, i.e., sum of (yi-mean(y))^2, should not be changed (=39434). Therefore, in the 2nd model, the Model SS = (Total SS)-(residual SS) = 39434-28427 = 11007 and the R2 should have been 0.2791, which is the answer I got from Matlab.
The curve will not be forced through the origin. The curve of the 1st model starts at (b0+b1=45.6) and decreases at an exponential rate. Similarly the curve of the 2nd model starts at (b1=44.5) and decreases at a similar exponential rate.
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Nick Cox
Sent: Thursday, June 30, 2011 4:05 PM
To: [email protected]
Subject: Re: st: FW: Model SS/R-square in nl
No, it is not a bug.
Your constant may not be significant by itself, but the model is
different. R-squares for different models are often difficult to
compare effectively.
Plot the fitted curves and the data to see what it is going on.
In my experience, especially with nonlinear models, it is far better
to rely on physical, biological, economic or other scientific
understanding to choose the better model and to compare fitted curves
with the data, rather than to rely blindly on a significance test.
Does it make sense to force the curve through the origin?
Nick
On Thu, Jun 30, 2011 at 6:06 PM, CJ Lan <[email protected]> wrote:
> I was using nl to run a 3-parameter NLS model estimation and got R2=0.28
> (see the first output). Since the parameter b0 is insignificant, I drop
> it and re-estimate it again. This time, I got the wrong R2 (=0.86 in
> the 2nd output). It is apparent that either the "Model SS" or "Total
> SS" is wrongly calculated. Is this bug? Thank you for help.
>
> (1)
> . nl exp3 : passby A in 1/152
> (obs =152)
> Iteration 0: residual SS =3D 29741.65
> Iteration 1: residual SS =3D 28448.53
> Iteration 2: residual SS =3D 28316.37
> Iteration 3: residual SS =3D 28315.61
> Iteration 4: residual SS =3D 28315.6
> Iteration 5: residual SS =3D 28315.6
> Iteration 6: residual SS =3D 28315.6
> Iteration 7: residual SS =3D 28315.6
> Source | SS df MS Number of obs =152
> -------------+------------------------------ F( 2, 149) =29.25
> Model | 11118.3472 2 5559.1736 Prob > F =0.0000
> Residual | 28315.6009 149 190.03759 R-squared =0.2819
> -------------+------------------------------ Adj R-squared =0.2723
> Total | 39433.9482 151 261.151975 Root MSE =13.78541
> Res. dev. =1225.905
> 3-parameter asymptotic regression, passby = b0 + b1*b2^A
> ------------------------------------------------------------------------
> passby | Coef. Std. Err. t P>|t| 95% Conf.Interval]
> -------------+----------------------------------------------------------
> b0 | 11.59292 10.68695 1.08 0.280 -9.52 32.71048
> b1 | 34.10476 9.433555 3.62 0.000 15.4 52.74559
> b2 | .998132 .0011685 854.19 0.000 .995 1.000441
> ------------------------------------------------------------------------
> * Parameter b0 taken as constant term in model & ANOVA table
> (SEs, P values, CIs, and correlations are asymptotic approximations)
>
> (2)
> . nl exp2 : passby A in 1/152
> (obs =3D 152)
> Iteration 0: residual SS =3D 29510.02
> Iteration 1: residual SS =3D 28427.14
> Iteration 2: residual SS =3D 28426.97
> Iteration 3: residual SS =3D 28426.97
> Source | SS df MS Number of obs =152
> -------------+------------------------------ F( 2, 150) =468.32
> Model | 177506.602 2 88753.3012 Prob > F =0.0000
> Residual | 28426.9672 150 189.513115 R-squared =0.8620
> -------------+------------------------------ Adj R-squared =0.8601
> Total | 205933.57 152 1354.82612 Root MSE =13.76638
> Res. dev. =1226.502
> 2-parameter exp. growth curve, passby =3D b1*b2^A
> ------------------------------------------------------------------------
> passby | Coef. Std. Err. t P>|t|[95% Conf.interval]
> -------------+----------------------------------------------------------
> b1 | 44.54536 2.038308 21.85 0.000 40.51785 48.57286
> b2 | .9988862 .0001727 5783.22 0.000 .9985449 .9992275
> ------------------------------------------------------------------------
> (SEs, P values, CIs, and correlations are asymptotic approximations)
>
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