Thanks Frank and Greg, 

This makes alot more sense to me now. I appreciate you are both very busy, but i was wondering if i could trouble you for one last piece of advice. As my data is a little complicated for a first effort at R let alone modelling!

The response is on a range from 1-6, which indicates extinction risk - 1 being least concern and 6 being critical - hence using a ordinal model

The factors (6) are categorical - FRUIT TYPE - fleshy/dry
						   	 HABITAT - terrestrial, aquatic, epiphyte

etc etc 

I am asking the question - How do different combinations of factors effect extinction risk.

Based on what you have both said i have called

> predict(model1, type="fitted")

Would this be the best way predicting the probability of falling into each response category  - 


            y>=2        y>=3         y>=4         y>=5         y>=6
1    0.502220616 0.410236021 0.2892270912 0.2191420568 0.1774250519
2    0.745221699 0.668501579 0.5412223837 0.4486151612 0.3847379442
3    0.720381333 0.639796647 0.5095814746 0.4174618165 0.3551631876
4    0.752321112 0.676811675 0.5505781183 0.4579680710 0.3937100283
5    0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
6    0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
7    0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
8    0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
9    0.526291649 0.433739868 0.3094355120 0.2360800803 0.1919312111

I have 100 species for which i have their factors and i want to predict their response, so if i do the above and use the newdata function, and present the probabilities  as above rather than trying to classify them?

I  tried polr and that "classified" each response as either 1 or 6 i.e no 2,3,4,5 - as did calling predict(model1, type="fitted.ind") which resulted in the probabilities of being 1 or 6 far outweighing 2,3,4,5 (Below) - this may just be that my model is not powefull enough to discrimate effectively as i know that is incorrect ( Brier score 2.01, AUC 66.9)?

     EXTINCTION=1 EXTINCTION=2 EXTINCTION=3 EXTINCTION=4 EXTINCTION=5     EXTINCTION=6
1       0.4977794 0.0919845942  0.121008930  0.070085034 0.0417170048			0.1774250519
2       0.2547783 0.0767201200  0.127279196  0.092607223 0.0638772170			0.3847379442
3       0.2796187 0.0805846862  0.130215173  0.092119658 0.0622986289			0.3551631876
4       0.2476789 0.0755094367  0.126233557  0.092610047 0.0642580427			0.3937100283
5       0.1756117 0.0604319173  0.109577572  0.088069011 0.0654116601			0.5008981585
6       0.1756117 0.0604319173  0.109577572  0.088069011 0.0654116601			0.5008981585
7       0.1756117 0.0604319173  0.109577572  0.088069011 0.0654116601			0.5008981585
8       0.1756117 0.0604319173  0.109577572  0.088069011 0.0654116601			0.5008981585
9       0.4737084 0.0925517814  0.124304356  0.073355432 0.0441488692			0.1919312111
10      0.2489307 0.0757263892  0.126424896  0.092614323 0.0641934484			0.3921102030																				    		 
Thanks very much for any advice given,

John


10   0.751069260 0.675342871 0.5489179746 0.4563036514 0.3921102030
On 1 Oct 2010, at 23:13, Frank Harrell wrote:


Well put Greg.  The job of the statistician is to produce good estimates
(probabilities in this case).  Those cannot be translated into action
without subject-specific utility functions.  Classification during the
analysis or publication stage is not necessary.

Frank

-----
Frank Harrell
Department of Biostatistics, Vanderbilt University
-- 
View this message in context: http://r.789695.n4.nabble.com/Interpreting-the-example-given-by-Frank-Harrell-in-the-predict-lrm-Design-help-tp2883311p2951976.html
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