About StatDave

StatDave · ‎07-03-2017

Statistical tests become more powerful at detecting small effects as the sample size increases. So, it is possible that these tests are detecting trivially small departures from fit for your practical purposes. You might want to assess how well the model does at the observation level by seeing how well it classifies observations. The graphical method you used, based on this note, is also a good way to assess the proportional odds assumption without using a test.

StatDave · ‎07-03-2017

If you pick a cutpoint on the predicted probabilities (such as the cutpoint that gave the sensitivity you mentioned), then you can use the value of that cutpoint to classify the observations as predicted events or nonevents. You can then use PROC FREQ to show the table of predicted by actual events and nonevents and get the statistics you want, along with confidence intervals, as shown in this note.

StatDave · ‎06-30-2017

Sounds like you simply need to change your predictors from binary variables, to continuous predictors that are estimates of the total light exposure and total flood (and/or ebb) exposure.

StatDave · ‎06-30-2017

Yes. GAMPL has the same DIST= option as GAM. See the GAMPL documentation for your release of SAS (9.4 TS1M3 | 9.4 TS1M4).

StatDave · ‎06-30-2017

A negative binomial or ZINB model is for a count response. A risk ratio is a ratio of event probabilities, so that assumes that the response is binary, not a count. Assuming your response is really a count, then you might want to estimate a RATE ratio. This note discusses estimating rates and rate ratios with a count model. This note talks about estimating rates using zero-inflated count models.

StatDave · ‎06-30-2017

If your data consist of a count of events and a count of total trials, then the proper syntax is the following to fit the model which is a logistic model for this binomial response. You can use either LOGISTIC or GENMOD with the same syntax. proc logistic data=DTI_mod; class grade DTI; model bad_accs/total_accs = Grade DTI grade*DTI; run;

StatDave · ‎06-30-2017

If you want an ordered logit, as you said, then you don't want the LINK=GLOGIT option. Remove it to get an ordinal logit model. See this note for examples.

StatDave · ‎05-11-2017

The generalized logit model you fit originaly is: l0=log(p0/p2)=int0+lang01+lang02 l1=log(p1/p2)=int1+lang11+lang12 You now fit the model l2=log(p2/(p0+p1))=int+lang1+lang2. When lang=1, under that first model: l0=int0+lang01 l1=int1+lang11 And in the second model: l2=int+lang1 Algebraically, I don't see that -(lang01+lang11) as you want to do, gets you to the lang1 parameter in the second model.

StatDave · ‎05-11-2017

The CONTRAST statements are meaningless since Drug is involved in interaction and so, by definition, its effect depends on the level of the interacting variable, Age. Your ODDSRATIO statement for Drug gives you estimates of the odds ratios comparing Drug levels at the mean of Age. If you want to make the Drug comparisons at a different level of Age, use the AT option in the ODDSRATIO statement.

StatDave · ‎05-11-2017

Sounds like you want to estimate so-called marginal effects which are the derivative of the event probability with respect to a predictor of interest. See this note.

StatDave · ‎05-11-2017

By including other potential predictors of Donor in the model, along with your BEC predictor, the resulting parameter estimate and odds ratio estimate for BEC tells you the effect of BEC after adjusting for the effects of the other predictors. If some of the other predictors in the model are associated with the Donor response, this gives you a better idea of what the separate effect of BEC is after accounting for those other predictors. As Rick says, the effect of BEC will obviously change if you change the other predictors that are being adjusted for by adding or deleting predictors from the model.

StatDave · ‎05-11-2017

That implies that you want to model a logit defined as log[ Pr(Edu3=2)/(1-Pr(Edu3=2) ]. To do this, simply create a new response variable: new=(edu3=2); and fit a binary logistic model using it as the response.

StatDave · ‎05-11-2017

Assuming that your response is binary (eligible vs not), you would be fitting a logistic model. You can use PROC LOGISTIC to fit the model and the SCORE statement in that procedure to get predicted probabilities of eligibility for you test data set. See the LOGISTIC documentation and examples there such as the "Scoring data sets" example. Basic code would be like: proc logistic data=train; model binary_response(event="Y") = predictor1 predictor2; score data=test out=mypreds; run;

StatDave · ‎05-11-2017

See the ROCPLOT macro. Specify the SCORE OUTROC= data set in the INROC= macro option, and the SCORE OUT= data set and its predicted probabilities in the macro's INPRED= and P= options. See the macro documentation for information on the various optimality criteria you can use.

StatDave · ‎04-26-2017

Assuming you are talking abou the area under the ROC curve being those amounts above the reference line (where AUC=0.5), you can do pretty well by using the following data set from one of the examples in the PROC LOGISTIC documentation and, as Rick suggests, using various models as shown: data Neuralgia; input Treatment $ Sex $ Age Duration Pain $ @@; datalines; Placebo Female 68 1 No TrtA Female 69 12 No Placebo Male 66 26 Yes TrtB Male 67 23 No TrtA Female 71 12 No TrtB Male 77 1 Yes TrtA Male 71 17 Yes Placebo Female 65 29 No TrtB Female 66 12 No TrtB Male 75 21 Yes TrtA Female 64 17 No Placebo Female 70 13 Yes Placebo Male 70 1 Yes Placebo Female 68 27 Yes TrtA Female 64 30 No TrtB Male 70 22 No TrtB Female 78 1 No TrtA Male 67 10 No TrtB Male 75 30 Yes TrtB Male 80 21 Yes TrtA Male 70 12 No Placebo Female 67 30 No TrtB Male 70 1 No TrtB Female 77 16 No Placebo Male 78 12 Yes TrtB Female 76 9 Yes Placebo Male 66 4 Yes TrtA Female 69 18 Yes TrtA Male 78 15 Yes Placebo Female 64 1 Yes Placebo Female 72 27 No TrtA Female 72 25 No TrtB Female 65 7 No TrtB Male 59 29 No Placebo Male 67 17 Yes TrtA Male 69 1 No Placebo Female 67 1 Yes TrtB Female 69 42 No TrtA Female 74 1 No Placebo Female 79 20 Yes TrtB Male 74 16 No TrtB Female 65 14 No TrtB Female 67 28 No TrtA Male 76 25 Yes TrtB Female 72 50 No TrtB Female 69 24 No TrtA Female 63 27 No Placebo Male 60 26 Yes TrtA Male 62 42 No TrtA Female 67 11 No Placebo Male 74 4 No TrtA Male 75 6 Yes TrtB Male 66 19 No Placebo Male 68 11 Yes TrtA Male 70 28 No TrtA Male 65 15 No Placebo Male 83 1 Yes Placebo Female 72 11 Yes Placebo Male 77 29 Yes TrtA Female 69 3 No ; ods select roccurve(persist); proc logistic data=Neuralgia plots(only)=roc; class Treatment Sex; model Pain(event="No") = Treatment Sex Treatment*Sex Age Duration ; run; proc logistic data=Neuralgia plots(only)=roc; class Treatment Sex; model Pain(event="No") = Treatment Sex Treatment*Sex ; run; proc logistic data=Neuralgia plots(only)=roc; class Treatment Sex; model Pain(event="No") = Treatment ; run; proc logistic data=Neuralgia plots(only)=roc; class Treatment Sex; model Pain(event="No") = Sex ; run; ods select all;

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