If I have two groups to compare. One group have 10 persons, and another group have 5 persons. Then some event happened for those persons, and the event may happen more than once. If I have 12 events happened for 10 persons group, and 0 event happened for 5 person group.  How can I compare the chance which occurring in those two groups? The problem is I cannot use the exact test because 12 exceeds 10, and I cannot have a negative value for one cell.  I don't know if proportion test is feasible, because one of the proportion exceeds 100% while the other is zero. And my data is the following.  a_event is the event occurring in a group, a_total is the total person in a group; the same for b group... data test; input a_event b_event a_total b_total; cards; 12 0 10 5 ; ... View more

Hi, I have a question about a comparison of two groups of data, which are small count.  The following are the code.  However, when I fit the model, the "contrast" p-value cannot be calculated.  And SAS always give me the warning: WARNING: The negative of the Hessian is not positive definite. The convergence is questionable. WARNING: The procedure is continuing but the validity of the model fit is questionable. WARNING: The specified model did not converge. WARNING: Negative of Hessian not positive definite. Is there any problem with my code?  The following is the data and the code: data test; input diet total; cards; 1 1 1 0 1 1 1 1 1 2 1 0 1 0 1 4 1 1 1 1 1 1 1 3 1 1 1 3 1 0 1 3 1 1 1 3 1 1 2 3 2 2 2 2 2 2 2 4 2 1 2 3 2 4 2 1 2 3 2 0 2 5 2 2 2 6 2 2 2 1 2 2 2 2 2 2 2 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 4 4 4 5 4 4 4 3 4 4 4 7 4 4 4 3 4 2 4 4 4 1 4 2 4 2 4 3 4 4 4 0 4 2 4 5 ; proc genmod data=test; class diet; model total=diet / dist=poisson link=log; contrast 'group, 1 vs. 2' diet 1 -1 0 0; contrast 'group, 1 vs. 3' diet 1 0 -1 0; contrast 'group, 1 vs. 4' diet 1 0 0 -1; run; ... View more

Hi, I have the data like the following. disease=1 means having the disease; disease=0 means no disease.  Can I draw the cumulative risk graph?  The x-axis is the response (continuous variable), y-axis should be the cumulative risk (probability??) of having the disease. How can I do it?  Thanks! data test; input response disease; cards; 45.6   1 190     0 34.5    1 89       0 56       1 45       1 58       0 ....... ........ ; ... View more