No failures in reference group when running a competing risk analysis.

[Bob Galbreath]

I have 268 prostate cancer patients and have placed them into four risk groups (low, favorable intermediate, unfavorable intermediate, and high; coded as 1, 2, 3, and 4, respectively). I want to compare the biochemical failure rates of the four groups using a competing risk regression with a competing risk being death. The analysis resulted in extremely high SHR values (3027318, 3674970 and 1.19e+07). All had P>|z| = 0.000. When I looked at the number of failures in the low-risk group, there was none. All of the other groups had failures (1, 4 and 7). Is it acceptable to report such high values for SHR or is there some other way I should analysis this data.

Thanks you,
Bob Galbreath

[Clyde Schechter]

Well, given that there are no failure in the reference group, I strongly suspect that if you look at the confidence intervals around those SHR values, they are very wide, reflecting the fact that the denominator is not very well determined. And, of course, while it is not uncommon to see people report central estimates such as SHRs along with p-values and no standard errors or confidence intervals, it is terrible statistical practice. I think if you look at the confidence intervals, you will probably be able to convince yourself that these SHRs are simply not meaningful or useful in this circumstance. Similarly, if you report the total number of failures in each group, along with the size of each group, it will probably, again, become clear that there just isn't enough information in this data set to reach clear-cut conclusions. If you are planning to write this up as the results of a pilot study or something like that, these kinds of results would be OK. If, however, you are planning to write this up as a validation of the risk-group classification, I don't think its results can be taken very seriously.

You might consider reducing the risk group classification to a dichotomy, high vs non-high, so that the numbers of failure would be 7 and 5, respectively, nearly equal but probably with rather different denominators. These results will likely be a bit more useful than one you have. But frankly, with only a total of 12 failures, I doubt this or any other maneuver will make more than a modest improvement.