Graham is mostly right, but slightly incorrect. In particular, suppose group A has the distribution f(x) and B has the distribution g(x).
If f(x) and g(x) are shaped significantly differently past the cutoff, then mean(H(x-C)f(x)) and mean(H(x-c)g(x)) might not agree even though there is no bias by construction. (Here H(x) is a step function and C the cutoff).
However, there is an easy fix: compute the minima of the support of the distribution rather than mean. min(H(x-C)f(x)) = min(H(x-C)g(x)) = C.
In practice, measure the weakest male and weakest female to be accepted in your sample set, or some similar approximation.
I'm pretty sure this is a valid frequentist hypothesis test. I've got half a proof worked out on paper already. It depends very weakly (and non-parametrically) on f(x) and g(x), but it works in basically the exact way Graham wants it to. Every counterexample I can think of is really pathological. My next blog post will probably be a proof of this.
All this negativity is really an overreaction. I know it's fun to totally debunk someone on details, but these are mostly fixable details.