All of the examples below use an equal number of invidivuals for sets M and W. They could differ, it doesn't matter much for the point I'm making.
“All M>W” – easily disproven
“All W>M” – easily disproven
“M and W are equal” (equality hypothesis predicts equal mean and variance)
M{2,4,6,8,10}, m=6, sd=3.16
W{2,4,6,8,10}, m=6, sd=3.16
To my knowledge, there is no evidence supporting this hypothesis.
“More M than W at the top and at the bottom” (‘pure’ variance hypothesis according to Saxmaan)
Equal n of M and W, most M stronger than W, but on average M=W, higher variance for M
M{2,2,8,8,9}, m=5.8, sd=3.5
W{3,3,6,8,9}, m=5.8, sd=2.8
If you want to argue that we see more men at the top because of higher variance, you have to assume (and demonstrate to others) that men and women perform similarly on average.
Why? Because the following is also possible:
“Most M at the top, most W at the bottom” (‘pure’ variance hypothesis counterexample)
Equal n of M and W, most M stronger than W, on average M>W, higher variance for W
M{3,5,7,9,10}, m=6.8, sd=2.86
W{2,3,4,8,10}, m=5.4, sd=3.44
“W are stronger than M (on average) but the top 10 are men so it won’t ever show”
M{1,1,1,1,1,1,1,1,1,1,6,6,7,7,8,8,9,9,10,10}, m=4.33, sd=3.64
W{4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5}, m=4.5, sd=0.51
For this argument to hold, you would have to show that M and W sample from extremely skewed distributions. Not supported by the evidence at my disposal.
“M=W (on average) but the top 10% are men so it won’t ever show”
M{2,2,4,4,6,6,8,8,10,10}, m=6, sd=2.98
W{3,3,4,4,4,8,8,8,9,9}, m=6, sd=2.58
Seems rather easy to show, if it was true.