Mann–Whitney U Test

Session lengths and order values are heavily skewed, so a t-test's normality assumption is shaky. The Mann–Whitney U test compares two groups using only ranks.

1. Pool both samples and rank them from 1, giving tied values the average of their ranks.
2. $R_a$ = the sum of the ranks belonging to sample a, and $U_a = R_a - \frac{n_a(n_a + 1)}{2}$.
3. For large samples, $z = \frac{U_a - \mu_U}{\sigma_U}$ with $\mu_U = \frac{n_a n_b}{2}$ and $\sigma_U = \sqrt{\frac{n_a n_b (n_a + n_b + 1)}{12}}$ (ignore the tie correction).

Implement rankdata(values) (average ranks for ties), and mann_whitney_u(a, b) returning (U_a, z). Also implement prob_superiority(a, b): $U_a / (n_a n_b)$, the probability that a random value from a beats a random value from b (ties count half). It's the same quantity as ROC AUC.