Why choose smaller U in Mann-Whitney U test? When performing a Mann-Whitney U hypothesis test between two samples of different sizes, why always use the lower $U$ value to compare to the critical value? Is this rule followed regardless of whether its a one-sided or two-sided test?
For reference:
$$U_1 = R_1 - \frac{n_1(n_1 + 1)}{2}$$
$$U_2 = R_2 - \frac{n_2(n_2 + 1)}{2}$$
$$U_1 +U_2 = n_1*n_2$$
 A: In practice, you typically use a computer; this isn't the twentieth century.
But back when people did use tables of critical values, the tables were for the smaller of $U_1$ and $U_2$ to save space and to save effort in checking the tables when printing them.
The tables are for a two-sided test. If you want a one-sided test, you take whichever $U_i$ you computed and check whether it is less than its expected null value, and convert the two-sided p-value accordingly
For example, suppose your one-sided alternative is that sample 1 has smaller values than sample 2. If you computed $U_1$ as 17 and the expected value $n_1n_2/2$ is 30, then the ranks in sample 1 are smaller than they would be under the null and your $p$-value will be less than 0.5 Then you can proceed as for the two-sided test to get the two-sided p-value, and halve it to get the one-sided value.  If, on the other hand, you had $U_1=43$, which is greater than the expected value, your p-value would be greater than 1/2; you would halve the two-sided p-value and subtract from 1 to get the one-sided p-value.
