Permutation p-Value for Difference in Medians

Problem

Run a two-sided permutation test on the difference in medians between x and y using n_perm label shuffles driven by the seeded Lehmer RNG. Return (count + 1)/(n_perm + 1), where count is the number of permuted absolute median differences at least as large as the observed one; return 1.0 for empty input or n_perm <= 0. Unlike the mean version, this statistic is robust to outliers.

Hint

Shuffle labels with Fisher-Yates, then recompute the median difference.

Test cases

case 1

in: [[1,2,3,4,5],[6,7,8,9,10],100,42]

expected: 0.09900990099009901

case 2

in: [[1,2,3,4,5],[3,4,5,6,7],100,7]

expected: 0.32673267326732675

case 3

in: [[5,5,5],[1,2,3],50,3]

expected: 0.3137254901960784

case 4

in: [[],[1],10,1]

expected: 1

case 5

in: [[1],[2],0,1]

expected: 1

Starter codePython

def permutation_p_value_medians(x, y, n_perm, seed):
    # Your code here
    pass

5 test cases · run your code against them in the editor