Holm-Bonferroni Step

Problem

Count rejections with the Holm-Bonferroni step-down procedure. Sort p-values ascending; reject while p_(i) <= alpha / (m - i) for 0-based i, stopping at the first failure.

Empty input returns 0.

Hint

Holm is uniformly more powerful than plain Bonferroni because thresholds loosen as you reject.

Test cases

case 1

in: [[0.001,0.01,0.04],0.05]

expected: 3

case 2

in: [[0.02,0.03,0.04],0.05]

expected: 0

case 3

in: [[0.01,0.5],0.05]

expected: 1

case 4

in: [[],0.05]

expected: 0

Starter codePython

def holm_count(p_values, alpha):
    # Your code here
    pass

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