The critical z after correcting for 50 tests

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A researcher runs 50 two-sided z-tests and holds the family-wise error rate at 5% with Bonferroni. A single test would clear at z = 1.96. What critical z must each of the 50 tests now clear?

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Shrink the per-test level. Bonferroni divides the 5% budget across 50 tests: \frac{\alpha}{m} = \frac{0.05}{50} = 0.001 \quad (	ext{two-sided}). Convert to a z-hurdle. A two-sided level of 0.001 places 0.001/2 = 0.0005 in each tail, so the critical value is the z with cumulative probability 1 - 0.0005 = 0.9995: z \approx 3.29. So each test must clear about 3.29 instead of 1.96. A useful landmark. The value 3.29 is worth remembering: it is the two-sided critical z for a 0.001 level, the "one in a thousand" bar. Lifting the hurdle from 1.96 to 3.29 pushes the finish line about 1.3 standard err…

The critical z after correcting for 50 tests

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