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Four independent experimental units have observed outcomes 1, 2, 4 and 5. Exactly two were chosen uniformly at random for treatment. They are the units with outcomes 4 and 5. Under the sharp null that treatment changes no unit’s outcome, use the absolute treatment-minus-control mean difference. What is the exact randomization p-value, as a percentage, counting ties as at least as extreme?
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Answer: 33.33% (1/3)
- The observed absolute difference is 3. The six signed differences are −3, −1, 0, 0, 1 and 3, so two of six assignments are at least as extreme and the exact p-value is 1/3.
Interview shortcut. List all C(4,2) = 6 assignments; 2 hit |diff| ≥ 3.
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