Permutation p-value with unequal group sizes

Question

You compare a small "new" group of 2 units against an "old" group of 3 units, five values in all: 2, \quad 4, \quad 6, \quad 8, \quad 20. The two "new" units scored 8 and 20. The test statistic is the sum of the new group (larger favors "new"). For an exact one-sided p-value, consider every way to pick which 2 of the 5 values form the "new" group. What is the one-sided permutation p-value?

Accepted Answer

Under the null, any 2 of the 5 values could be the "new" group. There are \binom{5}{2} = 10 ways. We only need to know how many of them give a sum at least as large as the observed 8 + 20 = 28. The two largest values are 20 and 8, so the only pair reaching a sum of 28 or more is \{8, 20\} itself (the next best, \{6, 20\}, sums to only 26). p-value. \frac{\{	ext{sums} \ge 28\}}{10} = \frac{1}{10} = 0.10. So the observed split is the single most extreme of the ten, giving the smallest possible p-value here, 0.10. Takeaway. With unequal groups the enumeration uses \binom{n}{k} splits; the exac…

Permutation p-value with unequal group sizes

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