Permutation p-value with three-versus-three groups

Question

You compare two groups of three on the values

$1, \quad 2, \quad 3, \quad 4, \quad 5, \quad 30,$

where the treatment group holds $4$ , $5$ , and $30$ . The statistic is the sum of the treatment group (larger favors treatment). Enumerate every way to split the six values into two groups of three.

What is the one-sided permutation p-value?

Accepted Answer

There are \binom{6}{3} = 20 ways to choose the treatment group. The observed treatment sum is 4 + 5 + 30 = 39. Any split whose sum is \ge 39 must include the outlier 30 (without it the max possible sum is 3+4+5 = 12). Given 30 is in, the other two must be the two largest remaining, 5 and 4, to reach 39; any other pair gives at most 30 + 5 + 3 = 38 < 39. So exactly one split of the twenty reaches 39 or more, the observed one. p-value. \frac{1}{20} = 0.05. With 20 arrangements the smallest attainable p-value is 1/20 = 0.05, and this data lands right on it, just barely significant at the usual t…

Permutation p-value with three-versus-three groups

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