 Solution

Assume the null distribution is a centrally symmetric probability distribution, so $X_i$ and $-X_i$ have the same law. A sign-flip randomization test draws signs $s_i\in \{-1,1\}$ independently and recomputes, for example,The exact p-value averages over all $2^n$ sign vectors:With $B$ random sign vectors, including the observed configuration, the standard Monte Carlo version is $(1+\#\{b:T_b\ge T_0\})/(B+1)$, where $T_0$ is the observed statistic. Joint sign invariance under the null makes this finite-sample valid.