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Dec 25, 2014 at 13:41 comment added David Z I suppose one could consider this as a probability distribution which is strictly zero except between $\frac{1}{n}\sum_1^n f(x_i)\pm V(f)D(\ldots)$, in which case it becomes equivalent to a confidence interval where the $p$-value can be arbitrarily close to $0$ (or the likelihood ratio arbitrarily close to $1$).
Dec 25, 2014 at 1:36 comment added Brendan McKay To help people like me unfamiliar with this: en.wikipedia.org/wiki/…
Dec 25, 2014 at 0:26 history answered Gerry Myerson CC BY-SA 3.0