Assuming no false negatives is just not an option :-) I think that can only happen if the situation is such that there can be no false positives either (i.e. the p-value when the null hypothesis is not true is always zero). EDIT: What I wrote is true only if the distributions under the null and the alternative are completely disjoint. You can actually have very low false negative rates if the distributions are not symmetric, and if you allow different distributions you can do even better: imagine the null hypothesis is x~Normal(0,1) and the alternative is x=C0=1.64 (exactly the cutoff value for 0.05 significance). If we get exactly p=0.05 then the probability of the null being true is 0%. I mean, we get p in [C0-epsilon C0+epsilon] with probability 1 under the alternative, but with probability-&gt;0 under the null as epsilon-&gt;0. Of course, this alternative is very unlikely and mixing continous and discrete distributions is always tricky. This is why it makes sense to make averages over prior distributions of the alternative.<p>As you can see in slide 14, there are multiple calibrations proposed under different assumptions. I agree it is misleading to give one as the &quot;real&quot; error I rate, but it&#x27;s interesting that all of them are giving rates well above the nominal alpha rate. EDIT: note as well that this is for the case where in 50% of the cases the null hypothesis is true(nowhere in the calculation of p-values do we consider how often the null hypothesis is true, but obviously if it&#x27;s always true 100% of the significant results will be false positives and if it&#x27;s never true 0% of the significant results will be false positives).<p>In slide 11 there are other calculations for the normal case, two-sided test this time. But instead of looking for the mu1 giving the lowest bound, they calculate the aggregate error rate making some assumptions about the distribution of the mu1. For example, if I understand correctly the results, assuming mu1 is normally distributed around mu0=0, if you get a p-value=0.05 (in the two-sided test, some modifications are required to the calculation we did) you should expect the null hypothesis to be true at least 32.1% (if the distribution of mu1 is very concentrated around 0, the 50% rejection rate on the left side of the chart dominates, if the standard deviation if very high the region of almost 100% rejection rate far from mu0 at the right of the chart dominates, for some intermediate standard deviation one will hopefully get the 32.1% lower bound).<p>Unfortunately, I think the assumption behind the nice result 1&#x2F;(1-1&#x2F;(e p log(p)))) is that p-values follow a beta distribution when the null hypothesis is not true and I don&#x27;t think there is a clear interpretation of that.