Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 / 2 / a / iii / Solution

The Bonferroni correction compares every p-value with $0.05/100000=5\times 10^{-7}$. Only the first two displayed values lie below that threshold; all undisplayed values are at least the tenth displayed value. HenceFor the Benjamini-Hochberg procedure at false discovery rate level $q=0.1$, the rank-$i$ threshold is $qi/100000=i\times 10^{-6}$. The first nine displayed ranks meet their thresholds, while rank ten does not: $1.1\times 10^{-5}>10^{-5}$. Consequently nine is the intended count if every later rank also fails its threshold. The rule is a step-up rule, however:It does not stop at the first failed comparison. The full rejection count is therefore not determined by ten order statistics. For example, completing the list with $99990$ values equal to $1$ gives $k=9$, whereas completing it with $99990$ values equal to $0.02$ gives $k=100000$, since $p_{(100000)}=0.02\le 0.1$. Both completions respect the printed ten smallest p-values. Thus at least nine are rejected; exactly nine requires an extra assumption about the undisplayed ranks. In the second completion even the tenth marker is rejected, despite failing its own rank threshold. The false discovery rate guarantee also requires the usual validity and independence or appropriate positive-dependence assumptions, unlike the arbitrary-dependence Bonferroni correction.