occurs with probability at most α . So in all three cases, τ a = τ b or τ a < τ b or τ a > τ $^{b}$, we have Pr ( I − v ⊇ τ$_{min}$ ) ≥ 1 − α , proving (ii). By a parallel argument, we obtain analogous 1 − α upper intervals, I + c and I + v , of the form ( −∞ ,$_{˜}$ τ ) for τ if τ a = τ b = τ or without restrictions for τ$_{max}$ . Taking the intersections, I c = I − c ∩ I + c and I v = I − v ∩ I + v , of two one-sided 1 − α/ 2 intervals yields analogous two-sided 1 − α intervals for τ if τ a = τ b = τ or without restrictions for the interval [ τ$_{min}$, τ$_{max}$ ]. In most cases, I v can be constructed by taking the union of I c and the two two-sided intervals constructed from the matched sets using each separate version of control. When this union is disjoint, I v is the shortest interval that contains all three intervals. In case (ii), the proof above that Pr ( I v ⊇ τ$_{min}$ ) ≥ 1 − α is similar to, but not quite iden- tical to, results in Lehmann (1952), Berger (1982) and Laska and Meisner (1989). These authors proposed tests that would invert to yield as a confidence interval the shortest inter- val I ∗ containing $^{{}$τ$_{0}$ : P a τ$_{0}$ > α or P b τ$_{0}$ > α $^{}}$, whereas I v is the shortest interval containing $^{{}$τ$_{0}$ : P$_{τ}$$_{0}$ > α or P a τ$_{0}$ > α or P b τ$_{0}$ > α $^{}}$, thereby ensuring I v ⊇ I $_{c}$. Of course, I v ⊇ I $_{∗}$, but unlike I $_{∗}$, our method ensures that I v and I c both simultaneously cover τ a = τ b = τ at rate 1 − α when there is actually only a single version of treatment. Because I c is built using all of the data and under stronger assumptions, it is unlikely that I ∗ will be much shorter than I $_{v}$; however, this logical possibility is the price for reporting the usual interval, I $_{c}$, without multiplicity correction. Why not report a single robust interval like I ∗ instead of two intervals, I c and I $_{v}$, that have slightly more nuanced coverage properties? A simple example may help illustrate the advantage of reporting I c and I v over I $_{∗}$. Suppose that I ∗ and I v both contain zero but I c does not. If we choose to report I ∗ then we have little additional information to determine why I ∗ contains zero – is Fisher’s sharp null true or is the smaller of the the two versions of 15