Proof. Let \(E_k\) be the event that the \(k\)th interval misses. By Boole’s inequality, \(P(E_1 \cup \dots \cup E_m) \le \sum _kP(E_k) = m(\alpha /m) = \alpha \). □

Proof. Let \(Z_i = (\bar y_{i\cdot } - \mu _i)/(\sigma /\sqrt r)\), which are independent \(N(0, 1)\), and \(V = (N - t)MSE/\sigma ^2 \sim \chi ^2_{N-t}\), independent of them. Every interval covers if and only if \(|(\bar y_{i\cdot } - \mu _i) - (\bar y_{k\cdot } - \mu _k)| \le q_\alpha \sqrt {MSE/r}\) for all pairs, that is, if and only if the largest such difference, which is \(\max _i(\bar y_{i\cdot } - \mu _i) - \min _i(\bar y_{i\cdot } - \mu _i)\), is at most \(q_\alpha \sqrt {MSE/r}\). Dividing by \(\sqrt {MSE/r}\), this is the event \((\max Z_i - \min Z_i)/\sqrt {V/(N - t)} \le q_\alpha \), which has probability \(1 - \alpha \) by definition. □

Proof. Let \(u_i = \bar y_{i\cdot } - \mu _i\) and \(\bar u = \sum n_iu_i/N\). For any contrast, since \(\sum c_i = 0\), \[ \hat C - C = \sum _ic_iu_i = \sum _ic_i(u_i - \bar u) = \sum _i\frac {c_i}{\sqrt {n_i}}\cdot \sqrt {n_i}(u_i - \bar u) . \] By the Cauchy–Schwarz inequality, \[ (\hat C - C)^2 \le \Bigl (\sum _i\frac {c_i^2}{n_i}\Bigr )\sum _in_i(u_i - \bar u)^2 , \] with equality when \(c_i \propto n_i(u_i - \bar u)\), which is itself a contrast. Hence \[ \max _{\text {contrasts}}\frac {(\hat C - C)^2}{MSE\sum c_i^2/n_i} = \frac {\sum _in_i(u_i - \bar u)^2}{MSE} . \] The numerator is the treatment sum of squares computed from the errors \(\varepsilon _{ij}\) in place of the \(y_{ij}\), so it is \(\sigma ^2\chi ^2_{t-1}\), independent of \(MSE\), and the right side is distributed as \((t - 1)F_{t-1,\,N-t}\). So every interval covers if and only if this maximum is at most \(S^2\), which has probability \(1 - \alpha \). □