The construction survives a difference of two weighted means

Coverage of δ̂ ± t√(S_D²/H) on b − 1 degrees of freedom, over 900 runs at a requirement of 0.3, where δ̂ is the block differences weighted by h_b = (1/m_A + 1/m_B)⁻¹ and H is their total. The theorem the one-mean field rests on goes through with h_b in place of the block size, and the reason is that the weights a weighted least squares decomposition needs are the inverse variances — which is exactly what h_b is. The stopping rule reads only within-arm within-block contrasts, so it is a function of nothing the interval reports, whatever it does with the block sizes. Each bar is within 2.9% of the level it claims.

Coverage of the same interval under three weightings. h_b is the inverse variance when the arms share a variance or the allocation is constant; equal weights are right when every block has the same two counts; the estimated precision weights are right in the limit and exact nowhere, because the decomposition needs the weights to be the constants they are only estimating. In the corner — two variances, changing sizes, changing allocation — the two exact estimators are the ones that miss, at 98.45% and 95.65%, and the one with no theorem behind it is at 95.05%. That is the whole statement: there is an exact estimator under either condition, and none under both.

Each point is one run. The one-mean field's identity is (b − 1) + (N − b) = N − 1, and every schedule moves along that line rather than off it. Two arms give the rule N − 2b and the interval b − 1, which come to N − b − 1 — short of the N − 2 two arms leave by exactly one per block, since a block's arm counts absorb one degree of freedom each and only one of the two directions carries the difference. The hollow points add what the block sums are worth, b − 1 more, and land on the total. The missing degrees of freedom are not lost; they are in a place the interval has to be shown it may read.