Proposition: No 0.1 acceptance at the reported sample count

For $n_j=40$, ${\alpha }_{\mathrm{s}}=1/20$, and $b_{\mathrm{cal}}=b=0$, no dataset can be accepted at $\epsilon =0.1$ by the three-setting binomial procedure. This remains true with outward conservative numerical brackets.

Proof. The binomial lower endpoint is nondecreasing in the pass count: the upper-tail probability decreases with $k$, so its inverse endpoint increases with $k$. Thus all-pass counts maximize $F_L$.

For exact all-pass endpoints, let $p_0=120^{-1/40}$. The maximal lower fidelity is

Acceptance would require $p_0\ge 599/600$. But Bernoulli's inequality gives

hence $p_0<599/600$, excluding acceptance. Outward lower brackets cannot increase $F_L$, and outward upper square-root brackets cannot decrease the error upper endpoint.