Power and sample size — Examples

Give the procedure the constant-false oracle $E_0(n)=0$. Correctness forces the output none, and universal termination makes its queried set $Q$ finite. Put $M=1+{\sum }_{q\in Q}q$, which is a positive integer greater than every queried positive size, including $M=1$ when $Q$ is empty, and define $E_M(n)=\mathbf{1}\{n\ge M\}$. Both oracles are nondecreasing and answer false at every query in the completed $E_0$ transcript.

Run the same deterministic procedure with $E_M$. Its initial internal state is the same, and whenever a query is reached along the original transcript, its answer is the same false value. Induction through the finite transcript therefore gives identical internal states, queries and terminal output. It returns none on $E_M$, although $M$ is the least feasible size. This contradicts the claimed correctness.

These oracles are realizable as exact power-feasibility inputs: use candidate plans $d_n$ with one allocation count equal to $n$, for every positive integer $n$; at that size take the measurable space $\{0,1\}$ with its full power set, parameters $0,1$, null law $P_{0,n}={\delta }_0$, target law $P_{1,n}={\delta }_{E(n)}$, and test $\phi (x)=x$. The full power set makes $\phi$ measurable, and the proved point-mass and measurable-rejection-function suppliers in [F2] make this an experiment and randomized test. Its null size is zero and target power is $E(n)$, so a size bound zero and target power $1\mathrm{/}2$ in [F1] make feasibility equivalent to $E(n)=1$. Use either $E_0$ or $E_M$; hence the distinction is compatible with explicit statistical laws, not a numerical approximation error.

Here $c_{\alpha }=z_{1-\alpha \mathrm{/}2}=z_{0.975}$ and $\lambda =\sqrt{4}\sigma \mathrm{/}\sigma =2$. Substitution into [F1] gives the displayed upper-tail term $1-\mathrm{\Phi }(z_{0.975}-2)$ and lower-tail term $\mathrm{\Phi }(-z_{0.975}-2)$.

Both arguments $z_{0.975}-2$ and $-z_{0.975}-2$ are finite, so [F2] gives $1-\mathrm{\Phi }(z_{0.975}-2)>0$ and $\mathrm{\Phi }(-z_{0.975}-2)>0$. Thus the displayed power has a positive contribution from each rejection tail.

Since $1\mathrm{/}2<\mathrm{\Phi }(1)<1$ by [F2], $\alpha ={\beta }_{\mathrm{I}\mathrm{I}}=1-\mathrm{\Phi }(1)$ belong to $(0,1)$. Thus $z_{1-\alpha }=z_{1-{\beta }_{\mathrm{I}\mathrm{I}}}=z_{\mathrm{\Phi }(1)}=1$, and the quantile sum in [F1] is $2$.

The theorem's formula gives $n_{\ast }=\max \{1,\lceil (\sigma \mathrm{/}\sigma {)}^2(2{)}^2\rceil \}=4$ by [F3]. At $n=3$, power is $\mathrm{\Phi }(\sqrt{3}-1)<\mathrm{\Phi }(1)$ because $\sqrt{3}<2$ and $\mathrm{\Phi }$ is strictly increasing. Therefore the target fails at $n=3$, while [F1] and the formula certify that $n=4$ is the least positive feasible size.

At $n=4$, halving $\delta$ changes $\sqrt{n}\delta \mathrm{/}\sigma$ from $2$ to $1$, so the power becomes $\mathrm{\Phi }(1-1)=\mathrm{\Phi }(0)=1\mathrm{/}2<\mathrm{\Phi }(1)=1-{\beta }_{\mathrm{I}\mathrm{I}}$.

If instead $\delta =\sigma$ and $n=4$ are fixed while the standard deviation becomes $2\sigma$, then the standardized effect also becomes $1$; the power is again $1\mathrm{/}2$, below target. This standard-deviation change quadruples the variance, so the example records variance sensitivity without treating the variance itself as the standard deviation.