ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A two-sided normal z-test retains its lower-tail power

Example

Let $α = 0.05$ , $n = 4$ , and $δ = σ > 0$ in the known-variance normal model, with simple null $μ = μ_{0}$ and alternative $μ = μ_{0} + δ$ . The two-sided test rejects for $∣ Z_{n} ∣ > z_{0.975}$ . Its exact power is $1 - Φ (z_{0.975} - 2) + Φ (- z_{0.975} - 2) .$ The second term is the lower-tail rejection probability under the positive shift; omitting it would compute only the upper-tail probability.

Facts & Assumptions

Given: AC; $α = 0.05$ , $n = 4$ , $δ = σ > 0$ , the simple null, and the two-sided rejection region in the Example.

[F1]

The known-variance two-sided test has exact power $1 - Φ (c_{α} - λ) + Φ (- c_{α} - λ)$ , where $c_{α} = z_{1 - α / 2}$ and $λ = \sqrt{n} δ / σ$ . Both tails determine power for the known-variance two-sided normal z-test

[F2]

$0 < Φ (c) < 1$ for every finite real $c$ , since $Φ$ is strictly increasing with limits $0$ and $1$ at the two infinite endpoints. Standard-normal CDF, tail, and quantile identities for power calculations

Verification

Given: The exact two-sided normal test and parameter values in the Example.

1.1F1givenalgebra

Here $c_{α} = z_{1 - α / 2} = z_{0.975}$ and $λ = \sqrt{4} σ / σ = 2$ . Substitution into [F1] gives the displayed upper-tail term $1 - Φ (z_{0.975} - 2)$ and lower-tail term $Φ (- z_{0.975} - 2)$ .

2.1F2step 1.1algebra∎

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

Depends on

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