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n≈2 p(1−p) (zα/2+zβ)2δ2n \approx \frac{2\,p(1-p)\,(z_{\alpha/2} + z_{\beta})^2}{\delta^2}

Why this formula appears here

Because a “pass” verdict is itself often a noisy measurement, not a fact, it is worth being explicit about how much noise a single run’s outcome carries before treating a change in the pass rate as real. If a task’s true pass rate is p under a baseline configuration and a candidate change is worth detecting only once it shifts that rate by at least δ\delta , the number of repeated trials needed per configuration to detect the shift reliably — at significance level α\alpha and statistical power 1-β\beta — is approximately n≈2 p(1−p) (zα/2+zβ)2δ2n \approx \frac{2\,p(1-p)\,(z_{\alpha/2} + z_{\beta})^2}{\delta^2} . Plugging in a fairly ordinary case — a baseline pass rate of 70 percent, a minimum shift worth caring about of 10 percentage points, a 5 percent…

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nn

Symbol n

n is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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zα/2z_{\alpha/2}

Symbol z_α/2

z_α/2 occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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zβz_{\beta}

Symbol z_β

z_β occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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δ2\delta^2

Symbol delta^2

delta2a^2 occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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2 p(1−p) (zα/2+zβ)22\,p(1-p)\,(z_{\alpha/2} + z_{\beta})^2

Numerator: 2p(1-p)(z_α/2 + z_β)^2

The complete quantity above the fraction bar.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Its accuracy depends on the assumptions and range of use described in the article.

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Published contexts (1)

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n≈2 p(1−p) (zα/2+zβ)2δ2.n \approx \frac{2\,p(1-p)\,(z_{\alpha/2} + z_{\beta})^2}{\delta^2} .

Equation 5 · Model Evaluation

Building a Custom Evaluation Suite for a Production Agent

This equation gives an approximation: it relates the quantities while allowing an approximation.

Because a “pass” verdict is itself often a noisy measurement, not a fact, it is worth being explicit about how much noise a single run’s outcome carries before treating a change in the pass rate as real. If a task’s true pass rate is p under a baseline configuration and a candidate change is worth detecting only once it shifts that rate by at least δ\delta , the number of repeated trials needed per configuration to detect the shift reliably — at significance level α\alpha and statistical power 1-β\beta — is approximately n≈2 p(1−p) (zα/2+zβ)2δ2n \approx \frac{2\,p(1-p)\,(z_{\alpha/2} + z_{\beta})^2}{\delta^2} . Plugging in a fairly ordinary case — a baseline pass rate of 70 percent, a minimum shift worth caring about of 10 percentage points, a 5 percent…

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