Power of a slippage-improvement test

Question

A team tests whether a new order-routing rule reduces mean slippage below the current level: $H_0: \mu = 0$ against $H_1: \mu > 0$ , where $\mu$ is the improvement in basis points. Fill-to-fill improvement has a known standard deviation of $\sigma = 3$ bps. They sample $n = 100$ fills and test at one-sided $\alpha = 0.05$ . Suppose the true improvement is $\mu = 1$ bp.

What is the power of this test?

Accepted Answer

Standard error. \sigma/\sqrt{n} = 3/\sqrt{100} = 3/10 = 0.3 bps. Effect in SE units. The true improvement is (1 - 0)/0.3 = 3.33 standard errors above the null. Power. 	ext{power} = \Phi(	ext{effect} - z\alpha) = \Phi(3.33 - 1.645) = \Phi(1.69) \approx 0.95. A 1 bp edge sounds tiny, but averaging over 100 fills shrinks the standard error to 0.3 bps, so the improvement is more than three standard errors clear of zero and the test catches it about 95% of the time. Takeaway. Small per-fill edges become detectable once you average enough observations: 100 fills turn a 1 bp improvement into a hig…

Power of a slippage-improvement test

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