Compute the power of a one-sample z-test

Question

A trading desk wants to test whether a new signal has a positive mean daily return: $H_0: \mu = 0$ versus $H_1: \mu > 0$ . Daily returns have a known standard deviation of $\sigma = 2.0\%$ . The desk will collect $n = 64$ days and run a one-sided test at $\alpha = 0.05$ . Suppose the signal's true mean is actually $\mu = 0.5\%$ per day.

What is the power of this test, the probability it correctly rejects $H_0$ ?

Accepted Answer

Step 1: standard error. The sample mean has standard error \sigma/\sqrt{n} = 2.0\%/\sqrt{64} = 2.0\%/8 = 0.25\%. Step 2: the rejection cutoff. A one-sided test at \alpha = 0.05 rejects when the sample mean exceeds \mu0 + z\alpha \cdot 	ext{SE} = 0 + 1.645 	imes 0.25\% = 0.411\%. Step 3: how far the truth sits above the cutoff. Measure the true mean in standard-error units: the "effect" is (0.5 - 0)/0.25 = 2.0 standard errors above the null. Step 4: the power. Power is the chance the sample mean lands past the cutoff when the true mean is 0.5\%: 	ext{power} = \Phi(	ext{effect} - z\alpha) =…

Compute the power of a one-sample z-test

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