From Brian Caffo’s Coursera biostatistics video:

Suppose that we wanted to detect an increase in mean RDI (respiratory disturbance index in the context of sleep apnea) of at least \(2\small \text{ events/hour}\) above \(30\). Assume normality and that the sample in question has a standard deviation of \(4\). What would be the power if we took a sample of \(16?\)



\[\frac{\mu_a - 30}{\sigma/\sqrt{n}}=\frac{2}{4/\sqrt{16}}=2\]



What \(n\) sample size would be required to get a power of \(80\,\%\) (a common benchmark in the sciences)?

For a one-sided test (\(H_a: \mu_a > \mu_o\)):

\[0.8=\Pr\left(Z> \, z_{1-\alpha} -\frac{\mu_a -\mu_o}{s/\sqrt{n}}\vert\mu=\mu_a\right)\]

We set \(z_{1-\alpha} - \frac{\mu_a -\mu_o}{s/\sqrt{n}} = z_{0.2}\) and solve for \(n\) for any value of \(\mu_a\). We pick \(\mu_a\) as the smallest effect that we would reasonably like to detect.

In the cases of \(H_a:\mu_a \neq \mu_o\) we can just take one of the sides but with \(\alpha/2\).

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