### SAMPLE SIZE:

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 $$16?$$

$Z_{1-\alpha}=1.645$

and

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

Therefore,

$\Pr(Z>1.645-2)=\Pr(Z>-0.355)=64\%$

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 $$\alpha/2$$.