# How can you derive the sample size to obtain a particular type II error rate?

I often see that the sample size for a $z$-test to achieve a particular type II error rate $\beta$ (at a given significance level, $\alpha$) is:

$$n = \frac{(Z_{1-\alpha/2}+Z_{1-\beta})^2\sigma^2}{\delta^2}$$

How do they reach that conclusion/determine that?

• Your question was a bit muddled/confusing. I have edited, but you'll need to review it to check it's asking what you want. – Glen_b -Reinstate Monica Feb 27 '14 at 23:26
• Such calculations are often called power calculations (power = $1-\beta$). I've removed one of your tags and added a couple of tags relating to power to (hopefully) help with finding related answers in the sidebar. – Glen_b -Reinstate Monica Feb 27 '14 at 23:31
• No doubt someone will write a good answer, but in the meantime, there's an outline of the calculation here. Keep in mind that they're doing a one tailed test calculation (hence the difference with $\alpha$ vs $\alpha/2$), and that $Z_{1-\beta}=-Z_\beta$, and their $d$ corresponds to $\delta/\sigma$. – Glen_b -Reinstate Monica Feb 27 '14 at 23:41

Here's an outline of the basic ideas:

The same critical value, $C$, applies whether $H_0$ is true or $H_1$ is true.

$C=\mu_0+\sigma Z_{1-\alpha/2}/\sqrt{n}$

$C=\mu_1+\sigma Z_\beta/\sqrt{n}$

$\mu_0+\sigma Z_{1-\alpha/2}/\sqrt{n}=\mu_1+\sigma Z_\beta/\sqrt{n}$

$(\mu_1-\mu_0)/(\sigma/\sqrt{n}) = (Z_{1-\alpha/2}- Z_\beta)$

$[(\mu_1-\mu_0)/\sigma]\sqrt{n} = (Z_{1-\alpha/2}+ Z_{1-\beta})$

$\sqrt{n} = (Z_{1-\alpha/2}+ Z_{1-\beta})/[(\mu_1-\mu_0)/\sigma]$

and then you just square both sides

$n = (Z_{1-\alpha/2}+ Z_{1-\beta})^2/[(\mu_1-\mu_0)^2/\sigma^2]$

$\quad = (Z_{1-\alpha/2}+ Z_{1-\beta})/\delta^2$

Note: (1) Here I'm doing just the case $\mu_1>\mu_0$. To cover both possibilities, we really need $|\mu_1-\mu_0|$ where I have $(\mu_1-\mu_0)$; the outcome is the same.

$\quad\quad$(2) This ignores a small piece of area to the far left of the second diagram.

$\quad\quad\quad\,\,\,$When $\delta$ is large, this is negligible.