# How to determine the sample size for a two sided $z$-test?

Let $$X_{1}, \ldots, X_{n}$$ be an iid sample from $$N(\mu,\sigma^2)$$ where $$\sigma$$ is known. We want to test a hypothesis $$H_{0}:\mu = \mu_{0} \quad \mbox{versus} \quad H_{1}: \mu \ne \mu_0$$ Now, assume that the values of $$\alpha$$ (probability of Type I Error) and $$\beta$$ (Probability of Type II Error) are fixed in advance.

Therefore, the problem now is to determine what should be the sample size to achieve the desired value of $$\beta$$?

Here is what I progressed:

The power function is given by $$w(\mu) = 1 + \Phi \left( k-z_{\alpha/2} \right) - \Phi \left( k+z_{\alpha/2} \right),$$ where $$k = \frac{\mu_0-\mu}{\sigma/\sqrt{n}}.$$ We also know that $$w(\mu) = 1 - \beta(\mu),$$ where $$\beta(\mu)$$ is the probability of making Type II error when the true parameter value is $$\mu$$.

Now, it is evident that in order to achieve the desired value of $$\beta$$, we need to set up the equation $$1-\beta = w(\mu),$$ and solve this equation for $$n$$.

But I am not sure how to solve this equation for $$n$$.

I just found on one of the textbooks without any work that the minimum sample size should be $$n \ge \left[ \frac{\sigma(z_{\beta} + z_{\alpha/2})} {\mu_0-\mu} \right]^2$$ as an approximated solution.

But again how do we get this approximated solution?

Thank you!

\begin{align*} & {\qquad} 1-\beta = w(\mu) \\ & {\qquad} = 1 + \Phi \left( k-z_{\alpha/2} \right) - \Phi \left( k+z_{\alpha/2} \right) \\ & {\qquad} = P(Z \ge z_{\alpha/2}-|k|) + P(Z \ge z_{\alpha/2}+|k|) \\ \Rightarrow & {\qquad} 1-\beta \approx P(Z \ge z_{\alpha/2}-|k|), \quad \mbox{assuming} \quad P(Z \ge z_{\alpha/2}+|k|) \approx 0 \\ \iff & {\qquad} z_{1-\beta} \approx z_{\alpha/2}-|k| \\ \iff & {\quad} -z_{\beta} \approx z_{\alpha/2}-|k| \\ \iff & {\quad} |k| \approx z_{\alpha/2}+z_{\beta}, \end{align*} this gives $$n \approx \left[ \frac{\sigma(z_{\beta} + z_{\alpha/2})} {\mu_0-\mu} \right]^2,$$ as desired.