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!
The approximated solution can be derived as below.
\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.