The only way to increase the power of a test without also increasing the significance level is to collect more data.
Failure of printed circuit boards
Consider the above printed circuit board example. Let us assume that the engineers have decided that it is acceptable for the probability of a Type I error (deciding that \(\pi\) < 0.06 when \(\pi\) is really 0.06) to be 5%:
significance level = P(Type I error) = 0.05
and that it is acceptable for the probability of a Type II error (deciding that \(\pi\) is 0.06) to be 10% if \(\pi\) is really 0.03.
With a sample size of \(n\) and decision rule to reject the null hypothesis if \(x \le k\), we therefore require
\[ \begin{align} \alpha \;&=\; P\left(\text {reject }H_0 \mid \pi = 0.06\right) \\[0.4em] &=\; P(X \le 7 \mid \pi = 0.06) \\ &=\; \sum_{x=0}^k {n \choose x} \; 0.06^x \; 0.94^{n-x} \\ &=\; 0.05 \end{align} \]and
\[ \begin{align} \beta \;&=\; P\left(\text {accept }H_0 \mid \pi = 0.03\right) \\[0.4em] &=\; 1 - P(X \le 7 \mid \pi = 0.03) \\ &=\; 1 - \sum_{x=0}^k {n \choose x} \; 0.03^x \; 0.97^{n-x} \\ &=\; 0.10 \end{align} \]The values of \(n\) (the sample size) and \(k\) (the cut-off for the decision rule) are the values that satisfy these two equations (or are as close as possible to this). The diagram below helps.
From trial and error with different \(n\) and \(k\), we can find that if \(n = 400\) and \(k = 16\), we get
\[ \alpha \;\;=\;\; 0.051 \spaced{and} \beta = 0.098 \]which is close to the required values.
The engineer should therefore test 400 circuit boards made with the new process and conclude that it is better than the old process if 16 or fewer fail.
For tests involving continuous distributions, it is often possible to find explicit solutions to the equations for the probabilities of Type I and II errors.