Hi @MichaelR1 : I think so. Your null and alternative hypotheses are as follows.

H0: p=p0

Ha: p<p0

Where p0 is the current failure rate. Ha is what you are trying to prove (the new failure rate, p, is less than the current rate p0).

To "prove" Ha, p must truly be somewhere in Ha (<p0). For example, if I assume p0=0.1, and expect p to be 0.05, and if I want my power to be 80%, and my type 1 error rate (alpha) to be 5%, then in JMP it looks like the following.

So, if you take a random sample of 199 units, there is 80% chance of "proving" Ha (via rejecting H0 in favor of Ha).

The type 1 error rate (Alpha) = 5% means that when H0 is true (p is not less than p0), there is a 5% chance of wrongly rejecting H0 in favor of Ha. Type 2 error = 1-Power= 0.2 (20%). Type 2 error then is the chance that you do not reject H0, even when Ha is true. You can choose Alpha and Power as you like, though 5% and 80% are common. And, I'm not a big fan of the Normal Approximation either; the Exact Test power plot looks weird due to the nature of pass/fail data. But normal approximations can be problematic especially when p is close to 0 or close to 1.

My two cents...