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.
![MRB3855_0-1721230064399.png MRB3855_0-1721230064399.png](/t5/image/serverpage/image-id/66257i5F95EBDA4974344E/image-size/medium?v=v2&px=400)
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...