Originally Posted by

**romsek** I think what you have to do here is note the following

a) for a given $n$, and $p$ the power of the tests are given by $power_p=1-(1-p)^n$

b) to get the test power you have to average that across p for the rejection range, in this case either (0.03, 1) or (0.1, 1)

c) for a general actual bad battery probability $p_0$, the formula for this average is given by

$power=\dfrac 1 {1-p_0} \displaystyle{\int_{p_0}^1} 1-(1-p)^n~dp$

$power =1 - \dfrac{(1-p_0)^{n}}{n+1}$

here you are treating p as a uniform$[p_0,1]$ random variable because you don't have any info about it other than it's range.

you can use this last formula to quickly evaluate the power of the the regions corresponding to $n$=1,2,3, with $p_0$=0.03, or 0.1