Proper power computation for one-sided binomial tests.

Am 23.09.2008 um 23:57 schrieb Peter Dalgaard:

For this kind of problem I'd go directly for the binomial 
distribution. If the actual probability is 0, this is essentially 
deterministic and you can look at

binom.test(0,99,p=.03, alt="less")

This means that you don't sample from the p=.03 population?
Note that there is a 5 per cent chance to have 0 failures in 99
trials with p=.03.

Yes, that's what I read the task as saying: Sample from p=0.00 when the 
hypothesis is p=0.03. Then rejection happens with probability 1 when n 
 >= 99. Actually, he said that we could assume the _sample_ rate to be 
0%, but that is only assured when p=0.0.

(You can continue the game by looking at the probability of getting 0 
failures, depending on the true p. E.g., if p=0.001, we have

 > dbinom(0, 99, 0.001)
[1] 0.9056978

i.e. 90% power to detect at 5% level. And further continue into a full 
power analysis where you calculate the probability of a failure rate 
that is significantly different from 0.03 depending on p and n.)

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