Suppose, I have done:

  • $n_1$ independent trials with an unknown success rate $p_1$ and observed $k_1$ successes.
  • $n_2$ independent trials with an unknown success rate $p_2$ and observed $k_2$ successes.

If, now $p_1 = p_2 =: p$ but still unknown, the probability $p(k_2)$ to observe $k_2$ for a given $k_1$ (or vice versa) is proportional to $\int_0^1 B(n_1,p,k_1) B(n_2, p, k_2) \text{d}p = \frac{1}{n_1+n_2+1}\binom{n_1}{k_1}\binom{n_2}{k_2}\binom{n_1+n_2}{k_1+k_2}^{-1}$, so if I want to test for $p_1 \neq p_2$, I only need to look in which quantile of the corresponding distribution my observations are.

So far for reinventing the wheel. Now my problem is that I fail to find this in literature, and thus I wish to know: What is the technical term for this test or something similar?

  • 2
    Why not use the two-proportion z-test (en.wikipedia.org/wiki/Statistical_hypothesis_testing) (If I understand your problem correctly). – Verena Haunschmid Dec 19 '13 at 15:04
  • @ExpectoPatronum: At a quick glance the biggest problem is that this test requires at least 5 successes and failures for each observation, which may not be given in my application and also indicates that (unneccessary) approximations are made. – Wrzlprmft Dec 19 '13 at 16:20
  • ok, that is a problem but most tests have similar requirements. – Verena Haunschmid Dec 19 '13 at 16:26
  • @ExpectoPatronum: Anyway searching for an exact alternative to the two-proportion z-test, I found Fisher’s exact test, which looks very similar at first glance (but I have yet to look into it in detail). – Wrzlprmft Dec 19 '13 at 16:31
  • 1
    @ExpectoPatronum: The division does not matter, since the large term is only proportional to $p(k_2)$ and $(n_1+n_2+1)$ is exactly the normalisation constant. Anyway, I have now confirmed that this is Fisher’s Exact Test, which I found thanks to you. – Wrzlprmft Dec 19 '13 at 21:41
up vote 4 down vote accepted

The test statistics $p(k_2)$ is that of Fisher’s Exact Test.

Since $$\sum_{k_2}^{n_2} \frac{1}{n_1+n_2+1}\binom{n_1}{k_1}\binom{n_2}{k_2}\binom{n_1+n_2}{k_1+k_2}^{-1} = \frac{1}{n_1+n_2+1},$$ normalisation can be obtained by multiplying with $n_1+n_2+1$ and thus: $$p(k_2) = \binom{n_1}{k_1}\binom{n_2}{k_2}\binom{n_1+n_2}{k_1+k_2}^{-1}.$$

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