p value bonferroni correction

http://www.aaos.org/news/aaosnow/apr12/research7.asp

For example, a researcher is testing 20 hypotheses simultaneously, with a critical P value of 0.05. In this case, the following would be true:

• P (at least one significant result) = 1 – P (no significant results)
• P (at least one significant result) = 1 – (1-0.05)²⁰
• P (at least one significant result) = 0.64

Thus, performing 20 tests on a data set yields a 64 percent chance of identifying at least one significant result, even if all of the tests are actually not significant. Therefore, while a given α may be appropriate for each individual comparison, it may not be appropriate for the set of all comparisons.


http://www.fon.hum.uva.nl/praat/manual/Bonferroni_correction.html


In general, if we have k independent significance tests at the α level, the probability p that we will get no significant differences in all these tests is simply the product of the individual probabilities: (1 - α)^k. For example, with α = 0.05 and k = 10 we get p = 0.95¹⁰ = 0.60. This means, however, we now have a 40% chance that one of these 10 tests will turn out significant, despite each individual test only being at the 5% level. In order to guarantee that the overall significance test is still at the α level, we have to adapt the significance level α′ of the individual test.
This results in the following relation between the overall and the individual significance level:

(1 - α′)k = 1 - α.

This equation can easily be solved for α′:

α′ = 1 - (1-α)1/k,

which for small α reduces to:

α′ = α / k

This is a very simple recipe: If you want an overall significance level α and you perform k individual tests, simply divide α by k to obtain the significance level for the individual tests.