**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

**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*= 0.95

^{10}= 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. |

α′ = 1 - (1-α)^{1/k}, |

α′ = α / k |

*k*individual tests, simply divide α by

*k*to obtain the significance level for the individual tests.