From Wikipedia:
In statistics, the Bonferroni correction is a method used to counteract the problem of multiple comparisons. It was developed and introduced by Italian mathematician Carlo Emilio Bonferroni. The correction is based on the idea that if an experimenter is testing n dependent or independent hypotheses on a set of data, then one way of maintaining the familywise error rate is to test each individual hypothesis at a statistical significance level of 1/n times what it would be if only one hypothesis were tested. So, if it is desired that the significance level for the whole family of tests should be (at most) α, then the Bonferroni correction would be to test each of the individual tests at a significance level of α/n. Statistically significant simply means that a given result is unlikely to have occurred by chance assuming the null hypothesis is actually correct (i.e., no difference among groups, no effect of treatment, no relation among variables).
The Bonferroni correction is derived by observing Boole's inequality. If n tests are performed, each of them significant with probability β, (where β is unknown) then the probability that at least one of them comes out significant is (by Boole's inequality) ≤ nβ. Our intention is for this probability to equal α, the significance level for the entire series of tests. By solving for β, we get β = α/n. This result does not require that the tests be independent.
http://en.wikipedia.org/wiki/Bonferroni_correction
Bonferroni校正法:
此方法是在进行两两比较时对检验水准进行调整的办法，但是该方法在比较的次数较多时，就不太适合，因为校正后的检验水准会过小。此时可采用sidark法进行多重比较（仍然是对检验水准进行调整）。
统计学中一般以小概率作为判断差异是否显著的标准，通常都以0.05或0.01作为判断标准。在多重比较中， bonferroni是以t分布作为检验分布的，但多重比较时若均以0.05作为小概率的话，每次比较就会有5%犯一型错误的可能。但如果有n次比较，如有4个组要做6次比较，则有C6（2）*5%一型错误发生的概率，不符合小概率判断的原则。因此，bonferroni中，将小概率0.05或0.01除以要比较的次数n，作为判断显著性的小概率，这样，多重比较总的一型错误发生的概率不会超过0.05或0.01。
控制累积Ⅰ类错误概率增大的方法
采用Bonferroni法,SNK法和Tukey法等方法
累积Ⅰ类错误的概率为α'
当有k个均数需作两两比较时,比较的次数共有c= = k!/(2!(k-2)!)=k(k-1)/2
设每次检验所用Ⅰ类错误的概率水准为α,累积Ⅰ类错误的概率为α',则在对同一实验资料进行c次检验时,在样本彼此独立的条件下,根据概率乘法原理,其累积Ⅰ类错误概率α'与c有下列关系:
α'=1-(1-α)c (8.6)
例如,设α=0.05,c=3(即k=3),其累积Ⅰ类错误的概率为α'=1-(1-0.05)3 =1-(0.95)3 = 0.143
一,Bonferroni法
方法:采用α=α'/c作为下结论时所采用的检验水准.c为两两比较次数, α'为累积I类错误的概率.
例8-1四个均值的Bonferroni法比较
设α=α'/c=0.05/6=0.0083,由此t的临界值为t(0.0083/2,20)=2.9271
Bonferroni法的适用性
当比较次数不多时,Bonferroni法的效果较好.
但当比较次数较多(例如在10次以上)时,则由于其检验水准选择得过低,结论偏于保守.