Bonferroni Correction Factor !!!

mp83

TS Contributor
#2
In statistics, the Bonferroni correction states that if an experimenter is testing nhypotheses on a set of data, then the statistical significance level that should be used for each hypothesis separately is 1/ dependent or independent n times what it would be if only one hypothesis were tested. Statistically significant simply means that a given result is unlikely to have occurred by chance.
For example, to test two independent hypotheses on the same data at 0.05 significance level, instead of using a p value threshold of 0.05, one would use a stricter threshold of 0.025.
The Bonferroni correction is a safeguard against multiple tests of statistical significance on the same data falsely giving the appearance of significance, as 1 out of every 20 hypothesis-tests will appear to be significant at the α = 0.05 level purely due to chance.
It was developed by Italian mathematician Carlo Emilio Bonferroni.
A uniformly more powerful test procedure is the Holm-Bonferroni method, however current methods for obtaining confidence intervals for the Holm-Bonferroni method do not guarantee confidence intervals that are contained within those obtained using the Bonferroni correction. A less restrictive criterion that does not control the familywise error rate is the rough false discovery rate giving (3/4)0.05 = 0.0375 for n = 2 and (21/40)0.05 = 0.02625 for n = 20.


from Wikipedia
 

Bala

New Member
#3
Thanks a lot........

I am using same data set to compare four different groups....so as per ur reply I have to consider the threshold p value as 0.00125..Is it right?


Thanks in Advance
Bala........:yup:
 
#5
Hi,
I am afraid I am getting confused with Bonferroni correction explained by Wikipedia so I hope to find some help here.
I have performed Wilcoxon's test on 12 variables of ten cases each, so I have 65 hypothesis to test. Which is the level of significance i have to choose using Bonferroni correction?
do i have to do 1/n so 1/65=0.015?
 
#6
Hi,
I am afraid I am getting confused with Bonferroni correction explained by Wikipedia so I hope to find some help here.
I have performed Wilcoxon's test on 12 variables of ten cases each, so I have 65 hypothesis to test. Which is the level of significance i have to choose using Bonferroni correction?
do i have to do 1/n so 1/65=0.015?
Wilcoxon's Test is non-parametric and more conservative than parametric methods already. Bonferroni correction adjusts parametric methods to account for more liberal use. Essentially, you just divide your significance level by the number of tests you run so that any one test is sufficiently stringent such that all your tests as a group are right (1-significance level)% of the time.

e.g. I want to be certain that my 65 hypotheses about a data set are correct 95% of the time, so each test should be tested at the 5%*1/65 or .077% confidence level to ensure that the failure of all the tests as whole is 65*.077% or 5%.
 

ad19

New Member
#7
Does anybody know where this formula came from for the FDR:

(3/4)0.05 = 0.0375 for n = 2 and (21/40)0.05 = 0.02625 for n = 20

I would really like to know as it would help greatly with my statistics that I'm conducting.
 
#8
@mad.casual

Do you mean, we can not use Bonferroni correction for a non parametric analysis?. For example in a Friedman's test.
 
#9
Wilcoxon's Test is non-parametric and more conservative than parametric methods already. Bonferroni correction adjusts parametric methods to account for more liberal use. Essentially, you just divide your significance level by the number of tests you run so that any one test is sufficiently stringent such that all your tests as a group are right (1-significance level)% of the time.

e.g. I want to be certain that my 65 hypotheses about a data set are correct 95% of the time, so each test should be tested at the 5%*1/65 or .077% confidence level to ensure that the failure of all the tests as whole is 65*.077% or 5%.
thanks for your answer, just to be sure: you ment 0.05 divided 65 comparisons? and not 5/65? otherwise you increase the level of significance with 0.077...