6
$\begingroup$

I'd just like to get clarification on something. When you perform a bonferroni adjustment (dividing the alpha level by the amount of tests you want to do, if say you're doing multiple ANOVAs) do you just check the assumptions as recommended in whatever guide you're following with no adjustments/alterations, and ONLY apply the bonferroni adjusted alpha level to your main outcomes sig values?

Or, do you have to do/apply anything to the assumption test sig values in your analysis?

For example, for the assumption tests for ANOVA of Levene’s Test of Equality of Error Variances (wanting them to be non-significant / over .05), or Box’s M statistic ( wanting them to be non-significant / over .001) do you use the divided alpha for them? Or do you divide them by the amount of tests your doing if they're already lower than the usual .05 (as in Box's M's 0.001)? Or do you multiply them by the amount of tests your doing? I can't remember if there are any statistical assumption tests that necessitate a significant result (in my case I don't think there are, but for those who may be wondering a similar thing in that instance), but the same question applies to those too.

If you could provide links to simple explanations, or simple explanations, and ideally links to text/references/sources to cite, that would be much appreciated.

No specific degrees in maths or stats, so the simpler the better.

$\endgroup$
1
$\begingroup$

When one analyses the proof that Bonferroni controls the type I error ''family-wise'' then you see that no assumptions are needed; it basically uses only the inequality of Boole. So Bonferroni does not need e.g. an independence assumption.

However, the analysis of the proof learns that the probability of a type I error is at most $\alpha$, i.e. the Bonferroni method can have a type-I error probability that is stricly smaller than $\alpha$ (and this will result in a loss of power).

The cases where the probability of a type I error probability is strictly smaller than $\alpha$, (one says that in these cases Bonferroni is conservative) occur when the tests are dependent or when the p-values of the indiviual tests are themselves conservative. The latter can be the case for discrete random variables (in a univariate test for a Binomial variable e.g. the ''observed'' type I error probability may be strictly smaller than $\alpha$). .

Note that Holm's method also controls type I error probability and its power is at least as good as the one of the Bonferroni method. For discrete random variables, like the Binomial, other multiple test correction methods methods have been shown to be more powerful (e.g. minP).

So to summarise: the Bonferroni method does not need any additional assumptions to show that the type-I error probability is controled, however, it can be conservative.

$\endgroup$
1
$\begingroup$

The Bonferroni (and similar corrections like Bonferroni-Holm etc.) assumes that the p-value you provide it follows a uniform distribution under the null hypothesis and would under the null hypothesis only be below 0.05 in 5% of the time, if you repeated your experiment again and again. Pre-tests such as for normality of residuals lead to a violation of this assumption and thus, lead to an inflation of the familywise type I error rate even if your use Bonferroni. I am not aware of any way of avoiding that (e.g. reducing the significance level of the pre-test) and any procedures with such a pre-test tend to lead to type I error inflation (if you are lucky they are small though).

It seems strange to so strictly control it with e.g. Bonferroni (or probably better at least Bonferroni-Holm) and then to get your p-values in such a way that a supposed level $\alpha$ test ist not actually a level $\alpha$ test so that even with a Bonferroni correction you cannot have type I error control. If you truly want strict familywise type I error rate control, then either being pretty sure that the residuals are sufficiently close to normal to use AN(C)OVA or using non-parametric (permutation) tests from the start seems more logical.

$\endgroup$
-1
$\begingroup$

Adjusting for multiple comparisons generally applies to your actual hypotheses, not to tests of assumptions.

$\endgroup$
  • $\begingroup$ Not sure why this answer was downvoted, given that it appears to be the only answer that directly addresses the question. $\endgroup$ – Bonferroni Apr 18 '17 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.