Re: Bonferroni correction and number of comparisons

Posted by Ware, William B on Jan 19, 2011; 9:14pm
URL: http://spssx-discussion.165.s1.nabble.com/Bonferroni-correction-and-number-of-comparisons-tp3348337p3348617.html

I agree with what Ryan has said, but I believe that the "straight-up Bonferroni" correction is overly conservative.  What about considering a sequential approach such as the Holm modification of the Bonferroni correction?

wbw


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-----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of R B
Sent: Wednesday, January 19, 2011 3:58 PM
To: [hidden email]
Subject: Re: Bonferroni correction and number of comparisons

You asked about  applying a Bonferroni correction in post hoc tests-->
"Is the Bonferroni adjustment for 6 comparisons and the corrected
p-value (.05/6=.008333)"

Answer. No. You have presumably adjusted the alpha level, not the
p-values. Multiply the each p-value by the number of post hoc tests
performed (6 in the example you provided). Then compare the adjusted
p-values to the alpha level you set (e.g., .05).

Ryan

On Wed, Jan 19, 2011 at 1:56 PM, J McClure <[hidden email]> wrote:

> Hi,
> I am doing a post hoc analysis comparing column proportions and
> adjusting (Bonferroni) for multiple comparisons (CTables, test
> statistics option).
> I've been asked to provide the corrected p-value.
> I'd like to know if I am correct in reporting the corrected value as .008.
> The row variable has two levels (yes and no) and the column variable has
> 4 levels, so I am making 6 comparisons for 'no' and 6 for 'yes'.  n*(n-1)/2
> The output shows the comparison for the 'no'  and for the yes level of
> the row variable. Is the Bonferroni adjustment for 6 comparisons and the
> corrected p-value (.05/6=.008333)
> Also, I have 19 row variables for which I am running column proportion
> comparisons. I think that the comparisons for each of the 19 variables
> are considered independent from each other so no further adjustment is
> made by SPSS.  Is this the way reviewers look at the question?
> Thanks for any help,
> Jan
>
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