Sample size with correction for multiple testing

[la volta statistics]

Hi all

I want to do a sample size calculation for comparing counts of two
groups (Poisson distribution)
I found the following calculator in the net giving me 365 cases per
group for lamda1 = 0.15 and lambda2 = 0.24:

http://www.stattools.net/SSiz2Counts_Pgm.php


Probability of Type I Error (alpha) = 0.05
Power (1 - beta) = 0.8
Mean count in group 1 = 0.15
Mean count in group 2 = 0.24
Tail = 2
Sample size per group = 365


Since it is planned to additionally analyze 4 subgroups within the
main group, I have to adjust the sample size for multiple testing.

If I do a Bonferroni correction by dividing the alpha by 4 (= 0.0125)
I get:

Probability of Type I Error (alpha) = 0.0125
Power (1 - beta) = 0.8
Mean count in group 1 = 0.15
Mean count in group 2 = 0.24
Tail = 2
Sample size per group = 518

Am I correct that I need 518 cases per group to analyze the results at
the alpha = 0.05 level, corrected for multiple testing? Somebody told
me I additionally need to multiply this sample size by 4 (4 x 518)
resulting in 2590 cases. What is correct?

TIA, Christian

=====================
To manage your subscription to SPSSX-L, send a message to
[hidden email] (not to SPSSX-L), with no body text except the
command. To leave the list, send the command
SIGNOFF SPSSX-L
For a list of commands to manage subscriptions, send the command
INFO REFCARD

[Rich Ulrich]

"What is correct" is hard to say.  You need a clearer statement
of the problem, and what you are doing with the "4 subgroups."

Let's say your first two groups are A and B. 

Are these extra tests between B1 to B4, comparing to A?
 - That would be a design where the best power for the
smallest N would use more cases in A.  (See" Dunnett's test")

Are these extra tests between all the possible contrasts of four
groups, where the groups are drawn from the pool of A and B?
 - There are 6 tests.  Bonferroni correction may be conservative
since the tests are not independent.  A much easier criterion
would be to consider the hypothesis that *some*  contrast is
going to be significant. 

Do you really consider all these tests (1+6) to be *equal*
in importance to you, so that they make up one "family" that
should be corrected by Bonferroni (or whatever)?
 - If "Test 1" is as important as the set in "Test 2", then one
proper partition of the family-wise 5% risk might be to assign
half that risk to Test 1: using a nominal test of 2.5%;  and to
assign the other 2.5% to the set of other tests.  - Of course,
if the sample is already going to be huge in order to test
the 4 subgroups, you don't have to worry so much about the
N for A versus B.  But the logical ordering of the test is what
properly comes first, and then you follow that. 

4*518 is not 2590, so I don't know what your friend
was asking or suggesting.  Did they really mean to say,
4 * 4 * 518   for some reason, for the total N for the
second set of tests?

--
Rich Ulrich

> Date: Tue, 26 Jun 2012 10:07:42 +0200

> From: [hidden email]
> Subject: Sample size with correction for multiple testing
> To: [hidden email]
>
> Hi all
>
> I want to do a sample size calculation for comparing counts of two
> groups (Poisson distribution)
> I found the following calculator in the net giving me 365 cases per
> group for lamda1 = 0.15 and lambda2 = 0.24:
>
> http://www.stattools.net/SSiz2Counts_Pgm.php
>
>
> Probability of Type I Error (alpha) = 0.05
> Power (1 - beta) = 0.8
> Mean count in group 1 = 0.15
> Mean count in group 2 = 0.24
> Tail = 2
> Sample size per group = 365
>
>
> Since it is planned to additionally analyze 4 subgroups within the
> main group, I have to adjust the sample size for multiple testing.
>
> If I do a Bonferroni correction by dividing the alpha by 4 (= 0.0125)
> I get:
>
> Probability of Type I Error (alpha) = 0.0125
> Power (1 - beta) = 0.8
> Mean count in group 1 = 0.15
> Mean count in group 2 = 0.24
> Tail = 2
> Sample size per group = 518
>
> Am I correct that I need 518 cases per group to analyze the results at
> the alpha = 0.05 level, corrected for multiple testing? Somebody told
> me I additionally need to multiply this sample size by 4 (4 x 518)
> resulting in 2590 cases. What is correct?
>
> TIA, Christian
> ...