On 3 Jun 2001, Bekir wrote, in part:
> My aim was to compare groups 2, 3, 4, 5 with control (group 1)". ...
>
> The rewiever had written me: "Accordingly, a "statistical penalty"
> needs to be paid in order to account for the increased risk of a Type
> 1 error due to multiple comparisons. The easist way to achieve this
> goal to adjust the P value require to declare significance using the
> Bonferroni correction."
>
> 1. What is the correct meaning of the last sentence? What must I do?
> As you wrote, must I find the adjusted p values or declare the
> adjusted significance level alpha?
Your choice: the two ways of approaching the problem are equivalent.
Either divide the criterion significance level (alpha) by the number
of comparisons, as Duncan Smith recommended, and compare the p-values
reported by your statistical routine to this adjusted value; or
adjust the p-values by multiplying the reported values by the number of
comparisons. Thus p = 0.02 > adjusted alpha = 0.125 for one of your
comparisons, or adjusted p = 0.08 > nominal alpha = 0.05. If I
understand your reviewer correctly, (s)he seems to be requesting the
latter: adjusting the p-value.
> 2. There are apparently and exactly three groups; groups 1, 3, 5 that
> had the same proportions of translocation.
Wasn't it groups 1, 4, 5 that had the same proportions?
> Therefore, to compare only the group 2 and 3 with the control can be
> appropriate, can it be?
Such a comparison may be appropriate (but see below); but this does not
change the situation. Had groups 4 and 5 NOT had proportion equal to
group 1, you would surely have wanted to make those two comparisons also.
The question is not, how many comparisons were useful or significant;
but how many comparisons would you have chosen to consider before you
observed the results of this particular experiment. By your description,
you certainly considered AT LEAST the four comparisons mentioned in your
first paragraph above.
> Thus, there would be two comparisons and the p valus 0.008 (0.008 x 2
> = 0.016) and 0.02 (0.02 x 2 = 0.04)would be significant. Is it right?
As explained above, and as Duncan Smith responded, No.
Duncan mentioned Dunnett's test. This might indeed be appropriate for
your design, but not for the analyses you have so far done. Dunnett's
test would normally follow the finding of a significant F value in a
one-way analysis of variance (testing the formal hypothesis that the
true (population) proportions in the five groups are all identical.
Such an analysis could be undertaken with your data, but some persons
(possibly including your reviewer? I don't know) would object to
carrying out an analysis of variance (ANOVA) with dichotomous data.
One advantage to ANOVA is the possibility of drawing conclusions more
complex, and possibly more interesting, than the pairwise comparisons
that you had originally envisioned. In particular, you could test the
contrast between Groups 2 and 3 combined, with Groups 1, 4, and 5
combined; since it seems clear that this is the only thing that is
going on in your data. Testing that contrast by the Scheffe' method,
which offers experimentwise protection against the null hypotheses for
any imaginable contrast, might be useful: that contrast, involving all
100 cases, is more powerfully tested than the series of pairwise
comparisons, and may well be significant even against the conservative
Scheffe' criterion.
Whether that is useful _for_your_purposes_ is another matter entirely.
If there is some useful meaning and interpretation to be gained in
observing that only groups 2 and 3 differ from the control group and
that groups 4 and 5 are indistinguishable from the control group, then
this contrast would be useful to test formally. If that outcome does
not lend itself to useful interpretation (and the advance of knowledge
in the field), you would probably be better off staying with the four
pairwise comparisons you started with.
------------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 603-535-2597
184 Nashua Road, Bedford, NH 03110 603-471-7128
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