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multiple comparison of proportions?

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jacques...@gmail.com

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Sep 6, 2007, 9:00:32 AM9/6/07
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Dear stat.consult folk

I have 2 samples (before/after) of a nominal/categorical variable
which assumes 6 possible values.
The before and after samples have different sizes.

I would like to check which categories changes with 95% confidence.

What I did was to calculate the Wald interval for each proportion
(before and after) and those whose interval had no intersection were
significantly different..

Then I realized

1) This looks similar to a multiple comparison procedure (for
averages) and I am using the t-test for each pair of comparison. Is
there some better way to compare the proportions?

2) I realized I could have used a X-square on the contingency table,
but that would only tell me whether the whole table is contingent or
not. How to determine which lines of the table are responsible for the
"non-contingency".

To summarize, can someone point me to some literature on multiple
comparison of proportions?


Thanks

jacques

Ray Koopman

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Sep 9, 2007, 4:30:00 PM9/9/07
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On Sep 6, 6:00 am, "jacques.wai...@gmail.com"

I take it that you have two independent sets of multinomial
observations, (x1,...,x6) and (y1,...y6), with parameters
(m; p1,...,p6) and (n; q1,...,q6).

In general, testing for differences by looking for non-overlapping
confidence intervals is valid but weak. That is, differences that
are declared significant would also be declared significant by a
proper test, but some differences that are declared nonsignificant
would be declared significant by a proper test.

The proper test for category j would be based on the usual single-df
chi-square for the 2 x 2 contingency table ((xj,m-xj),(yj,n-yj)).

For post-hoc tests, probably the most powerful procedure is the
stepwise Bonferroni, which compares the p-value for the largest
chi-square to alpna/6, then the p-value for the next-largest
chi-square to alpha/5, etc, stopping the testing as soon as a
nonsignificant result is obtained.

Scheffe's procedure, which is generally less powerful, would treat
all the chi-squares as if they had 5 df, and compare their p-values
to alpha.

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