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Re: st: How does -mcc- compute the CI for the difference in proportions?


From   Steven Samuels <[email protected]>
To   [email protected]
Subject   Re: st: How does -mcc- compute the CI for the difference in proportions?
Date   Wed, 18 Apr 2007 11:24:49 -0400

The formula should obviously be:
 Var (pc -pd) =  [Var(pc) + Var(pd) -2 Cov(pc,pd)]
              = [Pc (1 - Pc) + Pd (1 - Pd) + 2 Pc Pd]/n

The numbers are still okay.

Steve
On Apr 18, 2007, at 11:11 AM, Steven Samuels wrote:

Here are some paired data:
| test_2
test_1 | 0 1 | Total
-----------+----------------------+----------
0 | 1 2 | 3
| 10.00 20.00 | 30.00
-----------+----------------------+----------
1 | 3 4 | 7
| 30.00 40.00 | 70.00
-----------+----------------------+----------
Total | 4 6 | 10
| 40.00 60.00 | 100.00

mcc test_1 test_2 gives:
Proportion with factor
Cases .7
Controls .6 [95% Conf. Interval]
--------- --------------------
difference .1 -.4338565 .6338565

The CI is symmetric. We can recover the the standard error for the difference as the CI length divided by 2 x 1.96. This gives SE = . 27237578

Let the population values of the proportions in the table be Pa, Pb, Pc, Pd, with sample values pa=.1, ,pb=.2, pc=.3, pd=.4. The theory for a multinomial distribution gives:

n x Var(pc -pd) = Var(pc) + Var(pd) -2 Cov(pc,pd)
= Pc (1 - Pc) + Pd (1 - Pd) + 2 Pc Pd

With this formula, the sample standard error for pc-pd would be . 2213594 and the 95% CI [-.3338645 .5338645]. Why the difference?

Steve


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