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

[Steven Samuels <ssamuels@albany.edu>]

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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