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

[Steven Samuels <ssamuels@albany.edu>]

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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