🚀 go-pugleaf

Bob Parker (PARKERRA@VUCTRVAX.BITNET) wrote:
: For p1 = 0.01, p2 = 0.005, n1 = 2 * n2, Systat/Design gives n1=7888,
: n2=3944.
Greg Maislin (FGM@ASC.UPENN.EDU)  asked, in particular
>
>          I realize that the specifics for this problem will result in
>          unfeasible required samples sizes but I am still interested
>          in trying to solve the general problem.  Any help will be

If you have a pilot study, or a constructed example of the 
expected outcome, then estimating power is easy, for simple tests 
based on the normal z or on a 1 d.f. chi squared (which is just 
the square of a normal variate).  The only book or table you need
is the cumulative distribution of the normal curve, i.e., z-scores.

If you have a pilot sample, or model example, with z (or t) 
calculated, then that is the "noncentrality" (NC) that you say you 
expect to observe, based on so many subjects.  The noncentrality
increases when increasing either the sample sizes or distance 
between the means.

What "noncentrality" do you need to obtain?  For the normal, this 
is computed quite simply as the sum of the z-scores describing your 
desired power, and (1-alpha)  [ (1-alpha/2) for a two-tailed test].

Thus, noncentrality for a two-tailed, 5% alpha, at 80% power, is

        z(.975)+ z(.80)        
         (1.96 +   .842) = 2.802 = NC, noncentrality parameter.

What change in sample size does it take to rescale your 
observed t or z into NC?  Take the ratio of NC to z and square it,
and that is what you need to multiply your example-sample size by.
The computing is even easier for a chisquared test.

The noncentrality for a 1 d.f. chisquared is simply the square of
the corresponding NC, or,   2.802 x 2.802 = 7.8512  for the example.
Since chisquared is proportionate to sample size, the ratio of 
7.8512 to the computed chisquare is what you multiply the example-
size by.

      Examples of 10000 vs 5000

             x  not-x                  x  not-x
      1.5%  150  9850           1.0%  100  9900
       .5%   25  4975            .5%   25  4975
    
    chisquared= 28.909        chisquared= 10.084
    7.8512/28.909= .2716      7.8502/10.084= .7775
    -> 2716 vs. 1358          -> 7786 vs. 3893
      (2700 vs. 1350)           (7800 vs. 3900)

These samples are smaller than what another user reported to 
Usenet, using a computer program doing computations based on the
arc-sine transformation, which is also what Cohen's classic book 
uses for proportions.  I stopped using that chapter years ago, 
after noting, I think I remember, that the test was far too generous 
when comparing moderate proportions to tiny ones.  The reported
results suggest that it is also far too conservative when comparing
among these tiny one.

		Rich Ulrich, wpilib@vms.cis.pitt.edu
		Research Assistant Professor of Psychiatry
		Western Psychiatric Institute and Clinic
		University of Pittsburgh
		Pittsburgh, Pa.   15213