sample size
Richard F Ulrich
: For p1 = 0.01, p2 = 0.005, n1 = 2 * n2, Systat/Design gives n1=7888,
: n2=3944.
: Bob Parker
Richard F Ulrich
: For p1 = 0.01, p2 = 0.005, n1 = 2 * n2, Systat/Design gives n1=7888,
: n2=3944.
>
> 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, wpi...@vms.cis.pitt.edu
Research Assistant Professor of Psychiatry
Western Psychiatric Institute and Clinic
University of Pittsburgh
Pittsburgh, Pa. 15213