At the risk of oversimplifying the study design, this sounds like a two
sample comparison of proportions, in which case power.prop.test() would
be the function of interest. This could also be done via Monte Carlo
simulation, which would not be difficult to implement.
Note that I am also presuming that we are not in fact talking about the
*rate* of events in the two groups, as there was no mention of the time
to the events being a consideration.
I am going to make the further assumption that the phrase "significant
difference" as used here means a target chi-square test p value of
<=0.05, where the null is that the two proportions are equal and the
alternative is that they are not equal.
That being the case, then:
> power.prop.test(p1 = .06, p2 = .10, power = 0.8)
Two-sample comparison of proportions power calculation
n = 720.9169
p1 = 0.06
p2 = 0.1
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
See ?power.prop.test for more information.
If my assumptions are not correct, please post back with further
information about the study design.
HTH,
Marc Schwartz
on 07/31/2008 02:11 AM Moshe Olshansky wrote:
Hello Jason,
You are not specific enough. What do you mean by "significant
difference"? Let's assume that indeed the incidence in A is 6% and in
B is 10% and
we are looking for Na and Nb such that with probability of at least 80%
the mean of Nb sample from B will be at least, say, 0.03 (=3%) above the
mean of Na sample from A.
The solution is not unique. If Mb is the mean of the sample from B
and Ma is the one from A, using
Normal approximation we get the Mb is approximately normal with mean
0.10 and variance 0.1*0.9/Nb and Ma is approximately normal with mean
0.06 and variance 0.06*0.94/Na, so Mb - Ma is approximately normal with
mean 0.04 and variance 0.09/Nb + 0.0564/Na. So let V be the maximal
variance for which the probability that a normal rv with mean 0.04 and
variance V is above 0.03 equals 0.80 (finding such V is
straightforward). Then you must choose Na and Nb which satisfy 0.09/Nb +
0.0564/Na <= V. One such choice is Nb = 2*0.09/V, Na = 2*0.0564/V.
As I said, this solution is only approximate and probably not
optimal,
so see what other people say.
Regards,
Moshe.
--- On Thu, 31/7/08, Iasonas Lamprianou <[EMAIL PROTECTED]> wrote:
From: Iasonas Lamprianou <[EMAIL PROTECTED]>
Subject: [R] stats question
To: r-help@r-project.org
Received: Thursday, 31 July, 2008, 2:46 PM
Dear friends,
I am not sure that this is the right place to ask, but
please feel free to suggest an alternative discussion
group.
My question is that I want to do a comparative study in
order to compare the rate of incidence in two populations. I
know that a pilot study was conducted a few weeks ago and
found 8/140 (around 6%) incidence in population A.
Population B was not sampled. Assuming this is (about) the
right proportion in the Population A what is the sample
size I need for population A and B in the main study, in
order to have power of 80% to idenitfy
significant differences? I would expect the incidence in
population B to be around 10% compared to the 6% of the
Population A.
Any suggestions? How can I do this in R?
Jason
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