Yep, it is 20.000 per arm, sorry. The reference it's about an application of
the method, and I cannot download the paper with the main algorithm, so I don't
know exactly how they did.
Thanks everybody for the rich and interesting suggestions. Through free web
software (PS, others) I found also an N around 47.000 per arm. I guess these
are the values (also seen Marc's Monte Carlo).
Maybe the Poisson models approach suggested by David can be an alternative,
even if I guess at this point I won't get big differences in numbers. Would I?
Thanks a lot everybody again for your suggestions,
if anybody has other comments, they are always welcome.
Best,
Giulio
> Subject: Re: [R] Sample size calculation for differences between two very
> small proportions (Fisher's exact test or others)?
> From: marc_schwa...@me.com
> Date: Mon, 8 Nov 2010 11:13:12 -0600
> CC: perimessagg...@hotmail.com; r-h...@stat.math.ethz.ch
> To: mmal...@gmail.com
>
> Hi,
>
> I don't have access to the article, but must presume that they are doing
> something "radically different" if you are "only" getting a total sample size
> of 20,000. Or is that 20,000 per arm?
>
> Using the G*Power app that Mitchell references below (which I have used
> previously, since they have a Mac app):
>
> Exact - Proportions: Inequality, two independent groups (Fisher's exact test)
>
> Options: Exact distribution
>
> Analysis: A priori: Compute required sample size
> Input: Tail(s) = Two
> Proportion p1 = 0.00154
> Proportion p2 = 0.00234
> á err prob = 0.05
> Power (1-â err prob) = 0.8
> Allocation ratio N2/N1 = 1
> Output: Sample size group 1 = 49851
> Sample size group 2 = 49851
> Total sample size = 99702
> Actual power = 0.8168040
> Actual á = 0.0462658
>
>
>
>
> Using the base R power.prop.test() function:
>
> > power.prop.test(p1 = 0.00154, p2 = 0.00234, power = 0.8)
>
> Two-sample comparison of proportions power calculation
>
> n = 47490.34
> p1 = 0.00154
> p2 = 0.00234
> sig.level = 0.05
> power = 0.8
> alternative = two.sided
>
> NOTE: n is number in *each* group
>
>
>
> Using Frank's bsamsize() function in Hmisc:
>
> > bsamsize(p1 = 0.00154, p2 = 0.00234, fraction = .5, alpha = .05, power = .8)
> n1 n2
> 47490.34 47490.34
>
>
>
> Finally, throwing together a quick Monte Carlo simulation using the FET, I
> get:
>
> TwoSampleFET <- function(n, p1, p2, power = 0.85,
> R = 5000, correct = FALSE)
> {
> MCSim <- function(n, p1, p2)
> {
> Control <- rbinom(n, 1, p1)
> Treat <- rbinom(n, 1, p2)
> fisher.test(cbind(table(Control), table(Treat)))$p.value
> }
>
> # Run MC Replicates
> MC.res <- replicate(R, MCSim(n, p1, p2))
>
> # Get p value at power quantile
> quantile(MC.res, power)
> }
>
>
> # 50,000 per arm
> > TwoSampleFET(50000, p1 = 0.00154, p2 = 0.00234, power = 0.8, R = 500)
> 80%
> 0.04628263
>
>
>
> So all four of these are coming back with numbers in the 48,000 to 50,000
> ***per arm***.
>
>
> HTH,
>
> Marc Schwartz
>
>
> On Nov 8, 2010, at 10:16 AM, Mitchell Maltenfort wrote:
>
> > Not with R, but look for G*Power3, a free tool for power calc,
> > includes FIsher's test.
> >
> > http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
> >
> > On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni
> > <perimessagg...@hotmail.com> wrote:
> >>
> >>
> >> Hi,
> >> I'm try to compute the minimum sample size needed to have at least an 80%
> >> of power, with alpha=0.05. The problem is that empirical proportions are
> >> really small: 0.00154 in one case and 0.00234. These are the estimated
> >> failure proportion of two medical treatments.
> >> Thomas and Conlon (1992) suggested Fisher's exact test and proposed a
> >> computational method, which according to their table gives a sample size
> >> of roughly 20000. Unfortunately I cannot find any software applying their
> >> method.
> >> -Does anyone know how to estimate sample size on Fisher's exact test by
> >> using R?
> >> -Even better, does anybody know other, maybe optimal, methods for such a
> >> situation (small p1 and p2) and the corresponding R software?
> >>
> >> Thanks in advance,
> >> Giulio
>
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