From: Marc Schwartz <marc_schwartz_at_comcast.net>

Date: Thu, 31 Jul 2008 07:48:51 -0500

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.

so see what other people say.

>> From: Iasonas Lamprianou <lamprianou@yahoo.com>

*>> Subject: [R] stats question
*

*>> To: r-help_at_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
*

*>>
*

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Thu 31 Jul 2008 - 13:07:49 GMT

Date: Thu, 31 Jul 2008 07:48:51 -0500

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 <lamprianou_at_yahoo.com> wrote: >

>> From: Iasonas Lamprianou <lamprianou@yahoo.com>

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Thu 31 Jul 2008 - 13:07:49 GMT

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