sample size calculation three-arm trial

[Sandra Alba]

Hi all,

I have been asked to calculate the sample size for a 3-armed trial to assess which type of delivery method leads to higher ownership of mosquito nets in an African (clustered) community. All I have found on the internet and in books refers to 2 armed trials and I can't find any information on calculations for multi armed trials. Can anybody suggest anything? (open access references more welcome of course!)

Thank you for any help,

Sandra

[Bruce Weaver]

What kind of model is it? Logistic regression (owns nets: Yes/No)?
What are the 3 arms? Can you boil things down to an important
contrast (or contrasts) between two of them?

--
Bruce Weaver
bwe...@lakeheadu.ca
http://sites.google.com/a/lakeheadu.ca/bweaver/Home
"When all else fails, RTFM."

[Sandra Alba]

Hi,

Yes, it's a binary outcome - ownership of nets yes or no. Therefore I want to compare the prevalence in each arm.

arm 1: distribute new long lasting insecticide treated nets (LLITN) to everyone  without re-impregnating the old ones
arm 2: distribute new  (LLITN) to everyone and re-impregnation of the old ones
arm 3: only distribute exact number of nets requested by community members with no reimpregnation

Sandra

2009/9/14 Bruce Weaver <bwe...@lakeheadu.ca>

[Martin Holt]

Hi Sandra,

 

When there are more than two treatments there is no longer one clear alternative hypothesis. This can lead to multiple significance tests and so to misleading p-values. One answer is to conduct a global test such as an ANOVA, only looking at pairwise comparisons if the global test is significant. Or to use conventional tests and Bonferroni (eg).The simplest strategy is to look at the three-way comparison as three independent trials, and to use conventional significance tests without adjustment (Saville,1990). So the sample size is calculated as if three independent trials are to be tried, and then for each treatment group take the largest number as the sample size.

 

HTH,

Martin Holt (precised from Machin et al, sample size tables for clinical studies)

 

Saville , DJ (1990) Multiple comparison procedures: The practical solution. The American Statistician, 44, 174, -180

 

I know this sort of answer is like a red rag to a bull to Bendix Cartesen :) who might help you out with how to perform simulations of your situation. (I'm all for it, but don't have his ability to do it. To help people like me, he kindly put examples on the website)

[Steve Simon, P.Mean Consulting]

Sandra Alba wrote:

> Yes, it's a binary outcome - ownership of nets yes or no. Therefore I
> want to compare the prevalence in each arm.
>
> arm 1: distribute new long lasting insecticide treated nets (LLITN) to
> everyone without re-impregnating the old ones
> arm 2: distribute new (LLITN) to everyone and re-impregnation of the
> old ones
> arm 3: only distribute exact number of nets requested by community
> members with no reimpregnation

Martin Holt gives excellent advice, but let me elaborate a bit on it.
You can arrange your data from this experiment as a 3 by 2 table with
the 3 rows representing the three arms and the two columns representing
yes/no to ownership of nets.

You can test independence of the rows and columns using a chisquare
test, and Russ Lenth's power and sample size page

http://www.math.uiowa.edu/~rlenth/Power/

Is a nice place to start and the price is right. I believe that G*Power
will also do a power calculation for a chisquare test.

http://www.psycho.uni-duesseldorf.de/aap/projects/gpower/

G*Power is also free.

For a chisquare test, you need to estimate the non-centrality parameter
by creating some fake data that corresponds to the proportions of yes/no
in each group that you think represents a clinically relevant shift. For
exapmle, you might hypothesize that if arms 1, 2, and 3, have
proportions 0.2, 0.3, and 0.4, that you would have to have to look at a
chisquare statistic for a table that looks something like

20 80
30 70
40 60

The expected counts here would be

30 70
30 70
30 70

and the chisquare statistic for this artificial setting would be

10^2/30+10^2/70+0^2/30+0^2/70+10^2/30+10^2/70 = 9.52

That's your noncentrality parameter, assuming 100 patients per group.
For 50 patients per group, recalculate using this table

10 40
15 35
20 30

to get a non-centrality parameter of 4.76 (it's not just a coincidence
that this is half the size of the previous example).

Now this seems like a pain in the neck, and it is, so what would be a
simpler solution? One thing you might notice is that a 3 by 2 table
contains several 2 by 2 tables as subsets (three 2 by 2 tables to be
precise). If the sample size provides adequate power for each possible 2
by 2 table then surely that should be good enough to satisfy even the
pickiest grant reviewer.

I differ from Martin Holt in that I would apply a Bonferroni correction
(show adequate power for alpha = .05/3 = .0167). The price paid here is
small, but it adds credibility to your grant proposal.

So figure out power for a two by two table where the two proportions are
0.2 and 0.3, 0.2 and 0.4, and 0.3 and 0.4. This you already know how to do.

I hope this helps.
--
Steve Simon, Standard Disclaimer
Free statistics webinar, Wed, Oct 14, 10am CDT.
"P-values, confidence intervals, and the Bayesian alternative"
Details at www.pmean.com/webinars

[Barry Brown]

The  free program  STPLAN  handles the  k-sample  binomial case.   The
program can be downloaded from

biostatistics.mdanderson.org/SoftwareDownload

Here is  a transcript of the  answer provided for a  sample problem in
which I made up the values.  The power is against the null  hypothesis
that all probabilities  are equal -- many sided test.

 
                        ***** K-sample binomial *****                         


 All the groups are of equal size

 Probability of the first  2 groups ...... 0.300
 
 Group  3 probability .................... 0.500
 
 Significance ............................ 0.050
 
 Power ................................... 0.800
 
 The Number in group 1 is calculated to be 83.902
                                group  2 is 83.902
                                group  3 is 83.902



--
Barry W Brown
Professor, Biostatistics and Applied Mathematics
M. D. Anderson Cancer Center

[Björn]

On 14 Sep., 22:23, "Steve Simon, P.Mean Consulting" <n...@pmean.com>
wrote:

>
> I differ from Martin Holt in that I would apply a Bonferroni correction
> (show adequate power for alpha = .05/3 = .0167). The price paid here is
> small, but it adds credibility to your grant proposal.

The Bonferroni-Holm precedure is more powerful, so I would by default
apply that (for power calculations it's easiest to just simulate
things).

Regarding all the power calculations, it looks to me like some of the
answers are overlooking that this is (from what I read above) a
cluster randomized trial. I.e. I have the impression that the three
distribution methods would be assigned to different regions/
communities and then one would compare the proportions of owners of
nets between these regions/communities, which I fear means a higher
sample size, doesn't it?

[Sandra Alba]

Dear all,

thanks for very insightful responses to my query! I am certainly
learning a lot from all your comments.

As Björn pointed out, the data is indeed clustered, in villages and
"concessions" (extended families separated by a wall and surrounding
fields with average of 12 ppl). I decided to apply a correction factor
of 3 as I was once told that a correction factor between 1 and 2 for
each level of clustering is a reasonable rule of thumb when you have
no idea what the ICCs are. I would welcome people's thoughts on
this...

(I don't know what type of analysis will be carried out but my
impression is that they will not look at differences between
villages/concession but just look at net ownership overall)

Best,

Sandra

2009/9/15 Björn <bjoernh...@googlemail.com>

[Martin Holt]

Hi Sandra,

 

From your last statement below, I'm not sure that you are looking for help in a clustered environment. If you are, or if you're just interested, Pfof. Martin Bland's homepage has a number of useful references. I've selected one here:

 

http://www-users.york.ac.uk/~mb55/clust/clustud.htm

 

I'm confident this site will tell you what you need to know.

 

Best Wishes,

 

Martin Holt

----- Original Message -----

From: Sandra Alba

To: meds...@googlegroups.com

Sent: Tuesday, September 15, 2009 9:03 AM

Subject: {MEDSTATS} Re: sample size calculation three-arm trial