Power for a three arm experiment (created 2009-09-14)

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"I want to compute power for a three arm experiment. The outcome variable is binary (yes/no). I know how to compute power for a two-arm experiment already, but have no idea how to handle the third arm."

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 software, PiFace,


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.


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 example, 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, 300 patients total.

Here's how that data would be entered into PiFace program

The Chi2* represents the chisquare statistic computed on the idealized data set shown above and n* represents the row totals in the idealized data set.

Here's how the data would be entered into G*Power:

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). For PiFace, you don't have to do this second calculation. Just move the slider back from 300 total patients to 150 total patients.

For G*Power, enter 4.76 rather than 9.52.

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 would encourage you to apply a Bonferroni correction (show adequate power for alpha = .05/3 = .0167). The Bonferroni adjustment is controversial, but 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, then 0.2 and 0.4, and finally 0.3 and 0.4. This you already know how to do.