On Apr 4, 2011, at 01:10 , Timothy Spier wrote:

I've been searching for an answer to this for a while but no joy. I have a simple 2-way ANOVA with an interaction. I'd like to determine the power of this test for each factor (factor A, factor B, and the A*B interaction). How can I do this in R? I used to do this with "proc Glmpower" in SAS, but I can find no analogue in R.

They're not massively hard to do by hand, if you know what you're doing (which, admittedly is a bit hard to be sure of in this case). The basic structure can be lifted from power.anova.test and the name of the game is to work out the noncentrality parameter of the relevant F tests. E.g., lifting an example from the SAS manual:

twoway <- cbind(expand.grid(exúctor(1:2),varúctor(1:3)),x=c(14,10,16,15,21,16))

with(twoway,tapply(x,list(ex,var),mean))

1 2 3

1 14 16 21

2 10 15 16

Now, you have 10 replicates of this with a specified SD of 5. If we do a "skeleton analysis" of the above table, we get

anova(lm(x~ex*var,twoway))

Analysis of Variance Table

Response: x

Df Sum Sq Mean Sq F value Pr(>F)

ex 1 16.667 16.6667

var 2 42.333 21.1667

ex:var 2 4.333 2.1667

Residuals 0 0.000

Warning message:

In anova.lm(lm(x ~ ex * var, twoway)) :

ANOVA F-tests on an essentially perfect fit are unreliable

In a 10-fold replication, the SS would be 10 times bigger, and the residual Df would be 54; also, we need to take the error variance of 5^2 = 25 into account. The noncentrality for the interaction term is thus 43.333/25 and you can work out the power as

pf(qf(.95,2,54),2,54,ncpC.333/25,lower=F)

[1] 0.1914457

Similarly, the main effect powers are

pf(qf(.95,2,54),2,54,423.333/25,lower=F)

[1] 0.956741

pf(qf(.95,1,54),1,54,166.66667/25,lower=F)

[1] 0.7176535

(whatever that means in the presence of interaction, but that is a different discussion)

--

Peter Dalgaard

Center for Statistics, Copenhagen Business School

Solbjerg Plads 3, 2000 Frederiksberg, Denmark

Phone: (+45)38153501

Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com