Will,
Your SAS input indicates that within standard deviation is 9, not the
variance. If you use within.var=81 in your R statement you will get the
answer matching SAS.
Cheers,
Andy
__________________________________
Andy Jaworski
518-1-01
Process Laboratory
3M Corporate Research Laboratory
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Tel: (651) 733-6092
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"Will Holcomb"
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Subject
[R] one-way anova power
02/19/2008 10:19 calculations
AM
I have been attempting some basic power calculations using R and I am not
getting the results I expect. I had a homework assignment in SAS, but I
want
to learn R as well, so I was attempting to reproduce my result. (No one
else
in the class is doing R, so there's no need to obsfucate the answer, the
SAS
code is what I get my grade for.) The code I am using is:
# You assume that the within-population standard deviations all equal 9.
You
set the Type 1 error rate at
# alpha = .05. You presume that the population means will have the
following
values: 17.5, 19, 25, 20.5.
# You intend to run 80 subjects in all, with equal n's across all 4 groups.
Compute your power to reject
# the null hypothesis under these conditions.
means = c(17.5, 19, 25, 20.5)
power.anova.test(groups = length(means), n = 80 / length(means),
between.var = var(means), within.var = 9,
sig.level = 0.05)
The result I'm getting is:
Balanced one-way analysis of variance power calculation
groups = 4
n = 20
between.var = 10.5
within.var = 9
sig.level = 0.05
power = 1
NOTE: n is number in each group
I've already done the calculation in SAS with the following code:
data Dep;
Input cue $ mean weight;
datalines;
A 17.5 1
B 19 1
C 25 1
D 20.5 1
;
proc glmpower;
class cue;
model mean = cue;
weight weight;
power
stddev = 9
alpha = 0.05
ntotal= 80
power = .;
run;
This produces a power of 0.616 which is the answer I was supposed to get.
Any idea what I need to change?
Will
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