Displaying 1 result from an estimated 1 matches for "homework_2".
Did you mean:
homework_1
2008 Feb 21
1
anova power calculations
...2)
}
noncentral.param = noncentral.param / within.var
# Probability of central quantile in noncentral distribution
noncentral.p = pf(central.quant, J - 1, N - J, noncentral.param,
lower.tail= FALSE)
noncentral.p
The logic behind this is in my assignment at:
http://odin.himinbi.org/classes/psy304b/homework_2.xhtml#p2b
This works for a balanced ANOVA and gives the same result as
power.anova.test (and SAS). For the unbalanced ANOVA though it is giving me
a different result though than SAS, 0.8759455 versus 0.680.
So is there a straightforward way to compute the power of an unbalanced
ANOVA? If there is...