clear all
close all
%
disp('Power Calculation for ANOVA')
lw = 3;
set(0, 'DefaultAxesFontSize', 16);
fs = 16;
msize = 10;
%
%Suppose $k=4$ treatment
%means are to be compared at a significance level of $\alpha = 0.05.$
%The experimenter is to decide how many replicates $n$ to run at each
%level, so that the null hypothesis is rejected with a
%probability of at least 0.9 if $f^2 = 0.0625,$
%or if $\sum_i \alpha^2_i$ is equal to the overall $\sigma^2.$
1-ncfcdf( finv(1-0.05, 4-1, 4*60-4), 4-1, 4*60-4, 15) %0.9122
%%
% Assume two factor fixed effect ANOVA. Test for factor A
% at a=4 levels is to be evaluated for its power.
% Factor B has b=3 levels.
% Assume balanced design with 4 x 3 cells, with n=20 subjects in
% each cell. Total number of subjects is N = 3 x 4 x 20 = 240.
%
% For alpha = 0.05, medium effect size f^2 = 0.0625, and
% lambda = N f^2 = 20*4*3*0.0625 = 15, the power is 0.9075.
%
1-ncfcdf( finv(1-0.05, 4-1, 4*3*(20-1)),...
4-1, 4*3*(20-1), 15) %0.9075
%%
close all
k=4; %number of treatments}
alpha = 0.05; % significance level}
y=[]; %set values of power for n
for n=2:100
y =[y 1-ncfcdf(finv(1-alpha, k-1, k*(n-1)), ...
k-1, k*(n-1), n/4)];
end
plot(2:100, y,'b-','linewidth',3)
hold on
xlabel('Group sample size n'); ylabel('Power')
print -depsc 'C:\BESTAT\ANOVA\ANOVAeps\powermat.eps'