On Mon, 11 Apr 2005, Isaac Neuhaus wrote:
Here are the basic data:
f1 f2 response
S1 1 1 7
S6 1 2 9
S8 1 2 10
S2 2 1 1
S3 3 1 4
S5 3 2 5
S7 3 2 5
S4 4 2 5
X = [ 1, 1, 0, 0, 1, 0;
1, 1, 0, 0, -1, 0;
1, 1, 0, 0, -1, 0;
1, 0, 1, 0, 1, 0;
1, 0, 0, 1, 1, 1;
1, 0, 0, 1, -1, -1;
1, 0, 0, 1, -1, -1;
1, -1, -1, -1, -1, 1];
Y = [7;9;10;1;4;5;5;5]
I was looking at residual sums of squares (RSS) like this:
b=X\Y; e=Y-X*b; e'*e
octave:48> b=X\Y; e=Y-X*b; e'*e
ans = 0.500000000000000
That is the correct error SS and it is equal to the answer given by
SAS (it is on 2 df, so MSE = 0.50/2 = 0.25). See below.
octave:49> b=X(:,1:5)\Y; e=Y-X(:,1:5)*b; e'*e
ans = 1.25000000000000
That is the RSS for the interaction term. Subtract the error SS and
you'll have 0.75, which is again the SAS answer.
But then I run into a problem computing Type III SS for the other
effects, probably because I'm doing it the wrong way. I've really got
to return to some other work now. Maybe someone else on the list can
figure this out. We're trying to get this:
Source DF Type III SS Mean Square F Value Pr > F
R 3 36.30000000 12.10000000 48.40 0.0203
T 1 4.08333333 4.08333333 16.33 0.0561
R*T 1 0.75000000 0.75000000 3.00 0.2254
From these data:
Y = [7;9;10;1;4;5;5;5]
X = [ 1, 1, 0, 0, 1, 0;
1, 1, 0, 0,-1, 0;
1, 1, 0, 0,-1, 0;
1, 0, 1, 0, 1, 0;
1, 0, 0, 1, 1, 1;
1, 0, 0, 1,-1,-1;
1, 0, 0, 1,-1,-1;
1,-1,-1,-1,-1, 1]
Where the first column of X is for the intercept, the next three
columns are for R, the 5th column is for T and the last column is for
R*T.
Mike