Posted by B. Douglas Ward on July 18, 2001 at 17:36:01:
In Reply to: Re: 3dDeconvolve -glt posted by Phoebe on July 18, 2001 at 16:54:32:
Hello Phoebe:
I assume that the maxlags were 4 (and not 2):
1) The two different GLT matrices are performing two different tests.
The first GLT matrix:
0 0 1 1 1 1 1 -1 -1 -1 -1 -1 0 0 0 0 0 0 0 0 0 0
is performing a test of the hypotheses:
Ho: hA[0] + hA[1] + hA[2] + hA[3] + hA[4] = hB[0] + hB[1] + hB[2] + hB[3] + hB[4] vs.
Ha: hA[0] + hA[1] + hA[2] + hA[3] + hA[4] <> hB[0] + hB[1] + hB[2] + hB[3] + hB[4]
This is equivalent to testing whether the area of event A IRF = area of event B IRF.
The second GLT matrix:
0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0
is performing a test of the hypotheses:
Ho: hA[i] = hB[i] for all i=0,1,2,3,4, vs.
Ha: hA[i] <> hB[i] for some i.
This is equivalent to testing whether event A IRF == event B IRF at all points.
So, the answer to your question is: Yes, the sets of active voxels will be different,
since the GLT's are performing different tests.
2) The partial F-test for event A provides a comparison between Event A and fixation.
The Event A partial F-test is equivalent to the following GLT:
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
However, it would be redundant to calculate this GLT, since the partial F-statistics
for Events A, B, C, and D are already included in the bucket dataset output.
For more details, and for examples of the -glt option, see Sections 1.2.11 and 1.4.4
of 3dDeconvolve.ps.
Doug Ward