coherence again
Jan Hirschmann
Jan.Hirschmann at MED.UNI-DUESSELDORF.DE
Fri Aug 13 19:11:12 CEST 2010
Dear fieldtrip experts,
I have yet another question about the statistical testing of coherence.
Again, it relates to the group level, this time to the testing procedure
used in "Imaging the motor system's beta-band synchronisation during
isometric contraction", Schoffeleln et al., NeuroImage, 2008.
There it says: " The null hypothesis that was tested at the voxel level
states that the Z-transformed coherence between the EMG and that voxel,
accumulated across subjects, is not different from 0. Under this null
hypothesis, flipping the sign of the Z-transformed coherence values of a
random subset of the subjects before accumulating leads to an
alternative observation that belongs to the null distribution."
Now, as I understand it, the Z-transformation described here is
z=sqrt(-(df-2)*log(1-abs(C)^2)), i.e. z takes only positive values
(there is no comparing of conditions here)
And accumulation, it later says, means summing up the z-values of all
subjects and dividing by sqrt(n) at each voxel. I suppose n is the
number of subjects.
So "flipping the sign" means multiplying the z-value of some subjects by
-1 before summing up? As all z-values were originally positive, this
will inevitably decrease the sum. So the original state will inevitably
be the extreme end of the distribution.
Sorry, I must have some serious error in my thinking here (hope it's not
too obvious). Can you help me?
Many thanks,
Jan
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