Hi Eric,
Thanks for you reply but I am still unclear about this. There are only two conditions A and B in our current experiment. One could take the resting state as condition C but what if you do not have that.
By inhibited I do mean decrease from resting state.
Amit
-----Original Message-----
From: Eric Zarahn [mailto:[log in to unmask]]
Sent: Sun 5/29/2005 4:21 PM
To: Anand, Amit; [log in to unmask]
Cc:
Subject: Re: [SPM] Inference fo contrast
Hi Amit,
----- Original Message -----
From: "Amit" <[log in to unmask] <mailto:[log in to unmask]> >
To: <[log in to unmask] <mailto:[log in to unmask]> >
Sent: Sunday, May 29, 2005 4:57 PM
Subject: [SPM] Inference fo contrast
> Dear SPMers,
>
> The experimental contrasts (a difference between two conditions A and B)
> is interpreted in usual SPM analysis as indicating that condition A shows
> a larger response than condition B. However, difference measures,
> as conceptualized in SPM could have three potential ways in which
> conditions may differ: positive activity in A may be greater than positive
> activity in B, positive activity in A may be greater than negative activity
> in B, and finally, negative activity in B may be greater than negative
> activity in A. All lead to a positive difference between conditions.
Yes, absolutely. But implicit in your explanations is a third conditiion C to which you are comparing A and B.
>
> Is there a way of teasing out these three differences?
If you have a meaningful C in your experimental design, then you can compute 2 new contrasts to get your answers:
(1) A minus C
and
(2) B minus C.
>
> Also, how does one get a contrast for areas which are getting inhibited by
> a particular task - does one just reverse the contrast for the active
> versus control condition in a block-design experiment or is there a more
> sophisticated way for doing this?
The use of the word "inhibited" has connotations that I am not sure you intended to convey. But if you simply meant "how do see where the signal in B is greater than that in A?", then yes just compute the contrast B minus A.
Eric
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