Dear Takanori,

say we have N observations that we assume normally distributed, and we wish
to
test if a new observation departs from the mean of the first observations
enough to reject the null hypothesis of belonging to the same distribution.
Let m be the estimated mean of N signals in a voxel, distributed normally
with variance s2. We need the distribution of the quantity y - m, where y is
a single new observation in the voxel. Under the null hypothesis, y is
normally distributed with variance s2. m will be distributed with variance
(1/N)s2. Hence, the variate y - m (remember m here is random) is normally
distributed with variance s2 + (1/N)s2 == s2*[(N+1)/N]. To reject the null
hypothesis, we may therefore use the score

z = (y - m)/[s*sqrt((N+1)/N)]

z is distributed normally if s is known, or estimated globally from an image
of uniform variance. If the variance is not uniform and s is estimated
voxel-by-voxel, then z has a t distribution with N - 1 degrees of freedom.
(To see this, divide both numerator and denominator by the true standard
deviation, and check what distributions you have in numerator and
denominator).

If you apply the two-sample t test to the situation above, when one group
has only one subject (the group to which y belongs), you will see that you
obtain an expression that is identical to the expression above giving z (for
variance estimated voxel-by-voxel).

The difficulties using a two-sample t test in this situation is that it is
not clear at all that the test is appropriate until you carry out the
derivation above. The notion of 'pooling the variance of the two samples' is
difficult to interpret when one of the samples has size one.

R. Viviani
Dept. of Psychiatry III
University of Ulm
Ulm, Germany

----- Original Message -----
From: "Will Penny" <[log in to unmask]>
To: <[log in to unmask]>
Sent: Monday, January 17, 2005 3:18 PM
Subject: Re: [SPM] How to use contrast?


> Dear Takanori,
>
> I think you could make sensible arguments for both approaches
> eg.
>
> 1. The two-sample t-test; because we really do have two
> different populations. Just because we have N=1 for one
> population, this does'nt change.
>
> 2. The one-sample t-test; because we want to compare
> the control mean to a particular value.
>
> However, the one-sample t-test is usually used to
> compare eg. a control mean to a *fixed known value*
> (eg zero) - not to a *random variable* (eg. patient
> scan). So we need to adjust the test to account for this.
>
> Perhaps others would like to comment (Tom ?)
>
> Anyway, in practice the different approaches will
> produce very similar results.
>
> For a 12 subject study the difference in t-values
> is only 4% (sqrt(13)/sqrt(12)). For 24 subjects its 2%.
>
> Best,
>
> Will.
>
> Takanori Kochiyama wrote:
>
> > Dear Will and SPMers,
> >
> > I would like to ask you about the 2 sample t test if one of the samples
> > contains a single measurement because I face the similar problem.
> >
> > We agree on the point that GLM can deal with the two-sample t-test
design
> > including a single measurement in one of the samples. However,
> > I think that conventinal one-sample t-test design is suitable in this
case.
> > (although we need some modification in spm.)
> >
> > Based on your previous Email, we think about the following:
> >
> > The formula of T statistics is
> >
> > t = (Mc-Mp)/SE.
> >
> > Here,
> > Mc(p): mean of the contol (or patient) groups, and
> > SE: standard error.
> > and also,
> > Vc(p): variance of the contol (or patient) groups, and
> > Nc(p): Number of sub. in the contol (or patient) groups.
> >
> > The denominator for one-sample t-test is
> > SE = sigma*sqrt(c'*inv(X'X)*c)
> > where
> > sigma = Vc
> > sqrt(c'*inv(X'X)*c)  = sqrt(1/Nc)
> >
> > The denominator for two-sample t-test is
> > SE = sigma*sqrt(c'*inv(X'X)*c)
> > where
> > sigma = {(Nc-1)Vc + (Np-1)Vp}/{Nc+Np-2}
> >         = Vc for Np = 1
> > sqrt(c'*inv(X'X)*c)  = sqrt(1/Nc+1/Np)
> >                            = sqrt(1/Nc+1) for Np = 1
> >
> > The numerator {Mc-Mp} and df {Nc-1} of both tests are same.
> >
> > As a result, we have the following relationship in T value
> > between 1sample and 2sample T-test:
> > T_(2sample) = {1/sqrt(Nc+1)}*T_(1sample)
> >
> > i.e. T_(2sample) is smaller than T_(1sample).
> >
> > This seems to affect the confidence interval.
> > If we want to check e.g. a 95% confidence interval of the "control data
mean",
> > I think, 1sample T is preferable.
> >
> > And I am worried about the equal variance assumption
> > between control and patient group which is required by  2 sample T test,
> > because we never can measure the variance in the patient group with
single subject.
> > In this point, I think, 1sample T is safe approach
> > Please correct me if I am wrong.
> >
> > Thanks in advance for any clarification.
> >
> > -------------------------------------------------------------
> > Takanori Kochiyama
> > Faculty of Engineering
> > Kagawa Univ., Hayashi-cho 2217-20,Takamatsu, JAPAN
> > Phone: +81-87-864-2337,Fax: +81-87-864-2369
> > e-mail: [log in to unmask]
> > -------------------------------------------------------------
> >
> >
> >
>
> --
> William D. Penny
> Wellcome Department of Imaging Neuroscience
> University College London
> 12 Queen Square
> London WC1N 3BG
>
> Tel: 020 7833 7475
> FAX: 020 7813 1420
> Email: [log in to unmask]
> URL: http://www.fil.ion.ucl.ac.uk/~wpenny/
>