[BioC] F-test vs.T-test-on-differences

Naomi Altman naomi at stat.psu.edu
Wed Nov 1 15:11:24 CET 2006


Actually, since Benjamin took  abs(x-xbar) the means are not the 
same.  abs(x-sbar) should be centered roughly on SD(x).

--Naomi

At 04:15 AM 11/1/2006, Claus Mayer wrote:
>Hello Benjamin!
>
>I think there is some misunderstanding here. The t-test is a test for
>the differences between the means of two distributions. If you center
>your data like you propose the difference is 0 so the t-statistic will
>always behave very much like under the nullhypothesis (not exactly as
>the distributions might differ in variances and other properties, but
>the t-test is NOT meant to detect those).
>The F-test on the other hand specifically tests for difference in
>variances, so it is clearly the more appropriate test in your case (and
>if you are worrried about non-normality you might determine p-values by
>a resampling method like bootstrap).
>I think what might have confused you is that there are TWO F-tests:
>a) the one for testing differences between variances (lets call that F1)
>b) the F-test that is being used in Analysis of Variance (ANOVA) (lets
>call it F2)
>Despite its name ANOVA is a method to compare MEANS not VARIANCES. With
>two groups you have the trivial case of a one-way ANOVA and if you
>calculate the F-statistic F2 for that it is just a transformation of the
>usual t-statistic, i.e. the test will yield the same p-values.
>So F1 and F2 are very different statistics for very different things,
>but both have a F-distribution under normality assumptions so their
>names are the same (there are plenty of chi-square tests out there as well!)
>
>Hope this helps
>
>Claus
>
>Benjamin Otto wrote:
> > Dear community,
> >
> >
> >
> > That might be a stupid statistical question but I'm really not sure about
> > the answer:
> >
> >
> >
> > Suppose I have two groups of numeric values x11-x19  and y11-y19. The
> > conventional way to check for difference in variance here is performing an
> > F-test with say
> >
> >
> >
> >> g1 <- c(x11:x19)
> >
> >> g2 <- c(y11:y19)
> >
> >> var.test( g1, g2)
> >
> >
> >
> > and looking at the resuting p.value. A second possibility is calculating
> > some adjusted values first like
> >
> >
> >
> >> g1.adj <- abs(g1 - mean(g1))
> >
> >> g2.adj <- abs(g2 - mean(g2))
> >
> >
> >
> > And afterwards performing a T-test on those values. Should that give me the
> > same result? I tried to solve it mathematically and the statistic doesn't
> > seem to be the same. But then, why is the F-test calculated as it is AND is
> > it really better for a comparison than the second version?
> >
> >
> >
> > Regards,
> >
> >
> >
> > benjamin
> >
> >
> >
> > --
> > Benjamin Otto
> > Universitaetsklinikum Eppendorf Hamburg
> > Institut fuer Klinische Chemie
> > Martinistrasse 52
> > 20246 Hamburg
> >
> >
> >
> >
> >       [[alternative HTML version deleted]]
> >
> > _______________________________________________
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> >
> >
> >
> >
> >
>
>--
>***********************************************************************************
>   Dr Claus-D. Mayer                    | http://www.bioss.ac.uk
>   Biomathematics & Statistics Scotland | email: claus at bioss.ac.uk
>   Rowett Research Institute            | Telephone: +44 (0) 1224 716652
>   Aberdeen AB21 9SB, Scotland, UK.     | Fax: +44 (0) 1224 715349
>
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Naomi S. Altman                                814-865-3791 (voice)
Associate Professor
Dept. of Statistics                              814-863-7114 (fax)
Penn State University                         814-865-1348 (Statistics)
University Park, PA 16802-2111



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