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

Claus Mayer claus at bioss.ac.uk
Wed Nov 1 10:15:38 CET 2006


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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