Hi,
Okey... let me commit that for the first time I am coming across this kind of problem… but let me try
First the question is your F1, F2, and F3 are independent? If the answer is NO then I don't think there is any other way except to make judgement of 'SENSITIVITY' based on the magnitude of F
Now if F1, F2 and F3 are independent then we can perform two sample t-test of equality of means.
Mean [F(1,n)] = n/(n-1)
Var[F(1,9)] = 2*n^2(n-1)/{(n-2)^2(n-4)}
Now using Central Limit Theorem,
F(1,n) follows Normal[n/(n-1), 2*n^2(n-1)/{(n-2)^2(n-4)}]
And thus we can perform two sample t-test on any pair of F1, F2 and F3.
Hope this helps.
SUBSCRIBE allstat Anonymous <[log in to unmask]> wrote: Considering three comparisons with the following F values
1. F(1, 43) = 25.05, p < .01
2. F(1, 43) = 55.05, p < .01
3. F(1, 43) = 26.05, p < .01
I have concluded in my paper that since all three show a significant
difference their effect is the same. However, a reviewer has come back
with the comment that no. 2 shows 'more sensitivity' as the F value is
higher than nos 1 and 2. Does anyone know of any stats which compares F
values to show if they are significantly different from another? or is this
an odd way of looking at things?
thanks
--------------
Madan Gopal Kundu
Biostatistician I, i3 Statprobe
Gurgaon, Haryana
India
Web: http://www.freewebs.com/madanstata
mobile: 91-9868788406
e-mail: [log in to unmask]
Click to join Statisticians_group
---------------------------------
Check out what you're missing if you're not on Yahoo! Messenger
|