Re: F distribution

[Donald Burrill]

On 17 Nov 2001, Myles Gartland wrote:

 In an F distribution, the critical value for the lower tail is the
 reciprocal of the the critical value of the upper tail (with the
 degrees of freedom switched).
 
 Why?  I understand how to calculate it, but do not get why the math
 works.

Essentially for the same reason that in the normal distribution the 
critical value for the lower tail is the negative of the critical value 
for the upper tail.

Thnke about it.  For F = V1/V2, where V1 and V2 are two variance 
estimates with numbers of degrees of freedom n1 and n2 respectively, 
the relevant F distribution is said to have n1 and n2 degrees of 
freedom, naming the numerator first and then the denominator. 

For F = V2/V1, the relevant F distribution has n2 and n1 d.f. (hence the 
interchange of the numbers of degrees of freedom to which you allude).

Notice that V2/V1 is the reciprocal of V1/V2.  If V1/V2 is sufficiently 
larger than 1 that the hypothesis of equal variances in the populations 
can be rejected, then V2/V1 must be sufficiently smaller than 1 to permit 
rejection.  Hence the critical value for V2/V1 must be the reciprocal of 
the critical value for V1/V2, and the d.f. are interchanged simply by the 
choice of which direction to divide.

 
 Donald F. Burrill [EMAIL PROTECTED]
 184 Nashua Road, Bedford, NH 03110  603-471-7128



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F distribution

[Myles Gartland]

In an F distribution, the critical value for the lower tail is the
reciprical of the the critical value of the upper tail (with the
degrees of freedom swithced).

Why? I understand how to calculate it, but do not get why the math
works.


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Help : Integrating an F-Distribution

[Scott Nivens]

I am trying to incorporate an integration routine to calculate the area
under an F-Distribution given a score and the numerator and denominator
DOFs. I've made similar routines for the normal and T distributions. The
F-distribution is giving me some problems, especially when my DOFs are
small. I get good results using DOF of 5 or greater, but if I use 2 or 1,
I'm way off.

Is there some trick or alternate way to integrate this distribution? I am
just summing slices under the curve now.
Thanks.
-Scott





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