Attachment 21653

Verify that {F}_{\alpha,n,m} = \frac{1}{{F}_{ \alpha,n,m } }

I am very frustrated with this verification. Attached is a PDF of the question because I don't know how to really work the formatting on this site. Please Help!

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- Jun 12th 2011, 03:54 PMtech819Confidence Interval Using F
Attachment 21653

Verify that {F}_{\alpha,n,m} = \frac{1}{{F}_{ \alpha,n,m } }

I am very frustrated with this verification. Attached is a PDF of the question because I don't know how to really work the formatting on this site. Please Help! - Jun 12th 2011, 06:24 PMtheodds
An F distribution is equal in distribution to a ratio of chi-squares divided by their degrees of freedom. Hence, the distribution of 1/F is equal is also equal in distribution to a ratio of chi-squares divided by their degrees of freedom; you just end up switching the degrees of freedom. Use that fact to get the result.

- Jun 12th 2011, 06:28 PMbryangoodrich
First, use [tex] ... latex here ... [/tex] tags to post latex. Second, do you know

*what*the value $\displaystyle F_{\alpha, n, m}$ indicates? I'll give you a hint: it's a certain percentile of the F distribution. With that in hand, think about what percentage of the F distribution alpha and 1-alpha are specifying, and see how they relate according to the provided equality. Your proof will derive from that understanding. - Jun 12th 2011, 07:45 PMtech819
To Clarify,

does

$\displaystyle

F_{\alpha,n,m} = (1-\alpha) 100?

$ - Jun 12th 2011, 09:19 PMtheodds
No. $\displaystyle F_{\alpha, n, m}$ is the number such that if $\displaystyle X$ has an F distribution with n and m degrees of freedom, then

$\displaystyle P(X \ge F_{\alpha, n, m})= \alpha$.