Will somebody please help answer this question?
Use the result of F(3,4) = 1/F(4/3) about F distribution to show that the quantile relation f'subscript'(0.95,3,4) = 1/f'subscript'(0.05,4,3).
Using upper percentile points....
Which means that $\displaystyle a=P(F_{3,4}>F_{3,4,a})$
we have with a=.95............
$\displaystyle .95=P(F_{3,4}>F_{3,4,.95})$
(Note that $\displaystyle F_{3,4}$ is a rv and $\displaystyle F_{3,4,a}$ is a percentile point.)
Taking the reciprocal inside the probability we have
$\displaystyle .95=P(F_{3,4}>F_{3,4,.95})$
$\displaystyle =P(F^{-1}_{3,4}<F^{-1}_{3,4,.95})$
$\displaystyle =P(F_{4,3}<F^{-1}_{3,4,.95})$
(Which follows by the definition of an F, it's a ratio of chi-squares divided by their dfs.)
And note that
$\displaystyle P(F_{4,3}<F_{4,3,.05})=.95$ also.
Thus by the continuity of the densities, these percentile points must be equal....
Hence $\displaystyle F^{-1}_{3,4,.95}$ must equal $\displaystyle F_{4,3,.05}$
and this is true for any probability, not just .05.