Showing Rejection Region Equality with Fisher Distribution

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Homework Statement

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For reference:
Book: Mathematical Statistics with Applications, 7th Ed., by Wackerly, Mendenhall, and Scheaffer.
Problem: 10.81

From two normal populations with respective variances $\sigma_1^2$ and $\sigma_2^2$, we observe independent sample variances $S_1^2$ and $S_2^2$, with corresponding degrees of freedom $\nu_1=n_1-1$ and $\nu_2=n_2-1$. We wish to test $H_0: \sigma_1^2=\sigma_2^2$ versus $H_a: \sigma_1^2 \neq \sigma_2^2$.

(a) Show that the rejection region given by
$$\{F > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space F < (F_{\nu_1, \space \alpha/2}^{\nu_2})^{-1}\}$$
where $F=S_1^2/S_2^2$, is the same as the rejection region given by
$$\{S_1^2/S_2^2 > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space S_2^2/S_1^2 > F_{\nu_1, \space \alpha/2}^{\nu_2}\}.$$

(b) Let $S_L^2$ denote the larger of $S_1^2$ and $S_2^2$ and let $S_S^2$ denote the smaller of $S_1^2$ and $S_2^2$. Let $\nu_L$ and $\nu_S$ denote the degrees of freedom associated with $S_L^2$ and $S_S^2$, repectively. Use part (a) to show that, under $H_0$,
$$P(S_L^2/S_S^2 > F_{\nu_S, \space \alpha/2}^{\nu_L})=\alpha.$$
Note that this gives an equivalent method for testing the equality of two variances.

Homework Equations

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The Attempt at a Solution

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(a) $$\{F > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space F < (F_{\nu_1, \space \alpha/2}^{\nu_2})^{-1}\}$$
$$ = \{S_1^2/S_2^2 > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space S_1^2/S_2^2 < (F_{\nu_1, \space \alpha/2}^{\nu_2})^{-1}\}$$
$$ = \{S_1^2/S_2^2 > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space (S_1^2/S_2^2)^{-1} > F_{\nu_1, \space \alpha/2}^{\nu_2}\}$$
$$ = \{S_1^2/S_2^2 > F_{\nu_2, \space \alpha/2}^{\nu_1} \space or \space S_2^2/S_1^2 >(F_{\nu_1, \space \alpha/2}^{\nu_2})^{-1}\}$$

(b) I have no idea on, as I'm not entirely certain how the statement could be true in the first place. Because, let's assume that $S_1^2 = S_L^2$ and $S_2^2 = S_S^2$. Then the problem is saying to show $P(S_1^2/S_2^2 > F_{\nu_2, \space \alpha/2}^{\nu_1}) = \alpha$, but since this gives the tail probability of the Fisher distribution, and $F_{\alpha/2}$ is defined as the value of F such that the tail probability is $\frac{\alpha}{2}$, how can $P(F > F_{\nu_2, \space \alpha/2}^{\nu_1}) = \alpha$ when it by definition equals $\frac{\alpha}{2}$?

 

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Wait I THINK I may have figured it out. So we have, assuming $S_1^2 = S_L^2$ and $S_2^2 = S_S^2$:
$$P(F \in Rejection \space Region) = \alpha$$
$$P(S_L^2/S_S^2 > F_{\nu_S, \space \alpha/2}^{\nu_L} \space or \space S_S^2/S_L^2 > F_{\nu_L, \space \alpha/2}^{\nu_S}) = \alpha$$
$$P(S_L^2/S_S^2 > F_{\nu_S, \space \alpha/2}^{\nu_L})+P(S_S^2/S_L^2 > F_{\nu_L, \space \alpha/2}^{\nu_S}) = \alpha$$ since they are mutually exclusive
$$P(S_L^2/S_S^2 > F_{\nu_S, \space \alpha/2}^{\nu_L})+0 = \alpha$$ since $\frac{S_S^2}{S_L^2} < 1$ and F-values are greater than 1 (at least as far as I can tell looking at this table anyway)
$$P(S_L^2/S_S^2 > F_{\nu_S, \space \alpha/2}^{\nu_L})= \alpha$$

Is this right? And if so, could someone give a more intuitive reasoning to this? Because it still feels weird that the tail area is $\alpha/2$ but the probability of being the rejection region is $\alpha$ when it's not possible to be in one tail.