- #1
transmini
- 81
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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
N/A
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}##?