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Assuming perfect normality, independence, etc. and testing for equality of variance between two population (where we know that $\sigma_x^2 $ is not lower) and (thus) use a one-sided test.

$$H_0: \sigma_x^2 = \sigma_y^2 $$ $$H_1: \sigma_x^2 - \sigma_y^2 > 0 $$

A sample of size $n=40$, gives $s_x^2 =20 $ and $s_y^2=25$. Using $\alpha = 0.05 $ I will reject the hypothesis iff my test statistic exceeds the critical value $F>F_{0.05,39,39} \approx 1.69$

My test statistic is $F=S_x/S_y = 0.8 $ => I do not reject $H_0$.

Have I made any mistake? A colleague suggests I switch the numerator and denominator but I can't see why.

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    $\begingroup$ Suppose, just to get some intuition, that it turned out $s_x^2=2$ instead of $20$. This gives you two choices of $F$ statistic: $2/25=0.08$ and $25/2=12.5$. The latter clearly exceeds the critical value you computed. In this case would it make any kind of sense to conclude that $\sigma_x^2$ exceeds $\sigma_y^2$? $\endgroup$ – whuber Nov 12 '14 at 16:09
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    $\begingroup$ Right. It wouldn't! $\endgroup$ – snoram Nov 13 '14 at 19:13
  • $\begingroup$ Note that the alternative really relates to the ratio, not the difference. $\endgroup$ – Glen_b May 10 '17 at 12:52
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As you are doing an upper one-tailed test, you are correct. If fact it is probably best practice to keep the numerator and denominator in the same location. Your colleague is likely thinking of the lower one-tailed test.

Your alternative hypothesis can be rewritten as:

$$H_1: \sigma_x^2 > \sigma_y^2 $$

In which case, the corresponding F test is: $$F>F_{α,N1-1,N2-1}$$

If the alternative hypothesis is the lower one-tailed test:

$$H_1: \sigma_x^2 < \sigma_y^2 $$

Then the corresponding F test is: $$F<F_{α,N1-1,N2-1}$$ Or, in this case, you could reverse the numerator and denominator (as your colleague suggests) and use the former F-test to get the same answer.

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