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I am familiar with $F$-tests in which the alternative hypothesis is defined as $H_a=\{\sigma^2_1/\sigma^2_2>C\}$ (the sign "$>$" can be either "$<$" or "$\ne$" as well), where $C=1$.

If I wanted to know whether there is strong support for the idea of $\sigma^2_1$ being $C>1$ times as large as $\sigma^2_2$, can I calculate a $p$-value as

$p=P\left(F_{m,n}>\frac{s^2_1/s^2_2}{C}\right)?$

where $F_{m,n}$ is the $F$-distribution with $m$ (numerator) and $n$ (denominator) degrees of freedom.

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  • $\begingroup$ If you're calculating p-values, you're performing a hypothesis test -- in which case, what are the explicit null and alternative here? $\endgroup$ – Glen_b May 9 at 11:03
  • $\begingroup$ @Glen_b In this case, $H_0=\{\sigma^2_1/\sigma^2_2\le C\}$ and $H_a=\{\sigma^2_1/\sigma^2_2>C\}$ $\endgroup$ – Cromack May 9 at 11:06
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No.

Given suitable assumptions*, you could indeed calculate a p-value in the indicated manner, but that does not provide "strong support for the idea of $\sigma^2_1$ being $C$ times as large as $\sigma^2_2$"

If you didn't reject, you'd conclude that there's no clear evidence that it's more than $C$ times as large, but failure to reject the null is in no sense "strong support" for the null. (Absence of evidence of an effect is not the same thing as evidence of absence of an effect.), so there's not even a good basis for strongly asserting what the null says (which is also not what your statement says)

If you rejected you'd conclude that it's more than $C$ times as large, so for either rejection or non rejection the statement you're talking about doesn't work. (Indeed, even if you had a rejection, I don't think it's generally reasonable to claim that's "strong support" for the alternative.)


* However, be aware that this test can be pretty sensitive to deviations from the assumption of normality; it's not 'robust'.

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  • $\begingroup$ Oh, I understand the "more than" thing! Thank you very much for clarification. $\endgroup$ – Cromack May 9 at 11:31

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