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There is a following answer to an exercise in a text book:

The hypotheses are \begin{align*} H_0 &: Var_1 = Var_2\\ H_1 &: Var_1 \neq Var_2 \end{align*} Computation: $f = 78.800/913.333 = 0.086$.

Since $\text{P-value} = 2P(f < 0.086) = (2)(0.0164) =0.0328$ for $4$ and $6$ degrees of freedom, the variability of running time for company $1$ is significantly less than, at level $0.0328$, the variability of running time for company $2$. So the question is why do they multiple $\text{P-value}$ by $2$? I saw they did that when they were using $T$ or Normal distributions but the $F$ distribution is not symmetrical.

Gilles
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Nikolai Felix
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  • See the discussion here (which mostly covers it), and the last part of this answer, which gives some links, and this which gives some more discussion. Also see the comments under this answer. – Glen_b Aug 24 '14 at 06:40
  • Short version: it's a common approximation (dare one say 'so common to seems many people don't know how to do it any other way'?). None of those links are quite a duplicate - though the third one perhaps comes closest to a duplicate, the first one (which really isn't) probably comes closer to answering it. – Glen_b Aug 24 '14 at 06:43
  • Some of what's here might also help. – Glen_b Aug 24 '14 at 06:51

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