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When comparing variances, very often a F-test is used.

$$F=\frac{s_1^2}{s_2^2}$$

We then compare $F$ to an F-distribution. The assumption of this test is that the two samples (which variances are $s_1^2$ and $s_2^2$) are normally distributed.

Can you please explain why (given that the two samples are normally distributed) $\frac{s_1^2}{s_2^2}$ is F-distributed?

If I am not mistaken, if $U_1$ and $U_2$ are chi-squared distributed with degrees of freedom $df_1$ and $df_2$ respectively, then $\frac{U_1/df_1}{U_2/df_2}$ is F-distribution distributed. I would really appreciate an answer that also give some words between the link between the chi-squared distribution and the F-distribution.

Remi.b
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    I would say fairly rarely is an F-test used for this purpose. It's very sensitive to the assumption of normality. Less sensitive alternatives are more often used. – Glen_b Jun 24 '14 at 02:00
  • There are two parts to it. The first part relates to $\frac{s^2}{\sigma^2}\sim\chi^2_\nu/\nu$ where $\nu$ is the degrees of freedom in the estimate of $\sigma^2$, and the second part relates to the ratio of independent $\chi^2_{\nu_i}/\nu_i$ random variables having an F distribution. Which part do you need clarified? – Glen_b Jun 24 '14 at 02:03

2 Answers2

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If we have a normal sample of size $n$, then the random variable $U_1=(n-1)s_1^2/\sigma^2_1$ follows a $\mathcal \chi^2_{(n-1)}$. This follows from the fact that a chi-square is a squared standard normal, and that $(n-1)s_1^2/\sigma^2_1$ is the sum of standardized normal random variables (write it out to see that).

Re-arrange and consider the random variable

$$s_1^2 = \sigma^2_1[U_1/(n-1)]$$

Now consider another normal sample of size $m$, and the associated magnitude

$$s_2^2 = \sigma^2_2[U_2/(m-1)]$$

We form the statistic

$$T = \frac {s_1^2}{s_2^2} = \frac{\sigma^2_1}{\sigma^2_2}\cdot \frac {[U_1/(n-1)]}{[U_2/(m-1)]}$$

The second term is a random variable that follows an $F$-distribution. So if the null hypothesis is true, $H_0 : \sigma^2_1 = \sigma^2_2$ the statistic will be a random variable following the $F$-distribution:

$$T|_{Η_0} = \frac {[U_1/(n-1)]}{[U_2/(m-1)]} \sim F(n-1, m-1)$$

As for the link between the ratio of chi-squares and the F-distribution, is one of those "algebra of random variables" results that should make us very grateful to those people that actually devoted their time to derive them.

Alecos Papadopoulos
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The variance is defined as

$V(X)=E((X-E(X))^2)$,

so for a standardnormal distribution we have

$V(X)=E(X^2)$ (since $\mu=0$).

This shows that the variance is $\chi^2$-distributed, because the $\chi^2$-distribution is defined as the distribution of (df-variate) $\sum X^2$, $X\sim N(0,1)$.

Since $F$ is as you defined the ratio of two normal variances, it is also the ratio of two $\chi^2$-variables.

This is how the Normal, $\chi^2$, and $F$ distribution are linked.

emcor
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