Index: The Book of Statistical Proofs ▷ Statistical Models ▷ Univariate normal data ▷ Univariate Gaussian ▷ F-test for equality of variances

Theorem: Let

\[\label{eq:ug} \begin{split} y_{1i} &\sim \mathcal{N}(\mu_1, \sigma_1^2), \quad i = 1, \ldots, n_1 \\ y_{2i} &\sim \mathcal{N}(\mu_2, \sigma_2^2), \quad i = 1, \ldots, n_2 \end{split}\]

be two univariate Gaussian data sets representing two groups of unequal size $n_1$ and $n_2$ with unknown means $\mu_1$ and $\mu_2$ and unknown variances $\sigma_1^2$ and $\sigma_2^2$. Then, the test statistic

\[\label{eq:F} F = \frac{s_1^2}{s_1^2} = \frac{\frac{1}{n_1-1} \sum_{i=1}^{n_1} (y_{1i}-\bar{y}_1)^2}{\frac{1}{n_2-1} \sum_{i=1}^{n_2} (y_{2i}-\bar{y}_2)^2}\]

with sample means $\bar{y}_1$ and $\bar{y}_2$ and samle variances $s_1^2$ and $s_2^2$ follows an F-distribution with numerator degrees of freedom $n_1-1$ and denominator degrees of freedom $n_2-1$

\[\label{eq:F-dist} F \sim \mathrm{F}(n_1-1, n_2-1)\]

under the null hypothesis that the two variances are equal:

\[\label{eq:fev-h0} H_0: \; \sigma_1^2 = \sigma_2^2 \; .\]

Proof: We know that, for a sample of normal random variables, the sample variance is following a chi-squared distribution:

\[\label{eq:norm-chi2} X_i \sim \mathcal{N}(\mu, \sigma^2), \; i = 1, \ldots, n \quad \Rightarrow \quad V = (n-1) \frac{s^2}{\sigma^2} \sim \chi^2(n-1) \; .\]

Thus, we have:

\[\label{eq:V-dist} \begin{split} V_1 &= (n_1-1) \frac{s_1^2}{\sigma_1^2} \sim \chi^2(n_1-1) \quad \text{and} \\ V_2 &= (n_2-1) \frac{s_2^2}{\sigma_2^2} \sim \chi^2(n_2-1) \; . \end{split}\]

Moreover, by definition, the ratio of two chi-squared random variables, divided by their degrees of freedom, is following an F-distribution:

\[\label{eq:chi2-f} X_1 \sim \chi^2(d_1), \; X_2 \sim \chi^2(d_2) \quad \Rightarrow \quad Y = \frac{X_1/d_1}{X_2/d_2} \sim F(d_1, d_2) \; .\]

Thus, we have:

\[\label{eq:F-dist-qed} \begin{split} F &= \frac{V_1/(n_1-1)}{V_2/(n_2-1)} \\ &= \frac{\left. (n_1-1) \frac{s_1^2}{\sigma_1^2} \middle/ (n_1-1) \right.}{\left. (n_2-1) \frac{s_2^2}{\sigma_2^2} \middle/ (n_2-1) \right.} \\ &= \frac{s_1^2 / \sigma_1^2}{s_2^2 / \sigma_2^2} \\ &\overset{H_0}{=} \frac{s_1^2}{s_2^2} \; . \end{split}\]

This means that the null hypothesis of equal variances can be rejected when $F$ is as extreme or more extreme than the critical value obtained from the F-distribution with degrees of freedom $n_1-1$ and $n_2-1$ using a significance level $\alpha$.

∎
Sources:

Metadata: ID: P460 | shortcut: ug-fev | author: JoramSoch | date: 2024-07-05, 10:42.