Inference for Variance and Comparing Spread

Definition A pivot is a function of the data and an unknown parameter whose distribution is completely known, the same for every value of the parameters. Let $X_1, \dots, X_n$ be independent observations from one normal population $N(\mu, \sigma^2)$, with $\sigma > 0$ and $n \ge 2$, and let $S^2$ be the sample variance with denominator $n - 1$. Then

whatever the values of $\mu$ and $\sigma$. Write $\chi^2_{\nu,\,p}$ for the $p$ quantile of $\chi^2_\nu$. For $0 < \alpha < 1$ the pivot lies between $\chi^2_{n-1,\,\alpha/2}$ and $\chi^2_{n-1,\,1-\alpha/2}$ with probability exactly $1 - \alpha$. The result needs independence and normality. It holds for every $n \ge 2$ under that model, but a large sample does not in general rescue it when the population is not normal, as the last card explains.

Definition Under the same model, solving the pivot inequality for $\sigma^2$ gives the $100(1 - \alpha)$ percent confidence interval $[L, U]$ for $\sigma^2$, with

The larger quantile gives the lower endpoint, because dividing by a larger positive number gives a smaller result. The interval for $\sigma^2$ is in squared units. The square root is increasing on the nonnegative numbers, so $\sigma^2$ lies in $[L, U]$ exactly when $\sigma$ lies in $[\sqrt{L}, \sqrt{U}]$. Hence $[\sqrt{L}, \sqrt{U}]$ is a confidence interval for $\sigma$ with the same confidence level, in the original units. The intervals are not symmetric about the sample values $s^2$ and $s$. The confidence level describes the procedure. For a 95 percent interval this means that, in repeated samples from the normal model, 95 percent of the intervals built this way contain $\sigma$, which is not the same as a 95 percent probability that $\sigma$ lies in one computed interval.

Definition Suppose two independent samples of sizes $n_1 \ge 2$ and $n_2 \ge 2$ consist of independent observations from normal populations $N(\mu_1, \sigma_1^2)$ and $N(\mu_2, \sigma_2^2)$. Then

To test $H_0\colon \sigma_1^2 = \sigma_2^2$ against $H_1\colon \sigma_1^2 \ne \sigma_2^2$, use the statistic $F = S_1^2 / S_2^2$, which under $H_0$ has the $F_{n_1 - 1,\, n_2 - 1}$ distribution. For an observed value $f$, the equal-tailed two-sided p-value is $2\min\{P(F \le f), P(F \ge f)\}$. Putting the other sample on top swaps the degrees of freedom and gives the same two-sided p-value. Inverting the same pivot gives a confidence interval for the ratio $\sigma_1^2 / \sigma_2^2$.

Definition Let $k \ge 2$ independent groups have observations $Y_{ij}$, where $j = 1, \dots, k$ labels the group and $i = 1, \dots, n_j$ the observation, and let $N = n_1 + \dots + n_k$. Levene’s test replaces each observation by its absolute deviation from a group center $c_j$,

where the original version uses the group mean, $c_j = \bar{Y}_j$. It then computes the one-way ANOVA F statistic of the deviations,

with between-group sum of squares $\text{SS}_{\text{B}} = \sum_j n_j (\bar{Z}_j - \bar{Z})^2$ and within-group sum of squares $\text{SS}_{\text{W}} = \sum_j \sum_i (Z_{ij} - \bar{Z}_j)^2$, where $\bar{Z}_j$ is the mean of the deviations in group $j$ and $\bar{Z}$ the mean of all $N$ deviations. It compares $W$ with $F_{k-1,\, N-k}$. The null hypothesis is that the groups have equal population mean absolute deviations from their centers. This matches equal variances when the group distributions have the same shape apart from scale. The F reference is approximate, even for normal data, and the test requires independent observations within and between groups.

Definition The Brown-Forsythe test for spread is Levene’s procedure with each group’s median as the center, $Z_{ij} = \lvert Y_{ij} - \tilde{Y}_j \rvert$, where $\tilde{Y}_j$ is the median of group $j$. The statistic $W$ and its approximate $F_{k-1,\, N-k}$ reference are as before. Median centering makes the test less sensitive to skewness and to extreme observations, and simulation studies have found that it keeps the actual error rate closer to the nominal level than mean centering when populations are skewed. It is not free of assumptions. It still needs independent observations, its F reference is still an approximation, least reliable for tiny groups, and like Levene’s test it compares mean absolute deviations rather than variances directly. The same names are also attached to a different procedure, a test of equal means that allows unequal variances, so a report should say which one it used.

Definition A procedure is sensitive to nonnormality when its actual coverage or error rate moves away from the stated level once the population is not normal. The chi-square interval and the F test for variances are highly sensitive in this sense. For $n \ge 2$ independent observations from a population with mean $\mu$, variance $\sigma^2 > 0$ and finite fourth moment,

where $\kappa = E\bigl((X - \mu)^4\bigr)/\sigma^4 - 3$ is the excess kurtosis, equal to $0$ for every normal population. The chi-square model builds in $\kappa = 0$. When $\kappa > 0$, as for heavy-tailed populations, the sample variance varies more than the model says, so in large samples the intervals cover $\sigma^2$ less often than claimed and the F test rejects true null hypotheses too often. When $\kappa < 0$ the errors run the other way. Both terms shrink at the same rate, so the discrepancy does not go away as $n$ grows. By contrast, the t interval for a mean has approximately correct coverage in large samples from any population with finite variance.