SPSS uses the Levene test to evaluate homogeneity of variances in the independent group t-test procedure.

Why is the Levene test better than a simple F ratio of the ratio of the variances of the two groups?


1 Answer 1


You could use an F test to assess the variance of two groups, but the using F to test for differences in variance strictly requires that the distributions are normal. Using Levene's test (i.e., absolute values of the deviations from the mean) is more robust, and using the Brown-Forsythe test (i.e., absolute values of the deviations from the median) is even more robust. SPSS is using a good approach here.

Update In response to the comment below, I want to clarify what I'm trying to say here. The question asks about using "a simple F ratio of the ratio of the variances of the two groups". From this, I understood the alternative to be what is sometimes known as Hartley's test, which is a very intuitive approach to assessing heterogeneity of variance. Although this does use a ratio of variances, it is not the same as that used in Levene's test. Because sometimes it is hard to understand what is meant when it is only stated in words, I will give equations to make this clearer.

Hartley's test: $$ F=\frac{s^2_2}{s^2_1} $$ Levene's test / Brown-Forsythe test: $$ F=\frac{MS_{b/t-levels}}{MS_{w/i-levels}} $$

In all three cases, we have ratios of variances, but the specific variances used differ between them. What makes Levene's test and the Brown-Forsythe test more robust (and also distinct from any other ANOVA), is that they are performed over transformed data, whereas the F ratio of group variances (Hartley's test) uses the raw data. The transformed data in question are the absolute values of the deviations (from the mean, in the case of Levene's test, and from the median, in the case of the Brown-Forsythe test).

There are other tests for heterogeneity of variance, but I'm restricting my discussion to these, as I understood them to be focus of the original question. The rationale for choosing amongst them is based on their performance if the original data are not truly normal; with the F test being sufficiently non-robust that it is not recommended; Levene's test being slightly more powerful than BF if the data really are normal, but not quite as robust if they aren't. The key citation here is O'Brien (1981), although I could not find an available version on the internet. I apologize if I misunderstood the question or was unclear.

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    $\begingroup$ Because Levene's statistic is a ratio of squares constructed from those absolute residuals, and is referred to an F distribution, it is not immediately apparent that it should be any more robust than other tests based on ratios of squares! You may be thinking of more robust variants, such as the Brown-Forsythe test. See a good discussion by @chl at stats.stackexchange.com/questions/2591/…. $\endgroup$
    – whuber
    Mar 2, 2012 at 22:54
  • $\begingroup$ @whuber, thanks for the comment & link. There is too much to respond to in a comment, so I edited my answer. I believe what I am trying to get at should be clearer now. However, if I misunderstood, or am simply wrong, I can delete this answer. $\endgroup$ Mar 3, 2012 at 7:20
  • $\begingroup$ The (new) last paragraph makes your point well (+1). $\endgroup$
    – whuber
    Mar 3, 2012 at 15:39

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