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I am doing Levene's mean test and Levene's median test (Brown-Forsythe). I want to compare the p-values of these two tests to see which is better. I get large p-values for both tests which are 0.562 (Levene mean) and 0.611 (Levene median) for normal distribution.

  • Which test shows the better type I error rate?
  • does Levene's mean test perform best when the data follows a normal distribution?
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    $\begingroup$ I think you might want to read about type I error rates and hypothesis testing in general. Non-significant p-values are inconclusive and have anyway do not lead to a type I error. Also, what is "a better result"? Both tests fail to reject the zero hypothesis, meaning that you cannot conclude that the variances you are comparing are different. $\endgroup$
    – January
    Commented Apr 26, 2013 at 8:40
  • $\begingroup$ What is the purpose of your comparison? What is your definition of "better"? $\endgroup$
    – Peter Flom
    Commented Apr 26, 2013 at 10:12
  • $\begingroup$ i just want to know which test is more robust according to p values, couldn't i? $\endgroup$
    – Alice
    Commented Apr 26, 2013 at 10:56
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    $\begingroup$ because both tests show acceptance of null hypothesis which is equal variances (p value> alpha=0.05). but i want to know which test perform better in different distributions-normal, moderate skewed and extremely skewed distribution. can i make comparison based on p values? or any other ways for me to make comparison between these two tests? $\endgroup$
    – Alice
    Commented Apr 26, 2013 at 12:23
  • $\begingroup$ @January Don't odds ratio tests compare probabilities? I think comparison of probabilities can have meaning depending on the circumstances. I think I understand what you are saying, but it may be a half truth. It may be that a better test often has more extreme results, e.g. closer to $p=0$ or $p=1$. What do you think? $\endgroup$
    – Carl
    Commented Mar 15 at 13:58

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NIST & Wikipedia both cite Brown & Forsythe's 1974 paper in saying that the version of Levene's test using the median performs better for skewed distributions.

You can't infer that the test performed well or badly from the p-value you get unless you know whether the samples did in fact come from populations with unequal variances, & then you'd have to repeat many times to find the distribution of the p-value. Which is just what Brown & Forsythe did to justify their claim.

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  • $\begingroup$ thanks for your info. can i do it in SPSS or do you know any coding to do it in R? $\endgroup$
    – Alice
    Commented Apr 27, 2013 at 6:21
  • $\begingroup$ For R it's in the car package as leveneTest $\endgroup$
    – Scortchi ♦
    Commented Apr 27, 2013 at 12:50
  • $\begingroup$ i'm sorry but one more question. i would like to apply levene's test using both mean and median on sample data with different distributions. i wish to do comparison between these two tests on their effectiveness on different distributions. since you said i can't draw conclusion by using p values, then is there any other way that i can use to show the comparison? my variances are assumed equals for all distributions. $\endgroup$
    – Alice
    Commented Apr 27, 2013 at 21:56
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The Conover test, A.K.A., the squared ranks test is the only nonparametric equivalent that I know about of a test of difference of dispersion of data that works under non-normal conditions. That is, the Levene's test is somewhat sensitive to non-normal conditions.

The Brown-Forsythe test statistic is the F statistic resulting from an ordinary one-way analysis of variance on the absolute deviations from the median. This may reasonably be expected to be related to variance for symmetric distributions like the Cauchy density function (which statement concerns the pdf, not the random values, as the Cauchy moments do not exist). However, it is likely more powerful just to use the Conover test for non-normal conditions.

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