I would like to compare the means across three groups of equal sizes (equal sample size is small, 21). The means of each group are normally distributed, but their variances are unequal (tested via Levene's). Is a transformation the best route in this situation? Should I consider anything else first?
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1$\begingroup$ What happens if you do the ANOVA despite the unequal variance? $\endgroup$– BehacadCommented Apr 23, 2013 at 15:01
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2$\begingroup$ The result is significant. I am particularly cautious in my interpretation, however, due to the increased chance of incorrectly reporting a significant difference in the means when none exists. As I understand, that chance of a significant result is greater when the population variances are very different from each other. In the case of these data, one of the populations has a variance that is about half as large as the other two. $\endgroup$– Diana ECommented Apr 23, 2013 at 15:10
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5$\begingroup$ That's not a huge difference in the variance, and if your sample sizes are equal, it doesn't matter. $\endgroup$– Jeremy MilesCommented Apr 23, 2013 at 16:08
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11$\begingroup$ This might not need saying, but unequal variances can be something interesting in its own right, & not just a nuisance when trying to compare means. $\endgroup$– Scortchi ♦Commented Apr 23, 2013 at 19:42
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3 Answers
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@JeremyMiles is right. First, there's a rule of thumb that the ANOVA is robust to heterogeneity of variance so long as the largest variance is not more than 4 times the smallest variance. Furthermore, the general effect of heterogeneity of variance is to make the ANOVA less efficient. That is, you would have lower power. Since you have a significant effect anyway, there is less reason to be concerned here.
Update:
- I demonstrate my point about lower efficiency / power here: Efficiency of beta estimates with heteroscedasticity
- I have a comprehensive overview of strategies for dealing with problematic heteroscedasticity in one-way ANOVAs here: Alternatives to one-way ANOVA for heteroscedastic data
answered Apr 23, 2013 at 18:20
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1$\begingroup$ Thanks - I wasn't aware of the rule of thumb you mentioned. Very helpful. $\endgroup$– Diana ECommented Apr 23, 2013 at 20:54
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1$\begingroup$ The point in @JeremyMiles' answer is equality of sample sizes. $\endgroup$ Commented Apr 23, 2013 at 20:55
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$\begingroup$ Great answer. Do you have a reference for the rule of thumb? Thank you $\endgroup$– J.ConCommented Sep 6, 2018 at 6:07
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$\begingroup$ @J.Con, no. You might find it in an introductory applied stats book. It's not a formal thing. $\endgroup$ Commented Sep 6, 2018 at 11:12
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$\begingroup$ "First, there's a rule of thumb that the ANOVA is robust to heterogeneity of variance so long as the largest variance is not more than 4 times the smallest variance" is not correct. According to Blanca (2017), the rule of thumb is that the variance ratio (VR) above 1.5 can be considered a threat to the robustness of the F-test w/ unequal sample size. Thus, usage of ANOVA should be taken with serious caution. There are many potential alternatives to the ANOVA w/ unbalanced sample size eg: Kursal-Wallis test, Welch ANOVA..Reference: link.springer.com/article/10.3758/s13428-017-0918-2. $\endgroup$– SimonCommented Oct 3, 2019 at 8:58
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(1) "The means of each group are normally distributed" - on what basis can you make such an assertion?
(2) your difference in variance sounds pretty small, and if sample sizes are nearly equal would cause little concern, as others have mentioned,
(3) Welch-type adjustments* for degrees of freedom exist for ANOVA just as with two-sample t-tests; and just as with their use in two sample t-tests, there's little reason not to use them as a matter of course. Indeed, the oneway.test function in R does this by default.
*B. L. Welch (1951), On the comparison of several mean values: an alternative approach.
Biometrika, 38, 330–336.
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I suggest to employ Bayesian ANOVA, which does not assume the variances are necessarily the same across groups. John K. Kruschke has made an excellent example, available here: http://doingbayesiandataanalysis.blogspot.mx/2011/04/anova-with-non-homogeneous-variances.html
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4$\begingroup$ Welcome to the site, @Luis. In general, we are cautious about answers that are primarily composed of links to external sources, because linkrot is so common on the internet. Would you mind expanding on this idea & including the most important parts here? $\endgroup$ Commented Jul 9, 2014 at 18:49