I do not know why testing homogeneity of variance is so important. What are the examples that require homogeneity of variance?

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    $\begingroup$ I think the premise of the question is faulty -- I don't think that it's important specifically to test it (who says otherwise, and on what basis?); indeed, some papers clearly indicate that that approach may not be a good idea at all. It is often important when you assume it that it be approximately true -- that's not at all the same thing as being important to carry out hypothesis tests relating to it to justify some procedure. Obvious examples of procedures that rely on it are OLS and related procedures (ANOVA, two-sample t-tests) - for standard errors of coefficients and hypothesis tests $\endgroup$
    – Glen_b
    Jan 11, 2014 at 8:46
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    $\begingroup$ Or, perhaps, @variant is interested in consequences of its violation? $\endgroup$
    – Peter Flom
    Jan 11, 2014 at 11:43
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    $\begingroup$ You need to add what kind of procedure you're using. You've tagged this "anova", but it's not clear. $\endgroup$
    – Wayne
    Jan 11, 2014 at 14:20

3 Answers 3


It is about a year since you have asked this question, @variant, and I assume you hopefully passed whatever exam you where studying for or passed your stats course. Homogeneity of variance is a standard assumption of ANOVA and most statistical tests. It is usually touched on quickly in most stats class. Most people have no understanding of what their prof is talking about and, frankly, most profs do not have the best handle on it as well. Homogeneity of variance (HOV) has a history and it is often helpful to understand that history if you want to understand statistics.

Variance, as a term, was first coined by Fisher in 1918. (hopefully you already have a good understanding of variance) What Fisher was interested in was decomposing differences among organisms into their genetic and environmental influences. You know, the idea that people are a combination of nature and nurture. Fisher also felt that most natural phenomenon was normally distributed -- or had a bell curve shape.

Before Fisher was Pearson and Galton, and they had a strong influence on Fisher. The normal distribution has an interesting history in statistics and Galton found it almost magical. Galton explored the normal distribution using a device called the Quincunx. There is a great demonstration of Galton's device on the MathisFun website. Basically it was a board with nails pounded in it. The nails were arranged in a triangular shape. Galton dropped a series of beans or marbles at the top of the triangle and watched them plink, plink, plink down through the nails until they reached the bottom. What he observed was that the objects would arrange themselves at the bottom in a normal(ish) distribution pattern.

Now imagine you are using Galton's game and you repeat the above experiment with a 1000 marbles. Then you measure the mean and variance of those 1000 marbles. You write that down. Then you repeat your experiment with the same marbles but first you only use 750 marbles (write down the resulting mean and variance of the 750 marbles) and then empty the game and then use the remaining 250 marbles (again write down the mean and variance of the 250 marbles).

If you add the variance of the 750 marbles with the variance of the 250 marbles you will get the exact variance (more or less) of the original 1000 marble distribution.

Now, repeat the above experiment but, this time, imagine that on the last 250 marble trial you slightly tilt the game so that one side is higher than the other. This will cause the marbles to slightly skew off-center and collect more to one side of the game then the other side. If you calculate the mean and variance of this skewed sample, and add it to the non-skewed 750 marble sample, you will find that it no longer adds up correctly to the original 1000 marble population variance.

This because your 250 sample is skewed and has a different distribution than the 750 marble sample. Also when a sample is skewed the mean may no longer be the best measure of central tendency and variance relies on the mean.

ANOVA is a special case of the general linear model of statistics. It is linear in that you are adding things up. It assumes that the distributions of those things you are adding up are the same. If they are not then your conclusion or estimate might be off or biased.

And this why HOV is important. Hope this helps.

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    $\begingroup$ The answer posted by Jeff is a good one, though it is important to note that homogeneity of variance does not simply apply to skew (asymmetry) of the distribution, but also variance within the distribution as reflected by the height of the distribution. To use the same marble and peg analogy, if we varied the width of the initial spout from which the marbles fall, a broader spout would result in more marbles landing in the peripheral locations than a narrow spout. $\endgroup$
    – B C
    Apr 13, 2019 at 16:00
  • $\begingroup$ Provided both spouts were centrally located, this would produce two distributions with identical skew with different variances - violating the homogeneity of variance assumption. $\endgroup$
    – B C
    Apr 13, 2019 at 16:00

When we conduct an ANOVA test, we examine the plausibility of a null hypothesis, a straw-man hypothesis that we may end up rejecting. Under this hypothesis we assume not only that all group means are equal, but that we have a certain data-generating process. This is a process in which 1) our observations come to us randomly sampled from the population and 2) there really aren't multiple groups: all observations come from what is the same underlying group, with the same degree of variability.

The usefulness of temporarily adopting, and evaluating, such an hypothesis breaks down if, by their shape, our data clearly don't look as if they came from such a data-generating process. If the variability between groups is radically different, p-values computed based on the probability of certain results occurring under the null hypothesis are no longer accurate. If the data-generating process is clearly different for the different groups, what point is there, and what validity is there, in assessing the probability occurring under the null of the obtained mean differences? The situation we face bears no resemblance to the situation described by such a null.

The good news is twofold. First, you'll see authors routinely referring to the robustness of ANOVA in the face of violations to assumptions--not to the random sampling assumption, mind you, but to assumptions of homogeneity of variance and of normal distributions within groups. Second, resourceful researchers such as Harwell have gleaned from many Monte Carlo simulations some helpful guidance in making adjustments in the face of such violations.

Harwell, M. (2003). Summarizing Monte Carlo results in methodological research: The single factor, fixed-effects ANCOVA case. Journal of Educational and Behavioral Statistics, 28:1, 45-70.

  • $\begingroup$ Great explanation! The idea that, under the null hypothesis, the same data-generating process is behind all the groups being inter-compared is the best foundation I found to understand this question. However, the 'good news' that someone can violate that assumption and still use the test is quite puzzling. So, paradoxically, the good understanding one gets from the 'data-generation process' idea seems to fade when learning that, well, we can also use the test even if the data-generation process is different for different groups... $\endgroup$
    – terauser
    Mar 1 at 10:50

Within regression models homogeneity of variance of the residuals relative to the estimates, referred to as homoskedasticity, is a key underlying assumption of linear regression. If such residuals are not deemed homoskedastic but heteroskedastic (variance changes over observations instead of remaining roughly constant), the calculated statistical significance of your independent variables are invalid (t stat and P value are inaccurate). So, your whole selection of independent variables is questionable.

So, you have to test whether such residuals are homoskedastic or heteroskedastic using the Breusch-Pagan test or the White test.

If they are heteroskedastic, you have to calculate White standard errors (robust standard errors of regression coefficients adjusted for heteroskedasticity). Now, using those robust standard errors you can recalculate the statistical significance (p value) of those regression coefficients.

Another alternative is to rerun your multiple regression model using a weighted least square regression (WLS) that will resolve the heteroskedasticity issue.

From what I gather, calculating robust standard errors (White) is somewhat preferred to WLS regression because with the former you maintain the original coefficient and the causal meaning of your model. With the latter you don't. The meaning of your model can be somewhat altered and much more complicated to interpret.

In short, homogeneity of variance is key because otherwise you just don't know if the independent variables you have selected within your multiple regression model are statistically significant.


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