What does it mean for variances to be equal? Is this the same as the variance to follow a normal distribution?

  • $\begingroup$ It means that they are equal. No catch there, except that a hypothesis that (population) variances are equal can be consistent with (sample) variances not being exactly the same. Variance following a normal distribution would not be the same. As it happens, it would rarely be expected that variances follow a normal distribution; that's not how they behave, even under ideal or idealised circumstances. $\endgroup$
    – Nick Cox
    Dec 4, 2013 at 20:56
  • $\begingroup$ 'variances equal' simply means that the population variance for one thing is the same as the population variance for some other thing or things. The distribution of the variance is restricted to the non-negative half of the real line - so variances can't be normal, except in a limiting sense (a variance is a kind of average, and the CLT will apply to it if the usual CLT conditions apply, so I think you'd need at least 4th moments to exist, for example) $\endgroup$
    – Glen_b
    Dec 4, 2013 at 21:43
  • $\begingroup$ Is there a reason why we test for equality of variance and not equality of standard deviation? Does equality of variance imply equality of standard deviation and viceversa? $\endgroup$
    – MSIS
    Oct 14, 2020 at 21:08

2 Answers 2


If the variances of two random variables are equal, that means on average, the values it can take, are spread out equally from their respective means.

  • 1
    $\begingroup$ You mean that the mean is the same for both variables? And what does "spread out equally from the mean" really mean? $\endgroup$ Dec 4, 2013 at 22:52
  • $\begingroup$ @Dilip 'spread out equally from the mean' is just an intuitive way ive always thought about variance. If we drew the pdf of a normally distributed RV, the general shape of the bell curve shows us how, the possible values the RV can take are spread about the mean. The smaller the variance , the lesser the spread and this would mean the RV would most likely take a value that is very close to the mean. $\endgroup$ Dec 4, 2013 at 23:14
  • $\begingroup$ Your answer and response to my comment are mostly gobbledygook. A $N(2/3, 1/18)$ random variable has its mass spread evenly about the mean $2/3$, and mostly in the interval $(-0.04, 1.37)$. How is this comparable to a random variable $X$ with density $2x$ for $0 \leq x \leq 1$, mean $E[X]= \int_0^1 2u^2 du = 2/3$ and variance $E[X^2] - (E[X])^2 = \int_0^1 2u^3 du -\left(\frac{2}{3}\right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}$?? Now the mass is only between $0$ and $1$ and is clearly not "spread out equally" from the mean in any sense of the words. $\endgroup$ Dec 5, 2013 at 3:51
  • $\begingroup$ @Dilip, thanks for your comment. If I reworded to reflect that I am referring to two normally distributed RVs, would my answer and previous comment make sense? I chose to use 2 Normal RVs because I found it easier to think about variances and mean. Or do you think I am still confused about what equal variances mean? $\endgroup$ Dec 5, 2013 at 9:23
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    $\begingroup$ @Dilip Although you are correct, your approach seems overly harsh. After all, people do refer to the variance as a "measure of spread about the mean." If you read this answer, then, as saying "variance is one way to measure the spread of a variable about its mean, so equality of variances means those measures of spread are equal," then it makes perfect sense and does not require the assumption of Normality. Moreover, although such an answer is not terribly informative--it's practically circular--it does directly address some of the confusion expressed in the question. $\endgroup$
    – whuber
    Dec 5, 2013 at 16:39

I am only a 2nd year psychology degree student, but I think the below demonstration is a likely approximation to assume equal variances. If I am not likely to be accurately correct, your explanation would be appreciated. From - Ms.J.E.Sabharwal

Example scenario (put simply): Let’s say, a researcher hypothesised that more than 21% of public service employees currently work more than 40 hours per week. He conducted the appropriate hypothesis test and obtained a p-value of 0.143. Based on this result, he concludes that exactly 21% of public service employees currently work more than 40 hours per week.

However, if the data shows the distribution is not significantly different, p=.143, greater than alpha (α=0.05), 143≥α=0.05=Ho. We assume equal variance (measures of spread about the mean is equal) between p̂=.2 (21%) and the sample ‘n ‘ (group of p-hats). This is enough evidence not to reject the value stated in the null, H0: μ ≤ p̂=.2=21% (type 1 error – false positive). Enough evidence that 21% is the average mean of the model (expected value), Mu=p̂=.2=21%. Therefore, this example shows the point estimate is p̂=.2 (21%), suggesting it is likely the null hypothesis is: ‘On average, 21% of public service employees currently work less than 40 hours per week’, instead of the researcher's above conclusion.

I hope this helped.

  • $\begingroup$ I was asking for feedback...which is part of the terms of conduct. $\endgroup$
    – user231386
    Dec 17, 2018 at 5:28

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