I have a series of vectors, which are measurements from one sample at two time points.

First time point:
1.[5,3,2,4,4,3,6,5] (mean=4.000)
2.[5,6,3,3,4,3,4,5] (mean=4.125)
3.[6,3,4,2,5,3,5,7] (mean=4.375)
... etc.

grand mean=4.17

Second time point:
1.[1,2,1,2,1,3,4,5] (mean=2.375)
2.[2,2,3,1,1,3,3,5] (mean=2.500)
3.[1,3,1,2,2,3,5,4] (mean=2.625)
.... etc.

grand mean=2.5

I want to see if the variance for each measurement/vector is significantly different between the two time points.

However, the second measurement has a lower mean, which can therefore drive overall variance. How do you compare the variance of two conditions when the means differ?

  • 3
    $\begingroup$ Could you elaborate on how you think a lower mean could "drive overall variance"? What kinds of measurements are these? Are they perhaps bounded below by 1 (and above by some other integer such as 7)? Counts? Responses on a Likert scale? Digital instrument readings? These three circumstances would indicate three different answers to the question! $\endgroup$
    – whuber
    Dec 12, 2011 at 22:38
  • $\begingroup$ The values are % signal change from a baseline condition -- they are roughly between 1 and 10 and can be negative as well, but no bounds. What I meant by "drive overall variance" is that in playing with these data, if the vector was highly correlated at 1st and 2nd timepoint, but experienced a mean decrease, its variance also decreased. $\endgroup$
    – Amyunimus
    Dec 13, 2011 at 0:34

2 Answers 2


It is also important to keep in mind that a strong assumption in the usual F test for equality of variances is that of normality -- the test is very sensitive to the assumption of normality so the resulting p-vlaues can be very distorted. Levene suggested a simple alternative and Brown-Forsythe followed with a test statistic that aims at increasing robustness. You can start in Wikipedia http://en.wikipedia.org/wiki/Brown%E2%80%93Forsythe_test

  • $\begingroup$ Does this work with non-independent samples? $\endgroup$
    – Amyunimus
    Dec 13, 2011 at 17:32
  • $\begingroup$ @AlanForsythe good points. You know this as well as anyone. $\endgroup$ Sep 8, 2012 at 3:17
  • $\begingroup$ @Amyunimus These robustness studies refer to both samples being iid. If there is serial correlation, you are in the realm of comparing two time series. Then for it to make sense to talk about a "population" variance you need to assume stationarity. $\endgroup$ Sep 8, 2012 at 3:28
  • $\begingroup$ Also you need to keep in mind that sample means are not likely to be exactly equal even when the population means are equal. If you are interested in the variance of the sample mean the correlation affects the rate of decrease. For independent observations the variance of the mean decrease at the rate of 1/√n but positively correlated data such as an AR(1) process with a positive coefficient on the lag 1 coefficient the variance decrease at a slower rate. $\endgroup$ Sep 8, 2012 at 3:39

However, the second measurement has a lower mean, which can therefore drive overall variance.

Do you know this for sure? If that's so, you need to figure out what the relationship is and transform the variance to a more stable scale (e.g. log).

Otherwise, try the naive approach and use a folded F test, dividing the larger sample variance by the smaller one, and find the p-value for the resulting F statistic. This approach assumes the samples are independent, which might be reasonable if your overall sequence is stationary or the sample times are widely separated. If not, you probably will need to fit some sort of time series model to the data (ARIMA maybe), and compare the variance of the residuals.

  • $\begingroup$ The samples aren't independent -- they're taken from the same individual, same place, and separated by only about 48 seconds. I'd also like to compare the variance across other conditions with different means, not just time points. $\endgroup$
    – Amyunimus
    Dec 13, 2011 at 17:23

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