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Let be two samples collections A and B. Mean of the first is 5 and the second is 15. Standard deviation of the first is 2 and the second is 5.

Can we conclude something even though the two datasets have a different mean ?

Can we consider the ratio (standard deviation / mean) and conclude that A (0.4) is more widespread than B (0.33) ?

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As long as the distributions have the same shape, I guess it makes sense... –  Arthur Sep 17 '12 at 14:36

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up vote 2 down vote accepted

Your proposed ratio (std / mean) is called the coefficient of variation (http://en.wikipedia.org/wiki/Coefficient_of_variation). Since this is independent of the unit of measure of a sample/distribution, it is perfect for comparison of the dispersion.

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Certainly you can compare the spread of two distributions in dimensionless measures like the coefficient of variation as miladydesummer points out so nicely. But I think the important question is why? These two distributions have different means and variances. Are they measured in the same units? Do they represent items that you would be interested in comparing? –  Michael Chernick Sep 17 '12 at 15:15
    
From experience, this can be useful, for example, when comparing data from experiments that have different time scales (e.g., one experiment lasts 10 months, while a similar experiment can only be done for a week)... –  miladydesummer Sep 17 '12 at 15:19
    
@MichaelChernick In my case I am interested in the amount of data sent by different detectors during a set of experiments. The detectors are very different and thus their data size, however I am interested to know if their standard deviation is the same, i.e. they behave the same way amongst the experiments. –  Barth Sep 18 '12 at 6:37
    
Okay then if you can assume a paramtric form you can compare the sample estimates of variance. If both distributions are normal there is the F test for equality of variance. In other situations there are robust and nonparametric tests for determining if there is a difference in scale between two distributions, This was discussed recently in another question. –  Michael Chernick Sep 18 '12 at 12:20

You can think of them the same way you normally would, unless the standard deviation is linked to the mean in some way (such as with the poisson or exponential distribution). For more intuition, note that if you center a variable by its mean, you haven't changed anything fundamentally about the variable, and the standard deviation is unchanged.

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