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I have multiple sets of measurements: set1: 5 6 10 20 15 set2: 18 55 62 51 12 set3: 150 180 210 100 200

The three sets are not supposed to have the same mean, but the dispersion of the data is supposed to be similar.

What I do is I calculate the coefficient of variation, by dividing the standard deviation by the mean, which gives 0,562570857 0,578222175 0,264194536 for set1 set2 and set3 respectively.

How to test the significance in differences between the CV? In other words how to test the significance of differences in dispersion of the data between these 3 sets? Thank you

Youcef
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  • Because the CV is not a pure measure of dispersion (and would not even be applicable if some of your measurements could have negative means), and yet you have computed and compared CVs in an effort to address this question, one wonders what you actually mean by "dispersion." Could you clarify that? – whuber Jul 25 '15 at 13:44
  • If your data sets are supposed to have the same dispersion (defined however it is defined) then you would not test the null that they are the same. You could look into equivalence testing or you could just forgo testing and present the numbers and say "Look!" – Peter Flom Jul 25 '15 at 13:50
  • @whuber by dispersion I mean the distance of the measuerments from the mean. For example the set 10, 20, 30, 40 has a mean value of 25. The set 0, 10, 20, 70 also has a mean value of 25, but the different points are farther from their mean in comparison to the first set. And I can obviously not compare viariances or standard deviations since my different sets are not supposed to have the same mean value, but supposed to have the same dispersion, that's why I chose to divide the standard deviation which is a measure of dispersion by the mean. – Youcef Jul 25 '15 at 15:36
  • @PeterFlom But how to prove that the differences in coefficient of variation are significant? I might not even present these data, but even then I would like to understand how to compare data dispersion between various sets that are not supposed to have same means? Let's take set1: 10 55 44 30 17 and set2: 12 25 33 21 1506. how to prove that the dispersion of data between these two sets is significantly different? – Youcef Jul 25 '15 at 15:41
  • Your latest comment to me seems a little confused in its self-contradictions. If you want to compare typical distances to the mean, then do so--but don't alter them by then dividing by the mean itself! Moreover, you can compare variances, because they are separate from the mean and independent of it. This possibility is integral to basic statistical techniques like Analysis of Variance. – whuber Jul 25 '15 at 17:27
  • You don't want to prove they are significant. You want to show they are similar. That's completely different. Also, see @whuber 's comment: You can compare (or do equivalence testing) on variances. – Peter Flom Jul 25 '15 at 21:23

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