# comparing two sets of coefficient of variation values

I have two sets of coefficient of variations (cv):

ste1 = c(cv1, cv2, ..., cv15)

set2 = c(cv1, cv2, ..., cv15)

how can I compare set1 with set2? which test should I apply?

I know that if I want to compare only two CVs I can use one of the followings:

Feltz, C. J., & Miller, G. E. (1996). An asymptotic test for the equality of coefficients of variation from k populations. Statistics in Medicine, 15(6), 647-658.

Krishnamoorthy, K., & Lee, M. (2014). Improved tests for the equality of normal coefficients of variation. Computational Statistics, 29(1-2), 215-232.

but how can I compare two vectors of CVs?

Is it fine to only find the distribution of set1 and set2 and if they follow normal distributions, then I use any test which can be applied to normal distributions?

Thanks a lot and looking forward,

________________________________Update____________________________________

I have two populations, each includes 15 individuals. For each individuals I have measurements overtime, I summarized each individuals with its corresponding CV value.

Now I have two sets including 15 CVs per each population and I want to compare two populations to see which shows higher changes in comparison to the other one.

• Some more information about the underlying data would be welcome, because ordinarily (a) comparing statistics like a CV requires knowing the amounts of data involved in each one and (b) they are unlikely to have a Normal distribution. Moreover, how you proceed might depend on the purpose of your comparison: what is your hypothesis? – whuber Dec 7 '18 at 16:23
• I have two populations, each includes 15 individuals. For each individuals I have measurements overtime, I summarized each individuals with its corresponding CV value. – sbmm Dec 10 '18 at 10:12
• Now I have two sets including 15 CVs per each population. – sbmm Dec 10 '18 at 10:13
• and I want to compare two populations to see which shows higher changes in comparison to the other one. – sbmm Dec 10 '18 at 10:13