3
$\begingroup$

Can we use a normalized standard deviation to represent a large or small variation when we have data sets with different scales? For example, the first data set has numbers between 1,000,000 and 1,200,000 and the standard deviation is 3000. On the other hand the second data set has numbers between 0 and 200 and the standard deviation is 30.

Comparing these two, we roughly say that the variation in the first data set is smaller than the second data set. I would like to know how can I normalize the std values? Diving by the mean value? Or dividing by the max value? Any suggestion about that?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

"Coefficient of variation" is a statistic that seems to get at what you're describing, where you divide the standard deviation by the mean. However, for your task of saying which group has more variability, it seems straightforward: one group has $100$-times as high a standard deviation of the other. Given that one variable is spread over a range of $200$ and the other over a range of $200,000$, this is what I would expect, and I would be comfortable disagreeing with your assessment that the second group is more variable than the first.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Anyone new to this measure should perhaps look at some cautionary literature first e.g. stats.stackexchange.com/questions/118497/… before committing themselves to extensive use; $\endgroup$
    – Nick Cox
    Commented Dec 6, 2021 at 17:24
  • $\begingroup$ I didn't get the point about your reason. Let me rephrase my understanding. In the first data set, the raw values is in the order 10^6 while the variation is in the order of 10^3. In the second data set, the raw values are in the order of 10^2 and the variation is in the order of 10^1. So, I would say the variation in the second data set is larger than the first data set. I refer to the scale of data. $\endgroup$
    – mahmood
    Commented Dec 6, 2021 at 17:26
  • $\begingroup$ Sure, the SD is larger, but is it surprisingly larger given the mean? Using a coefficient of variation is equivalent to thinking on a logarithmic scale, which often seems entirely natural -- but is far from the only choice. $\endgroup$
    – Nick Cox
    Commented Dec 6, 2021 at 18:02
  • $\begingroup$ Perhaps a ratio of the variances or the standard deviations would represent what you are interested in. $\endgroup$
    – Geoffrey Johnson
    Commented Dec 6, 2021 at 18:24
2
$\begingroup$

There are two issues to consider here: scale (i.e., what counts as a lot and a little) and spread (i.e., whether values are similar or different from each other). Unfortunately, the standard deviation is often used to define both, which means it is challenging to distnguish between scale and spread when comparing two distributions. It is clear your distributions have different scales, but using the standard deviation alone to determine the scale means that you cannot use it to determine the spread. You will have to use some other criteria to do so.

The coefficient of variation is one way to compute the spread separately from the scale, but the mean (which is in its denominator) doesn't always represent the scale of the data well. Sometimes, the (possible) range of the data is more useful, e.g., when comparing the spread of results of two tests with different numbers of questions. Or there may be an external criterion, like the typical standard deviation of the type of data you are observing, that you can use to create a measure of spread in comparison. There is no single statistically valid way to this and your ultimate choice will depend on the substantive nature of your problem. You can use any strategy as long as you justify it; some strategies will be more justified than others, but there is no perfect solution to this problem.

Improve this answer
$\endgroup$
1
  • $\begingroup$ "There are two issues to consider here: scale and spread." Please provide general definitions, as at least some people wouldn't distinguish between them. $\endgroup$
    – Nick Cox
    Commented Dec 7, 2021 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.