I have 2 data sets. The first data set, let's call it $X$ has an average value of ($\bar X$) and standard deviation of ($STD_X$), the second set of data also has the average value of ($\bar Y$) and standard deviation of ($STD_Y$). I want to find out the standard error or standard deviation of a percentage change of data set 2 compared to data set 1. So I have $((\bar Y-\bar X)/\bar X)*100$. Now my question is, how do you take into account the standard deviations for this percentage value?
Solved – Standard deviation of a ratio (percentage change)
mathematical-statisticsstandard deviationstandard error
Best Answer
If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables
(You have to recognize that the two sample means are themselves random variables to apply it.)
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Edit:
The formula is directly relevant for your case because $Var(100(y-z)/z) = 100^2 Var(\frac{y}{z} -1) = 100^2 Var(y/z)$.
I don't know of a specific book reference off the top of my head, it feels a bit like asking for a reference for how to do long division.
It's an absolutely standard technique for approximating means and variances, based quite directly (and in a fairly obvious way) off Taylor series, which have been around for 300 years now. It's certainly mentioned in books, but I've never learned it from a book, in spite of encountering it many times - it's always 'expand this transformation in a Taylor series' (usually, but not always about the mean) and 'take expectations' or 'take variances' (or whatever, as necessary).
Once you learn how to do Taylor series (standard early-undergrad mathematics) and know a few properties of expectations and variances (standard early mathematical statistics), you're done; it's something undergrad students are given as an exercise.
I'll see if I can dig up a reference; there's sure to be something in a standard old reference like Cox and Hinkley or Kendall and Stuart or Feller or something (none of which I have to hand at the moment).