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?

  • $\begingroup$ Dou you know how $X$ and $Y$ are distributed? $\endgroup$ – Sven Hohenstein Feb 6 '13 at 13:33
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    $\begingroup$ I'm too afraid of the hardcore statisticians here to post a real answer, so I post it here: I was told that if you want to normalize your data $y \pm \Delta y$ to $z \pm \Delta z$ $$x = 100 \times \frac{y}{z},$$ you have to calculate the standard deviation $\Delta x$ as follows: $$\Delta x = x \sqrt{\left ( \frac{\Delta y}{y} \right )^2 + \left (\frac{\Delta z}{z} \right )^2}$$ Maybe you can take it from there, but I'm not sure if this helps. $\endgroup$ – Eekhoorn Feb 6 '13 at 15:12

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



(You have to recognize that the two sample means are themselves random variables to apply it.)



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).

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  • $\begingroup$ Sven Hohenstein: I know how X and Y are distributed. $\endgroup$ – Lucy Feb 6 '13 at 16:44
  • $\begingroup$ zenbomb: Thank you very much. Can you help me to find out the book source of this? Can this formula be applied for my case in which (x=100*(y-z)/z)? $\endgroup$ – Lucy Feb 6 '13 at 16:47
  • $\begingroup$ see the edit to my answer $\endgroup$ – Glen_b Feb 6 '13 at 23:09
  • $\begingroup$ Lucy, if you know the joint distribution of X and Y (or are prepared to assume that they're independent), then you shouldn't need the Taylor expansion approximation. You may well be able to compute the distribution of the ratio you want - and then get variances - either exact, or to a much better approximation - from that. $\endgroup$ – Glen_b Feb 6 '13 at 23:17

The Taylor series method yields an estimator for the variance which can then be used to estimate a symmetric confidence interval based upon a crude normality assumption of your ratio.

A more general approach which directly yields a (possibly unsymmetric) confidence interval for the ratio of two random variables is MOVER-R by Donner and Zhou:

Donner, Zhou: Closed-form confidence intervals for functions of the normal mean and standard deviation. Statistical Methods in Medical Research, 21(4), 347–359 (2012)

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