6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Dou you know how $X$ and $Y$ are distributed? $\endgroup$ – Sven Hohenstein Feb 6 '13 at 13:33
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.)

---

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).

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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