I faced an estimation problem of compute the mean of a random variable (rv) $X$. What I know is the coefficient of variation of this rv, namely its standard deviation divided by its mean: $$c_{v}(X) = {\sigma_{x}\over \mu_{x}}$$ Surely, I know that the more samples I have, the more accurate my estimation of $\mu$, but I want to use as small number of samples as possible (while keeping my estimation reasonably sound).

If we look the sample mean $\bar{X}=\sum_{i=1}^N{X_i}/N$, we know that $$c_{v}(\bar{X}) = {1\over \sqrt{N}} c_{v}(X)$$ My plan is to choose the number of samples $N$ with respect to $c_{v}(\bar{X})$, but I am not sure what a good $c_{v}(X)$ should be. Is there any rule of thumb?

PS: $c_{v}(X)>1$.

  • $\begingroup$ Do you know anything beyond the CV? Do you at least know the mean or have an estimate of it? Ordinarily the CV is less than relevant. What matters is how accurately do you want to estimate the mean and what you need to know, or estimate, is the standard deviation rather than the CV. $\endgroup$
    – whuber
    Sep 25, 2013 at 17:45
  • $\begingroup$ I could estimate both mean and variance for N samples. Maybe I should ask for what N is reasonably good enough. I actually want to balance the following two things: 1) N should be large enough to ensure a reliable mean estimation; and 2) N should be small enough to keep my cost low. $\endgroup$
    – pitfall
    Sep 25, 2013 at 18:01
  • 1
    $\begingroup$ That's the right way to think about it. To make this answerable, though, you need to quantify both (1) and (2). How would you measure reliability--relative error, absolute error, or something else? What is your sampling budget? $\endgroup$
    – whuber
    Sep 25, 2013 at 18:58
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    $\begingroup$ Section 2.2 of Gerald van Belle's book, "Statistical Rules of Thumb" is called "Calculating Sample Size Using the Coefficient of Variation". This is in the context of knowing only the CV and the % change in the mean you expect between test and control, and a standard power. The sample size (per group) = 16(CV^2) / (ln(mean1)-ln(mean2)^2. So if the CV is 30% and the expected difference in means is 20% each group should have >=29 in it. This is the two sample case, and you're asking about one sample, so some further algebra is needed. $\endgroup$
    – zbicyclist
    Sep 26, 2013 at 4:29
  • $\begingroup$ @zbicyclist Thanks. I think I could use some historical data as one sample and the actual observed ones as another, and choose the sample size as you suggested. $\endgroup$
    – pitfall
    Sep 26, 2013 at 18:53


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