1
$\begingroup$

A colleague does replication on a quite coslty experiments. There are four different conditions, each one duplicated. The outcome with $4\times 2 = 8$ points is illustrated below:

Replication dataset

The analysis is standardized: each pair of colored points yields an average $\mu$ associated with its dispersion $d$, to assess the variability in the given condition. Black diamonds apparently exhibit higher variation, for instance. The colleague directly computes the standard deviation on each pair of points, and uses $d=2\sigma$. I know statistics on such small dataset can be touchy. I would like to provide advices to keep numbers as clean as possible on this condition.

My questions are the following (I am not a trained statistician):

  1. The variance is computed directly, and not corrected by the $\frac{1}{n-1}$ factor (Bessel's correction), which I believe would be the standard with sample estimation (see here). In fact, this is the same here, since $n=2$, for each pair, so $n-1 = \sqrt{n-1} = 1$. Am I correct in insisting on mentioning this factor in an explicit manner, although it does not change the results? By "explicit manner", I mean writing in documents, or even in Excel macros (in case somebody use them in a different context), that $d=2\sqrt{\frac{(x_1-\mu)^2+(x_2-\mu)^2}{n-1}}$, instead of $d=2\sqrt{(x_1-\mu)^2+(x_2-\mu)^2}$, which the colleague uses, even if it does not make a difference. Some people are afraid of formulas in some fields.
  2. Sometimes, the colleague gathers the red and black dots. Am I correct that in this case the correction on $\sigma$ is mandatory, as now $n =4$ for one quadruplet, and $n=2$ for each of the two remainging couples?
  3. I feel that if the experiments could be triplicated, one could expect a "huge" gain in estimating sample dispersion, since one morally would expect to divide the estimate by $\sqrt{3-1}$ instead of $\sqrt{2-1}$. Does that sounds sound enough?

I know two points is not much, and we are close to the singularity limit of one-sample statistics. I have heard a colleague say: "I do not replicate, to avoid dispersion".

$\endgroup$
  • 1
    $\begingroup$ I'm hesitant to answer because I'm not a publishing academic, but I certainly think it's important to mention when nonstandard computations are used. By "directly," do you mean "dividing by $n$" instead of "dividing by $n-1$?" $\endgroup$ – shadowtalker Apr 30 '16 at 13:52
  • $\begingroup$ @ssdecontrol Do not hesitate, I am pretty such datasets rarely appear in the academic statistical literature. I meant that I would like the colleague to write the correct formula, even if it gives the same results as a casual one $\endgroup$ – Laurent Duval Apr 30 '16 at 13:56
1
$\begingroup$
  1. Yes, you are correct. That calculation is standard for a reason.if your colleague feels the need to use a different calculation, it is their responsibility explain why, and to enumerate the different calculation they use. That formula is not "casual," it is wrong. If anything, what would be "casual" would be to use the standard calculation from any statistics software, which in most cases will use $n-1$ in the denominator.
  2. Errors propagate. In this case, using the wrong calculation will yield the wrong results even on a relative scale, so your colleague's wrongness will just be compounded.
  3. Keep in mind that you will be increasing the numerator as well: there's no way to tell (without some prior knowledge) whether the third point will increase dispersion or decrease it. However, adding a point will always improve the precision of the dispersion estimate: with $n=3$ the variance of the variance estimate would be $\frac{\mu^4}{3}$, whereas with $n=2$ is would be $\frac{\mu^4}{2} + \frac{\sigma^4}{2}$. The latter is always larger than the former.
$\endgroup$
  • $\begingroup$ Could you please provide me with a pointer for the variance of the variance estimate? $\endgroup$ – Laurent Duval May 11 '16 at 18:28
  • $\begingroup$ @LaurentDuval it's in that link $\endgroup$ – shadowtalker May 11 '16 at 20:25

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.