When averaging independent results, the type A uncertainty can be estimated as the standard error of the mean, $$\frac{\sigma}{\sqrt{n}}$$ where $\sigma$ and $n$ are the standard deviation and the number of values being averaged.

If each value corresponds to a mean and has its own variance, then a better estimate would be the DerSimonian Laird estimator, which I understand calculates a weighted mean of the means and provides a standard error.

In that case, the type A uncertainty would also be the standard error divided by $\sqrt{n}$?

Example: I have the following means and the sampling variances (code in R)

means  <- c(1004.068, 1002.567, 1002.448)
samVar <- c(0.1449915, 0.1623947, 0.4040860)

The mean of means and standard error would be

mean(means); sd(means)/sqrt(length(means))
# > [1] 1003.027
# > [1] 0.5212864

while the DerSimonian estimator would be

metafor::rma(yi = means, vi = samVar, method = 'DL')
# > (trimmed)
# > 1003.0785  0.5648

The concrete question here is if shall I divide that 0.5648 by $\sqrt{3}$ to get the type A uncertainty estimator.

  • $\begingroup$ No, 0.5648 is already the standard error, so you should not divide this by $\sqrt{3}$. $\endgroup$
    – Wolfgang
    Jul 1 at 6:11
  • $\begingroup$ Thank you. So would you say that the 0.521 that results from $S(x)/\sqrt{3}$ is underestimating the type A uncertainty? $\endgroup$
    – Crparedes
    Jul 2 at 13:18

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