I am conducting a meta analysis using the metafor package in R and am trying to calculate the Standard Error (if possible) of the Tau statistic.

The S.E of the tau-squared is obviously available easily from output of the metafor::rma function for example:

Random-Effects Model (k = 6; tau^2 estimator: REML)

tau^2 (estimated amount of total heterogeneity): 0.0331 (SE = 0.0384)
tau (square root of estimated tau^2 value):      0.1820
I^2 (total heterogeneity / total variability):   56.53%
H^2 (total variability / sampling variability):  2.30

Test for Heterogeneity: 
Q(df = 5) = 12.8056, p-val = 0.0253

Model Results:

estimate      se    zval    pval   ci.lb   ci.ub     
0.3463  0.1012  3.4239  0.0006  0.1481  0.5446  ***

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I know from this this answer that creating CI's for Tau is is straight forward (simply square root the upper and lower bounds) can the same be applied to the standard error? So in the above case the standard error would be:

> sqrt(0.0384)
[1] 0.1959592 

Is this the correct method and if not what would the correct method be?


The problem with what you suggest is that the confidence interval for $\tau$ or, for that matter, $\tau^2$ is not symmetric about the central value so the standard error is not really useful for establishing it. The package does provide a confint() function which gives you the appropriate confidence interval for either of them. In fact I am not really sure why the standard error is provided in that output.


The Q-profile method is directly implemented in the metafor package for calculating confidence intervals of tau and other statistics. Standard error is not useful and I should also mention that it is rarely important in practice. This is one of the compelling reason for the use of I-square as heterogeneity statistics. Check this material from the package author http://www.metafor-project.org/doku.php/analyses:viechtbauer2007a


You want to use the delta method. Based on this, the standard error of $\hat{\tau}$ would be $$\mbox{SE}[\hat{\tau}] = \mbox{SE}[\hat{\tau}^2] / (2\tau).$$ To actually compute this in practice, one would substitute $\hat{\tau}$ for $\tau$, so $$\mbox{SE}[\hat{\tau}] = \mbox{SE}[\hat{\tau}^2] / (2\hat{\tau}).$$

As @mdewey already mentioned, the usefulness of this SE (and that of $\hat{\tau}^2$) is rather limited. It would only be useful when $k$ and $\tau^2$ are both large.

  • $\begingroup$ Thanks for the answer - I really appreciate it! I want to use the S.E (and Tau value or Tau-squared for that matter) in a z-test. Would this be appropriate? Thanks. $\endgroup$ – Mwatch Nov 10 '18 at 0:42
  • 1
    $\begingroup$ No, Wald-type test of variance components have terrible performance. $\endgroup$ – Wolfgang Nov 10 '18 at 15:15
  • $\begingroup$ Would the same apply in a 2 sample z test for checking difference in population means? Thanks. $\endgroup$ – Mwatch Nov 10 '18 at 18:38
  • $\begingroup$ That's something entirely different (if you really mean the difference between means and not the difference between two $\tau^2$ values). For testing (differences of) means, Wald-type tests are usually fine (the Knapp & Hartung method is even better, but this is a refinement of the Wald-type tests, not an entirely different procedure). $\endgroup$ – Wolfgang Nov 11 '18 at 9:06
  • $\begingroup$ @Michael That seems a different question from the original one you asked. The crude approach would be to convert your single value to a $z$-score using the mean and standard deviation of the values of $\tau$ you have in your comparison group. $\endgroup$ – mdewey Nov 12 '18 at 11:55

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