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I have results of meta-analysis, which have been carried out and reported. These analyses report tau-squared as a measure of heterogeneity (and the other details from the analysis). I've been asked to report the I-squared.

Is it possible to calculate I-squared from the reported results, or would I need to track down the original files to rerun the meta-analysis?

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    $\begingroup$ It would be helpful to tell us what tau-squared & I-squared are, apart from Greek & Latin letters. And don't you mean "Calculating I-squared [...] given tau-squared" in the title? $\endgroup$ – Scortchi - Reinstate Monica Aug 23 '13 at 23:54
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    $\begingroup$ They are measures of heterogeneity that are used in meta-analysis. I-squared is the percentage of total variation across studies that is due to heterogeneity rather than chance. It's calculated using 100%×(Q - df)/Q where Q is the Cochran's heterogeneity statistic, which is chi-square distributed. Tau-squared as an absolute measure of heterogeneity, it's square root is a measure of the standard deviation of effect sizes across studies. My assumption was that someone unfamiliar with the terms as used in meta-analysis would not know answer the question, but that might have been incorrect. $\endgroup$ – Jeremy Miles Aug 24 '13 at 0:14
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    $\begingroup$ Most reports of meta-analyses would include the value of $Q$ and its $df$. So why don't you compute $I^2$ based on that? $\endgroup$ – Wolfgang Aug 24 '13 at 14:54
  • $\begingroup$ I don't have the values of Q. $\endgroup$ – Jeremy Miles Aug 26 '13 at 18:08
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    $\begingroup$ Also you might want to take a look at: - Multivariate random-effects meta-regression: Updates to mvmeta (2011) STATA Journal by Ian R. White, MRC Biostatistics Unit, Cambridge, UK $\endgroup$ – user55310 Sep 5 '14 at 22:46
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In the random-effects model the method-of-moments (DerSimonian–Laird) estimator of the between-study variance for $n$ studies is given by

$$ \hat{\tau}^2=\frac{Q-(n-1)}{\sum{w_i}-\frac{\sum{w_i^2}}{\sum{w_i}}}$$

where Cochran's $Q=\sum{w_i}\left(y_i-\frac{\sum{w_i y_i}}{\sum{w_i}}\right)^2$ (with $y_i$ & $w_i$ as the effect size & the reciprocal of the within-study variance respectively, estimated from the $i$th study).

The total proportion of variance owing to heterogeneity is

$$I^2=\frac{Q-(n-1)}{Q}$$

So from $\hat{\tau}^2$ alone you can't calculate $I^2$; you also need the number of studies, the sum of the weights, & the sum of the squared weights.

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