The metafor Package

A Meta-Analysis Package for R

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faq [2020/06/03 23:54]
Wolfgang Viechtbauer
faq [2020/06/26 06:56] (current)
Wolfgang Viechtbauer
Line 75: Line 75:
 !!! The pseudo $R^2$ statistic (Raudenbush, 2009) is computed with $$R^2 = \frac{\hat{\tau}_{RE}^2 - \hat{\tau}_{ME}^2}{\hat{\tau}_{RE}^2},$$ where $\hat{\tau}_{RE}^2$ denotes the estimated value of $\tau^2$ based on the random-effects model (i.e., the total amount of heterogeneity) and $\hat{\tau}_{ME}^2$ denotes the estimated value of $\tau^2$ based on the mixed-effects model (i.e., the residual amount of heterogeneity). It can happen that $\hat{\tau}_{RE}^2 < \hat{\tau}_{ME}^2$, in which case $R^2$ is set to zero. Again, the value of $R^2$ will change depending on the estimator of $\tau^2$ used. Also note that this statistic is only computed when the mixed-effects model includes an intercept (so that the random-effects model is clearly nested within the mixed-effects model). You can also use the ''anova.rma.uni()'' function to compute $R^2$ for any two models that are known to be nested. !!! The pseudo $R^2$ statistic (Raudenbush, 2009) is computed with $$R^2 = \frac{\hat{\tau}_{RE}^2 - \hat{\tau}_{ME}^2}{\hat{\tau}_{RE}^2},$$ where $\hat{\tau}_{RE}^2$ denotes the estimated value of $\tau^2$ based on the random-effects model (i.e., the total amount of heterogeneity) and $\hat{\tau}_{ME}^2$ denotes the estimated value of $\tau^2$ based on the mixed-effects model (i.e., the residual amount of heterogeneity). It can happen that $\hat{\tau}_{RE}^2 < \hat{\tau}_{ME}^2$, in which case $R^2$ is set to zero. Again, the value of $R^2$ will change depending on the estimator of $\tau^2$ used. Also note that this statistic is only computed when the mixed-effects model includes an intercept (so that the random-effects model is clearly nested within the mixed-effects model). You can also use the ''anova.rma.uni()'' function to compute $R^2$ for any two models that are known to be nested.
  
-??? For random-effects models fitted with the rma() function, how is the credibility/prediction interval computed by the predict() function?+??? For random-effects models fitted with the rma() function, how is the prediction interval computed by the predict() function?
  
 !!! By default, the interval is computed with $$\hat{\mu} \pm z_{1-\alpha/2} \sqrt{\mbox{SE}[\hat{\mu}]^2 + \hat{\tau}^2},$$ where $\hat{\mu}$ is the estimated average true outcome, $z_{1-\alpha/2}$ is the $100 \times (1-\alpha/2)$th percentile of a standard normal distribution (e.g., $1.96$ for $\alpha = .05$), $\mbox{SE}[\hat{\mu}]$ is the standard error of $\hat{\mu}$, and $\hat{\tau}^2$ is the estimated amount of heterogeneity (i.e., the variance in the true outcomes across studies). If the model was fitted with the Knapp and Hartung (2003) method (i.e., with ''test="knha"'' in ''rma()''), then instead of $z_{1-\alpha/2}$, the $100 \times (1-\alpha/2)$th percentile of a t-distribution with $k-1$ degrees of freedom is used. !!! By default, the interval is computed with $$\hat{\mu} \pm z_{1-\alpha/2} \sqrt{\mbox{SE}[\hat{\mu}]^2 + \hat{\tau}^2},$$ where $\hat{\mu}$ is the estimated average true outcome, $z_{1-\alpha/2}$ is the $100 \times (1-\alpha/2)$th percentile of a standard normal distribution (e.g., $1.96$ for $\alpha = .05$), $\mbox{SE}[\hat{\mu}]$ is the standard error of $\hat{\mu}$, and $\hat{\tau}^2$ is the estimated amount of heterogeneity (i.e., the variance in the true outcomes across studies). If the model was fitted with the Knapp and Hartung (2003) method (i.e., with ''test="knha"'' in ''rma()''), then instead of $z_{1-\alpha/2}$, the $100 \times (1-\alpha/2)$th percentile of a t-distribution with $k-1$ degrees of freedom is used.
  
-Note that this differs from Riley et al. (2001), who suggest to use a t-distribution with $k-2$ degrees of freedom for constructing the interval. Neither a normal, nor a t-distribution with $k-1$ or $k-2$ degrees of freedom is correct; all of these are approximations. The computations in metafor are done in the way described above, so that the credibility/prediction interval is identical to the confidence interval for $\mu$ when $\hat{\tau}^2 = 0$, which could be argued is the logical thing that should happen. +Note that this differs slightly from Riley et al. (2001), who suggest to use a t-distribution with $k-2$ degrees of freedom for constructing the interval. Neither a normal, nor a t-distribution with $k-1$ or $k-2$ degrees of freedom is correct; all of these are approximations. The computations in metafor are done in the way described above, so that the prediction interval is identical to the confidence interval for $\mu$ when $\hat{\tau}^2 = 0$, which could be argued is the logical thing that should happen.
  
 ??? How is the Freeman-Tukey transformation of proportions and incidence rates computed? ??? How is the Freeman-Tukey transformation of proportions and incidence rates computed?
faq.txt ยท Last modified: 2020/06/26 06:56 by Wolfgang Viechtbauer