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This is the formula I would need to use to perform a meta-analysis in R using the package metafor:

REmodel <- rma(m1i=elev.mean,m2i=amb.mean,sd1i=?,
               sd2i=?,n1i=?,n2i=?,data=NPPsiteWide,
               measure="ROM")

Note that sd and n refers to standard deviations and sample size. The problem is that the dataset I am using (data gathered by others) contains se values, but not sd and n.

Is there any way I can still do the meta-analysis using the data I have without sd and n? Thanks

EDIT:

After following Wolfgang’s suggestion below, it seems I am doing something wrong because rma can’t calculate vi (all NA’s), and returns this error:

REmodel <- rma(m1i=elev.mean,m2i=amb.mean,sd1i=elev.se,
               sd2i=amb.se,n1i=1,n2i=1,data=db,
               measure="ROM",subset=Myc=="AM”)


Error in rma(m1i = elev.mean, m2i = amb.mean, sd1i = elev.se, sd2i = amb.se,  : 
          Processing terminated since k = 0.
        In addition: Warning message:
        In rma(m1i = elev.mean, m2i = amb.mean, sd1i = elev.se, sd2i = amb.se,  :
          Studies with NAs omitted from model fitting.
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  • $\begingroup$ rma() can also take the estimate and its standard error as arguments. You don't need to use the mean and sd. $\endgroup$ – Jeremy Miles May 13 '15 at 17:11
  • $\begingroup$ The standard error of a mean is equal to the standard deviation divided by the square-root of the sample size. So, if you know the standard error and the sample size, it's easy to calculate the SD. $\endgroup$ – Wolfgang May 13 '15 at 22:32
  • $\begingroup$ What I said is that I do know se, but I do not know sd and n, hence my question $\endgroup$ – fede_luppi May 14 '15 at 7:08
  • $\begingroup$ @JeremyMiles how can I include SE as an argument considering the structure of my dataset? (see EDIT). Thanks $\endgroup$ – fede_luppi May 14 '15 at 7:19
  • $\begingroup$ @fede_luppi Ah, okay, it wasn't clear to me whether you knew the n's (in which case it would have been easy to compute the SDs). But for the (log-transformed) ratio of means as the outcome measure, there is a simple solution. I'll provide a proper answer. $\endgroup$ – Wolfgang May 15 '15 at 9:26
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The outcome measure used is the (log-transformed) ratio of means (often called the response ratio in the ecology literature), which is given by $$y = \ln\left[\frac{\bar{x}_1}{\bar{x}_2}\right] = \ln[\bar{x}_1] - \ln[\bar{x}_2].$$ The large-sample approximation to the sampling variance of $y$ is given by $$Var[y] = \frac{SD_1^2}{n_1 \bar{x}_1^2} + \frac{SD_2^2}{n_2 \bar{x}_2^2}$$ (see, for example, Hedges et al., 1999). Since $$SE[\bar{x}_1] = \frac{SD_1}{\sqrt{n_1}} \quad \mbox{and} \quad SE[\bar{x}_2] = \frac{SD_2}{\sqrt{n_2}},$$ it follows that $$Var[y] = \frac{SE_1^2}{\bar{x}_1^2} + \frac{SE_2^2}{\bar{x}_2^2}.$$ So, you could easily compute this by hand based on the information you have.

But there is an even simpler trick. All that you have to do is feed the SEs to the escalc() or rma() functions and at the same time set the sample sizes to 1. An example:

library(metafor)
escalc(measure="ROM", m1i=15.6, m2i=12.2, sd1i=3.82, sd2i=3.22, n1i=15, n2i=20, digits=6)
escalc(measure="ROM", m1i=15.6, m2i=12.2, sd1i=3.82/sqrt(15), sd2i=3.22/sqrt(20), n1i=1, n2i=1, digits=6)

Both give you:

          yi         vi
1 0.24583496 0.00748055

Hedges, L. V., Gurevitch, J., & Curtis, P. S. (1999). The meta-analysis of response ratios in experimental ecology. Ecology, 80(4), 1150-1156.

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  • $\begingroup$ Dear Wolfgang, thanks for your through answer. I would like to share a further though with you: Is it OK for a paper if I weight each experiment based on the number of years each experiment was carried out? I can found #years in the literature. In that case, could I calculate SD as SE/sqr(#years)? The next step would be to perform the meta-analysis for each Mic * N level so I would have 4 groups: AM-LowN, AM-HighN, ECM-LowN, ECM-HighN, but to be honest I am not sure how I should structure the dataset. Thanks and sorry for my lack of skills. $\endgroup$ – fede_luppi May 15 '15 at 11:04
  • $\begingroup$ The reason is that the longer the experiment, the more important its results should be within the meta-analysis $\endgroup$ – fede_luppi May 15 '15 at 11:19
  • $\begingroup$ This seems to be a separate question/issue. Somewhat related, but still different. I would suggest to ask a new question. $\endgroup$ – Wolfgang May 15 '15 at 22:17
  • $\begingroup$ please see me edit above about the error after following your advice. $\endgroup$ – fede_luppi May 16 '15 at 11:14
  • $\begingroup$ I will ask a new question with the related topic. Many thanks $\endgroup$ – fede_luppi May 16 '15 at 11:15

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