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I want to run a meta-analysis over several studies reporting one-sample data (e.g. comparing participants' scores against a baseline score of zero). I calculated Cohen's d by dividing the difference of the sample mean and the baseline score by the sample standard deviation (as reported in the first answer here). How do I get the sampling variance of this effect size? (needed in the meta-analysis for calculating inverse variance weights)

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  • $\begingroup$ I'm using the R package 'metafor' $\endgroup$
    – Johanna
    Aug 2, 2016 at 9:57

1 Answer 1

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This is an interesting question because (so far as I know) there is no widely used formula for computing the variance in this situation. Some time ago, I did some simulations to examine the performance of different formulas to estimate the sampling variance of Cohen's d in case of a one-sample t-test.

I was aware of three different formulas:

The formula used in the Comprehensive Meta-analysis Software:

(1/sqrt(ni))*sqrt(1+di^2/2)^2,

with ni being the sample size per study and di the observed Cohen's d.

Other people use the standard formula for the dependent samples t-test (e.g., Borenstein, 2009) with correlation between pre- and posttest (r) equal to 0.5:

(1/ni)+di^2/(2*ni)

Another formula I have seen is one that was used in a paper by Koenig et al. (2011). This formula is obtained by personal communication with B. Becker.

(1/ni)+di^2/(2*ni*(ni-1)) 

I did a very small simulation study to examine the performance of these three formulas with sample sizes ranging from 10 to 500 and effect sizes in the population ranging from 0 to 0.8. The differences between the formulas were most observable for a population effect size of 0.8.

enter image description here

Using the formula of the dependent samples t-test with r=0.5 yielded the least biased estimates. However, there may be other formulas with better properties. I am curious what other people think about this.

Code:

rm(list = ls()) # Clean workspace

k <- 10000 # Number of studies

thetais <- c(0, 0.2, 0.5, 0.8) # Effect in population 

nis <- c(10,15,20,30,50,75,100,250,500) # Sample size in primary study
sigma <- 1 # Standard deviation in population

### Empty objects for storing results
vi.ac <- vi.beck <- vi.comp <- vi.dep <- matrix(NA, nrow = length(nis), 
                                                ncol = length(thetais), 
                                                dimnames = list(nis, thetais))

############################################
for(thetai in thetais) {
  for(ni in nis) { 

    ### Actual variance Cohen's d
    sdi <- sqrt(sigma/(ni-1) * rchisq(k, df = ni-1))
    mi <- rnorm(k, mean = thetai, sd = sigma/sqrt(ni))
    di <- mi/sdi

    vi.ac[as.character(ni),as.character(thetai)] <- var(di)

    ############################################

    ### Suggestion by Becker in Koenig et al.
    vi <- (1/ni)+di^2/(2*ni*(ni-1))
    vi.beck[as.character(ni),as.character(thetai)] <- mean(vi)

    ############################################

    ### Comprehensive meta-analysis software
    vi <- (1/sqrt(ni))*sqrt(1+di^2/2)^2
    vi.comp[as.character(ni),as.character(thetai)] <- mean(vi)

    ############################################

    ### Dependent sample t-test with r=0.5
    vi <- (1/ni)+di^2/(2*ni)
    vi.dep[as.character(ni),as.character(thetai)] <- mean(vi)

  }
}

plot(x = nis, y = vi.ac[ ,1], type = "l", main = "theta = 0", ylab = "Variance")
lines(x = nis, y = vi.beck[ ,1], type = "l", col = "red")
lines(x = nis, y = vi.comp[ ,1], type = "l", col = "blue")
lines(x = nis, y = vi.dep[ ,1], type = "l", col = "green")
legend("topright", legend = c("Actual variance", "Becker", "CMA", "Dep. samples"), 
       col = c("black", "red", "blue", "green"), lty = c(1,1,1,1))

plot(x = nis, y = vi.ac[ ,2], type = "l", main = "theta = 0.2")
lines(x = nis, y = vi.beck[ ,2], type = "l", col = "red")
lines(x = nis, y = vi.comp[ ,2], type = "l", col = "blue")
lines(x = nis, y = vi.dep[ ,2], type = "l", col = "green")
legend("topright", legend = c("Actual variance", "Becker", "CMA", "Dep. samples"), 
       col = c("black", "red", "blue", "green"), lty = c(1,1,1,1))

plot(x = nis, y = vi.ac[ ,3], type = "l", main = "theta = 0.5")
lines(x = nis, y = vi.beck[ ,3], type = "l", col = "red")
lines(x = nis, y = vi.comp[ ,3], type = "l", col = "blue")
lines(x = nis, y = vi.dep[ ,3], type = "l", col = "green")
legend("topright", legend = c("Actual variance", "Becker", "CMA", "Dep. samples"), 
       col = c("black", "red", "blue", "green"), lty = c(1,1,1,1))

plot(x = nis, y = vi.ac[ ,4], type = "l", main = "theta = 0.8")
lines(x = nis, y = vi.beck[ ,4], type = "l", col = "red")
lines(x = nis, y = vi.comp[ ,4], type = "l", col = "blue")
lines(x = nis, y = vi.dep[ ,4], type = "l", col = "green")
legend("topright", legend = c("Actual variance", "Becker", "CMA", "Dep. samples"), 
       col = c("black", "red", "blue", "green"), lty = c(1,1,1,1))

data.frame(vi.ac[,1], vi.beck[,1], vi.comp[,1], vi.dep[,1])

References:

Borenstein, M. (2009). Effect sizes for continuous data. In H. Cooper, L. V. Hedges & J. C. Valentine (Eds.), The Handbook of Research Synthesis and Meta-Analysis (pp. 221-236). New York: Russell Sage Foundation.

Koenig, A. M., Eagly, A. H., Mitchell, A. A., & Ristikari, T. (2011). Are leader stereotypes masculine? A meta-analysis of three research paradigms. Psychological Bulletin, 137, 4, 616-42.

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  • $\begingroup$ Thank you, this is extremely helpful! If I want to include your simulation in a paper or presentation, how should I cite you? $\endgroup$
    – Johanna
    Aug 2, 2016 at 15:18
  • $\begingroup$ I am happy to help ;-). I think you should just cite this webpage. This simulation is not a an article or book. $\endgroup$
    – User33
    Aug 4, 2016 at 7:13

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