I am trying to do a quick and dirty meta-analysis based on published values

The paper I'm reading has a table showing two independent population, with means ($m_1$ and $m_2$), stdev ($std_1$ and $std_2$) and N ($N_1$ and $N_2$) for each population. I do not have the exact p-values for the t-test (only a range 0.05>p>0.01 for example).

To compare to my data, I would like to calculate the effect size (Hedges' $g$) ± 95%CI.

From my research, I think I found how to calculate Cohen's $d$, and from there Hedges' $g$ (formulas from here)

$$ d = \frac{m_1-m_2}{\sqrt{\frac{((N_1-1)*std_1+(N_2-1)*std_2)}{(N_1+N_2-2)}}} $$

and $$ g = d * (1-\frac{3}{4*(N_1+N_2-2)-1)} $$

However, I'm not sure how to calculate the CI of the effect size. The document explains how to calculate the variance $V_d$ and $V_g$ but I don't know how to go from variance to CI.


1 Answer 1


Find the standard error by taking the square root of the sampling variance for which you say you have the formula. Then find the normal deviate corresponding to the confidence coefficient you require, for 95% this is 1.96. Multiply the standard error by that value and take that either side of the central estimate which you already have.

  • $\begingroup$ Thank you. I have read before that 95%CI = 1.96*SE. But doesn't that assume a normal distribution for ... (I'm not sure what actually, for g?)? And is that a problem? $\endgroup$ Apr 14, 2020 at 18:01
  • $\begingroup$ Why would you assume its' not a normal distribution? You can always contact the authors, but generally speaking, most authors who report a mean/ SD do so because the data isn't normally distributed. $\endgroup$
    – abousetta
    Apr 15, 2020 at 1:23

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