[R-meta] Effect size calculation
James Pustejovsky
jepu@to @end|ng |rom gm@||@com
Thu Dec 23 00:12:29 CET 2021
Hi Tharaka,
There are formulas in the literature for converting between a
standardized mean difference (of which Hedges' g is an estimate) and
the Pearson correlation, as well as from other test statistics (F, t,
X^2); See Jacobs & Viechtbauer (2017) and Pustejovsky (2014). However,
the formulas are only valid under specific distributional assumptions
about the variables involved (e.g., that a continuous predictor has
been artificially dichotomized, and that interest is in the
correlation between the continuous predictor) and the outcome. For the
study you attached, it's not at all clear why the authors would use
the Pearson correlation coefficient as the effect size measure. If I
understand correctly, they are interested in the effects of
contrasting treatment conditions (AES vs. control) so the assumptions
for converting from SMD to r would not really make sense here.
James
Jacobs, P., & Viechtbauer, W. (2017). Estimation of the biserial
correlation and its sampling variance for use in meta‐analysis.
Research synthesis methods, 8(2), 161-180.
Pustejovsky, J. E. (2014). Converting from d to r to z when the design
uses extreme groups, dichotomization, or experimental control.
Psychological methods, 19(1), 92.
On Tue, Dec 21, 2021 at 5:32 AM Tharaka S. Priyadarshana
<tharakas.priyadarshana using gmail.com> wrote:
>
> Dear all,
>
> I recently came across this paper (kindly see the attached pdf), and I am a bit confused about how these authors have measured the effect size for the MA. They have measured Pearson’s correlation coefficient in two different ways (see section 2.3, page 3).
>
> (i). Hedges’ g has been transformed to a Pearson’s correlation coefficient.
> (ii). Or, Pearson’s r has been calculated from F, t, or χ2 data.
>
> Is this a correct way of measuring the effect size? if yes, how this can be done??
> Then, can we really pool the effect sizes calculated from these two methods for the final analysis???
>
> Thank you,
> Tharaka
>
>
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