I have study-level means/SDs reflecting depression symptom severity from multiple single-group studies. I do not have access to participant-level data. The studies all use different measures, and I have the sample size and the minimum and maximum possible scale scores for each. I would like to be able to aggregate the means using meta-analysis to indicate the average severity of depression across these studies.

I was thinking about rescaling the means to a 0-100 scale so they roughly represent % of maximum severity but I'm not sure what I could use as the variance, because I can't compute a rescaled SD without participant-level data. Is there any other way I could go about this?

I'm using metafor to calculate other effect sizes for this project but could also calculate by hand. I'm assuming it would be inappropriate to treat the rescaled means as if they were percentages or event counts with ni or ti = 100, respectively, because that wouldn't take into account the actual variation in estimates, but if I'm wrong I'd love to know.


2 Answers 2


Ideally, one would want to use proper test equating methods to properly link the scores (and hence means and SDs) of different measures to each other. But since you only have the means, SDs, and possible ranges of the different studies, you are limited in what you can do. As you suggested, you could rescale each measure, so that it has the same range (0 to 1) with: $$y_i = \frac{\bar{x}_i - \mbox{min}_i}{\mbox{max}_i - \mbox{min}_i},$$ where $\bar{x}_i$ is the observed mean in study $i$ and $\mbox{max}_i$ and $\mbox{min}_i$ are the maximum and minimum possible scores on the measure used in the study (note that this must be the range of the possible scores, not the observed scores!). Then the sampling variance of $y_i$ is given by: $$\mbox{Var}[y_i] = v_i = \frac{\mbox{SD}^2_i}{n_i (\mbox{max}_i - \mbox{min}_i)^2},$$ where $\mbox{SD}^2_i$ is the observed SD in the study and $n_i$ is the sample size.

If you prefer a 0 to 100 range, multiply $y_i$ by 100 and $v_i$ by $100^2$.

You can then feed these estimates and corresponding sampling variances (or their square-root if the standard errors are the required input) into the meta-analysis software of your choice.

  • $\begingroup$ Do you have a reference for the equation of the variance? The first one is min-max normalisation. $\endgroup$
    – RikH
    Jun 16, 2020 at 13:53
  • $\begingroup$ It's trivial to derive this. If random variable $x$ has $\mbox{Var}[x] = s^2$, then $\mbox{Var}[a + b \times x] = b^2 \times s^2$ for any constants $a$ and $b$. $\endgroup$
    – Wolfgang
    Jun 16, 2020 at 13:58

I would like to extend the answer by Wolfgang on rescaling the measure. Consider some study. Let $X$ be the averaged mean, $l$ be the lower bound for the Likert scale, $u$ be the upper bound for the Likert scale (both bounds over all the answers) and $n_q$ be the number of items. We can split the lower and upper bound up into the number of items and the lower and upper bound for individual answers, denoted by $k_l$ and $k_u$ respectively. Then, the transformed mean $m_t$ can be obtained by using

$$ m_t = \frac{X \cdot n_q - l}{u - l} = \frac{X \cdot n_q - (n_q \cdot k_l)}{(n_q \cdot k_u) - (n_q \cdot k_l)} = \frac{(X - k_l) \cdot n_q}{(k_u - k_l) \cdot n_q} = \frac{X - k_l}{k_u - k_l}. $$

This is known as min-max normalisation.

To scale the standard deviations, we can use the equation for a linear transformation of the variance. According to Hogg et al. (2005):

Let $X$ be a random variable with finite mean $\mu$ and variance $\sigma^2$. Then for all constants $a$ and $b$, $$ Var(aX + b) = a^2 \cdot Var(X). $$

So, for the variance of $\frac{X - k_l}{k_u - k_l}$, $$ \begin{aligned} Var(\frac{X-k_u}{k_u - k_l}) &= Var((k_u - k_l)^{-1} \cdot (X - k_u)) \\ &= (k_u - k_l)^{-2} \cdot Var(X - k_u) \\ &= (k_u - k_l)^{-2} \cdot 1^2 \cdot Var(X) \\ &= \frac{Var(X)}{(k_u - k_l)^2}, \end{aligned} $$

and for the standard deviation

$$ \begin{aligned} sd(\frac{X-k_l}{k_u - k_l}) &= \sqrt{Var(\frac{X-k_l}{k_u - k_l})} \\ &= \sqrt{\frac{Var(X)}{(k_u - k_l)^2}} \\ &= \frac{sd(X)}{k_u - k_l}. \end{aligned} $$

Note that I have not taken Bessel's correction into account which will be an issue for studies having small sample sizes. I'll add that later if I understand how I can take the correction into account.


Hogg, R. V., McKean, J., & Craig, A. T. (2005). Introduction to mathematical statistics. Pearson Education.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.