# How to transform a negative effect into positive? How to reverse the effect direction?

I am conducting a meta-analysis on the effectiveness of a learning intervention. In the studies I analyse, there are multiple outcome measures: the majority of them have a positive direction, i.e. we expect them to increase (e.g., accuracy, proficiency, vocabulary-size, etc.), but some have a negative direction, i.e. we expect them to decrease (e.g., number of errors/sentence).

outcome_negative <- data.frame(Timing = c(1, 2), Mean = c(.067, .036), SD = c(.251, .187))
# outcome_negative is reduced in ±half between pre and post-test: it is very positive!
# let's just copy it to the target variable before adapting the values
outcome_positive <- outcome_negative


I am computing the effect sizes for all these values, through Standardized Mean Difference (d), but I could not find references on how to manage negative direction effects.

• $$-\bar{x}$$
• outcome_positive$Mean <- -outcome_negative$Mean
• What about the standard deviations? $$-s$$ ? or $$s$$ ?
• $$\dfrac{1}{\bar{x}}$$
• outcome_positive$Mean <- 1/outcome_negative$Mean
• What about the standard deviations? $$\dfrac{1}{s}$$ ? $$s$$ ?
• Reversing pre- and post-test values (but it seems even more wrong)
• outcome_positive$Mean[outcome_positive$Timing == 1] <- outcome_negative$Mean[outcome_negative$Timing == 2] outcome_positive$Mean[outcome_positive$Timing == 2] <- outcome_negative$Mean[outcome_negative$Timing == 1]

Most of all, I wonder what is statistically sound. Don't hesitate to send me to a reference document.

In general, you should use $-\bar x$ as the effect size to account for the change in polarity of the outcome, but leave $s$ unchanged.
• A heuristic argument for leaving $s$ unchanged would be that negative standard errors don't make any sense.
$$\sigma(-x) = \sqrt{\sigma^2(-x)} = \sqrt{(-1)^2\sigma^2(x)} = \sqrt{\sigma^2(x)} = |\sigma(x)| = \sigma(x)$$ according to the basic rules governing variances (i.e. if $c$ is a constant, $\sigma^2(cx) = c^2 \sigma^2(x)$).