4 ms·

No, that's just wrong. See the original source: https://journals.sagepub.com/doi/abs/10.1111/1467-9280.t01-1-01430 https://journals.sagepub.com/doi/abs/10.1111/

• drawkward 3y ago
  Nice link to a paywalled source.
• icegreentea2 3y ago
  Thanks to scihub, we get the paragraph: > Point estimates for effect size on all cognitive tasks were 0.164 > (SE = 0.028, n = 96, p = .05) for control groups and 0.478 (SE = > 0.029, n = 101, p = .01) for exercisers, suggesting that both groups > improved between Times 1 and 2. However, the control groups’ im- > provement was about 1/8 a standard deviation, whereas the exercise > groups’ improvement was nearly 1/2 a standard deviation, on average. > Both values are significantly different from zero, and from each other > (see Table 1). Earlier in the paper they define their effect size as: g = (M_Post - M_Pre )/SD_P , where M_Pre is the preintervention mean task performance, M_Post is the postintervention mean, and SD_P is the pooled standard deviation. I think the standard deviation mentioned is relative to overall population spread of cognitive performance, given that standard errors of the effect size are given separately?