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The standard deviation referred to in the article is the effect size, and not the standard error of that estimate, which is what would be used to determine its
by drawkward 3y ago
The standard deviation referred to in the article is the effect size, and not the standard error of that estimate, which is what would be used to determine its significance.
- sigmoid10 3y agoNo, 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/10.1111/1467-9280.t01-1...
- drawkward 3y agoNice link to a paywalled source.
- icegreentea2 3y agoThanks 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?