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...in the elderly. And across all tasks it's only half a standard deviation, meaning the result is far from significant. There are lots of good studies on posit

...in the elderly. And across all tasks it's only half a standard deviation, meaning the result is far from significant. There are lots of good studies on positive effects of exercise in all sorts of ages and populations, but this is not one of them. Edit: Jesus, this thread is proof than HN techies think they know math but actually have no clue about statistics. They are easy prey for this kind of journalism.

JoeAltmaier 3y ago
More complicated than that? Across all tasks, each was half a std dev higher. That's very significant?
mathgradthrow 3y ago
Half a standard deviation is 8 IQ, and effect size has nothing to do with significance.
- sigmoid10 3y ago
  What are you talking about? Statistical significance is directly calculated from effect size and standard deviation. https://en.wikipedia.org/wiki/Statistical_significance https://en.wikipedia.org/wiki/Statistical_significance
  - 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 ago
      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/10.1111/1467-9280.t01-1...
      - 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?
  - rfrey 3y ago
    > What are you talking about? Statistical significance is directly calculated from effect size and standard deviation. But it's not about how big the effect is. It's how likely that the effect was or was not due to chance. A very small effect can be statistically significant. You missed "sample size" in your list above - a big sample size can help you show a small effect is probably not chance.
oldpersonintx 3y ago
your response is proof once again, HN will go through any amount of mental gymnastics to avoid common sense on diet and exercise
- networkchad 3y ago
  I would love to know the waist size of the average HN user.
  - deleted 3y ago
    [deleted]
  - paulcole 3y ago
    My money is on bimodal distribution.
- Etheryte 3y ago
  This isn't really constructive regardless of whether you're right or not. It's completely valid to e.g. both agree that exercise is good as well as criticize a specific study on the matter.
- lazide 3y ago
  Anyone telling me something I don't want to hear is clearly wrong, and probably a malicious asshole to boot. /s
- Der_Einzige 3y ago
  I'm pretty sure that HN skews far fatter than average, which is insane to think about since the average tech bro is far LESS fat than the average person. I imagine the average HN user to look like Richard Stallman.
  - FartyMcFarter 3y ago
    Why do you think HN skews fatter than average?
rfrey 3y ago
Half a standard deviation means going from the 50th percentile of IQ to the 69th. I'm assuming you didn't mean statistical significance since size of movement has nothing to do with statistical significance, so you must have meant significant in the sense of important. (Just saw your rebuttal about "significance" below. Statistical significance is not how big a movement is, it is the chance that such a movement could be due to random chance. A very small movement could be statistically significant if your sample is big enough to show it's probably (e.g. 95% probably) not due to chance.)
- drawkward 3y ago
  Or, they are confusing the way that standard deviation is used in the article as the size of the effect, with the standard error of that estimate, which would b3 used to determine the significance.
deleted 3y ago
[deleted]
enoch_r 3y ago
> And across all tasks it's only half a standard deviation, meaning the result is far from significant. I can't read the original article because it's paywalled, but assuming that the original claim is correct ("increased performance by approximately .5 of a standard deviation across all tasks tested") and that by "significant" you mean "statistically significant" then you are spectacularly and confidently wrong. Imagine we have a normal distribution of IQs around 100 with a standard deviation of 15. Imagine there were some intervention which resulted in a uniform increase in IQ by half a standard deviation across the board. Perhaps the intervention is "literally add 7.5 points to every single score in the treatment group." You seem to be asserting that because the effect size is "only" half a standard deviation, we have no way of determining whether or not the intervention actually increased scores, no matter how much data is collected. We could collect 100 million IQ scores in the control group, and 100 million IQ scores in the "treatment" group, and see that the average score in the control group is 100 and the average score in the treatment group is 107.5, and we'd have to say "nope, the null hypothesis looks completely plausible here, because the effect size was just half a standard deviation." Does that seem like a reasonable claim?
ramblenode 3y ago
> Edit: Jesus, this thread is proof than HN techies think they know math but actually have no clue about statistics. They are easy prey for this kind of journalism. And yet you are the one confusing the standard deviation of an effect size with the standard error of a point estimate...

JoeAltmaier 3y ago

More complicated than that? Across all tasks, each was half a std dev higher. That's very significant?

mathgradthrow 3y ago

Half a standard deviation is 8 IQ, and effect size has nothing to do with significance.

sigmoid10 3y ago
What are you talking about? Statistical significance is directly calculated from effect size and standard deviation. https://en.wikipedia.org/wiki/Statistical_significance https://en.wikipedia.org/wiki/Statistical_significance
- 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 ago
    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/10.1111/1467-9280.t01-1...
    - 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?
- rfrey 3y ago
  > What are you talking about? Statistical significance is directly calculated from effect size and standard deviation. But it's not about how big the effect is. It's how likely that the effect was or was not due to chance. A very small effect can be statistically significant. You missed "sample size" in your list above - a big sample size can help you show a small effect is probably not chance.

oldpersonintx 3y ago

your response is proof once again, HN will go through any amount of mental gymnastics to avoid common sense on diet and exercise

networkchad 3y ago
I would love to know the waist size of the average HN user.
- deleted 3y ago
  [deleted]
- paulcole 3y ago
  My money is on bimodal distribution.
Etheryte 3y ago
This isn't really constructive regardless of whether you're right or not. It's completely valid to e.g. both agree that exercise is good as well as criticize a specific study on the matter.
lazide 3y ago
Anyone telling me something I don't want to hear is clearly wrong, and probably a malicious asshole to boot. /s
Der_Einzige 3y ago
I'm pretty sure that HN skews far fatter than average, which is insane to think about since the average tech bro is far LESS fat than the average person. I imagine the average HN user to look like Richard Stallman.
- FartyMcFarter 3y ago
  Why do you think HN skews fatter than average?

rfrey 3y ago

Half a standard deviation means going from the 50th percentile of IQ to the 69th. I'm assuming you didn't mean statistical significance since size of movement has nothing to do with statistical significance, so you must have meant significant in the sense of important. (Just saw your rebuttal about "significance" below. Statistical significance is not how big a movement is, it is the chance that such a movement could be due to random chance. A very small movement could be statistically significant if your sample is big enough to show it's probably (e.g. 95% probably) not due to chance.)

drawkward 3y ago
Or, they are confusing the way that standard deviation is used in the article as the size of the effect, with the standard error of that estimate, which would b3 used to determine the significance.

deleted 3y ago

[deleted]

enoch_r 3y ago

> And across all tasks it's only half a standard deviation, meaning the result is far from significant. I can't read the original article because it's paywalled, but assuming that the original claim is correct ("increased performance by approximately .5 of a standard deviation across all tasks tested") and that by "significant" you mean "statistically significant" then you are spectacularly and confidently wrong. Imagine we have a normal distribution of IQs around 100 with a standard deviation of 15. Imagine there were some intervention which resulted in a uniform increase in IQ by half a standard deviation across the board. Perhaps the intervention is "literally add 7.5 points to every single score in the treatment group." You seem to be asserting that because the effect size is "only" half a standard deviation, we have no way of determining whether or not the intervention actually increased scores, no matter how much data is collected. We could collect 100 million IQ scores in the control group, and 100 million IQ scores in the "treatment" group, and see that the average score in the control group is 100 and the average score in the treatment group is 107.5, and we'd have to say "nope, the null hypothesis looks completely plausible here, because the effect size was just half a standard deviation." Does that seem like a reasonable claim?

ramblenode 3y ago

> Edit: Jesus, this thread is proof than HN techies think they know math but actually have no clue about statistics. They are easy prey for this kind of journalism. And yet you are the one confusing the standard deviation of an effect size with the standard error of a point estimate...