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I'm writing a review in which I need to find the change in mean score of a treatment group before and after treatment. The standard deviation of the mean is known for pre and post treatment seperately. Is it possible to calculate the standard deviation for the change in score?

Example data:

Number of participants = 29

Pre-treatment mean and SD = 68.07, 25.43

Post-treatment mean and SD = 58.31, 21.94

Mean change in score = 68.07 - 58.31 = 9.76

P value = 0.001

What is the standard deviation of this change?

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  • $\begingroup$ You have a pre- and post-measurement for each participant, right? You subtract pre- from post- so you have a single number for each participant that represents their change, then take the standard deviation of that. $\endgroup$
    – rishi-k
    Jul 11, 2021 at 18:22
  • $\begingroup$ Sorry about the confusion. I'm writing a review and that data is not available to me. $\endgroup$ Jul 11, 2021 at 18:28
  • $\begingroup$ Ah. That is going to be tough without actually looking at the data. Do you have access to the standard error of the change in mean? $\endgroup$
    – rishi-k
    Jul 11, 2021 at 18:33
  • $\begingroup$ Unfortunately no, but the p value is mentioned. $\endgroup$ Jul 11, 2021 at 18:37
  • $\begingroup$ Oh, you can use the p-value and the number of participants to figure out value of the t-statistic, which is the change in mean over the standard error of the change in mean. Since you know the value of the change in mean, you can get the standard error. $\endgroup$
    – rishi-k
    Jul 11, 2021 at 18:39

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It would be best if you had access to the raw data, but it is possible to back-calculate what you want given the mean difference, the number of observations, and the p-value. This experiment is a within-subjects design, so the p-value should have been calculated with a paired t-test.

First, we can use the p-value and the number of observations to determine the value of the t-statistic. Because this is a paired design, there are 28 degrees of freedom. I'm going to assume the p-value you have is two-tailed, so we can use a t-distribution calculator to figure out the t-value corresponding the the 99.95th percentile of the distribution, which is $ t = 3.674 $.

Remember that the t-statistic is just an expression of the mean difference divided by the standard error of the mean difference:

$$ t = \dfrac{\bar{X} - \bar{Y}}{S.E.(\bar{X} - \bar{Y})} $$

In the case of the paired t-test, this is really just:

$$ t = \dfrac{(\bar{X} - \bar{Y}) * \sqrt{n}}{S.D.(\bar{X} - \bar{Y})} $$

You know t, n, and the mean difference, so you can solve this equation to find that the standard deviation of the mean difference is:

$$ S.D.(\bar{X} - \bar{Y}) = 14.3 $$

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  • $\begingroup$ Thank you! I have just one more doubt. There are a few cases in which p value is only mentioned as non significant. Is there any way I can find the standard deviation of change in this case? $\endgroup$ Jul 11, 2021 at 19:18
  • $\begingroup$ No, you need the actual value. You can set a lower bound, but if you need these standard deviations for a meta-analysis, you need the actual numbers. $\endgroup$
    – rishi-k
    Jul 11, 2021 at 19:20
  • $\begingroup$ Well then, I'll exclude those values from the meta-analysis. Thanks again! $\endgroup$ Jul 11, 2021 at 19:22
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    $\begingroup$ Be careful with that - you might overestimate your treatment effect if non-significant studies get excluded because you can't back-calculate their standard error. $\endgroup$
    – rishi-k
    Jul 11, 2021 at 19:28
  • $\begingroup$ The studies that I'm including also have controls with significant p value which will be analysed. And I'm only excluding few results that have non significant p values for both treatment and control groups. So, I don't think it'll cause a significant change. $\endgroup$ Jul 11, 2021 at 19:38

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