Very frequently in papers, devoted to parallel clinical trials, I face a situation, when a calculated SD for an effect in each group is approximately equal to an SD of effect difference between the two groups.
An example may be found, e.g. in the following paper (Table 2): Article Comparison of the efficacy and safety profiles of a pelubipr...
Pelubiprofen achieved an efficacy in VAS scale of 26.2 with SD = 19.5. Celecoxib achieved efficacy of 21.2 with SD = 20.8. However, a difference is 5.0 with SD = 20.1! I was expecting SD ~sqrt(2) more, since the samples are independent and has approximately equal size.
Probably, it is related to the way how t-test are calculated. t-value is usually calculated through dividing the variance between groups (which is the means difference in your cases) by the overall variance within the groups, (rather than the arithmetic in variance between two groups). The later is generated from standard error of the means (this is calculated from S.D and sample size).
So practically, the more clear separation between samples in outcome, you will see larger means differences, which is of course relevant when it come to clinical interpretation e.g. the drug X increases patient satisfaction by a mean of 50 points (mean difference).
One another hand, the better the study, you should never detect any difference in variance or S.D within groups! What is actually important when it come to S.D and variance, is equality of variance between between each group and in both groups combined. So, in some how, if we manged to calculate the S.D difference, it should be equal or near Zero.
This is necessary important because the t-test and ANOVA (the bigger brother) depend on the assumption of equal variance between groups. if the groups don't have equal variance (S.D), then these tests couldn't be used at all; you should use non-parametric approaches.
Also, the variance and S.D would gives you an idea about the sampling methods e.g. if there is unequal variance, i would assume the presence of selection bias and bias with randomization process; the individuals characteristics are not matched between two groups. Moreover, the smaller the S.D, the smaller the Standard error of the mean (SEM) and the high the power of the test.
I think you should review the statistics behind each test to interpret how to deal of test results. One of the good sources to use is statistics at glance.
Amr Muhammed , thank you for your reply! Seems, that I understand a meaning of S.D. and SEM, but I don't understand, why the variance of difference between two independent random numbers (effects in the 1st and the 2nd groups) is not equal to a sum of variances. One should expect to see something like SD = sqrt(19.5^2 + 20.8^2) = 28.5.
Variance is calculated depending on sum of( Variables - Mean )/N-1 equation.
if you collected the two groups into one group and do the calculation, the mean and S.D will differ. Probably, if need to figure out the S.D for both groups, you should sum up the S.D for group one and two, then divide by two to get the average S.D for both group. (19.5 + 20.8)/2 = 20.15 ~ quite equal to output 20.1
I think this would be the explanation.
Amr Muhammed , yes, to be more precise the weights will be not exactly 1/2 but will depend on the groups sizes. But in the mentioned table (and many other similar papers) this SD is referred to a difference: difference mean = 5.0, difference SD = 20.1.... Probably it is some kind of conventional notation for a united sample SD but it seems to be too confusing.
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