I am calculating Standard error of mean(SEM), from Standard error of difference in mean(SED). Everytime it gives me almost 70% of SED why ?
Thats always true for iid random variables:
https://en.wikipedia.org/wiki/Variance#Propagation
Assuming independent random variables X and Y, i.e. Cov(X, Y)=0, the variance of X-Y is equal to Var(X-Y)=Var(X)+Var(Y). If X and Y have the same variance, this simplifies to Var(X-Y)=2 Var(X) and therefore SD(X-Y)=sqrt(2)*SD(X). Thus SD(X)=1/sqrt(2)*SD(X-Y) and 1/sqrt(2) is almost exactly 0.7
The standard error of the mean (SEM) is typically smaller than the standard error of the difference in means (SED) because they measure different quantities. SEM estimates how much a sample mean is likely to deviate from the actual population mean, while SED measures the variability between the means of two samples. The SEM is derived from the standard deviation (SD) of a single sample divided by the square root of the sample size, which tends to be a smaller value, especially as the sample size increases. On the other hand, SED takes into account the SD of two samples and the sample sizes of both groups. The fact that you're consistently finding SEM to be about 70% of SED could be due to the relative sizes of your samples or the inherent variability in the data. It's important to ensure that the calculations are done correctly and that the sample data accurately represents the population. If the SEM is consistently a specific proportion of the SED, it might be worth reviewing the formulas and the data to ensure that there are no systematic errors in your methodology.
Note: This answer refers in part to another answer that was subsequently deleted
~~that pretty much looks like a chatgpt answer, doesn't it?~~
Dividing SED by sqrt(2) is virtually the same as multiplying it by 0.7 (up to rounding of the 2nd significant digit), so chatgpt comes to the wrong conclusion that this is a mistake
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