I have a question regarding variance, paired testing and minimum detectable difference (MDD).
Paired samples: $$ MDD (δ) = \sqrt{ \frac{σ^2}{n} (t_{(α/2,n-1)} + t_{(1-β, n-1)})} $$ I have a set of baseline samples ($A$) sampled at time 0. Because acquiring those samples is destructive, I’m unable to re-sample the same sample twice (similar to for example a soil sample). Therefore I cannot assume that a future re-sampling will have the same variance as the baseline sampling.
To measure how big this error is I have taken a second set of samples ($B$) in close proximity to the first samples ($A$), which gives me a number ($n = 15$) of $A-B$ pairs. Sample $B$ is also taken at time 0. If I run a paired t-test on these two sample populations I fail to reject the null hypothesis ($H_0: A\neq B$).
I should probably mention that both sample population $A$, $B$ and the difference ($A-B$) are normally distributed.
Q1: If I want to estimate the MDD for the baseline mean, how do I integrate this sample error into the equation?
Q2: Can I simply take the variance from the baseline sample and add the variance of the differences ($A-B$)?
Q3: What if the samples don't reject the null hypothesis? Can I then assume that the samples are independent and use a pooled variance?
