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 (H0: A≠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?
