Follow-up Comment #3, task #7180 (project relax):
> C:
> Your variance averaging would work as follows:
> loop over the first 3 replicates, calculate variance for each peak
> loop over the 2 replicates, calculate variance for each peak
> finally take the average over 6 variances and assign this as a systematic
error to all peaks at all mixing times
> Is this correct ?
This is correct. Using 100 peaks, duplicates for 1 time point (ignoring the
multiple relaxation periods for now), and M = 1e6 for some stats on the error
estimate, the results are:
ave(error) = 0.998
sd(error) = 0.071
This is for Gaussians of mu = 20 and sigma = 1. As you can see, the standard
deviation of 1 is nicely estimated (plus and minus 0.071 for different
attempts). This is the best error estimate of them all.
(file #11440)
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