Involving validation/mathematics

Basic questions from students; resources for projects and reports.

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The goal of my internship was to perform a validation on a GCMS method. One of the performance characteristics i had to determine was the intralaboratory-repeatability (Srw and VCrw).

According to the dutch NEN 7777 norm, VCrw can be determined by extracting and analyzing n samples in duplicate on n different days. VCrw is then calculated with the 2nd formula:


Srw: reproducibility standard deviation
VCrw = Reproducibility coefficient of variation
xi1 = measurement result of the first duplicate sample for measurement i.
xi2 = measurement result of the second duplicate sample for measurement i.
n = the number of samples pairs measured in duplicate to determine the VCrw

Because of an error in the research design, the samples were not processed in duplicate. In which case Srw is calculated as follows:


xi2 is replaced with the mean result of n measurements

Now to conclude: a formula to calculate the VCrw based on the 3rd formula was not given. Would it be correct to also substitute xi2 with the sample mean in the 2nd formula?
I'm really sorry, but I'm not sure you can do what you've done already.
The original method quantifies the differences between duplicate runs, so it's looking at the variation between two measurements of the same sample on two different days. The number should be the same, but it's different because the method isn't behaving the same on the two days.

If you subtract the mean, you're quantifying something different. It's still related to the repeatability of the measurements, but it's not the same as doing paired measurements. What it does depends on what your samples were - the nature of the samples and when they were run.

The original method will work even if all the samples are completely different; the data might be 2.1, 2.0; 45.2, 45.3; 10.5, 10.3; 13.2, 13.2
The method of subtracting the mean will only work if you have used only samples that are precisely the same (or injected the same sample n times), so your data are more of the form 10.1, 10.0, 10.1, 10.2, 10.1

Since the Dutch norm could have specified the method you've used, but chose not to, you have to think about why the authors of the norm chose not to. If they had specific statistical reasons for wanting to work with paired data, then you're not going to satisfy the Dutch norm. If they were merely trying to come up with a way to make the measurement when faced with n uniquely different samples (i.e. business as usual for the analytical lab) then your measure may be sufficient.

But there's another caveat. The variability depends on what you did. If a single analyst ran all the samples on the same day, the results will be less variable than if two different analysts ran the samples on successive days. This will again be less variable than if two analysts ran the samples, one in Jan 2018, the other in July 2019.

Intra-laboratory repeatability to me implies the reliability with which your lab can produce the same results, given the same samples, within the reasonable time-course of the work I might submit to you. So if I expect to submit samples over the course of years, and your lab employs 6 analysts, I would expect you to measure the repeatability using different analysts on different days. If you use two nominally identical instruments, it'd be quite nice to check they both give the same results too!

Sorry to be awkward!

(but to answer your actual question, no, it wouldn't be appropriate to use the mean in place of Si2 in the second equation, because 0.5(Si1+Si2) is actually the mean for that sample. Since your method actually only works if all samples are the same anyway, you should be dividing all your differences by the mean (i.e use the mean in place of 0.5(Si1+Si2)), which also means you could simplify the whole thing.)
(and note that your formula is arithmetically different to the Dutch formula too. In its simplest case, if you ran two samples (that must be identical or it's meaningless), and the Dutch method was run with a single pair of samples, you'd be doing the same thing. But yours subtracts the mean, while theirs looks at the whole difference, so theirs gives twice the value that yours would).
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