- #1
Doctor_G
- 10
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This is an issue I've seen asked and answered before on this forum some years ago. However, the answer doesn't quite make sense to me, so I want to see if I can get either a more satisfactory answer or a better explanation of the original.
Suppose that I have some cells that produce a luminescent protein, and I have a drug that curtails this process. To quantitatively assay how much the drug impacts protein production as a function of dose, I take the cell culture supernatant from each condition and measure it under a luminometer. I make each measurement in triplicate, and I repeat the experiment three times. For simplicity's sake, we'll refer to each triplicate measurement in each experiment as a Rep, and each experiment as a Set. So, three Sets, each with three Reps.
Now, each Set is going to have an average value (Set AVG) with a degree of uncertainty about the mean (using SEM). I want to take the average of these three set averages. This Meta-Average will have its own standard deviation (SD). To propagate the measurement uncertainty of each set average to the SD of the meta-average, I use this formula:
STDEVmeta-average=SQRT[(STDEV(avg_set1, avg_set2, avg_set3))^2+(STDEV(sem_set1, sem_set2, sem_set3))^2]. This, as I understand it, is the method that was advocated by user viraltux in a thread back in 2012. You can find it here: https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/
I like this formula, as it always generates a variance that is always a little bit bigger than simply taking the SD of the three Set averages. This makes sense, as the calculated variability should always be a little bit compounded by the initial uncertainty in my measurements. My problem, though, is that when I subtract the SD from the values generated by this formula, they do not correspond linearly with the sum of the SEMs from each Set. This doesn't make sense to me. If the propagated error formula takes the SD and then adds, essentially, a function of the SEM to it, then once you subtract the SD, the difference should depend entirely on the initial variance in my measurement. Can anyone explain why this is not the case?
Thank you.
Suppose that I have some cells that produce a luminescent protein, and I have a drug that curtails this process. To quantitatively assay how much the drug impacts protein production as a function of dose, I take the cell culture supernatant from each condition and measure it under a luminometer. I make each measurement in triplicate, and I repeat the experiment three times. For simplicity's sake, we'll refer to each triplicate measurement in each experiment as a Rep, and each experiment as a Set. So, three Sets, each with three Reps.
Now, each Set is going to have an average value (Set AVG) with a degree of uncertainty about the mean (using SEM). I want to take the average of these three set averages. This Meta-Average will have its own standard deviation (SD). To propagate the measurement uncertainty of each set average to the SD of the meta-average, I use this formula:
STDEVmeta-average=SQRT[(STDEV(avg_set1, avg_set2, avg_set3))^2+(STDEV(sem_set1, sem_set2, sem_set3))^2]. This, as I understand it, is the method that was advocated by user viraltux in a thread back in 2012. You can find it here: https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/
I like this formula, as it always generates a variance that is always a little bit bigger than simply taking the SD of the three Set averages. This makes sense, as the calculated variability should always be a little bit compounded by the initial uncertainty in my measurements. My problem, though, is that when I subtract the SD from the values generated by this formula, they do not correspond linearly with the sum of the SEMs from each Set. This doesn't make sense to me. If the propagated error formula takes the SD and then adds, essentially, a function of the SEM to it, then once you subtract the SD, the difference should depend entirely on the initial variance in my measurement. Can anyone explain why this is not the case?
Thank you.