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I am new to this so I don't have enough reputation to comment on the original post here. I have a similar problem with equal weighting, and have n = m = 3.

I believe this means I need to use sample standard deviation, which with equal weighting I believe that gives a similar equation, where I ended up multiplying each standard deviation by n-1 and each mean by n (for their respective sets) and then dividing by the total n-1. Is this the proper way to go about this problem?

My exact problem is I have three sets of data (so I have a third set of terms as well), each with their own noise associated with each value (also based on three subsets of data, so 9 total). Their mean values are slightly distinct, and the noise of the subsets is quite small. So it looks something like: 1, 1.01, 1.005, 1.51, 1.50, 1.505, 1.20, 1.21, 1.19. | first set | second set | third set | | --------- | ---------- | --------- | | 1.00 | 1.51 | 1.20| | 1.01 | 1.5 | 1.19| | 0.99 | 1.52| 1.18|

The data is measured in groups of three (the noise) so I believe I should pool these first, so something like 1.00 ± SD1, 1.51 ± SD2, 1.19 ± SD3. I then think I should pool those together, accounting for the variance between the means and the variance of the means.

cmay
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  • The thread you reference fully answers your questions. But remember that the SD must first be *squared* to obtain the variance. Whether you multiply by $n-1$ or $n$ depends on how you computed the SD in the first place. – whuber Jul 29 '21 at 16:43
  • See [Bessel's correction](https://en.wikipedia.org/wiki/Bessel%27s_correction) for why $n-1$ is used for sample variance and standard deviation. – DifferentialPleiometry Jul 29 '21 at 17:18

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