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I have two related questions:

Assuming (as the extreme case) I measured a quantity (A) just twice, but I know that the values would "become" normally distributed if I measured the quantity repeatedly. I can calculate the CI for the mean by multiplying the standard deviation of the mean by 12.71 (see here).

I also measured another quantity (B), this time with a higher number of repeats (e.g. 1000x), so that I don't have to correct for bias and can use the usual factor of 1.96 to get the CI.

What about the CI of a derived quantity A+B (or in general f(A,B))? I know how to propagate the standard deviations, but when I calculate the CI from this propagated standard deviation (let's assume the derived quantity is also normally distributed), what's the sample number I have to use for the Student t to calculate the CI? 2? 1000? 10002? Something else? Do I have to propagate the CIs in another way in such a case (how?)? Is the Student's t method to get the CIs not appropriate in this case?

The second question: If I measured two quantities 1000x, and both are sampled from normal distributions with the same mean but different standard deviations. It's easy enough to calculate the weighted mean & corresponding variance/standard deviation as it is done in fixed effects meta analysis (see here). Again the same question - what sample number do I have to use to calculate the CI from the weighted mean standard deviation? 2? 1000? 2000?

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