Error on the Mean: How to Compute w/ Individual Error

[DEvens]

The buzz phrase is "least squares fit with error bars." Here are two examples of working it out. If you don't like these, please Google some more.

https://young.physics.ucsc.edu/242/lsfit.pdfhttps://www.phas.ubc.ca/~oser/p509/Lec_09.pdf
Basically, you have $\{\frac{x_i - \bar{x}}{\sigma_i}\}^2$ in the least-squares instead of the usual thing. Then there's a formula to estimate the net error in the slope and intercept you get.

 

[mfb]

Figuring out what the question is is the most important part.

It is completely meaningless to say that they come from the same distribution and then give different variances for the two measurements. If they come from the same distribution then they must have the same variance.

That is not true. The uncertainty doesn't have to be the variance of the underlying distribution of the numbers. Toy example: You measure radioactive decays in 1 minute and estimate the true decay rate based on that. A measurement that has 5 decays will come with a different uncertainty than a measurement with 10 decays, even though they are identical repetitions of the experiment.

Thank you for this! I actually found this: https://ned.ipac.caltech.edu/level5/Leo/Stats4_5.html I think this is what I was looking for.

That is the approach to get a best estimate for the "true" parameter and the uncertainty for your estimate.

However if I am trying to approximate the real distribution with my measurements (which I assume it is why experimentalists are trying to do in general), I need to add on top of the error the standard deviation of the samples themselves?

Your measured variance will be the sum of the variance from the underlying distribution itself and your measurements. The above formula will give you the variance coming from your measurement uncertainties. Subtract that from the sample variance to estimate the variance of the underlying distribution.

Note: That will only give a best estimate, which even might be negative (e.g. you measure 1000+-100 and 1030+-60, the spread is smaller than you expect from the measurement uncertainties). If your measurement uncertainties are large compared to the width of the underlying distribution you will need many measurements to make this approach viable.
If you need a confidence interval for your estimate of the variance of the distribution run toy MC, doing that analytic won't work well.

 

[Dale]

That is not true. The uncertainty doesn't have to be the variance of the underlying distribution of the numbers.

Oh, you are right. Good point.

If you need a confidence interval for your estimate of the variance of the distribution run toy MC, doing that analytic won't work well.

I have also suggested this a couple of times here. Monte Carlo is so flexible and useful that it should be a standard tool anyone doing statistics uses.