Error propagation in an average of two values

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To calculate the error for the average value V from two runs, one approach is to use the addition/subtraction propagation formula on the sum of the two values. This method allows for the determination of the fractional error in the average V. While using standard deviation is common for larger datasets, it may not be suitable for just two values. A weighted average, where weights are based on the inverse of the squared errors, can also be considered, especially if the errors differ significantly. Ultimately, given the similarity of the two values and their errors, the choice of method may not greatly impact the final result.
beth92
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I'm writing up an experiment I did for a lab course and I am calculating the error in quantity V. I have two runs and have ended up with a value of V for each one, as well as an error. Ie, I have

V = 0.1145±0.0136 for Run 1
V= 0.1146± 0.0134 for Run 2

I got my errors through some tedious propagation which I won't go into, but what I'm wondering is what's the best way to calculate the error for my final V? (which will be the average V for the two runs) I have looked around and can't seem to find anything which gives a straight answer.

Would it be ridiculous to use the fact that Vaverage=(V1+V2)/2 and then propagate the error in (V1+V2) using the addition/subtraction propagation formula, then equate this quantity's fractional error to the fractional error in Vaverage? This seems a little over complicated.
Normally I would take the error in an average using Standard Deviation but that doesn't seem appropriate for just two values.
 
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beth92 said:
Would it be ridiculous to use the fact that Vaverage=(V1+V2)/2 and then propagate the error in (V1+V2) using the addition/subtraction propagation formula, then equate this quantity's fractional error to the fractional error in Vaverage? This seems a little over complicated.
That is how I would do it. It is just one of the simplest examples of propagation of error.
 
I find these figures fascinating. Is it possible to explain what you measured and what instruments were used.
I would like to know about the tedious propagation you used to arrive at the errors.
The explanation may be there.
 
Been many years since I did much statistics but I think you normally use a weighted average where the weighting is 1/Δ2. That way the result is biased towards the value with the lowest error.

In this case the two values and their error are virtually the same so it won't make much difference.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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