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Uncertainty of the Average

  • Thread starter LCHL
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Homework Statement


This isn't a specific problem. It's more of something that I have encountered before and I have been unable to find an answer to by looking online or through notes. Hopefully this thread might serve others with similar problems in the future.

Say for example, you want to measure the speed of sound experimentally, and you get four values from four attempts:

340 ± 1 m/s
345 ± 5 m/s
341 ± 2 m/s
335 ± 20 m/s

It seems sensible to average the data to get a value which will hopefully be a good estimate. That is not too difficult, but how would one deal with uncertainty in this case?

Homework Equations



I'm not really sure.

The Attempt at a Solution



The mean of these values gives a speed of sound of 340 m/s correct to one decimal place.

I have been told that the error of the average is not the average of the error, so that would eliminate using ± 7 m/s as the uncertainty. The standard deviation can be used to give an uncertainty of ± 4 m/s but that ignores the uncertainty in each of the measured quantities.

Another way of obtaining uncertainty is to subtract the smallest obtained value from the largest and divide it by the number of samples, giving (345-335)/4 = ± 2.5 m/s but this also ignores the measurement uncertainty.

What is the appropriate course of action in this case? Thanks in advance. :oldsmile:
 

Answers and Replies

  • #3
haruspex
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gneill
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That thread only deals with assessing the error range for the average (as far as I could see), but there is a question before that: how to find the average. It does not make sense to give equal weight to a measurement with greater uncertainty.
See for example http://labrad.fisica.edu.uy/docs/promedios_ponderados_taylor.pdf
I think @D H's post #7 in that thread covered the more precise weighted average adequately (the "One last item" towards the end).
 
  • #5
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Oh wow the formula in D H's post is really useful. Thank you so much for pointing that out. :biggrin: From inspection, that would imply that if you took a measurement a large number of times and measurement uncertainty is constant like in that previous example, the error would be quite small.

Taking the example in the opening post as an example, if the mean after 100 measurements was 3.31 and all uncertainties were ± 0.01, the uncertainty would be reduced to ± 0.001. Obviously that uncertainty is extremely low, but it makes some sense given that if you measure something properly a large number of times, it should converge on the correct answer.

I've taken a look at that very useful document and the method used is the same as the one mentioned above. They are just phrased differently.

Therefore, the answer to the question would be 340 ± 1 m/s rounded appropriately. Thanks all! :cool:
 
  • #6
haruspex
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I think @D H's post #7 in that thread covered the more precise weighted average adequately (the "One last item" towards the end).
Ah yes, I missed it, thanks.
 

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