Uncertainty of the Best Value from a Multiple-Trial Experiment

AI Thread Summary
The discussion centers on calculating the best value from multiple trial measurements with associated uncertainties. The initial method used an average of the best estimates and uncertainties, resulting in a value of 5.39 ± 0.03. However, it was pointed out that a more accurate approach involves weighting the individual measurements by their uncertainties, leading to a better estimate and a standard deviation calculation. The correct method emphasizes that trials with lower uncertainties should carry more weight in the final average. A recommended resource for further understanding is John R. Taylor's "An Introduction to Error Analysis," which covers the method of weighted averages.
WannaLearnPhysics
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Homework Statement
Uncertainty of the best value in a Multiple-Trial Experiment in which each trial has its own uncertainty.
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I took the average of the best estimates in each trial to get the best estimate of the best value. I also did that with the uncertainties, however, I'm wondering if this is wrong.
For Example:
Trial 1: 5.36 ± 0.03
Trial 2: 5.42 ± 0.04
Trial 3: 5.35 ± 0.01
Trial 4: 5.38 ± 0.03
Trial 5: 5.45 ± 0.02

What I did was take the average of the best estimates and the uncertainties.
Best Value 5.39 ± 0.03

(0.03+0.04+0.01+0.03+0.02)/5=0.026=0.03
 
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WannaLearnPhysics said:
I'm wondering if this is wrong.
Wrong is a strong term. But you must agree you don't do justice to your best measurement that has ##\pm\;##0.01.

If your uncertainties are reasonably well established (*), the proper way to do this is to weigh the individual measurements by weight ##w_i = 1/\sigma_i^2## and to calculate$$\bar x = {\sum w_i\,x_i\over \sum w_i}$$ as your best estimate. The estimate of the standard deviation ##\sigma _{\bar x}## follows from $$\sigma_{\bar x} = \sqrt{1\over {\sum w_i}}$$ (*) Since weights are ##1/\sigma_i^2##, your trial 3 gets 16 times the weight of trial 2 !
 
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Thanks for this! I really appreciate this. Oh yeah, it seems like wrong is kinda strong. I don't understand everything now but from what I understood, what I did seems fine if all of the uncertainties have the same value. I really appreciate your help! I've been reading my lab manual and books for hours but this wasn't mentioned.
 
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BvU said:
The estimate of the standard deviation ##\sigma _{\bar x}## follows from $$\sigma_{\bar x} = \sqrt{1\over {\sum w_i}}$$
That's the standard error of the mean, yes?
 
Right. I get 5.372 ##\pm## 0.008 as internal error
 
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WannaLearnPhysics said:
I've been reading my lab manual and books for hours but this wasn't mentioned.

An excellent resource is John R. Taylor's An Introduction to Error Analysis. I didn't find it until after I was done with college, but I like the balance he strikes between informality and rigor. There is a short chapter on the method of weighted averages described above.
 
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brainpushups said:
An excellent resource is John R. Taylor's An Introduction to Error Analysis. I didn't find it until after I was done with college, but I like the balance he strikes between informality and rigor. There is a short chapter on the method of weighted averages described above.
Thank you very much for this! : D
 
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