Uncertainty of the Best Value from a Multiple-Trial Experiment

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SUMMARY

The discussion centers on calculating the best value from multiple trial experiments, specifically using the method of weighted averages. The trials presented yielded values with associated uncertainties, leading to a calculated best value of 5.39 ± 0.03. Participants emphasized the importance of weighing individual measurements by their uncertainties, using the formula $$\bar x = {\sum w_i\,x_i\over \sum w_i}$$ for the best estimate and $$\sigma_{\bar x} = \sqrt{1\over {\sum w_i}}$$ for standard deviation. A recommended resource for further understanding is John R. Taylor's "An Introduction to Error Analysis."

PREREQUISITES
  • Understanding of statistical concepts such as mean and standard deviation
  • Familiarity with the method of weighted averages
  • Basic knowledge of experimental uncertainty and error analysis
  • Ability to interpret and manipulate mathematical formulas
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  • Study the method of weighted averages in detail
  • Learn about calculating uncertainties in experimental measurements
  • Explore statistical analysis tools like R or Python for data analysis
  • Read John R. Taylor's "An Introduction to Error Analysis" for comprehensive insights
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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.
Relevant Equations
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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