Uncertainty of the Best Value from a Multiple-Trial Experiment

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 2K views
WannaLearnPhysics
Messages
4
Reaction score
2
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
 
Physics news on Phys.org
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 !
 
  • Like
  • Love
Likes   Reactions: brainpushups and WannaLearnPhysics
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.
 
  • Like
Likes   Reactions: BvU
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
 
  • Like
Likes   Reactions: WannaLearnPhysics
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.
 
  • Like
Likes   Reactions: WannaLearnPhysics
Tha
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
 
Last edited: