Getting an average figure with ±

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Homework Help Overview

The discussion revolves around calculating an average value from a series of measurements that include uncertainty, specifically in the context of a lab assignment involving slope figures from a v(t) graph.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore methods for averaging values with uncertainties, questioning the appropriateness of using a common mean versus a weighted mean. There is also discussion about the implications of measurement precision and the independence of measurements.

Discussion Status

The conversation is ongoing, with participants providing insights into the calculation methods and raising questions about the nature of the measurements. Some guidance has been offered regarding the use of weighted means, but confusion remains about the correct approach to take.

Contextual Notes

Participants note the ambiguity in the original measurements and the need for more details to provide effective assistance. There is also mention of the precision of individual measurements affecting the final average.

Monocerotis
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Homework Statement



Quick question, working on a lab

Just wondering how I could go about getting an average value for a series of figures with +- values.

For instance
1) 1.5 ± 0.02
2) 1.7 ± 0.84
3) 1.7 ± 0.08

The Attempt at a Solution



I tried enclosing the figures within parenthesis and then trying to find an average with wolfram alpha but no success.

I was just reading something on the net

would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?
 
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Monocerotis said:
would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?
Incorrect, and very much so. Assuming the measurements are statistically independent (are they? There's no telling from the original post) then

1. That 1.5 measurement is considerably more precise than either of the other two measurements. You should get something closer the 1.5 than 1.7 for your final estimate.

2. You combined error is larger than the largest error. It should be smaller than any of individual errors.Google "weighted mean".
 
k I've been reading and it seems as though what I'm dealing with is a common mean rather than a weighted mean

I'm dealing with slope figures of a v(t) graph and trying to find g avg +- delta g avg as to compare it with my theoretical results

the values I gave above were ambigous but the figures I'm actually dealing with from my perspective appear to have equal weight so

still confused as to what to do however because even the common mean will only give me a single value
 
What exactly are you measuring here? Without details it is a bit hard to help you.
 

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