Uncertainty For Multiple Measurements

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Homework Statement

We needed to measure the temperature of a bucket of water using a thermometer which can be read to 0.1°C. Each group would do a single measurement, the data from five groups (including our own) was to be averaged with the uncertainty stated.

The data collected was:
19.5
19.5
19.8
20.0
20.3
(all in °C)

I'm having difficulty with calculating the uncertainty.

Homework Equations

ΔZ = ΔX + ΔY

The Attempt at a Solution

[/B]
I have a feeling that I shouldn't be using this equation. Taking multiple measurements and averaging them should give better results, but... if I take more measurements then I have more uncertainties to add which will lead to a greater total error and an overall worse result.I end up with ±0.5°C which doesn't seem right.

 

[gneill]

Summing the uncertainties is at best an upper bound approximation on the uncertainty of the sum. A better estimation is given by summing them in quadrature (a fancy term for the square root of the sum of the squares -- think Pythagoras).

$ΔS = \sqrt{ΔX_1^2 + ΔX_2^2 + ΔX_3^2 +... + ΔX_n^2}$

Now, not only are you summing the individual temperature values but you're also dividing that sum by the number of terms in order to find the average, right? So what do you do with the uncertainty of the sum when you divide the sum by a constant value?

 

[Login]

I probably should have said that this is a section of a prac, for the entire prac we are only supposed to use; mean, standard deviation, error on a slope and the error combinations:
\Delta{Z}=\Delta{X}+\Delta{Y} For quantities which are added/subtracted
\frac{\Delta{Z}}{\left|Z\right|}=\frac{\Delta{X}}{\left|X\right|}+\frac{\Delta{Y}}{\left|Y\right|} For quantities which are multiplied/divided
\frac{\Delta{Z}}{\left|Z\right|}=\left|r\right|\frac{\Delta{X}}{\left|X\right|} For quantities which are raised to a power

I don't think that summing the quadrature is meant to be used for this prac

Now, not only are you summing the individual temperature values but you're also dividing that sum by the number of terms in order to find the average, right? So what do you do with the uncertainty of the sum when you divide the sum by a constant value?

ah, so I have to divide the uncertainty by the constant as well? ±0.1°C seems a lot more reasonable for the average.

Could I extend this method to measurements where the uncertainties differ? Or would that require a different process?
.

 

[gneill]

Could I extend this method to measurements where the uncertainties differ? Or would that require a different process?

Yes, it would apply also when the uncertainties on the individual quantities differ.

 

[BvU]

Dear Loq,

The averaging process doesn't improve the accuracy when adding the errors straight, but it doesn't make it worse either:
Adding up gives you +/- 0.5 degrees on approximately 100 degrees, so 0.5%
And per your multiplication/division equation (with zero uncertainty in the number 5 -- I hope ) the relative uncertainty in the average is also 0.5%.

But there's something else: if the data is in chronological order, you need to worry if the teams have measured the same thing! They haven't, if the water heats up in the mean time.
However, if the data has just been sorted in ascending order, I haven't said a thing. Except perhaps: why do that ?

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If uncertainties differ, you want to weigh the measurements. That is easy with quadratic addition of uncertainties: weight = 1/sigma². For straight addition you could use weight = 1/sigma.