Uncertainty in group of measurements, given single measurement uncertainty.

1. Jul 21, 2012

EricVT

Hello! I have a question regarding measurement uncertainties. This is not a homework problem.

Let's say that I want to measure some quantity, and I want to measure it multiple times using multiple identical but separate instruments. That is,

-First measurement taken using equipment 'A'
-Second measurement taken using equipment 'B'
-Third measurement taken using equipment 'C'

'A', 'B', 'C' are all the same type of equipment and somehow it is known that they each have the same 'X' % uncertainty in their individual measurements.

Once I have the set of measurements I can look at them individually with their 'X' % uncertainty, or I can look at them as a group. How do I propagate the 'X' % uncertainty in each individual measurement into a group uncertainty?

Of course you could take the results and find a mean value and associated standard deviation, but how does the 'X' % uncertainty come into the picture?

2. Jul 21, 2012

HallsofIvy

Staff Emeritus
After you have "measured it several times" what do you do with them? Average them? It is easily shown that if add a number of meaurements, you add their errors. If you then divide by the number of measurements, you divide the error also so that averaging the measurements your average the errors.

3. Jul 21, 2012

haruspex

It can help to think about what the X% uncertainty means physically. E.g. if it comes from the error in reading exact values off a scale, the uncertainty will be (pretty much) a uniform distribution across a known range (but not really a percentage in this case). In other situations, it may be someone else's (a manufacturer's?) representation of, say, a Gaussian error distribution; in this case, it might represent a certain number of standard deviations, but how many?
The ideal way to combine multiple such errors will depend on both the underlying distribution and on the use to which the result will be put. In an engineering context, you might care much more about hard limits on errors (max possible error) than about standard deviations.

4. Jul 21, 2012

chiro

Hey EricVT.

Typically what happens is that for a measurement, you assign a distribution for that particular realization of the measurement. You already have specified a variance and now you need to supply an assumed underlying model for the distribution. Many applications use a normal, but don't just default to this without thinking about the context of your data and experiment.

Then what you do after this is that you supply the transformation of your measurements in the way corresponding to how you calculate something. For example if you are averaging, then you sum all the measurement distributions and divide by the number of measurements.

In the averaging case, you can also find the variance of this sum by using the properties of the variance operator and see how the variance gets affected as you apply a function of your sample.

Then if you want to do actual statistics, you can derive an estimator for the parameters of your distribution (like the mean or variance) or some other property that might be distribution-independent (like say the median) and obtain a distribution that corresponds to an estimator for that particular property and possibly make an inference.

In this regard, it would help if you gave a mathematical expression with what you are doing with respect to your samples (i.e. describe the function of your sample and what calculation you are looking at: average? sample variance? other?)

5. Jul 22, 2012

Stephen Tashi

You don't have a specific question until you define what "uncertainty" means in your situation. Saying it is a "percent uncertainty" doesn't define it adequately.

One type of uncertainty is due to limited precision in the measurement read-out. For example a digital multi-meter may give you only 4 digits. Another type of uncertainty is due to "errors" in measurement and might be specified as a certain number of standard deviations of a Gaussian distribution. (How many standard deviations is a question that must be answered and perhaps only the manufacturer of the equipment can tell you.) Another type of uncertainty is due to variation in the thing measured. For example, you might be measuring many different Thompson seeldless grapes and trying to estimate the average weight per grape of the entire population of Thompson seedless grapes.

6. Jul 22, 2012

EricVT

Maybe if I rephrase this as a specific problem it will be clearer what I am hoping to learn.

A manufacturer sends me 5 radiation dosimeters. I use all 5 of them -- one at a time -- to measure the same unknown quantity of radiation dose. I send the devices back to the manufacturer and I am provided with 5 result reports.

Each report gives a quantity of dose measured and claims accuracy of the device for measuring the "true quantity of radiation dose" as within +/- 6%. Whether this is one sigma or two sigma or something else is not clear, but for the sake of moving forward let's say it is a two sigma uncertainty.

Assuming the distribution is Gaussian, if I average the 5 dose measurements and want an uncertainty for the average result, how do I combine the 6% uncertainties associated with the individual measurements?

Result 1: 200 +/- 12
Result 2: 198 +/- 11.9
Result 3: 201 +/- 12.1
Result 4: 204 +/- 12.2
Result 5: 193 +/- 11.6

Average Result : 199.2 +/- ?

Sorry if I am making this more complicated than it is. I know I could compute the standard deviation from the 5 results, but I don't understand how to incorporate the individual measurements uncertainties into an uncertainty in the mean result.

Thanks.

7. Jul 22, 2012

Parlyne

The simple answer is that you can't. To be able to combine errors, you need to know something about whether they are correlated. If the 6% error means that the design of the dosimeters is such that their results overall may be miscorrelated by as much as 6%, then the average would have a 6% miscorrelation as well. On top of that, there would also be a random error, which can be approximated using the standard deviation of the five measured values. On the other hand, if the 6% is meant to mean that the dosimeter readings will typically have fluctuations of around 6% away from the correct dose value at random, then the best combination is actually a weighted average; and, the uncertainty will be what you get from error propagation through the resulting formula.

Of course, it's also possible that the 6% is meant to include both random and systematic effects. If this is the case, you can't really do a valid combination without knowing how much of that 6% is random, and how much is systematic.

8. Jul 22, 2012

Stephen Tashi

Even if you assume away all Parlyne's objections by saying that the measurements are independent and unbiased, you still have the problem of saying exactly what you are trying to accomplish. (Applying mathematics to real world problems demands more information than most people care to contemplate.)

Taking the straightforward average of all the measurements gives them all equal "weight", but if some are more precise than others there is an intuitive appeal to giving more weight to the ones that are more precise. I think the exact mathematics of that can be worked out if you define the objective for your estimator. Is it to be the maximum likelihood estimator? The least squares estimator? An unbiased estimator? etc.

Last edited: Jul 22, 2012
9. Jul 23, 2012

Stephen Tashi

EricVT, this is not the question, as you have phrased it, but perhaps what you want to know is:

If $X_i, i = 1,2,..n$ are uncorrelated random variables with respective means $\mu_i$ and variances $\sigma^2_i$ then the random variable $X = \frac{\sum_{i=1}^n X_i}{n}$ has mean $\mu = \frac{ \sum_{i=1}^n \mu_i }{n}$ and variance $\sigma^2 = \frac{\sum_{i=1}^n \sigma^2_i}{n^2}$

However, this result applies to population parameters, not to one specific numerical value of a sample parameter like 199.2. It doesn't show that using the formula $\hat{\mu} = \frac{\sum_{i=1}^n \hat{\mu_i}}{n}$ (where the $\hat{\mu_i}$ are the sample means) is ithe "best" estimator for $\mu$.

10. Jul 24, 2012

EricVT

Thanks for the help, everyone.