I have $N$ rulers that each have a different but known level of accuracy, e.g. one's a meterstick, one's a yard stick, etc.

I measure the length of my table using each ruler.

How do I combine those $N$ measurements? If I simply take the mean and standard deviation of all the measurements, does that neglect the (known) systematic uncertainty of each measurement?

I would assume it does, so I would assume I would want to take a weighted mean and a weighted standard deviation with weights 1/$\sigma$, where $\sigma$ is the uncertainty from each ruler. Is that right?

If each ruler has a different uncertainty, is it possible to create a histogram or density plot of the results? How would I obtain, say, the 86% percentile?

  • 2
    $\begingroup$ Under the assumption of Normal errors the Maximum Likelihood estimator of the true mean is a weighted sum of measurements (with inverse variance weights, i.e. $1/\sigma_i^2$). The variance of this estimator is another one. See en.wikipedia.org/wiki/… for the details. $\endgroup$ – conjugateprior Mar 22 '16 at 4:18

Suppose you have three rulers which have accuracies with variances $\sigma_1^2, \sigma_2^2, \sigma_3^2$.

We can assume that the error has a normal distribution with $0$ mean.

Let the measurements iven by each ruler be $X_1, X_2, X_3$

If we take the sample mean $\frac{1}{3}(X_1+X_2+X_3)$ that is the sum of three normal distributions.

The sum of many normally distributed variables has a normal distribution with a variance equal to the sum of the individual variances.

Therefore the distribution of $X_1+X_2+X_3$ has a mean of $0$ and a variance of $\sigma_1^2+ \sigma_2^2+ \sigma_3^2$

The variance of the sample mean $\frac{1}{3}(X_1+X_2+X_3)$ is $(\frac{1}{3})^2(\sigma_1^2+ \sigma_2^2+ \sigma_3^2)$

You can use this variance to construct a confidence interval like you normally would for the sample mean.

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  • 3
    $\begingroup$ You can, although an inverse variance weighted estimator will be more efficient. $\endgroup$ – conjugateprior Mar 22 '16 at 4:20
  • $\begingroup$ I you have a very accurate and a very inaccurate instrument, this method would make the combined variance worse than what you'd get by using just the single more accurate measurement. $\endgroup$ – relatively_random Jan 16 at 7:25

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