I have some measurements of the same quantity, using instruments of different resolutions. So I have some data of the form $x_i \pm \delta x_i$. I want to report the value obtained from these measurements as the mean of them, but I am not sure what to use for the error. I guess I should include both the error on each measurement, but also the spread (i.e. standard deviation) of the measurements and I am not sure how to do it. Just to give a concrete example to avoid confusion: Let's say I have a machine which produces cubes of size coming from a normal distribution of mean 1000 and standard deviation 100 (this is the true distribution of the length of the cubes). I pick one cube from this distribution and measure it with a ruler of resolution 10 and I get a value of $900 \pm 10$. Then I pick a second one and measure it with a ruler of resolution 20 and I get a value of $1100 \pm 20$. The mean is just $\frac{1100+900}{2}=1000$. What is the error associated to this value?


1 Answer 1


What you're looking for is a mechanism for computing how errors propagate. So, you have a bunch of data, and each individual measurement has its own error.

Then the question is: What's the error on the reported average? That depends a bit on how you compute the average. From your description, it sounds like you use the simple arithmetic mean, i.e. you sum up all your measurements, then divide by the total number.

In this case, we start with the error for adding numbers, which is the square-root of the sum of the squares of the error. (The rule is that when adding or subtracting number, you have to add the squares of the absolute errors). Then you divide by a constant (the total number of measurements) and that just goes directly into the error.

So in your example, the error would be $\frac{\sqrt{10^2 + 20^2}}{2} \approx 11.2$.

Well, that's no good because that makes your error bigger than if you had just used the first ruler alone. Often when you deal with measurements with different errors, it makes sense to compute a weighted mean, where you give a measurement more weight if it has lower error and less weight if it has bigger error.

See https://ned.ipac.caltech.edu/level5/Leo/Stats4_5.html for a discussion. In this case, we'd use 1/100 and 1/400 for the weights, respectively:

$\frac{900 * 1/100 + 1100 * 1/400}{1/100 + 1/400} = 940$.

So basically the ruler that's more accurate gets more weight. In this case, the error will be $\frac{1}{\sqrt{1/100 + 1/400}} \approx 8.9$. So in this case, your error is smaller than that of just one single ruler.

  • $\begingroup$ Thank you so much! This is what I needed. Just an aside note, I am a bit confused about the fact that the actual values of the measurements (i.e. without errors) don't appear at all in the error formula. If I had the same errors, but 90 and 110 instead, this formula would give the same error, but intuitively I feel like in this case the error should be smaller, as the range of my measurement is smaller, so the result is "constraint" to a smaller range than before. $\endgroup$ Commented Sep 10, 2019 at 0:29
  • $\begingroup$ What I am trying to say is, shouldn't I add on top of the error on the mean, the actual standard deviation of the sample? I.e. if I try to approximate the real distribution using my measurements, the error on the mean seems to be to small (i.e. the sample I have would be very unlikely). $\endgroup$ Commented Sep 10, 2019 at 2:01
  • $\begingroup$ The actual values of the measurements do not appear in the calculation of the errors because the absolute errors do not care about the values to which they are attached. A three foot snake is still three feet long regardless that you measure it with a millimeter ruler or a football field-marked ruler. The former ruler has a smaller absolute error (a higher measurement precision). What you are asking next is to appreciate the magnitude of the RELATIVE uncertainty. Use those same two rulers to measure an inch worm and you may understand absolute versus relative errors. $\endgroup$ Commented Sep 10, 2019 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.