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I am very confused as to what the correct way is to calculate the uncertainty of the average of values ($x_{avg}$) in a data set of measurements $(x_1 ... x_N)$. I have found at least four different ways of doing it around the internet, as follows:

  • Method 1: Uncertainty is the average of the deviations from the mean. That is, $$\Delta x_{avg} = \frac{(|x_{avg} - x_1| + ... + |x_{avg} - x_N|)}{N}$$ (as described in this Youtube video)

  • Method 2: $\Delta x_{avg} = \frac{R}{2}$, where $R$ is the range of the values (from this Youtube video)

  • Method 3: $\Delta x_{avg} = \frac{R}{2\sqrt{N}}$ from this document

  • Method 4: $\Delta x_{avg} = \frac{\sigma}{\sqrt{N}}$, $\sigma$ being the standard deviation of the data set (from here)

Which is the correct way?

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  • $\begingroup$ If the errors of individual measurements are normally distributed around zero, the most useful thing to know is the second parameter of this normal distribution which is what the fourth method gives. $\endgroup$ Commented Sep 4, 2021 at 15:41

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You evaluate the mean and express the uncertainty in the mean as the standard deviation of the mean.

Your estimate of mean of the sample with $n$ values is $m = {1 \over n} \sum_{j = 1}^{n} y_{j}$ where $y_{j}$ is the $j^{th}$ value in the sample. The standard deviation of the mean for the sample is $S = \sqrt{s^2 \over n}$ where $s = \sqrt{{\sum_{j= 1}^{n} (y_{j} - m)^2} \over {n - 1} }$ is the standard deviation for the sample. You report $m \pm S$.

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  • $\begingroup$ Yes, I corrected it. Thanks! $\endgroup$
    – John Darby
    Commented Mar 1, 2023 at 18:41
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First, you intuitively want a measure that will decrease as $N$ increases, because the more (independent) measurements you make, the smaller you expect your uncertainty to be. So Method 2 is clearly wrong since $\frac R 2$ will not decrease as $N$ increases. If anything, it will tend to increase as $N$ increases, because you are more likely to get outliers in a larger set of measurements.

Second, you need to distinguish between the expected uncertainty in a single new measurement $x_{N+1}$ and the expected uncertainty in the average of a whole new set of $N$ measurements. Method 1, the average deviation, is a measure of the uncertainty in a single new measurement. An alternative would be to use the standard deviation of the first $N$ measurements, $\sigma$. But you want to know the uncertainty in the average of $N$ measurements, $x_{avg}$. So Method 1 is incorrect for this.

This leaves Method 3 and Method 4. If you only know the range of measurements and the number of measurements then use Method 3. However, if you know all of the $N$ individual measurements then you can calculate their standard deviation, in which case you should use Method 4.

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  • $\begingroup$ Great explanation, thanks. $\endgroup$ Commented Sep 5, 2021 at 9:07
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None of the above if $n$ is small.

John Darby's answer is correct if $n$ is sufficiently large that the departure from a normal distribution caused by the fact that you are trying to estimate the mean and the standard deviation from the same set of data is small.

The (more) correct procedure is to use a Student's t-distribution with $n-1$ degrees of freedom to make a better, unbiased estimate of the population variance $s^2$ and then use $s/\sqrt{n}$ for your estimate of the standard error in the mean.

In practice the Bessell-corrected estimate of the population standard deviation given in John Darby's answer asymptotes to that obtained from the t-distribution with $n-1$ degrees of freedom when $n$ is large. It is probably sufficiently accurate for $n >5$ that there is unlikely to be any change in the most significant figure of the error estimate.

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Actually there is no best way. in you cited this document from pen they tell you the difference if you have many or just a few data. usually one assumes that the errors ar statistical and then the last formula is the correct one. $Δ𝑥𝑎𝑣𝑔=𝑅/2$ is for few data a good estimate. the first formula is for a quick estimate (mostly in school) but seldom used in scientific papers.

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