Relation between uncertainty in measurement of mean and standard error on the mean I have a series of individual (time) measurements with a certain uncertainty each per measurement, which is the same for all the measurements (±one frame). I have understood that the uncertainty on the arithmetic mean of these measurements will be equal to that same uncertainty of the measurements (±one frame in this case). I am completely lost on how this relates to the standard error from the mean (I suspect the standard error is an additional error on top on the uncertainty that comes from the underlying parent distribution and true value). How would one combine the uncertainty in the mean from measurement with the standard error, to give a "final" error (±) for the mean?
 A: In order to understand what is happening it is helpful to know the true distribution. Thus, I simulated $N=100$ data points from a normal distribution with mean value $\mu=21$ and standard deviation $\sigma=3$. The data looks like this

The red line is the average value, and the blue line is a Gaussian (least square) fit to the data.
Formula: y ~ k * exp(-1/2 * (x - mu)^2/sigma^2)

Parameters:
      Estimate Std. Error t value Pr(>|t|)    
mu    20.93766    0.35034  59.764 3.18e-16 ***
sigma  3.01581    0.35599   8.472 2.08e-06 ***
k      0.12922    0.01302   9.922 3.90e-07 ***

Note that the center of the Gaussian fit is not equal to the average value $\bar x$, because the fit does not weight the data points linearly.
Next, I repeat the calculation $nLoop = 1000$ times. In each iteration I draw $N=100$ samples from a normal distribution and calculate the average value. Hence, after the loop I obtained $\{\bar x_1, \bar x_2, \ldots, \bar x_{1000}\}$.
Now the key point for your question: The central limit theorem tells us that the average value $\bar{\bar x} = \frac{1}{1000} \sum_{i=1}^{1000} \bar x_i$ is  expected to be Gaussian, $N(\mu, \sigma/10)$, which is the red curve

Note that I use the center of the fit as estimator for the mean value and not the average value. Usually, the fit provides a more accurate estimate -- although this it is not true for this particular sample (my first $N=100$ data points).  If I include the 1000 sample averages in the plot I get

It is obvious that the standard deviation of $\bar x$ (red) is much smaller than the standard deviation of the original data (blue).
