# Standard deviation and single errors of measures

During my courses of laboratory, we used as a rule of thumb the following statement: If you have less than five measurements (each of them with its error), the error of the mean is simply the squared sum of the errors divided by the square root of N. If you have more measurements, you instead calculate the average and the standard error of the set.

Let's say that I have 100 measurements $\{x_i , \delta x_i\}$ and I want to consider the average and the error.

I should compute the average $\mu = \sum \frac{x_i}{N}$ and the formula for the sample standard deviation $S=\sqrt{\sum(x_i - \mu)^2/N}$. I could also compute the sum of the squares of the errors $\sqrt{\sum \delta x_i^2/N}$.

Why should I use the first formula instead of the second one? Why if I have many measures I can "throw away" the errors of each measure?

## 1 Answer

Generally speaking, your $\delta x_i$ will be a manual estimate of the error in measurement $x_i$ - e.g. this could be the smallest grid size on your ruler. This is a rather rough estimate since it presumes very little about your measurement methodology and is reliant on the experimenter's best judgement. $S$ (usually we use the symbol $\sigma$ here) on the other hand is measuring how consistent your measurements of $x$ are. The presumption here is that the less consistently you can measure $x$, the larger $S$ is and the larger the error is.

Maybe a concrete example will help here. So let's say you are measuring the length of a table while standing on a ship that rocks back and forth randomly. You have a ruler that is accurate down to the millimeter. One way an experimenter might report $\delta x_i$ is then $\delta x_i=1\text{mm}~\forall x_i$. The rms of this would again be $\bar{\delta x}=1\text{mm}$ . However, this is probably an underestimate of the error. How else should the experimenter report his error though? Eyeballing how far the ruler moves around during his measurements? Such efforts are difficult. This is when we introduce $S$. We let the experimenter measure the length of the table 100 times, we take the average of those measurements as the length of the table and then we take the measure of how consistently that experimenter will get some value - this is exactly what $S$ measures - as the error in that measurement. Perhaps even though the ruler the experimenter has has guides at every $1\text{mm}$, he gets values that are spread out such that $S=10\text{mm}$. Hopefully it is clear now that this latter value is a better estimate of the error.

Now, why isn't $\delta x_i$ even considered when computing $S$? Well, presumably $S$ includes contributions that $\delta x_i$ would give. If you have a ruler that is only accurate to $1\text{mm}$, presumably over many measurements, your measurements will be spread out so that $S\geq 1\text{mm}$. Of course, one can conceive of instances where this might fail. It is of course the experimenter's perogative to report the best estimate of error he can. And if in some instances $\delta x_i$ is a better measure, then he should report those. But, generally speaking, $S$ is just a better measure overall.

• I was wondering if the question has to do with the central limit theory. If you have many variables you know that they have to be distributed (if the variances are small enough) as a Normal with a certain width, well estimated by S. On the contrary, when you have a small set of measures, the average is not normally distributed, so the best thing you can do is to use the formula for the standard deviation of the sum. In theory, when N is large enough, both estimates should converge. Do you think it's right? – Francesco Di Lauro Jun 19 '18 at 22:30
• What's your precise definition of $\delta x$? I took it to be the $\delta x$ I described in my answer, which is more of a basic intro to laboratory physics definition, but it feels like you have another definition in mind. – enumaris Jun 19 '18 at 22:33
• For me $\delta x$ is a number that comes along with each measure. It could be the precision of the instrument as well as any estimate of the error in measurement. I take it to be an estimate of the standard deviation of a measure. – Francesco Di Lauro Jun 19 '18 at 22:38
• So, the answer to the question "when N is large enough, both estimates should converge" will obviously depend on how exactly you define $\delta x$ and how you find the value of $\delta x$. Go back to my example of measuring the length of a table while you're on an unsteady boat. If you report $\delta x$ as instrument precision, then you will not have $S$ converging to $\delta x$ for any $N$ since you have not accounted for the rocking of the boat in your report of $\delta x$. – enumaris Jun 19 '18 at 22:43