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This question already has an answer here:

I've been studying experimental physics on the book "The art of experimental physics" and on the chapter about error analysis there's something that has been bothering me. The author says:

Now that we have determined the "best value" for the measurement, that is, $\bar{x}$, we need to estimate the uncertainty or error in this value. We start with defining one way in which the spread of data about the mean value can be characterized.

The standard deviation $s$ is defined as

$$s = \sqrt{\dfrac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2}$$

If the standard deviation is small then the spread in the measured values about the mean is small; hence, the precision in the measurement is high. Note that the standard deviation is always positive and that it has the same units as the measured values.

The error or uncertainty in the mean value, $\bar{x}$, is the standard deviation of the mean, $s_m$, which is defined to be

$$s_m = \dfrac{s}{n^{1/2}}$$

where $s$ is the standard deviation and $n$ is the total number of measurements.

The result to be reported is then

$$\bar{x}\pm s_m.$$

Now, why to get the error on the quantity measured we must divide the standard deviation by $\sqrt{n}$ instead of just using the standard deviation?

Why things are done in this way?

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marked as duplicate by user10851, Gert, user36790, John Rennie, ACuriousMind Aug 31 '16 at 23:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Would Cross Validated be a better home for this question? $\endgroup$ – Qmechanic Aug 30 '16 at 19:41
  • $\begingroup$ @Qmechanic, I thought on posting it on Physics because that definition was made on the context of experimental physics. So I wanted to know exactly why in experimental physics one needs to divide $s$ by $\sqrt{n}$. Is it off topic here? $\endgroup$ – user1620696 Aug 30 '16 at 19:53
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    $\begingroup$ Possible duplicate of Multiple measurements of the same quantity - combining uncertainties $\endgroup$ – user10851 Aug 30 '16 at 20:02
  • $\begingroup$ @Qmechanic In a normal world, some SE users could simply decide to offer the question to the CV, and they could decide if they accept it. Fairly and fast. Territorial conflicts were unheard in the whole process. Yes, I know, it is not your mistake that we are quite far from this ideality. $\endgroup$ – peterh Aug 30 '16 at 20:33
  • $\begingroup$ So it seems the expression becomes $\sqrt(\frac{n}{n-1}\sum[(\frac{x_i-\bar x_i}{n})^2])$ if I didn't make mistakes again. This looks a bit more approachable to me. Looks a bit like the L2 norm of the mean error. $\endgroup$ – Emil Aug 30 '16 at 21:23
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One concept essential to understand this process is that whatever we are measuring in the lab is basically a sample out of the large number of experiments needed to determine the actual mean $\mu$. So, the mean obtained from the experiment, i.e. the mean of a value measured $N$ times is basically a sample mean.

We can get the actual mean by repeating the measurement infinite number of times, which gives us a normal curve with mean $\mu$ and variance $\sigma^2$. Since this is not feasible in practice, we take a finite number of measurements $X_1, X_2, ... , X_N$ in the lab, and take the sample mean $\bar{X}$, defined as: \begin{equation} \bar{X}= \frac{X_1+X_2+...+X_N}{N} \end{equation}

This is stated as the 'mean obtained in the experiment'. It is worth-noting that $\bar{X}$ is not a certain value but a random variable, and has a standard deviation ($\sigma_\bar{X}$) associated with it, which is of our interest.

Each of the individual measurements $X_1, X_2, ... , X_N$ are also normal random variables, independent and identically distributed with mean $\mu$ and variance $\sigma^2$. It should be made clear that this standard deviation ($\sigma$) is different from standard deviation of the (sample) mean ($\sigma_\bar{X}$).

Since all $X_i$'s are identical and independent, variance of $N\bar{X}\left(=X_1+X_2+...+X_N\right)$ is \begin{equation} \sigma^2_{N\bar{X}}=N\sigma^2 \implies N^2\sigma^2_{\bar{X}}=N\sigma^2 \end{equation}

\begin{equation} \therefore \sigma^2_{\bar{X}}=\frac{\sigma^2}{N} \text{, or } \sigma_{\bar{X}}=\frac{\sigma}{\sqrt{N}} \end{equation}

The key motivation behind this error-analysis is considering the whole set of measurements (needed for actual mean) to be the population, and the set of finite recorded observation (taken in the lab) as a sample. Then, this is nothing but finding the standard error in the sample mean.

One of the implication of this formula is that: "To decrease the error in the mean by $k$, the number of observation taken should be increased to $k^2$."

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