Reason for taking the mean of experimental observations

Whenever we do some experiment in Physics, e.g. measure Planck's constant, we take the mean of several experimental observations. This seems to be the most obvious way to report a result.

Is there a reasoning based on MLE or other statistics concepts that can be used to justify that the mean is the best way to measure the value over several observations?

• If you know the distribution of experimental observations then give a valid reason by MLE estimate of interested parameter. – love-stats Oct 8 '11 at 14:08
• Example: For Physics experiments done in a lab the different observations would vary due to lots of environment factors coming into play. Thus I think the disturbance in the actual value would be normal (lots of random variables influencing the actual value). Can we build upon this assumption? – Rohit Banga Oct 8 '11 at 14:54

If $n$ (independent) measurements are made of some quantity $\mu$, then, assuming that systematic errors (e.g. bias in the measuring instrument such that the measurement is always $0.3$ larger than the actual value) have been eliminated, a suitable model for the measurements is $n$ independent identically distributed random variables $X_1$, $X_2, \ldots, X_n$, with mean $\mu$ and variance $\sigma^2$.
If we ignore $X_2, \ldots, X_n$ and report just the observed value of $X_1$ as the value of $\mu$, the variance of the measurement is $\sigma^2$. If we report the sample mean $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ as the value of $\mu$, then the variance is $\sigma^2/n$. Why is the reduction in variance important? Well, the Chebyshev inequality says that for a random variable $X$ with mean $\mu$ and variance $\sigma^2$, $$P\{\vert X - \mu \vert \geq \alpha \sigma \} \leq \frac{1}{\alpha^2}.$$ Choosing $\alpha = 5$, we have that if $x_1$ is the observed value of $X_1$, the $96\%$ confidence interval for $\mu$ is of length $10\sigma$, that is, $(x_1 - 5\sigma, x_1 + 5\sigma)$ is the $96\%$ confidence interval. On the other hand, the much shorter interval $(\bar{x} - 5\sigma/\sqrt{n}, \bar{x} + 5\sigma/\sqrt{n})$ can be used if we report $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$ as the value of $\mu$. Note the reduction in length by a factor of $\sqrt{n}$.
There are theories going back to Gauss and perhaps even earlier that after things like systematic errors (e.g. bias in instruments that always give higher readings than the actual value) have been eliminated, the residual error in a measurement can be modeled as a zero-mean normal random variable. In this case, we get $(x_1 - 1.97\sigma, x_1 + 1.97\sigma)$ as the $95\%$ confidence interval for one measurement, and $(\bar{x} - 1.97\sigma/\sqrt{n}, \bar{x} + 1.97\sigma/\sqrt{n})$ as the $95\%$ confidence interval. Once again, we have the same reduction in the length of the confidence interval by a factor of $\sqrt{n}$ when the average of $n$ measurements is used instead of just one measurement.