# If I group my data the variance changes, what does this tell me?

Imagine that I measure 9 times a certain physical properties (which I know its exact value is 11) and the outcomes are:

$$X_i = 10, 11, 11, 10, 11, 12, 10, 11, 12$$

If I analyse $X_i$ as a random variable I can compute it's expected value and variance as it follows:

$$\mathbb{E}(X_i) = \frac{1}{9} \sum X_i = 11$$ $$Var(X_i) = \frac{1}{9} \sum (X_i-11)^2 =0.5$$

Hence the measurement can be written as $X = 10.9 \pm \sqrt{0.5}$

If I define a second random variable:

$$\bar X_j =\frac{1}{3} \sum_{\textrm{group of 3 X_i}} X_i$$

This means that:

$$X_i = 10.7, 11, 11$$ $$\mathbb{E}(\bar X_i) = \frac{1}{3} \sum \bar X_i = 10.9$$ $$Var(\bar X_i) = \frac{1}{3} \sum (\bar X_i-11)^2 = 0.05$$ Hence the measurement can be written as $X = 10.9 \pm \sqrt{0.05}$

The fact that the variance is smaller in the second case, follows directly by the sum of iid and goes like $1/\sqrt{N}$ where $N$ is the number of averaged value to obtain $\bar X_i$.

From a mathematical point of view it seems all clear but from a physical point of view what does it mean? If I consider averaged values the error I make on my measurement is smaller? The uncertainty of my measurement depends on how I treat my data?

• This analysis is a particularly clever way to motivate the concept of standard error, q.v.. – whuber Oct 3 '16 at 14:49
• Note that what you are computing are sample statistics rather than random variable mean and variance. The former depend on the realized values of the random variables (the 9 numbers you use) while the latter are properties of the random variable itself. Therefore, you should divide by $n-1$ when computing variances (i.e., by 8 and by 2 respectively). – A.G. Oct 3 '16 at 15:48

• Thank you for your reply, but what I can't see is how can the standard deviation account for statistical errors (which I assume have some physical origin), if they depend on how I play with my data. Imagine that I want to present an experiment showing the measurement above. What standard deviation should I use? I guess it doesn't matter as soon as everybody knows which one I am using right? And in any case a guy which makes the measurement of X will observe a spread $\sqrt{0.5}$ on its measurement. Right? – Worldsheep Oct 3 '16 at 15:13
• @Worldsheep. If you look at your data you will see that the spread of measurement error is from $-1$ ($X_1=10$) to $+1$ ($X_9=12$). – A.G. Oct 3 '16 at 15:54