chi squared distribution

Let \(Z_i\) be normal random variables with \(\mu = 0\) and \(\sigma^2 = 1\)

Then, the sum \(\sum_{i=1}^{k} Z_i^2\) is distributed according to a \(\chi^2\) distribution.

1. Notes

It turns out that the distribution of the sum of residuals (the squared error from the sample mean) follows a \(\chi^2\) distribution with \((n-1)\) degrees of freedom (see degrees of freedom).
On a very related note, think about the sample variance. Recall that the sample variance is given by \(S_n = \frac{1}{n}\sum_{i=1}^{n}(X_i - \bar{X}_n)^2\)
For \(X_1, ..., X_n \sim N(\mu, \sigma^2)\), if \(S_n\) is the sample variance, then \(nS_n/\sigma^2 \sim \chi^2_{n-1}\). Note that Cochran's theorem says that the sample mean and sample variance are independent.
see also student's t test