Anderson–Darling test

Remarks

With the order statistics $X_{(1)} \le \dots \le X_{(n)}$ and $U_i = F_0(X_{(i)})$, $$A^2 = -n - \frac{1}{n} \sum_{i=1}^{n} (2i - 1) \big( \ln U_i + \ln(1 - U_{n+1-i}) \big).$$ Under $H_0$ the law of $A^2$ does not depend on $F_0$. When parameters of $F_0$ are estimated from the data, for example in a test of normality, other tables of critical values are used.

Example: $n = 1$, $X_1 = 1/2$ and $F_0(x) = x$ on $[0, 1]$. The integral is $\int_0^{1/2} \frac{x}{1 - x} \, dx + \int_{1/2}^{1} \frac{1 - x}{x} \, dx = 2\big(\ln 2 - \tfrac12\big) = 2\ln 2 - 1$. The formula gives $-1 - 2\ln(1/2) = 2 \ln 2 - 1 \approx 0.386$, the same value.

Non-example: the Cramér–von Mises statistic $n \int (F_n - F_0)^2 \, dF_0$ lacks the weight $1/(F_0(1 - F_0))$, so it is a different statistic that gives less weight to the tails.