p = anderson_darling_cdf(A,n)

Return the CDF for the given Anderson-Darling coefficient A
computed from n values sampled from a distribution. For a
vector of random variables x of length n, compute the CDF
of the values from the distribution from which they are drawn.
You can uses these values to compute A as follows:

A = -n - sum( (2*i-1) .* (log(x) + log(1-x(n:-1:1,:))) )/n;

From the value A, anderson_darling_cdf returns the probability
that A could be returned from a set of samples.

The algorithm given in [1] claims to be an approximation for the
Anderson-Darling CDF accurate to 6 decimal points.

Demonstrate using:
n=300; reps=10000;
z=randn(n,reps);
x=sort((1+erf(z/sqrt(2)))/2);
i = [1:n]'*ones(1,size(x,2));
A = -n - sum( (2*i-1) .* (log(x) + log(1-x(n:-1:1,:))) )/n;
p=anderson_darling_cdf(A,n);
hist(100*p,[1:100]-0.5);
You will see that the histogram is basically flat, which is to
say that the probabilities returned by the Anderson-Darling CDF
are distributed uniformly.

You can easily determine the extreme values of p:
[junk,idx]=sort(p);
The histograms of various p aren't  very informative:
histfit(z(:,idx(1)),linspace(-3,3,15));
histfit(z(:,idx(end/2)),linspace(-3,3,15));
histfit(z(:,idx(end)),linspace(-3,3,15));
More telling is the qqplot:
qqplot(z(:,idx(1))); hold on; plot([-3,3],[-3,3],';;'); hold off;
qqplot(z(:,idx(end/2))); hold on; plot([-3,3],[-3,3],';;'); hold off;
qqplot(z(:,idx(end))); hold on; plot([-3,3],[-3,3],';;'); hold off;

Try a similarly analysis for z uniform:
z=rand(n,reps); x=sort(z);
and for z exponential:
z=rande(n,reps); x=sort(1-exp(-z));

[1] Marsaglia, G; Marsaglia JCW; (2004) "Evaluating the Anderson Darling
distribution", Journal of Statistical Software, 9(2).