This is not built into the system as far as I can tell. However, it's a really good example of how to construct a fairly sophisticated function using just a few lines of code. I've taken the definition directly from the Wikipedia page, and applied it to Sign[data] on the assumption that if you are testing residuals from a regression, you are interested in runs of positive and negative values.
runstest[x:{__?NumericQ}] :=
Module[{n = Length[x], res = Sign[x], n1, n2, runs, mu, sig2, y},
{n1, n2} = Last /@ Tally[res];
runs = Split[res];
mu = 2 n1 n2/n + 1;
sig2 = (mu - 1) (mu - 2)/(n - 1);
NProbability[y <= Length[runs],
y \[Distributed] NormalDistribution[N@mu, N@sig2]] ]
Some notable things about this code:
- The pattern
{__?NumericQ}. This ensures that the function only works on vectors of numeric data.
- The use of local variables via
Module, some of which can be set initially in the first argument of Module, while others can wait till later, perhaps because they depend on other local variables.
- The use of
Tally and Split, built-in functions that automatically divide lists into sublists of like elements.
- The use of
NProbability to work out the one-tailed test. Alternatively the last line could be something like N@CDF[NormalDistribution[mu, sig2], Length[runs]]: it gives the same answer except where the p-value is so tiny that numerical error creeps in (but seriously, who cares if it's $10^{-16}$ or $10^{-20}$?). It's possible that you want to express the test a different way, in which case that line is easy to modify.
Here is some test data that should definitely be rejected as random because of strong serial correlation:
testdata1 =
FoldList[0.98 #1 + #2 &, RandomReal[],
RandomVariate[NormalDistribution[0, 0.5], 100]];
And here is some data that should probably pass the test.
testdata2 = RandomVariate[NormalDistribution[0, 0.5], 100];
Sure enough:
runstest[testdata1]
(* -7.90479*10^-14 *)
runstest[testdata2]
(* 0.451757 *)
