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Is there a better alternative to using run tables for testing for randomness in a sample?

Currently, for statistical process control charting, I am testing the number of runs identified in a sample (essentially the number of times the sample crosses the average, plus one) against a standard table of significantly high or low run counts given a particular sample size:

  n     Low     High
  10     2        9
  11     2        9
  12     2        11
  13     2        11
  14     3        13
       .  .  .
  46     15       32
       .  .  .
  50     17       34

For example, if a sample of size 11 has less than 2 runs or more than 9 runs, then it can be said that that sample exhibits special cause variation (i.e. is not random).

I'm trying to approach this programmatically and having some formulaic approach rather than attempting to lookup values in a finite table would be a huge help.

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I use SPC frequently but we do not use this rule. Where can I find the whole table? – kaine Feb 27 '13 at 21:27
up vote 1 down vote accepted

Well, this test apears to check for autocorrelation of the variable. Rather than using a table like this, I am aware that most places have rules that state the following as our of control:

9 adjacent samples that are all increasing or decreasing

14 adjacent samples that alternate between increasing and decreasing

This would check for simular cyclic/autocorrelated behavior in a way easy to check but it is obviously not identical.

I believe that the table could be approximated by assuming that the samples are not autocorrelated and therefore have an equal probabilty (50%) of crossing the mean each sample (besided the first). I can then calculate the predicted number of runs (r) compared to the number of samples (n). I am still defining number of runs as number of times crossing the mean +1. The probability of there being r runs is $$P(r,n) = \frac{(n-1)!}{(n-r)!(r-1)!2^{(n-1)}}$$

You then can calculate the l (lower limit) and u (upper limit) values by picking a threshold probability ($P_t$) and finding the largest l and smallest u such that:

$$P_t < \underset{j=1}{\sum^{l-1}}\frac{(n-1)!}{(n-j)!(j-1)!2^{(n-1)}}$$ $$P_t < \underset{k=u-1}{\sum^{n-1}}\frac{(n-1)!}{(n-k)!(k-1)!2^{(n-1)}}$$

I get simular numbers to the table above using $P_t=.005$ but it doesn't work very well for n=11 and n=13. My calculations (and intuition) indicate that those upper limits should be 10 and 12 respectively instead of 9 and 11. I think that the long term trend would be closer but need more data to confirm this.

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That's great. I think that ultimately the table is simplifying the actual values a bit. I've seen multiple versions of this table, all slightly different. Perhaps the threshold was different... If I set the threshold to 0.05, how would I go about solving for the various values? If I'm trying to approach this programmatically (which I am), I'd like to have a formula for k and u given n and Pt. – mbeasley Feb 28 '13 at 18:23
$P_t$ is equal to the cumulative binomial distribution for p=.5. You only need to solve for l as $u = n-l+1$. The binomial distribution is approximated by the normal distribution at high values of n with $\mu=np$ and $\sigma=\sqrt(np(1-p))$so I will try that. Please try this and tell me if it works: $$x \approx \frac{n}{2}+(\sqrt{2}/2)erf^{-1}(1-2p) \sqrt(n)$$ where $erf^{-1}$ is the inverse error function. – kaine Mar 1 '13 at 14:42
it won't let me edit this for some reason, x should be replaced by $l-1$ – kaine Mar 1 '13 at 14:53
I still have to test this out a little further, but this is definitely the right direction. Thanks for the help! – mbeasley Mar 4 '13 at 12:59

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