Description

Performs a test whether the elements of x are serially independent - say, whether they occur in a random order - by counting how many runs there are above and below a threshold. If y is supplied a two sample Wald-Wolfowitz-Test testing the equality of two distributions against general alternatives will be computed.

Usage

Arguments

Details

The runs test for randomness is used to test the hypothesis that a series of numbers is random. The 2-sample test is known as the Wald-Wolfowitz test.

For a categorical variable, the number of runs correspond to the number of times the category changes, that is, where x_i belongs to one category and x_(i+1) belongs to the other. The number of runs is the number of sign changes plus one.

For a numeric variable x containing more than two values, a run is a set of sequential values that are either all above or below a specified cutpoint, typically the median. This is not necessarily the best choice. If another threshold should be used use a code like: RunsTest(x > mean(x)).

The exact distribution of runs and the p-value based on it are described in the manual of SPSS "Exact tests" http://www.sussex.ac.uk/its/pdfs/SPSS_Exact_Tests_21.pdf.

The normal approximation of the runs test is calculated with the expected number of runs under the null

μ_r=\frac{2 n_0 n_1}{n_0 + n_1} + 1

and its variance

σ_r^2 = \frac{2 n_0 n_1 (2 n_0 n_1 - n_0 - n_1) }{(n_0 + n_1)^2 \cdot (n_0 + n_1 - 1)}

as

\hat{z}=\frac{r - μ_r + c}{σ_r}

where n_0, n_1 the number of values below/above the threshold and r the number of runs.

Setting the continuity correction correct = TRUE will yield the normal approximation as SAS (and SPSS if n < 50) does it, see http://support.sas.com/kb/33/092.html. The c is set to c = 0.5 if r < \frac{2 n_0 n_1}{n_0 + n_1} + 1 and to c = -0.5 if r > \frac{2 n_0 n_1}{n_0 + n_1} + 1.

Wald-Wolfowitz with Ties. Ideally there should be no ties in the data used for the Wald-Wolfowitz test. In practice there is no problem with ties within a group, but if ties occur between members of the different groups then there is no unique sequence of observations. For example the data sets A: 10,14,17,19,34 and B: 12,13,17,19,22 can give four possible sequences, with two possible values for r (7 or 9). The "solution" to this is to list every possible combination, and calculate the test statistic for each one. If all test statistics are significant at the chosen level, then one can reject the null hypothesis. If only some are significant, then Siegel (1956) suggests that the average of the P-values is taken. Help for finding all permutations of ties can be found at: https://stackoverflow.com/questions/47565066/all-possible-permutations-in-factor-variable-when-ties-exist-in-r

However this solutions seems quite coarse and in general, the test should not be used if there are more than one or two ties. We have better tests to distinguish between two samples!

Value

A list with the following components.

Author(s)

Andri Signorell <andri@signorell.net>, exact p-values by Detlew Labes <detlewlabes@gmx.de>

References

Wackerly, D., Mendenhall, W. Scheaffer, R. L. (1986) Mathematical Statistics with Applications, 3rd Ed., Duxbury Press, CA.

Wald, A. and Wolfowitz, J. (1940): On a test whether two samples are from the same population, Ann. Math Statist. 11, 147-162.

Siegel, S. (1956) Nonparametric Statistics for the Behavioural Sciences, McGraw-Hill Kogakusha, Tokyo.

See Also

Run Length Encoding rle

Examples