The test can be seen as a special case of
Maurice Kendall’s more general method of
rank correlation and makes use of the Kendall's
S statistic. This can be computed in one of two ways:
The ‘direct counting’ method • Arrange the samples in the predicted order • For each score in turn, count how many scores in the samples to the right are larger than the score in question. This is
P. • For each score in turn, count how many scores in the samples to the right are smaller than the score in question. This is
Q. •
S =
P –
Q The ‘nautical’ method • Cast the data into an ordered
contingency table, with the levels of the
independent variable increasing from left to right, and values of the
dependent variable increasing from top to bottom. • For each entry in the table, count all other entries that lie to the ‘South East’ of the particular entry. This is
P. • For each entry in the table, count all other entries that lie to the ‘South West’ of the particular entry. This is
Q. •
S =
P –
Q Note that there will always be ties in the independent variable (individuals are ‘tied’ in the sense that they are in the same group) but there may or may not be ties in the dependent variable. If there are no ties – or the ties occur within a particular sample (which does not affect the value of the test statistic) – exact tables of
S are available; for example, Jonckheere
Normal approximation to S The
standard normal distribution can be used to approximate the distribution of
S under the null hypothesis for cases in which exact tables are not available. The
mean of the distribution of
S will always be zero, and assuming that there are no ties scores between the values in two (or more) different samples the
variance is given by :\operatorname{VAR}(S)=\frac{2(n^3-\sum t^3_i)+3(n^2-\sum t^2_i)}{18} Where
n is the total number of scores, and
ti is the number of scores in the ith sample. The approximation to the standard normal distribution can be improved by the use of a continuity correction:
Sc = |
S| – 1. Thus 1 is subtracted from a positive
S value and 1 is added to a negative
S value. The z-score equivalent is then given by :z =\frac{S_c}{\sqrt{\operatorname{VAR}(S)}}
Ties If scores are tied between the values in two (or more) different samples there are no exact table for the S distribution and an approximation to the normal distribution has to be used. In this case no continuity correction is applied to the value of
S and the variance is given by :\begin{align}\operatorname{VAR}(S)=&\frac{2\left(n^3-\sum t^3_i -\sum u^3_i\right)+3\left(n^2-\sum t^2_i -\sum u^2_i\right)+5n}{18} \\ &{}+\frac{\left(\sum t^3_i-3\sum t^2_i+2n\right)\left(\sum u^3_i-3\sum u^2_i+2n\right)}{9n(n-1)(n-2)} \\ &{}+\frac{\left(\sum t^2_i-n\right)\left(\sum u^2_i-n\right)}{2n(n-1)}\end{align} where
ti is a row marginal total and
ui a column marginal total in the contingency table. The
z-score equivalent is then given by :z =\frac{S}{\sqrt{\operatorname{VAR}(S)}} ==A numerical example==