Mosteller has recently [1,, 1948] proposed a $k$-sample slippage test and has given percentage points for selected $n, k$ and $R$ for the case of $k$ equal samples of size $n$. when the samples are of unequal size, exact significance levels can be calculated very quickly from $P_r = \frac{\sum n^{(r)}_i}{N^{(r)}} \text{where} x^{(r)} == x(x - 1) \cdots (x - r + 1),$ by the method explained in section 3 below. The significance values for $k$ equal samples of $n \geq 10$ are very well approximated by $P_r = \frac{1}{k^{r - 1}} e^{-r(r - 1)(K - 1)/2n}$ where $N = kn.$ A convenient rough approximation for unequal samples may be given in terms of $k^\ast,$ an "effective" number of samples, which is given by $k^\ast = \frac{(\sum_in_i)^2}{\sum n^2_i},$ the one-sided significance level will then be approximately given by $P_r = (k^\ast)^{-(r - 1)}.$ This approximation can be easily applied with the aid of Table 1. Thus, for example, with four samples of sizes 7, 5, 5, 2, we have $k^\ast = \frac{(7 + 5 + 5 + 2)^2}{49 + 25 + 25 + 4} = \frac{361}{103} = 3.50,$ whence from the table $r = 3$ lies at a one-sided level approximately between 5% and 10%, $r = 4$ approximately between 1% and 2.5%, $r = 5$ between 0.5% and 1%, $r = 6$ near 0.2%, and so on. Direct calculation yields 5.7%, 1.2%, 0.2% and 0.03%. The approximation is, in this example, quite satisfactory for moderate significance levels and conservative for more extreme significance levels.