## Show me the mean(ing)…

Well testing a bunch of samples for the largest population mean isn’t that common yet a simple test is at hand. Under the obvious title “*The rank sum maximum test for the largest K population means*” the test relies on the calculation of the sum of ranks under the combined sample of size , where is the common size of the k’s samples.

For illustration purposes the following data are used. They consist of 6 samples of 5 observations.

> data [1] 4.17143986 1.31264787 0.12109036 0.63031601 1.56705511 0.58817076 [7] 1.98011001 1.63226118 -0.03869368 1.80964611 4.80878278 0.67015153 [13] 2.07602321 1.52952749 1.68483297 2.00147364 9.30173048 0.58331012 [19] 2.49537140 1.31229842 1.40193543 0.11906268 4.76253012 1.26550467 [25] 0.69497074 -0.27612056 5.05751484 1.96589383 2.58427547 -0.36979229

Next we construct a convenient matrix

data.mat=expand.grid(x=rep(NA,5),sample=c("1","2","3","4","5","6")) data.mat$x=data data.mat$Rank=rank(data.mat$x)

and we compute the sample ranks

R=rep(NA,6) for (i in 1:6) { R[i]=sum(subset(data.mat,data.mat$sample==i)$Rank) }> rank(R) [1] 3 2 5 6 1 4

So we would test whether the 4th sample has the largest population mean. First we need critical values.

##Critical valus 115/119/127/134 for 10%,5%,1% and 0.1%> R[rank(R)==length(R)]>119FALSE

So, we cannot accept the hypothesis of the largest mean for the 4th sample.

**Look it up… **Gopal K. Kanji, 100 Statistical Tests , Sage Publications [google]