I have one variable that attains a certain value (between 0 and 1), about 1000 observations and each belongs to one (and only one) of fifteen groups. Now I would like to reduce the number of groups to two or three by combining them. It is important that all the data within a given group in the former division will stay together after the pooling.
The problem is, that the whole dataset has too large variance, so I would prefer several groups that are more focused, less varied.
Example: Say you have hundreds of data on incomes of people from different EU countries. From this, you estimate average salary for each country (just an example). You would like to now create a group of "rich countries" and "poor countries" — but where do you draw the division line? Or two division lines in case of three groups (adding, say, "average countries").
Since yet again I don't know the proper statistical approach, I at least give you what my idea was:
For the reduction to two, instead of trying all possible divisions, I had the idea of sorting the groups by their sample means (from $g_1$ to $g_{15}$ where means satisfy $m_1 \leq m_2 \leq \dots \leq m_{15}$) and creating pool combinations — first combination would be {[$g_1$]; [$g_2, ..., g_{15}$]}, second would be {[$g_1, g_2$]; [$g_3, ..., g_{15}]$} and so on -- this notation means that in the first case the first pool are only values from $g_1$ and the second pool are all other values. The second case the first pool are values from $g_1$ and $g_2$, the rest is in the other pool.
I would then compute the p-value for sample mean equivalence test of these pool combinations. The lowest p-value of these 14 combination tests would mark the "most different" division into pools.
The same could be applied to the three groups, though there would be a higher number of pool combinations and the "diversity test" would have to be constructed quite differently.
Feel free to ignore my scribbling... since I do not know statistics very well, I at least tried an approach that seemed most logical.
Thanks again!