2 sample rank sum test 
hi,
the above question is one of the past exam question of the course which i am studying.
The question i want to ask is: how can i answer part a)? I calculated the U=27 and mean = 15{(5*6)/2} , so i know that i can use normal approximation to do it. While the question need me to use the exact distribution to handle the problem. I know the no. of possible outcome is 11C6=462. If n is small , i can simply list out all the possibilities . but now i have 462 cases, it is obviously not to do in this way. So what should i do?  
 A: Since the OP has figured out how to get the answer with a table here's how to do it without a table.
You need a one-tailed test, so you simply need to find all the rank sets for sample 1 that have a lower rank sum.
Data

A 10.0  10.8   11.1    11.7    12.3
B                  11.5   12.1      12.8  13.6  13.8  15.5 

Ranks

A   1    2      3       5        7
B                    4       6        8     9    10    11  

Note that the sum of the ranks in sample A is 18.
Let's list all sets of sample-A ranks with a sum of 18 or less:
 Sample A    Rank
  ranks       sum
1 2 3 4 5     15
1 2 3 4 6     16
1 2 3 5 6     17
1 2 3 4 7     17
1 2 4 5 6     18
1 2 3 5 7     18
1 2 3 4 8     18

That's a total of 7 sets of ranks out of 462 that are at least as extreme (in the direction of $H_1$) as the observed set of ranks.
Consequently the p-value is  7/462 $\approx$ 0.01515
This can be done exactly, without tables, under exam conditions, in the space of a couple of minutes (with practice you can do it much faster); if the test statistic were higher it would take longer, but even with a 10% test, as soon as you write more than 46 sets of ranks for sample A you know you've exceeded the significance level and you can stop at that point.
