How does the 30% trimmed mean of this equal 6.167? [6, 3, 7, 11, 5, 3, 8, 7, 2, 6, 9, 13, 10, 4, 3]
So I'm going through some revision for stats, one of the questions asks to find the trimmed mean of the above data. I've worked it out but my answer does not match up with the Memo, my working is such:
Order in ascending order: [2, 3, 3, 3, 4, 5, 6, 6, 7, 7, 8, 9, 10, 11, 13]
Work out K:
K = np
= 15 x 30%
= 4,5
Then we remove 4 values from either side, and then find the mean of the remaining data:
(0.5 x 4) + 5 + 6 + 6 + 7 + 7 + (0.5 x 8)/7 = 5,28
Where am I going wrong?

Your weights sum to 6 not 7. In Stata, although any other program should suit, my version of your calculation is
. mata : ((0.5 * 4) + 5 + 6 + 6 + 7 + 7 + (0.5 * 8))/6
  6.166666667

which tallies with the reported correct answer. Some recipes, including the recipe you are using, imply fractional weights for some values. Regardless of that detail, a trimmed mean is just a weighted mean, in which values are assigned weights between 0 (totally excluded) and 1 (totally included), so that the calculation is $\sum$ (values $\times$ weights) / $\sum$ weights.
The simplest recipe boils down to trimming an integer number of values, so weights are either 0 or 1 and never fractional. Also in Stata, I get these results. Here number means number trimmed on each side and # is number of values averaged. Just about any answer between 6 and 6.5 is defensible.
. trimmean foo, number(0/7)

  +----------------------------+
  | number    #   trimmed mean |
  |----------------------------|
  |      0   15       6.466667 |
  |      1   13       6.307693 |
  |      2   11       6.181818 |
  |      3    9       6.111111 |
  |      4    7       6.142857 |
  |      5    5            6.2 |
  |      6    3       6.333333 |
  |      7    1              6 |
  +----------------------------+

More discussion can be found at this paper (tutorial review, historical notes, Stata details).
