Trimmed Mean – How to Calculate the Truncated or Trimmed Mean

Question

How can I calculate the truncated or trimmed mean? Let's say truncated by 10%?

I can imagine how to do it if you have 10 entries or so, but how can I do it for a lot of entries?

Answer 1

Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 38.64$

Source: Class notes on P percent trimmed mean