Solved – An X% trimmed mean means

trimmed-mean

Rand Wilcox in Fundamentals Of Statistical Methods, 1st. edition, gives a formula which says that for a 20% trimmed mean, you would trim away 20% of one end of the ranked data, and 20% of the other end, making 40% trimmed away in total.

But spreadsheets such as the Calc of LibreOffice5, would for a 20% trimmed mean only trim away 10% from one end and another 10% from the other end, making 20% trimmed away in total.

Which one is right?

The author also writes that a 20% trimmed mean is best for mixture distributions. Is this correct?

Best Answer

Neither is "right" or "wrong"; it's just that usage is not universal. However, I've seen Wilcox's definition used more than the other. Wikipedia agrees with him, as do several other sites I browsed to, and so do SAS, and R.

Related Solutions

Solved – Is R’s trimmed means function biased

In such cases it is a good idea to look at the source code. In this case the source code of mean.default

> mean.default
function (x, trim = 0, na.rm = FALSE, ...) 
{
    if (!is.numeric(x) && !is.complex(x) && !is.logical(x)) {
        warning("argument is not numeric or logical: returning NA")
        return(NA_real_)
    }
    if (na.rm) 
        x <- x[!is.na(x)]
    if (!is.numeric(trim) || length(trim) != 1L) 
        stop("'trim' must be numeric of length one")
    n <- length(x)
    if (trim > 0 && n) {
        if (is.complex(x)) 
            stop("trimmed means are not defined for complex data")
        if (any(is.na(x))) 
            return(NA_real_)
        if (trim >= 0.5) 
            return(stats::median(x, na.rm = FALSE))
        lo <- floor(n * trim) + 1
        hi <- n + 1 - lo
        x <- sort.int(x, partial = unique(c(lo, hi)))[lo:hi]
    }
    .Internal(mean(x))
}

From this the trimming is based on the lower and upper index bounds lo <- floor(n * trim) + 1 and hi <- n + 1 - lo, which are the design decisions implemented in R. Now, while writing this I see that Nick Sabbe posted more or less the same answer, but I leave the entire code posted here as well.

P.S. I think the clue to reading the help page is that it says a symmetrically trimmed mean, not the symmetrically trimmed mean.

Solved – Mean has lower standard error than 5% trimmed mean

If the underlying population is normally distributed without contamination then the sample mean is the best unbiased estimate (in the sense of the lowest mean square error) of the centre of the population distribution.

This is not always the case with other distributions, which might include those with contamination. So your observation depends on the particular distribution and contamination.

Best Answer

Related Solutions

Solved – Is R’s trimmed means function biased

Solved – Mean has lower standard error than 5% trimmed mean

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