# How to calculate a confidence interval for a trimmed mean

I try to get the sample size that presents the population, I averaged randomly the 10 and then the 20 readings then I used the formula CI = 1.96 * std/root n, that gave me the CI at any (n), but how can I get the CI if the (n) has been excluded the high and low value at say 10 or 20 readings?

• The trimmed mean is not a parameter of a distribution, so the question arises: what are you trying to estimate with it? If you are using it to estimate the mean (which may be questionable, because the trimmed mean could be a heavily biased estimator of the mean), then your question is "how do I obtain a confidence interval for the mean." The answer to that might depend on why you are trimming. Presumably that's because you are concerned about influential outliers: but do you have any particular expectations about the frequency and magnitudes of those outliers? – whuber May 28 '16 at 0:59
• Yaser -- If you are trimming in order to estimate a population mean, how are you estimating the standard deviation? Much about your post needs to be made explicit. What you're attempting to achieve (and why), and exactly what calculations you're doing (and why) might all be important. As whuber has already indicated, a CI is an interval for a population quantity, not a sample statistic, so you need to be clear what population quantity you're trying to produce an interval for and under what assumptions. – Glen_b -Reinstate Monica May 28 '16 at 1:04
• You may want to look at the book "Introduction to Robust Estimation and Hypothesis Testing", by Rand Wilcox. You are probably interested in estimating the population trimmed mean $\mu_t$; formulas and R functions for constructing confidence intervals for this parameter, based on the sample trimmed mean, are described in Chapter 4. – Brent Kerby May 28 '16 at 4:54
• @whuber: The trimmed mean is indeed not a parameter, but its expectation is (if it exists). – Michael M May 28 '16 at 9:39
• @Michael I believe there may be a subtle problem with your claim. We should be concerned that the expectation of the trimmed mean might depend on the sample size. In fact it will when the underlying distribution is not symmetric. Thus (in general) its expectation is not a property of the distribution itself. My original question stands: exactly what property of that distribution is the trimmed mean intended to estimate? (Brent Kerby proposed a reasonable one--but notice that it is not the mean.) – whuber May 28 '16 at 18:27