Trimmed mean in a specific column?

General
#1

One of the problems in my assignment for class is the following:

Calculate a so-called "trimmed mean" of column 9 of x_mat by trimming off 5 percent of the values from each end of x. Please do not store the result in an object.

I have to be making this harder than it actually is. Here is what I came up with:

mean(x_mat[, 9], trim = 0.1)

#2

From the help file on the mean function (emphasis mine)

trim the fraction (0 to 0.5) of observations to be trimmed from each end of x before the mean is computed. Values of trim outside that range are taken as the nearest endpoint.

#3

I am not too sure how to interpret this info into a coding format in R.. can this be explained differently?

#4

The question you presented asks that you "[trim] off 5 percent of values from each end of x". The help file says that the trim argument is " the fraction (0 to 0.5) of observations to be trimmed from each end of x". How much will be trimmed off each end of x_mat[,9] by the code you presented:

mean(x_mat[, 9], trim = 0.1)

I am being somewhat indirect to try to fulfill the spirit of the forum's homework policy.

#5

Ah fair enough. Thanks for the input. It is difficult knowing if the coding is actually doing what you expected when R gives you a result. Rather it just say the input was bad because then at least you know it is wrong!

#6

Let us create a simple example to get familiar with the function. Simple vector with 10 elements.

x<-c(2,3,1,1,1,1,1,1,3,2)
sum(x)
#> [1] 16
mean(x)
#> [1] 1.6

Created on 2020-09-26 by the reprex package (v0.3.0)

Then we go ahead and remove 1 element from each end. So we have element 2 to 9 (after ordering). 1 and 3 are removed. We are trimming 1/10=0.1 from each end.

x<-c(2,3,1,1,1,1,1,1,3,2)

x[order(x)][2:9]
#> [1] 1 1 1 1 1 2 2 3

sum(x[order(x)][2:9])/8
#> [1] 1.5

mean(x,trim=0.1)
#> [1] 1.5

Created on 2020-09-26 by the reprex package (v0.3.0)

Hope this helps.

#7

That makes total sense. I just need to do this for a data set of 1000. But I understand the steps and more importantly the logic.
Thank you!

#8

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