Question

A company samples the distribution of conductivity in isolation of receptacles and got the following values:

24.46 25.61 26.25 26.42 26.66 27.15 27.31 27.54 27.74 27.94 27.98 28.04 28.28 28.49 28.50 28.87 29.11 29.13 29.50 30.88

Trimming 10% mean when we discard the smallest and largest 10 % of the sample. Can we use that sample for to calculate the point estimate of population mean? What are the values for estimator of population mean with and without trim? Provide full reasoning.

Answer 1

Solution-:

A company samples the distribution of conductivity in isolation of receptacles and got the following values:

24.46,25.61,26.25,26.42,26.66,27.15,27.31,27.54,27.74,27.94,27.98,28.04,28.28,28.49,28.50,28.87,29.11,29.13,29.50,30.88.

To calculate the point estimate of population mean trimming 10% mean when we discard the smallest and largest 10 % of the sample;

By using R-Software:

>x=c(24.46,25.61,26.25,26.42,26.66,27.15,27.31,27.54,27.74,27.94,27.98,28.04,28.28,28.49,28.50,28.87,29.11,29.13,29.50,30.88)
> x
[1] 24.46 25.61 26.25 26.42 26.66 27.15 27.31 27.54 27.74 27.94 27.98 28.04
[13] 28.28 28.49 28.50 28.87 29.11 29.13 29.50 30.88
> n=length(x);n
[1] 20> sum(x)
[1] 555.86
> mean=(sum(x)/n);mean # mean of obervation without trimmed
[1] 27.793
># Yes, we use that sample for to calculate the point estimate of population mean.

># First we sort given observation:

> sort(x)
[1] 24.46 25.61 26.25 26.42 26.66 27.15 27.31 27.54 27.74 27.94 27.98 28.04
[13] 28.28 28.49 28.50 28.87 29.11 29.13 29.50 30.88

Here, n=20 , therefore 10% trimmed,

(We need to ignore the 2 highest obervation (29.50, 30.88) and 2 lowest observtion (24.46, 25.61 ). The mean of middle 16 observation is the required trimmed mean.

> x1=c(26.25,26.42,26.66,27.15,27.31,27.54,27.74,27.94,27.98,28.04,28.28,28.49,28.50,28.87,29.11,29.13)
> x1
[1] 26.25 26.42 26.66 27.15 27.31 27.54 27.74 27.94 27.98 28.04 28.28 28.49
[13] 28.50 28.87 29.11 29.13
> sum(x1) # sum of 10 % trimmed obervation
[1] 445.41
> n1=length(x1);n1
[1] 16
> mean1=sum(x1)/n1;mean1 # mean of 10% trimmed obervation
[1] 27.83812

Therefore, the value for estimator of population mean without trim is = 27.793 and

the value for estimator of population mean with trim is =27.83812