Question

According to the journal Chemical Engineering, an important property of a fiber is its water ab- sorbency. A random sample of 20 pieces of cotton fiber was taken and the absorbency on each piece was mea- sured. The following are the absorbency values:
18.71 21.41 20.72 21.81 19.29 22.43 20.17 23.71 19.44 20.50 18.92 20.33 23.00 22.85 19.25 21.77 22.11 19.77 18.04 21.12
(a) Calculate the sample mean and median for the above sample values.
(b) Compute the 10% trimmed mean.
(c) Do a dot plot of the absorbency data.
(d) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?

Answer 1

a) Sample mean $\bar{x}=\frac{\sum {x_i}}{N}=\frac{415.35}{20}=20.768;\\$

Sorted data Y=

18.040  18.710  18.920  19.250  19.290  19.440  19.770  20.170  20.330  20.500  20.720 21.120  21.410  21.770  21.810  22.110  22.430 

 22.850  23.000  23.710;

$Y_{10}=20,500\space Y_{11}=20.720,\\Me=\frac{Y_{10}+Y_{11}}{2}=\frac{20.500+20.720}{2}=20.610;$

b)

Let we trim Y by removing 10% or 2 largest and 1o% or 2 smallest samples:

Y'=18.920  19.250  19.290  19.440  19.770  20.170  20.330  20.500  20.720 21.120  21.410  21.770  21.810  22.110  22.430  22.850 

Sum(Y')=331.89;

$\bar{x}_{.10}=\frac{Sum(Y')}{16}=\frac{331.89}{16}= 20.743$

c)

d) if $|Me-\bar{x}|>k\cdot|Me-\bar{x}_{.10}|$

where k some weight for example k=2 then there are outliers with considerable reliability. For given data we have

$\frac{|Me-\bar{x}}{|Me-\bar{x}_{10}|}=\frac{20.768-20.610}{20.768-20/743}=\frac{0.158}{0.025}=6.32>2$

So we should conclude that outlers are.