absorbency

logistic_guy

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According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of \(\displaystyle 20\) pieces of cotton fiber was taken and the absorbency on each piece was measured. The following are the absorbency values:

\(\displaystyle 18.71 \ \ \ 21.41 \ \ \ 20.72 \ \ \ 21.81 \ \ \ 19.29 \ \ \ 22.43 \ \ \ 20.17\)
\(\displaystyle 23.71 \ \ \ 19.44 \ \ \ 20.50 \ \ \ 18.92 \ \ \ 20.33 \ \ \ 23.00 \ \ \ 22.85\)
\(\displaystyle 19.25 \ \ \ 21.77 \ \ \ 22.11 \ \ \ 19.77 \ \ \ 18.04 \ \ \ 21.12\)

\(\displaystyle \bold{(a)}\) Calculate the sample mean and median for the above sample values.
\(\displaystyle \bold{(b)}\) Compute the \(\displaystyle 10\%\) trimmed mean.
\(\displaystyle \bold{(c)}\) Do a dot plot of the absorbency data.
\(\displaystyle \bold{(d)}\) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?
 
According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of \(\displaystyle 20\) pieces of cotton fiber was taken and the absorbency on each piece was measured. The following are the absorbency values:

\(\displaystyle 18.71 \ \ \ 21.41 \ \ \ 20.72 \ \ \ 21.81 \ \ \ 19.29 \ \ \ 22.43 \ \ \ 20.17\)
\(\displaystyle 23.71 \ \ \ 19.44 \ \ \ 20.50 \ \ \ 18.92 \ \ \ 20.33 \ \ \ 23.00 \ \ \ 22.85\)
\(\displaystyle 19.25 \ \ \ 21.77 \ \ \ 22.11 \ \ \ 19.77 \ \ \ 18.04 \ \ \ 21.12\)

\(\displaystyle \bold{(a)}\) Calculate the sample mean and median for the above sample values.
\(\displaystyle \bold{(b)}\) Compute the \(\displaystyle 10\%\) trimmed mean.
\(\displaystyle \bold{(c)}\) Do a dot plot of the absorbency data.
\(\displaystyle \bold{(d)}\) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?
Please show us what you have tried and exactly where you are stuck.

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\(\displaystyle \bold{(a)}\) Calculate the sample mean
\(\displaystyle \text{Mean} = \frac{1}{20} (18.71 + 21.41 + 20.72 + 21.81 + 19.29 + 22.43 + 20.17 + 23.71 + 19.44 + 20.50 + 18.92 + 20.33 + 23.00 + 22.85 + 19.25 + 21.77 + 22.11 + 19.77 + 18.04 + 21.12) = \frac{415.35}{20} = \textcolor{blue}{20.77}\)
 
\(\displaystyle \bold{(a)}\) Calculate the sample median
\(\displaystyle 18.04,\ 18.71,\ 18.92,\ 19.25,\ 19.29,\ 19.44,\ 19.77,\ 20.17,\ 20.33,\ 20.50,\ 20.72,\ 21.12,\ 21.41,\ 21.77,\ 21.81,\ 22.11,\ 22.43,\ 22.85,\ 23.00,\ 23.71\)

Since the number is even, we take the average of the two numbers in the middle of the list.

\(\displaystyle \bold{median} = \frac{20.50 + 20.72}{2} = \textcolor{blue}{20.61}\)
 
\(\displaystyle \bold{(b)}\) Compute the \(\displaystyle 10\%\) trimmed mean.
\(\displaystyle \bold{trimmed \ mean} = \frac{1}{16} (21.41 + 20.72 + 21.81 + 19.29 + 22.43 + 20.17 + 19.44 + 20.50 + 18.92 + 20.33 + 22.85 + 19.25 + 21.77 + 22.11 + 19.77 + 21.12) = \frac{331.89}{16} = \textcolor{blue}{20.74}\)
 
\(\displaystyle \bold{(d)}\) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?
Since the mean, median, and trimmed mean are almost the same, there are no outliers.
 
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