According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber was taken and the absorbency on each piece was measured. 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?
Solution:
Part A. The sample mean is computed as follows:
\begin{align*} \bar x & = \frac{\Sigma x_{i}}{n} \\ \bar x & = \frac{18.71+21.41+20.72+\cdots +21.12}{20} \\ \bar x & = 20.77 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
We can solve for the sample median by arranging the data in increasing order first.
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 there are 20 measurements (even), the middle measurements are the (20/2) 10th and the (20/2 + 1) 11th measurement. The 10th measurement is 20.50 and the 11th measurement is 20.72. The median is the average of these two measurements.
\begin{align*} \tilde x & = \frac{20.50+20.72}{2} \\ \tilde x & = 20.61 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Part B. The 10% trimmed mean is calculated by removing 10% of the lowest data and 10% of the highest data. That is, removing the 2 lowest and 2 highest data. We are left with the following:
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
The 10% trimmed mean, \bar x _{tr10} is
\begin{align*} \bar x _{tr10} & = \frac{\Sigma x_{i}}{n} \\ \bar x _{tr10} & = \frac{18.92+19.25+\cdots+22.85}{16} \\ \bar x _{tr10} & = 20.74 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Part C. The dot plot is shown
Part D. Since the values of the mean, median, and trimmed mean are not actually far from each other, we can conclude that there are no outliers in the given measurements.
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