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:

(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) 10^th and the (20/2 + 1) 11^th measurement. The 10^th measurement is 20.50 and the 11^th 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.

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint.

Assume that the measurements are a simple random sample.
(a) What is the sample size for the above sample?
(b) Calculate the sample mean for these data.
(c) Calculate the sample median.
(d) Plot the data by way of a dot plot.
(e) Compute the 20% trimmed mean for the above data set.
(f) Is the sample mean for these data more or less descriptive as a center of location than the trimmed mean?

Solution:

Part A. Sample size, n is the total number of measurements.

n=15 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B. The sample mean, \bar x is computed as follows:

\begin{align*}
\bar x & = \sum_{i=1}^{n}\frac{x_{i}}{n} \\
& = \frac{3.4+2.5+4.8+2.9+3.6+2.8+3.3+5.6+3.7+2.8+4.4+4.0+5.2+3.0+4.8}{15} \\
& = \frac{56.8}{15} \\
& = 3.79 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The sample median, \tilde x is the number at the middle of the arranged measurements in increasing magnitude. There are 15 measurements, n=15. If we arranged the data in increasing magnitude, the median is the measurement in the middle.

2.5, \ 2.8, \ 2.8, \ 2.9, \ 3.0, \ 3.3, \ 3.4, \ \underset{\color{Blue} \text{middle number}}{3.6}, \ 3.7, \ 4.0, \ 4.4, \ 4.8, \ 4.8, \ 5.2, \ 5.6

The middle number is 3.6. That is

\tilde x= 3.6 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D. This dot plot has been created using the Statistical Software Rguroo

Part E. The 20% trimmed mean means the average of the measurements left after removing 20% highest and 20% lowest data. This means we remove the 3 highest and 3 lowest numbers. Therefore, the data becomes

\ 2.9, \ 3.0, \ 3.3, \ 3.4, \ 3.6,  \ 3.7, \ 4.0, \ 4.4, \ 4.8,

The 20% trimmed mean, \bar x _{tr20} is

\begin{align*}
\bar x _{tr20} & = \frac{2.9+3.0+ \cdots+4.8}{9} \\
\bar x _{tr20} & = \frac{33.1}{9} \\
\bar x _{tr20} & = 3.678 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part F. The sample mean for these data is \bar x = 3.79 \ \text{hours} while the 20% trimmed mean is \bar x _{tr20} = 3.678 \ \text{hours}. Seems like the two means are not really far from each other, but because of the elimination of the extreme values, we can treat the trimmed mean as a better descriptive mean.