Question

Fretting is a wear process that results from tangential oscillatory movements of small amplitude in machine parts. The article “Grease Effect on Fretting Wear of Mild Steel” (Industrial Lubrication and Tribology, 2008: 67–78) included the following data on volume wear (1024 mm3 ) for base oils having four different viscosities.

Viscosity Wear
20.4 58.8 30.8 27.3 29.9 17.7 76.5
30.2 44.5 47.1 48.7 41.6 32.8 18.3
89.4 73.3 57.1 66.0 93.8 133.2 81.1
252.6 30.6 24.2 16.6 38.9 28.7 23.6

a. The sample coefficient of variation 100s/x-bar assesses the extent of variability relative to the mean (specifically, the standard deviation as a percentage of the mean). Calculate the coefficient of variation for the
sample at each viscosity. Then compare the results and comment.
b. Construct a comparative boxplot of the data and comment on interesting features.

Short answer

a.

0.5601 or 56.01%

0.2966 or 29.66%

0.3230 or 32.30%

0.2786 or 27.86%

Relative variability of viscosity level (20.4) is highest among all the four samples while the relative variability of viscosity level (252.6) is lowest as compared with other samples.

b.

Two samples, 20.4 and 252.6, are positive skewed in the comparison boxplot.

The viscosity level of 89.4 is, on the other hand, negative skewed.

In all samples, there are no outliers. The volume wears corresponding to oil viscosity 252.6 have the least variation, as can be shown. The viscosity boxplot is almost symmetric (252.6).

Step-by-step solution

Given information

The data are provided that consists of 24 observations on volume wear for base oils having 4 different viscosities.

Compute sample means for each viscosity

a.

The sample mean is computed using the formula,

\(\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\)

Let \({\bar x_{vis\cos ity\left( {20.4} \right)}}\),\({\bar x_{vis\cos ity\left( {30.2} \right)}}\),\({\bar x_{vis\cos ity\left( {89.4} \right)}}\) and \({\bar x_{vis\cos ity\left( {252.6} \right)}}\) are the four sample means. Each sample size is\(n = 6\). Each sample mean can be calculated as,

\(\begin{aligned}{{\bar x}_{vis\cos ity\left( {20.4} \right)}} &= \frac{{\sum\limits_{i = 1}^6 {{x_i}} }}{6}\\ &= \frac{{\left( {58.8 + 30.8 + 27.3 + 29.9 + 17.7 + 76.5} \right)}}{6}\\ &= \frac{{241}}{6}\\ &= 40.1667\end{aligned}\)

\(\begin{aligned}{{\bar x}_{vis\cos ity\left( {30.2} \right)}} &= \frac{{\sum\limits_{i = 1}^6 {{x_i}} }}{6}\\ &= \frac{{\left( {44.5 + 47.1 + 48.7 + 41.6 + 32.8 + 18.3} \right)}}{6}\\ &= \frac{{233}}{6}\\ &= 38.8333\end{aligned}\)

\(\begin{aligned}{{\bar x}_{vis\cos ity\left( {89.4} \right)}} &= \frac{{\sum\limits_{i = 1}^6 {{x_i}} }}{6}\\ &= \frac{{\left( {73.3 + 57.1 + 66 + 93.8 + 133.2 + 81.1} \right)}}{6}\\ &= \frac{{504.5}}{6}\\ &= 84.0833\end{aligned}\)

\(\begin{aligned}{{\bar x}_{vis\cos ity\left( {252.6} \right)}} &= \frac{{\sum\limits_{i = 1}^6 {{x_i}} }}{6}\\ &= \frac{{\left( {30.6 + 24.2 + 16.6 + 38.9 + 28.7 + 23.6} \right)}}{6}\\ &= \frac{{162.6}}{6}\\ &= 27.1\end{aligned}\)

Compute sample standard deviations for each viscosity

The sample standard deviation is computed using the formula,

\(s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \)

Let \({s_{vis\cos ity\left( {20.4} \right)}}\),\({s_{vis\cos ity\left( {30.2} \right)}}\),\({s_{vis\cos ity\left( {89.4} \right)}}\) and \({s_{vis\cos ity\left( {252.6} \right)}}\) are the four sample means. Each sample size is \(n = 6\). Each sample standard deviation can be calculated as,

\(\begin{aligned}{s_{vis\cos ity\left( {20.4} \right)}} &= \sqrt {\frac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{6 - 1}}} \\ &= \sqrt {\frac{{{{\left( {58.8 - 40.1667} \right)}^2} + \ldots + {{\left( {76.5 - 40.1667} \right)}^2}}}{{6 - 1}}} \\ &= 22.4978\end{aligned}\)

\(\begin{aligned}{s_{vis\cos ity\left( {30.2} \right)}} &= \sqrt {\frac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{6 - 1}}} \\ &= \sqrt {\frac{{{{\left( {44.5 - 38.8333} \right)}^2} + \ldots + {{\left( {18.3 - 38.8333} \right)}^2}}}{{6 - 1}}} \\ &= 11.5193\end{aligned}\)

\(\begin{aligned}{s_{vis\cos ity\left( {89.4} \right)}} &= \sqrt {\frac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{6 - 1}}} \\ &= \sqrt {\frac{{{{\left( {73.3 - 84.0833} \right)}^2} + \ldots + {{\left( {81.1 - 84.0833} \right)}^2}}}{{6 - 1}}} \\ &= 27.1557\end{aligned}\)

\(\begin{aligned}{s_{vis\cos ity\left( {252.6} \right)}} &= \sqrt {\frac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{6 - 1}}} \\ &= \sqrt {\frac{{{{\left( {30.6 - 27.1} \right)}^2} + \ldots + {{\left( {23.6 - 27.1} \right)}^2}}}{{6 - 1}}} \\ &= 7.5493\end{aligned}\)

Construct a comparative box plot for the given data

Following are the steps to make comparative boxplot by hand:

1. Draw a plot line of range 0 to 240.
2. Draw three horizontal lines that consists of first quartile, second quartile and third quartile and make two vertical lines to make it in rectangular form like a box for Viscosity level (20.4).
3. Do the Step 2 again for Viscosity level (30.2), Viscosity level (89.4) and Viscosity level (252.6).
4. Draw whiskers on both sides of four boxplots and set the minimum and maximum value with respect to the obtained lower fence and upper fence.

From the comparative boxplot, Two samples are positively skewed i.e. 20.4 and 252.6. On the other hand, the viscosity level 89.4 is negative skewed. There are no outliers exist in all samples. It is observe that the volume wears corresponding to oil viscosity 252.6 have the smallest variability. The boxplot is almost symmetric for the viscosity (252.6).