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

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,

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

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

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

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

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,

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

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

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

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

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).