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The data are provided that consists of 16 observations on milk selenium concentration (mg/L) for samples of cows given a selenium supplement and a control sample given no supplement, both initially and after a 9-day period.

The following table represent the ordered Male and Female samples:

The sample mean is computed using the formula,

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

Let \({\bar x_M}\) and \({\bar x_F}\) are the two required sample means. The sample size of male group is 13 while the sample size of female group is 10. The mean of each sample can be determined as follows:

\(\begin{aligned}{{\bar x}_M} &= \frac{{\sum\limits_{i = 1}^{13} {{x_i}} }}{{13}}\\ &= \frac{{\left( {3.1 + 3.4 + \ldots + 4.0 + 4.0} \right)}}{{13}}\\ &= \frac{{47.3}}{{13}}\\ &= 3.638\end{aligned}\)

\(\begin{aligned}{{\bar x}_F} &= \frac{{\sum\limits_{i = 1}^{10} {{x_i}} }}{{10}}\\ &= \frac{{\left( {2.6 + 3 + \ldots + 3.8 + 4.3} \right)}}{{10}}\\ &= \frac{{32.8}}{{10}}\\ &= 3.28\end{aligned}\)

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 \({s_M}\)and\({s_F}\) are two standard deviations. Each sample standard deviation can be calculated as,

\(\begin{aligned}{s_M} &= \sqrt {\frac{{\sum\limits_{n = 1}^{13} {{{\left( {{x_i} - 3.638} \right)}^2}} }}{{13 - 1}}} \\ &= \sqrt {\frac{{{{\left( {3.1 - 3.638} \right)}^2} + \ldots + {{\left( {4 - 3.638} \right)}^2}}}{{13 - 1}}} \\ &= 0.27\end{aligned}\)

\(\begin{aligned}{s_F} &= \sqrt {\frac{{\sum\limits_{n = 1}^{10} {{{\left( {{x_i} - 3.28} \right)}^2}} }}{{10 - 1}}} \\ &= \sqrt {\frac{{{{\left( {2.6 - 3.28} \right)}^2} + \ldots + {{\left( {4.3 - 3.28} \right)}^2}}}{{10 - 1}}} \\ &= 0.47\end{aligned}\)

The five-number summary are smallest \({x_i}\), lower fourth, median, upper fourth and largest \({x_i}\).Since sample size of test is odd, the median is the\({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) ordered value when the data are in ascending order as below.

The smallest value is: 3.1 and the largest value is: 4.

Let \({\tilde x_M}\) be the required median. Use the formula to calculate median,

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}{{\tilde x}_M} &= {\left( {\frac{{13 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {7^{th}}\,ordered\,value\\ &= 3.7\end{aligned}\)

Therefore, the median of supplement group is 3.7

The lower fourth is the median of smallest half of the data as the median of the data is\({\tilde x_M} = 3.7\)so the lower half contains 7 values.

Since \(n = 7\)is odd, calculate the median using the formula:

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}\tilde x &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {4^{th}}\,ordered\,value\\ &= 3.4\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\({\tilde x_M} = 3.7\)so it contains 6 values.

Since \(n = 6\)is even, calculate the median using the formula:

\(\tilde x = \frac{{\left( {{{\left( {\frac{n}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value} \right)}}{2}\)

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{6}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{3^{th}}ordered\,value + {4^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {3.8 + 3.9} \right)}}{2}\\ &= 3.85\end{aligned}\)

Thus, the five-number summary to construct boxplot are as follows:

Smallest \({x_i}\): 3.1, lower fourth: 3.4, median: 3.7, upper fourth: 3.85,

Largest \({x_i}\): 4.0.

The five-number summary are smallest \({x_i}\), lower fourth, median, upper fourth and largest \({x_i}\).Since sample size of test is even, the median is the average of\({\left( {\frac{n}{2}} \right)^{th}}\)and\({\left( {\frac{n}{2} + 1} \right)^{th}}\) ordered value when the data are in ascending order as below.

The smallest value is: 2.6 and the largest value is: 4.3.

Let \({\tilde x_F}\) be the required median. Use the formula to calculate median,

\(\tilde x = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{{n + 1}}{2}} \right)}^{th}}ordered\,value\,}}{2}\)

\(\begin{aligned}{{\tilde x}_F} &= \frac{{\left( {\left( {\frac{{10}}{2}} \right)\,ordered\,value + {{\left( {\frac{{10}}{2} + 1} \right)}^{th}}\,ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{5^{th}}\,ordered\,value + {6^{th}}\,ordered\,value} \right)}}{2}\\ &= \frac{{3.1 + 3.2}}{2}\\ &= 3.15\end{aligned}\)

Therefore, the median of Female group is 3.15.

The lower fourth is the median of smallest half of the data as the median of the data is \({\tilde x_F} = 3.15\) so the lower half contains 5 values.

Since \(n = 5\)is odd, calculate the median using the formula:

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}\tilde x &= {\left( {\frac{{5 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {3^{rd}}\,ordered\,value\\ &= 3.0\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\({\tilde x_F} = 3.15\) so the upper half also contain 5 values.

Since \(n = 5\)is odd, calculate the median using the formula:

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}\tilde x &= {\left( {\frac{{5 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {3^{rd}}\,ordered\,value\\ &= 3.5\end{aligned}\)

Thus, the five-number summary to construct comparative boxplot are as follows:

Smallest \({x_i}\): 2.6, lower fourth: 3.0, median: 3.15, upper fourth: 3.5,

Largest \({x_i}\): 4.3.

Following are the steps to make comparative boxplot by hand:

1. Draw a plot line of range 8 to 12.
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 Supplement group.
3. Do the Step 2 again for Control group.
4. Draw whiskers on both sides of two boxplots and set the minimum and maximum value with respect to the obtained lower fence and upper fence.

From the comparative boxplot, it is observe that the variability is less for the Male sample than the Female group. From the box plot of Males, there is no outlier and the middle value line is near 3.7 that is the median of data. The distribution is slightly negative skewed. From the box plot of Females, there is no outlier and the middle value line is near 3.15 that is the median of data. The distribution is slightly positive skewed.

Let \(\bar x_{Tr\left( {10} \right)}^M\)be the required trimmed mean. The trimmed value is calculated by multiplying the sample size and given trimming percentage i.e. 10%.

\(13\left( {0.10} \right) = 1.3\)

Since 1.3 is not an integer so eliminate 1 and 2 values from the male sample and then interpolate between the resulted trimmed means. Therefore, the trimmed values are 1 and 2. Ignore one observation for first trimmed mean and two observations for second trimmed means. At last, calculate the mean of remaining data.

The trimming percentages are \(100 \times \frac{1}{{13}} = 7.692\% \)and \(100 \times \frac{2}{{13}} = 15.385\% \).

The remaining data after trimming one value from both sides are as follows:

The remaining data after trimming two value from both sides are as follows:

Compute \(\bar x_{Tr\left( {7.692} \right)}^M\)and\(\bar x_{Tr\left( {15.385} \right)}^M\)using the formula:

\({\bar x_{Tr}} = \frac{{\sum {{x_{Tr}}} }}{{{n_{Tr}}}}\)

Substitute the value of\({n_{Tr\left( {7.692} \right)}} = 11\)and\({n_{Tr\left( {15.385} \right)}} = 9\)in the above formula,

\(\begin{aligned}\bar x_{Tr\left( {7.692} \right)}^M &= \frac{{\sum {{x_{Tr\left( {7.692} \right)}}} }}{{{n_{Tr\left( {7.692} \right)}}}}\\ &= \frac{{\left( {3.4 + 3.4 + \ldots + 3.9 + 4.0} \right)}}{{11}}\\ &= \frac{{40.2}}{{11}}\\ &= 3.65\end{aligned}\)

\(\begin{aligned}\bar x_{Tr\left( {15.385} \right)}^M &= \frac{{\sum {{x_{Tr\left( {15.385} \right)}}} }}{{{n_{Tr\left( {15.385} \right)}}}}\\ &= \frac{{\left( {3.4 + 3.4 + \ldots + 3.8 + 3.9} \right)}}{9}\\ &= \frac{{32.8}}{9}\\ &= 3.64\end{aligned}\)

Compute \(\bar x_{Tr\left( {10} \right)}^M\) by interpolation method:
\(\bar x_{Tr\left( {10} \right)}^M = 0.7\bar x_{Tr\left( {7.692} \right)}^M + 0.3\bar x_{Tr\left( {15.385} \right)}^M\)

Substitute the values of\(\bar x_{Tr\left( {7.692} \right)}^M = 3.65\)and\(\bar x_{Tr\left( {15.385} \right)}^M = 3.64\) in the above formula to obtain the required trimmed mean for male group.

\(\begin{aligned}\bar x_{Tr\left( {10} \right)}^M &= 0.7\left( {3.65} \right) + 0.3\left( {3.64} \right)\\ &= 2.555 + 1.092\\ &= 3.647\end{aligned}\)

Thus, the 10% trimmed mean for Male group is 3.647

Let \(\bar x_{Tr\left( {10} \right)}^F\)be the required trimmed mean. The trimmed value is calculated by multiplying the sample size and given trimming percentage i.e. 10%.

\(10\left( {0.10} \right) = 1\)

Therefore, the trimmed value is 1. Ignore one observation on both sides and calculate the mean of remaining data.

The remaining data are as follows:

Compute the required trimmed mean by using the formula:

\(\bar x_{Tr\left( {10} \right)}^F = \frac{{\sum {{x_{Tr}}} }}{{{n_{Tr}}}}\)

Substitute the value of\({n_{Tr}} = 8\)in the above formula,

\(\begin{aligned}{{\bar x}_{Tr\left( {10} \right)}} &= \frac{{\left( {3 + 3 + \ldots + 3.5 + 3.5} \right)}}{8}\\ &= \frac{{25.9}}{8}\\ &= 3.237\end{aligned}\)

Thus, the 10% trimmed mean for Female group is 3.237.