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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 Init Se sample:

The following table represent the ordered Init Cont sample:

The sample mean is computed using the formula,

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

Let\({\bar x_{Init\,Se}}\)and\({\bar x_{Init\,Cont}}\)are the tworequired sample means. The sample size of initial selenium concentration for supplement group is 16 while the sample size of initial selenium concentration for control group is 14. The mean of each sample can be determined as follows:

\(\begin{aligned}{{\bar x}_{Init\,Se}} &= \frac{{\sum\limits_{i = 1}^{16} {{x_i}} }}{{16}}\\ &= \frac{{\left( {8.2 + 8.5 + \ldots + 11.4 + 11.8} \right)}}{{16}}\\ &= \frac{{163.7}}{{16}}\\ &= 10.231\end{aligned}\)

\(\begin{aligned}{{\bar x}_{Init\,Cont}} &= \frac{{\sum\limits_{i = 1}^{14} {{x_i}} }}{{14}}\\ &= \frac{{\left( {8.7 + 8.8 + 9.1 + \ldots + 10.9 + 11.6 + 12.3} \right)}}{{14}}\\ &= \frac{{145.2}}{{14}}\\ &= 10.371\end{aligned}\)

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\({s_{Init\,Se}}\)and\({s_{Init\,Cont}}\)are two sample means. Each sample standard deviation can be calculated as,

\(\begin{aligned}{s_{Init\,Se}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{16} {{{\left( {{x_i} - 10.231} \right)}^2}} }}{{16 - 1}}} \\ &= \sqrt {\frac{{{{\left( {8.2 - 10.231} \right)}^2} + \ldots + {{\left( {11.8 - 10.231} \right)}^2}}}{{16 - 1}}} \\ &= 1.0023\end{aligned}\)

\(\begin{aligned}{s_{Init\,Cont}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{14} {{{\left( {{x_i} - 10.231} \right)}^2}} }}{{14 - 1}}} \\ &= \sqrt {\frac{{{{\left( {8.7 - 10.231} \right)}^2} + \ldots + {{\left( {12.3 - 10.231} \right)}^2}}}{{14 - 1}}} \\ &= 1.03\end{aligned}\)

The five-number summary are smallest\({x_i}\), lower fourth, median, upper fourth and largest\({x_i}\).Since sample size of test is 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: 8.2 and the largest value is: 11.8.

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

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

\(\begin{aligned}{{\tilde x}_{Init\,Se}} &= \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\\ &= \frac{{{{\left( {\frac{{16}}{2}} \right)}^{th}} + {{\left( {\frac{{16}}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {{8^{th\,}}observation + {9^{th}}observation} \right)}}{2}\\ &= \frac{{\left( {10.3 + 10.3} \right)}}{2}\\ &= 10.3\end{aligned}\)

Therefore, the median of supplement group is 10.3

The lower fourth is the median of smallest half of the data as the median of the data is \(\tilde x = 10.3\) so the lower half contains 8 values.

Since \(n = 8\)is even, calculate the median using the formula:

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{8}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{8}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{4^{th}}ordered\,value + {5^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {9.6 + 9.8} \right)}}{2}\\ &= 9.7\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is \(\tilde x = 10.3\) so it contains 8 values.

Since \(n = 8\)is even, calculate the median using the formula:

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{8}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{8}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{4^{th}}ordered\,value + {5^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {10.8 + 10.9} \right)}}{2}\\ &= 10.85\end{aligned}\)

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

smallest \({x_i}\): 8.2, lower fourth: 9.7, median: 10.3, upper fourth: 10.85,

largest \({x_i}\): 11.8.

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 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: 8.7 and the largest value is: 12.3.

Let \({\tilde x_{Init\,Cont}}\) be the required median. Use the formula to calculate median when the sample size is even,

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{{14}}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{{14}}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{7^{th}}ordered\,value + {8^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {10.4 + 10.6} \right)}}{2}\\ &= 10.5\end{aligned}\)

Therefore, the median of control group is 10.5

The lower fourth is the median of smallest half of the data as the median of the data is \(\tilde x = 10.5\) so the lower half contains 7 values.

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

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

\(\begin{aligned}\tilde x &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}ordered\,value\\ &= {4^{th\,}}ordered\,value\\ &= 9.7\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 10.5\)so the upper half contains 7 values.

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

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

\(\begin{aligned}{{\tilde x}_{Init\,Cont}} &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}ordered\,value\\ &= {4^{th\,}}ordered\,value\\ &= 10.9\end{aligned}\)

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

Smallest \({x_i}\): 8.7, lower fourth: 9.7, median: 10.5, upper fourth: 10.9,

Larges t\({x_i}\): 12.3.

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 both boxplots are quite similar. Both data distributionare skewed to the left or negatively skewed.

As a result, the selenium and control group samples appear to be similar.

The following table represent the ordered Final Se sample:

The following table represent the ordered FinalCont sample:

The sample mean is computed using the formula,

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

Let\({\bar x_{Final\,Se}}\)and\({\bar x_{Final\,Cont}}\)are the two required sample means. The sample size of final selenium concentration for supplement group is 16 while the sample size of final selenium concentration for control group is 14. The mean of each sample can be determined as follows:

\(\begin{aligned}{{\bar x}_{Final\,Se}} &= \frac{{\sum\limits_{i = 1}^{16} {{x_i}} }}{{16}}\\ &= \frac{{\left( {88 + 89 + 93 + \ldots + 146.3 + 147.3 + 172.6} \right)}}{{16}}\\ &= \frac{{1839.4}}{{16}}\\ &= 114.96\end{aligned}\)

\(\begin{aligned}{{\bar x}_{Final\,Cont}} &= \frac{{\sum\limits_{i = 1}^{14} {{x_i}} }}{{14}}\\ &= \frac{{\left( {8.4 + 8.6 + 8.7 + \ldots + 10.4 + 12.4 + 12.5} \right)}}{{14}}\\ &= \frac{{135.5}}{{14}}\\ &= 9.678\end{aligned}\)

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 \({s_{Final\,Se}}\) and \({s_{Final\,Cont}}\) are two sample standard deviations. Each sample standard deviation can be calculated as,

\(\begin{aligned}{s_{Final\,Se}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{16} {{{\left( {{x_i} - 114.96} \right)}^2}} }}{{16 - 1}}} \\ &= \sqrt {\frac{{{{\left( {88 - 114.96} \right)}^2} + \ldots + {{\left( {172.6 - 114.96} \right)}^2}}}{{16 - 1}}} \\ &= 24.75\end{aligned}\)

\(\begin{aligned}{s_{Init\,Cont}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{14} {{{\left( {{x_i} - 9.678} \right)}^2}} }}{{14 - 1}}} \\ &= \sqrt {\frac{{{{\left( {8.4 - 9.678} \right)}^2} + \ldots + {{\left( {12.5 - 9.678} \right)}^2}}}{{14 - 1}}} \\ &= 1.30\end{aligned}\)

The five-number summary are smallest \({x_i}\), lower fourth, median, upper fourth and largest \({x_i}\). Since sample size of test is 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: 88 and the largest value is: 172.6.

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

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

\(\begin{aligned}{{\tilde x}_{Final\,Se}} &= \frac{{{{\left( {\frac{{16}}{2}} \right)}^{th}} + {{\left( {\frac{{16}}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {{8^{th\,}}observation + {9^{th}}observation} \right)}}{2}\\ &= \frac{{\left( {103.8 + 104} \right)}}{2}\\ &= 103.9\end{aligned}\)

Therefore, the median of supplement group is 103.9

The lower fourth is the median of smallest half of the data as the median of the data is\({\tilde x_{Final\,Se}} = 103.9\)so the lower half contains 8 values.

Since\(n = 8\)is even, calculate the median using the formula:

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{8}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{8}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{4^{th}}ordered\,value + {5^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {96.4 + 97.1} \right)}}{2}\\ &= 96.75\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\({\tilde x_{Final\,Se}} = 103.9\)so it contains 8 values.

Since\(n = 8\)is even, calculate the median using the formula:

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{8}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{8}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{4^{th}}ordered\,value + {5^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {122.3 + 138.3} \right)}}{2}\\ &= 130.3\end{aligned}\)

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

Smallest\({x_i}\): 88.0, lower fourth: 96.75, median: 103.9, upper fourth: 130.3,

Largest \({x_i}\): 172.6.

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 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: 8.4 and the largest value is: 12.5.

Let\({\tilde x_{Final\,Cont}}\)be the required median. Use the formula to calculate median when the sample size is even,

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

\(\begin{aligned}{{\tilde x}_{Final\,Cont}} &= \frac{{\left( {{{\left( {\frac{{14}}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{{14}}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{7^{th}}ordered\,value + {8^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {9.3 + 9.3} \right)}}{2}\\ &= 9.3\end{aligned}\)

Therefore, the median of control group is 9.3.

The lower fourth is the median of smallest half of the data as the median of the data is\({\tilde x_{Final\,Cont}} = 9.3\)so the lower half contains 7 values.

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

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

\(\begin{aligned}\tilde x &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}ordered\,value\\ &= {4^{th\,}}ordered\,value\\ &= 8.8\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 10.5\)so the upper half contains 7 values.

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

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

\(\begin{aligned}\tilde x &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}ordered\,value\\ &= {4^{th\,}}ordered\,value\\ &= 10.1\end{aligned}\)

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

Smallest\({x_i}\): 8.4, lower fourth: 8.8, median: 9.3, upper fourth: 10.1,

Largest \({x_i}\): 12.5.

Following are the steps to make comparative boxplot by hand:

1. Draw a plot line of range 0 to 180.
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 both boxplots are quite different. From the box plot of Final Se, there is no outlier and the middle value line is near 103.9 that is the median of data. The distribution is slightly negative skewed.

From the box plot of Final Cont, there is no outlier and the middle value line is near 9.3 that is the median of data. The distribution is slightly positive skewed. It is evident that the final selenium concentration for cows is much greater for the supplement group than for control group which indicate that the selenium supplement significantly increases selenium concentration in milk over the 9-day period.