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The data are provided that consists of 13 observations each on receptor binding measure for samples of individuals suffering from PTSD and a sample of healthy individuals

The sample mean is computed using the formula,

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

Let\({\bar x_{PTSD}}\)and\({\bar x_{Healthy}}\)are the two required sample means. The sample size of receptor binding measure for PTSD group is 13 while the sample size of receptor binding measure for Healthy group is 13. The mean of each sample can be determined as follows:

\(\begin{aligned}{{\bar x}_{PTSD}} &= \frac{{\sum\limits_{i = 1}^{13} {{x_i}} }}{{13}}\\ &= \frac{{\left( {10 + 20 + \ldots + 42 + 46} \right)}}{{13}}\\ &= \frac{{428}}{{13}}\\ &= 32.92\end{aligned}\)

\(\begin{aligned}{{\bar x}_{Healthy}} &= \frac{{\sum\limits_{i = 1}^{13} {{x_i}} }}{{13}}\\ &= \frac{{\left( {23 + 39 + \ldots + 69 + 72} \right)}}{{13}}\\ &= \frac{{679}}{{13}}\\ &= 52.23\end{aligned}\)

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

\(\begin{aligned}{s_{PTSD}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{13} {{{\left( {{x_i} - 32.92} \right)}^2}} }}{{13 - 1}}} \\ &= \sqrt {\frac{{{{\left( {10 - 32.92} \right)}^2} + \ldots + {{\left( {46 - 32.92} \right)}^2}}}{{13 - 1}}} \\ &= 9.93\end{aligned}\)

\(\begin{aligned}{s_{Healthy}} &= \sqrt {\frac{{\sum\limits_{n = 1}^{13} {{{\left( {{x_i} - 52.23} \right)}^2}} }}{{13 - 1}}} \\ &= \sqrt {\frac{{{{\left( {23 - 52.23} \right)}^2} + \ldots + {{\left( {72 - 52.23} \right)}^2}}}{{13 - 1}}} \\ &= 14.86\end{aligned}\)

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\({\left( {\frac{{n + 1}}{2}} \right)^{th}}\)ordered value when the data are in ascending order as below.

The smallest value is: 10 and the largest value is: 46.

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

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

\(\begin{aligned}{{\tilde x}_{PTSD}} &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {\left( {\frac{{13 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {7^{th}}\,observation\\ &= 37\end{aligned}\)

Therefore, the median of PTSD group is 37.

The lower fourth is the median of smallest half of the data as the median of the data is\({\tilde x_{PTSD}} = 37\)so the lower half contains 6 values.

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

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{6}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{3^{th}}ordered\,value + {4^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {25 + 28} \right)}}{2}\\ &= 26.50\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 37\)so it contains 6 values.

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

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{6}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{3^{th}}ordered\,value + {4^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {39 + 39} \right)}}{2}\\ &= 39\end{aligned}\)

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

smallest\({x_i}\): 10, lower fourth: 26.5, median: 37, upper fourth: 39,

largest\({x_i}\): 46.

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: 23 and the largest value is: 72.

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

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

\(\begin{aligned}{{\tilde x}_{Healthy}} &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {\left( {\frac{{13 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {7^{th}}\,observation\\ &= 51\end{aligned}\)

Therefore, the median of Healthy group is 51.

The lower fourth is the median of smallest half of the data as the median of the data is\({\tilde x_{Healthy}} = 51\)so the lower half contains 6 values.

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

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{6}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{3^{th}}ordered\,value + {4^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {40 + 41} \right)}}{2}\\ &= 40.50\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 51\)so it contains 6 values.

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

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

\(\begin{aligned}\tilde x &= \frac{{\left( {{{\left( {\frac{6}{2}} \right)}^{th}}ordered\,value + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {{3^{th}}ordered\,value + {4^{th}}ordered\,value} \right)}}{2}\\ &= \frac{{\left( {66 + 67} \right)}}{2}\\ &= 66.50\end{aligned}\)

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

smallest\({x_i}\): 23, lower fourth: 40.5, median: 51, upper fourth: 66.5,

largest\({x_i}\): 72.

Following are the steps to make comparative boxplot by hand:

1. Draw a plot line of range 10 to 80.
2. Draw three vertical lines that consists of first quartile, second quartile and third quartile and make two horizontal lines to make it in rectangular form like a box for Supplement group.
3. Do the Step 2 again for Healthy 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. The data distribution of PTSD group is skewed to the left or negatively skewed,while; it is right-skewed for the Healthy group.

The variability is greater in the PTSD group as compared to Healthy group.