91影视

The data are provided that consists of 11 observations on burst strength for test nozzle closure welds and 12 observations on burst strength for production cannister nozzle welds.

The five-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: 6100 and the largest value is: 8300.

Let\(\tilde x\)be the required median. Use the formula to calculate median,

\(\begin{aligned}\tilde x &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {\left( {\frac{{11 + 1}}{2}} \right)^{th}}\\ &= {6^{th\,}}observation\\ &= 7300\end{aligned}\)

The lower fourth is the median of smallest half of the data as the median of the data is\(\tilde x = 7300\)so lower half contains 7 values.

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

\(\begin{aligned}\tilde x &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {4^{th\,}}observation\\ &= 7300\end{aligned}\)

Since 7300 is the median value, so the lower fourth becomes 7200.

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 7300\)so it contains 4 values.

Since\(n = 4\)is even, calculate the median using the formula:

\(\begin{aligned}\tilde x &= \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\\ &= \frac{{{{\left( {\frac{4}{2}} \right)}^{th}} + {{\left( {\frac{4}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {8000 + 8000} \right)}}{2}\\ &= 8000\end{aligned}\)

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

Smallest\({x_i}\): 6100, lower fourth: 7200, median: 7300, upper fourth: 8000,

Largest \({x_i}\): 8300.

The five-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: 5250 and the largest value is: 6600.

Let\(\tilde x\)be the required median. Use the formula to calculate median,

\(\begin{aligned}\tilde x &= \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\\ &= \frac{{{{\left( {\frac{{12}}{2}} \right)}^{th}} + {{\left( {\frac{{12}}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{{6^{th\,}}observation + {7^{th}}observation}}{2}\\ &= \frac{{5875 + 5900}}{2}\\ &= 5887.5\end{aligned}\)

The lower fourth is the median of smallest half of the data as the median of the data is\(\tilde x = 5887.5\)so each half contains 6 values.

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

\(\begin{aligned}\tilde x &= \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\\ &= \frac{{{{\left( {\frac{6}{2}} \right)}^{th}} + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {5700 + 5800} \right)}}{2}\\ &= 5750\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 5887.5\)so it contains 6 values.

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

\(\begin{aligned}\tilde x &= \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\\ &= \frac{{{{\left( {\frac{6}{2}} \right)}^{th}} + {{\left( {\frac{6}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {6000 + 6050} \right)}}{2}\\ &= 6025\end{aligned}\)

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

Smallest\({x_i}\): 5250, lower fourth: 5750, median: 5887.5, upper fourth: 6025,

Largest \({x_i}\): 6600.

Following are the steps to make comparative boxplot by hand:

1. Draw a plot line of range 5000 to 8500.
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 Test nozzle closure welds.
3. Do the Step 2 again for Cannister nozzle welds.
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.
5. Mark the two outliers of Cannister nozzle welds.

From the comparative boxplot, one can see that both have different burst strengths. The distribution of cannister nozzle welds is more consistent and have little skewness to the right. The distribution of Test nozzle welds, on the other hand, is skewed to the right, resulting in more variability.