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Magazine Chapter 1: Q16E (page 25)
The article cited in Example 1.2 also gave the accompanying strength observations for cylinders:
6.1 | 5.8 | 7.8 | 7.1 | 7.2 | 9.2 | 6.6 | 8.3 | 7.0 | 8.3 |
7.8 | 8.1 | 7.4 | 8.5 | 8.9 | 9.8 | 9.7 | 14.1 | 12.6 | 11.2 |
a. Construct a comparative stem-and-leaf display(see the previous exercise) of the beam and cylinder data, and then answer the questions in parts(b)鈥�(d) of Exercise 10 for the observations oncylinders.
b. In what ways are the two sides of the display similar? Are there any obvious differences between the beam observations and the cylinder observations?
c. Construct a dotplot of the cylinder data.
Short Answer
a. A comparative stem and leaf display of the beam and cylinder data is
No. Both are rightly skewed.
Yes.
The proportion of strength that exceeds 10 MPa is 0.15.
b.Both are right-skewed. The cylinder observations are more spread out as compare to beam observations.
c.
Step by step solution
Given information
The data are provided in which there are 27 beam observations (data given in Example 1.2) and 20-cylinder observations.
Construct comparative stem-and-leaf plot for beam and cylinder data.
a.
Following are the steps to construct acomparative stem and leaf plot:
1. Arrange the given data in order.
2. Construct the stem and leaf display by using the same digits as stems.
3. List possible stem values vertically and record the leaf for each observation for every stem value.
4. Place the digits for leaves of cylinder on the right side and the digits of leaves of beam on the left side of the stem.
Answer to the problem 10-(b)
No, the data does not appear to be symmetric. In both cases, it appears that the data for beam and cylinder strengths are rightly skewed. It seems that the value 14.1 is an outlier.
Answer to the problem 10-(c)
From the plot, it is observed that most of the data lie between 6 and 9 for both the beam and cylinder. For cylinder data, there are three outliers i.e. 11.2, 12.6 and 14.1. It is due to the fact that the data for cylinder is more spread as compared to beam data.
Answer to the problem 10-(d)
In the given sample of 20-cylinder observations, there are 3 outliers that exceeds 10 MPa so, the proportion is calculated by dividing 3 from 20 using the formula:
\(P = \frac{{Number\,of\,observations\,that\,exceeds\,10\,{\rm{MPa}}}}{{Total\,number\,of\,observations}}\)
\(\begin{aligned}P &= \frac{3}{{20}}\\ &= 0.15\end{aligned}\)
Thus, the proportion of strength that exceeds 10 MPa is 15% or 0.15.
Given information
b.The data are provided in which there are total of 20 strength observations for cylinders.
6.1 | 5.8 | 7.8 | 7.1 | 7.2 | 9.2 | 6.6 | 8.3 | 7.0 | 8.3 |
7.8 | 8.1 | 7.4 | 8.5 | 8.9 | 9.8 | 9.7 | 14.1 | 12.6 | 11.2 |
State the reason
b.
The shape of the distribution of data for both beam and cylinder strengths are right-skewed.
The data for cylinder are more spread than the beam data as observed by the comparative stem and plot.
Given information
The data are provided in which there are total of 20 strength observations for cylinders.
6.1 | 5.8 | 7.8 | 7.1 | 7.2 | 9.2 | 6.6 | 8.3 | 7.0 | 8.3 |
7.8 | 8.1 | 7.4 | 8.5 | 8.9 | 9.8 | 9.7 | 14.1 | 12.6 | 11.2 |
Construct dot-plot of Cylinder Observations.
c.
Following are the steps to construct a dot-plot of cylinder observations:
1. Open Minitab and enter the given data into the worksheet.
2. Choose Graph and select 鈥淒ot-plot鈥�.
3. Choose the Simple dot plot from the list of One Y and then click 鈥淥k鈥�.
4. Double click on Cylinder observations to specify it in the graph variables box and click 鈥淥k鈥�.
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