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Poly(3-hydroxybutyrate) (PHB), a semicrystalline polymer that is fully biodegradable and biocompatible, is obtained from renewable resources. From a sustainability perspective, PHB offers many attractive properties though it is more expensive to produce than standard plastics. The accompanying data on melting point \(\left({ }^{\circ} \mathrm{C}\right)\) for each of 12 specimens of the polymer using a differential scanning calorimeter appeared in the article "The Melting Behaviour of Poly(3-Hydroxybutyrate) by DSC. Reproducibility Study" (Polymer Testing, 2013: 215-220). \(\begin{array}{llllll}180.5 & 181.7 & 180.9 & 181.6 & 182.6 & 181.6 \\ 181.3 & 182.1 & 182.1 & 180.3 & 181.7 & 180.5\end{array}\) Compute the following: a. The sample range b. The sample variance \(s^{2}\) from the definition [Hint: First subtract 180 from each observation.] c. The sample standard deviation d. \(s^{2}\) using the shortcut method

Short Answer

Expert verified
a. Range: 2.3 b. Variance: 0.471 c. Standard deviation: 0.686 For d, variance confirmed: 0.471

Step by step solution

01

Compute the Sample Range

The range of a set of data is the difference between the maximum and minimum values. First, identify the smallest observation (180.3) and the largest observation (182.6) within the data set. Then, calculate the range:\[\text{Range} = 182.6 - 180.3 = 2.3\]
02

Compute Sample Variance Using the Definition

Variance measures how much the data points differ from the mean on average. Follow the hint by subtracting 180 from each observation to simplify calculations. The adjusted data set is:\[0.5, 1.7, 0.9, 1.6, 2.6, 1.6, 1.3, 2.1, 2.1, 0.3, 1.7, 0.5\]Calculate the mean of these adjusted data points:\[\text{Mean} = \frac{(0.5 + 1.7 + 0.9 + 1.6 + 2.6 + 1.6 + 1.3 + 2.1 + 2.1 + 0.3 + 1.7 + 0.5)}{12} \approx 1.3667\]Now, calculate the squared deviations from the mean, sum these squared deviations, and then divide by \(n - 1\) (which is 11 for this sample of 12):\[\sum (x_i - \bar{x})^2 = (0.5 - 1.3667)^2 +(1.7 - 1.3667)^2 + \ldots + (0.5 - 1.3667)^2 \approx 5.18\]Variance:\[s^2 = \frac{5.18}{11} \approx 0.471\]
03

Compute the Sample Standard Deviation

The sample standard deviation is the square root of the sample variance. Since we found the variance \(s^2\) to be approximately 0.471, the standard deviation is:\[s = \sqrt{0.471} \approx 0.686\]
04

Use Shortcut Method for Variance

The shortcut formula for variance is:\[s^2 = \frac{n(\sum{x^2}) - (\sum{x})^2}{n(n-1)}\] Where \(x\) are the adjusted values (after subtracting 180). Calculate \(\sum{x^2}\) and \(\sum{x}\):\[\sum{x^2} = 0.5^2 + 1.7^2 + \ldots + 0.5^2 = 27.38\]\[\sum{x} = 0.5 + 1.7 + 0.9 + 1.6 + 2.6 + 1.6 + 1.3 + 2.1 + 2.1 + 0.3 + 1.7 + 0.5 = 16.4\]Plug these into the shortcut formula:\[s^2 = \frac{12(27.38) - (16.4)^2}{12(11)} = \frac{328.56 - 268.96}{132} \approx 0.471\]This confirms the variance calculated using the definition method.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Range
The sample range is a statistical measure that gives us an idea of the spread or dispersion in a data set. To calculate the sample range, you simply subtract the smallest value from the largest value in the data set. This range tells us how far apart the data points are from each other. In the context of the exercise we're looking at, which involves measurements of melting points of a biodegradable polymer, calculating the range helps to understand the variability in the melting temperatures across different samples.
The formula for the range is straightforward:
  • Identify the smallest observation in the data set.
  • Identify the largest observation in the data set.
  • Subtract the smallest observation from the largest observation to get the range.

In our given data, the smallest observation was 180.3°C and the largest was 182.6°C. Therefore, the sample range was calculated as:\[ \text{Range} = 182.6 - 180.3 = 2.3 \text{°C} \]
This tells us that the melting point of the polymer samples varied by 2.3°C.
Sample Standard Deviation
Sample standard deviation is a more refined measure of dispersion, building on the concept of variance. It tells us, on average, how far each data point is from the mean of the data set. Whereas variance gives us the squared differences, the standard deviation brings this back to the same units as the data by taking the square root.
The process to calculate it involves several steps:
  • Calculate the mean (average) of the data set.
  • Subtract the mean from each data point to find the deviations.
  • Square each deviation.
  • Find the average of these squared deviations (this is the variance).
  • Take the square root of the variance to get the standard deviation.

In our example, the sample variance was about 0.471, so we took its square root to find the sample standard deviation:\[ s = \sqrt{0.471} \approx 0.686 \]
This standard deviation measures how much the melting point temperatures vary around the average. It's particularly useful in contexts like quality control, where maintaining consistent product properties is crucial. Understanding how sample standard deviation works improves our ability to analyze and interpret data effectively, highlighting significantly different observations.
Biodegradable Polymers
Biodegradable polymers, like poly(3-hydroxybutyrate) (PHB), are materials that can break down naturally in the environment through the action of microorganisms. Unlike traditional plastics, they do not persist in the environment for long periods but instead decompose into natural substances like water, carbon dioxide, and biomass. This property makes them highly valuable from a sustainability perspective.
PHB, as mentioned in the problem, is derived from renewable resources and has characteristics such as biocompatibility and semi-crystallinity, which makes it suitable for various applications. However, despite these advantages, a significant downside is its cost of production. These materials are often more expensive to produce than conventional plastics, making widespread adoption challenging.
Some key characteristics of biodegradable polymers like PHB include:
  • Environmental Friendliness: They reduce plastic waste and are better for the environment.
  • Renewability: Often obtained from renewable resources, promoting sustainable usage.
  • Decomposition: They break down into harmless substances, which minimize environmental impact.

Research and advancements in this field aim to overcome the cost barrier, enhancing both the production processes and material properties. Understanding biodegradable polymers is crucial as we move towards a more sustainable future, focusing on reducing environmental pollution and conserving non-renewable resources.

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Most popular questions from this chapter

The accompanying summary data on \(\mathrm{CeO}_{2}\) particle sizes \((\mathrm{nm}\) ) under certain experimental conditions was read from a graph in the article "NanoceriaEnergetics of Surfaces, Interfaces and Water Adsorption" (J. of the Amer. Ceramic Soc., 2011: 3992-3999): \(\begin{array}{ccccc}3.0-<3.5 & 3.5-<4.0 & 4.0-<4.5 & 4.5-<5.0 & 5.0-5.5 \\ 5 & 15 & 27 & 34 & 22\end{array}\) \(\begin{array}{ccccc}5.5-<6.0 & 6.0-<6.5 & 6.5-<7.0 & 7.0-<7.5 & 7.5-<8.0 \\\ 14 & 7 & 2 & 4 & 1\end{array}\) a. What proportion of the observations are less than 5 ? b. What proportion of the observations are at least 6 ? c. Construct a histogram with relative frequency on the vertical axis and comment on interesting features. In particular, does the distribution of particle sizes appear to be reasonably symmetric or somewhat skewed? [Note: The investigators fit a lognormal distribution to the data; this is discussed in Chapter 4.] d. Construct a histogram with density on the vertical axis and compare to the histogram in (c).

The article "A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankease Lubricants" (Lubric. Engr., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145\end{array}\) \(\begin{array}{llllllll}153 & 152 & 138 & 87 & 99 & 93 & 119 & 129\end{array}\) a. Calculate the sample variance and standard deviation. b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

Observations on burst strength \(\left(\mathrm{lb} / \mathrm{in}^{2}\right)\) were obtained both for test nozzle closure welds and for production cannister nozzle welds (4proper Procedures Are the Key to Welding Radioactive Waste Cannisters, Welding \(J .\) Aug. 1997: 61-67). \(\begin{array}{lllllll}\text { Test } & 7200 & 6100 & 7300 & 7300 & 8000 & 7400 \\ & 7300 & 7300 & 8000 & 6700 & 8300 & \\ \text { Cannister } & 5250 & 5625 & 5900 & 5900 & 5700 & 6050 \\ & 5800 & 6000 & 5875 & 6100 & 5850 & 6600\end{array}\) Construct a comparative boxplot and comment on interesting features (the cited article did not include such a picture, but the authors commented that they had looked at one).

a. If a constant \(c\) is added to each \(x_{i}\) in a sample, yielding \(y_{i}=x_{i}+c\), how do the sample mean and median of the \(y_{i} \mathrm{~s}\) relate to the mean and median of the \(x_{i} \mathrm{~s}\) ? Verify your conjectures. b. If each \(x_{i}\) is multiplied by a constant \(c\), yielding \(y_{i}=c x_{p}\), answer the question of part (a). Again, verify your conjectures.

The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of \(n=129\) houses in Perth, Australia ("An Application of Bayes Methodology to the Analysis of Diary Records in a Water Use Study," J. Amer. Stat. Assoc., 1987: 705-711): \(\begin{array}{rrrrrrrrrr}4.6 & 12.3 & 7.1 & 7.0 & 4.0 & 9.2 & 6.7 & 6.9 & 11.5 & 5.1 \\ 11.2 & 10.5 & 14.3 & 8.0 & 8.8 & 6.4 & 5.1 & 5.6 & 9.6 & 7.5 \\\ 7.5 & 6.2 & 5.8 & 2.3 & 3.4 & 10.4 & 9.8 & 6.6 & 3.7 & 6.4 \\ 8.3 & 6.5 & 7.6 & 9.3 & 9.2 & 7.3 & 5.0 & 6.3 & 13.8 & 6.2 \\ 5.4 & 4.8 & 7.5 & 6.0 & 6.9 & 10.8 & 7.5 & 6.6 & 5.0 & 3.3 \\ 7.6 & 3.9 & 11.9 & 2.2 & 15.0 & 7.2 & 6.1 & 15.3 & 18.9 & 7.2 \\ 5.4 & 5.5 & 4.3 & 9.0 & 12.7 & 11.3 & 7.4 & 5.0 & 3.5 & 8.2 \\ 8.4 & 7.3 & 10.3 & 11.9 & 6.0 & 5.6 & 9.5 & 9.3 & 10.4 & 9.7 \\ 5.1 & 6.7 & 10.2 & 6.2 & 8.4 & 7.0 & 4.8 & 5.6 & 10.5 & 14.6 \\ 10.8 & 15.5 & 7.5 & 6.4 & 3.4 & 5.5 & 6.6 & 5.9 & 15.0 & 9.6 \\ 7.8 & 7.0 & 6.9 & 4.1 & 3.6 & 11.9 & 3.7 & 5.7 & 6.8 & 11.3 \\ 9.3 & 9.6 & 10.4 & 9.3 & 6.9 & 9.8 & 9.1 & 10.6 & 4.5 & 6.2 \\ 8.3 & 3.2 & 4.9 & 5.0 & 6.0 & 8.2 & 6.3 & 3.8 & 6.0 & \end{array}\) a. Construct a stem-and-leaf display of the data. b. What is a typical, or representative, flow rate? c. Does the display appear to be highly concentrated or spread out? d. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry? e. Would you describe any observation as being far from the rest of the data (an outlier)?

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