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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
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Exercise 34 presented the following data on endotoxin

Chapter 2, Problem 48E

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QUESTION:

Exercise 34 presented the following data on endotoxin concentration in settled dust both for a sample of urban homes and for a sample of farm homes: a. ?Determine the value of the sample standard deviation for each sample, interpret these values, and then contrast variability in the two samples. [Hint:?xi=237.?0 ?for the urban sample and =128.4 for the farm sample, and ?x2i = 10,079 for the urban sample and 1617.94 for the farm sample.] b?. ?Compute the fourth spread for each sample and compare. Do the fourth spreads convey the same message about variability that the standard deviations do? Explain. c. ?The authors of the cited article also provided endotoxin concentrations in dust bag dust: Construct a comparative boxplot (as did the cited paper) and compare and contrast the four samples.

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QUESTION:

Exercise 34 presented the following data on endotoxin concentration in settled dust both for a sample of urban homes and for a sample of farm homes: a. ?Determine the value of the sample standard deviation for each sample, interpret these values, and then contrast variability in the two samples. [Hint:?xi=237.?0 ?for the urban sample and =128.4 for the farm sample, and ?x2i = 10,079 for the urban sample and 1617.94 for the farm sample.] b?. ?Compute the fourth spread for each sample and compare. Do the fourth spreads convey the same message about variability that the standard deviations do? Explain. c. ?The authors of the cited article also provided endotoxin concentrations in dust bag dust: Construct a comparative boxplot (as did the cited paper) and compare and contrast the four samples.

ANSWER:

Answer: Step1: The given data on endotoxin concentration in settled dust both for a sample of urban homes and for a sample of farm homes: U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0 F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0 9.2 3.0 2.0 0.3 Step2: a). To determine the value of the sample standard deviation for each sample. U (u u) F (f f) 6 241.4916 4 20.7936 5 273.5716 14 29.5936 11 111.0916 11 5.9536 33 131.3316 9 0.1936 4 307.6516 9 0.1936 5 273.5716 8 0.3136 80 3417.5716 4 20.7936 18 12.5316 20 130.8736 35 181.1716 5 12.6736 17 20.6116 8.9 0.1156 23 2.1316 21 154.7536 9.2 0.4096 3 30.9136 2 43.0336 0.3 68.2276 2 2 u = (u u) = f = (f f) = 237 4972. 128.4 518.836 u f Mean u = n Mean f = n = 128.4 15 = 237 f = 8.56 11 u = 21.5454 2 (uu) (ff) Variance u = Variance f = n 1 n 1 = 4972 = 518.836 111 151 u2 = 497.2 f2 = 37.0598 u = 497.2 f = 37.0598 u = 22.30 f = 6.09 Here we conclude that the variability in endotoxin concentration is far greater in ur

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