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applied statistics and probability for engineers
Applied Statistics For Engineers And Scientists 3rd Edition Jay L. Devore, Nicholas R. Farnum, Jimmy A. Doi - Solutions
=+c. What is the probability that exactly one of the xi’s is less than ?d. What is P( , y2), where y2 denotes the second smallest xi
=+b. What is the probability that only x2 is smaller than the median and all other n21 observations exceed the median?
=+84. Consider the situation described in Exercise 83.a. What is P(x1 , , x2 . , x3 . ,…, xn . ), that is, the probability that only the first observation is smaller than the median and all others exceed the median?
=+g. Assuming that the data in part (f) was selected from a normal distribution (is this assumption justified?), calculate a confidence interval for (which for a normal distribution is identical to )using the same confidence level as in part (f), and compare the two intervals.
=+f. An experiment carried out to study the curing time (hr) for a particular experimental adhesive yielded the following observations:31.2 36.0 31.5 28.7 37.2 35.4 33.3 39.3 42.0 29.9 Referring back to part (e), determine the confidence interval and the associated confidence level.
=+d. With y1 5 min {x1,…, xn}, what is P( , y1)?e. Using the results of parts (c) and (d), what is P(y1 , , yn)? Regarding (y1, yn) as a confidence interval for , what is the associated confidence level?
=+c. Let yn 5 max {x1,…, xn}. What is P(yn , )?Hint: The condition that yn is less than is equivalent to what about x1,…, x2?
=+b. What is the probability that both the first and the second observations are smaller than the median?
=+a. What is P(x1 , ), the probability that the first observation is smaller than the median?
=+83. Suppose an investigator wants a confidence interval for the median of a continuous distribution based on a random sample x1,…, xn without assuming anything about the shape of the distribution.
=+82. It is important that face masks used by firefighters be able to withstand high temperatures. In a test of one type of mask, the lenses in 11 of the 35 masks popped out at a temperature of 250°F. Calculate a lower confidence bound for the proportion of all such masks whose lenses would pop
=+b. Estimate the true average difference in peak ER and IR velocities in a way that conveys information about reliability and precision.Interpret the resulting estimate.
=+a. Is it plausible that the differences came from a normally distributed population?
=+were determined for a sample of 15 female collegiate golfers during their swings. The following data was supplied by the article’s authors:Golfer ER IR diff z quan 1 2130.6 298.9 231.7 21.28 2 2125.1 2115.9 29.2 20.97 3 251.7 2161.6 109.9 0.34 4 2179.7 2196.9 17.2 20.73 5 2130.5 2170.7 40.2
=+81. Torsion during hip external rotation (ER) and extension may be responsible for certain kinds of injuries in golfers and other athletes. The article “Hip Rotational Velocities During the Full Golf Swing”(J. of Sports Sci. and Med., 2009: 296–299) reported on a study in which peak ER
=+c. Does the upper limit of the interval in part (a)specify a 95% upper confidence bound for the proportion being estimated? Explain.
=+b. What sample size is required if the desired width of the 95% CI is to be at most .04, irrespective of the sample results?
=+a. Calculate and interpret a 95% CI for the proportion of all American households in 2010 that owed student loan debt.
=+a record 19% owed student loan debt in 2010 (a sharp increase from the 15% that owed such debt in 2007).
=+80. As reported by the Pew Research Center’s Social and Demographic Trends Project in September 2012, a survey of 6500 American households revealed that
=+true average abrasion resistances for the two types of fabric. Does your interval provide convincing evidence that true average resistances differ for the two types of fabric? Why or why not?
=+in the filling direction of the yarn was 3975.0, with a sample standard deviation of 245.1. Another sample of 40 softener-plus specimens gave a sample mean and sample standard deviation of 2795.0 and 293.7, respectively. Calculate a confidence interval with confidence level 99% for the difference
=+611). One particularly important characteristic of fabric is its durability, that is, its ability to resist wear.For a sample of 40 softener-only specimens, the sample mean stoll-flex abrasion resistance (cycles)
=+79. An experiment was carried out to compare various properties of cotton/polyester spun yarn finished with softener only and yarn finished with softener plus 5% DP-resin (“Properties of a Fabric Made with Tandem Spun Yarns,” Textile Res. J., 1996: 607–
=+b. Estimate the difference between true average breaking force in a dry medium at 37° and true average force at the same temperature in a wet medium, and do so in a way that conveys information about precision and reliability. Then interpret your estimate.
=+a. Estimate true average breaking force in a dry medium at 37° in a way that conveys information about reliability and precision. Interpret your estimate.
=+bone structure. The paper “Validation of the SmallPunch Test as a Technique for Characterizing the Mechanical Properties of Acrylic Bone Cement”(J. of Engr. In Med., 2006: 11–21) gave the following data on breaking force (N):Temp Medium n x s 37° Dry 6 325.73 34.97 37° Wet 6 306.09
=+78. Acrylic bone cement is commonly used in total joint arthroplasty as a grout that allows for the smooth transfer of loads from a metal prosthesis to
=+deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.348 chapter 7 Estimation and Statistical Intervals thenar muscle volume (mm3)
=+77. The article “Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population” (Ergonomics, 1997: 642–649) reported that (2473.3, 1691.9) was a large-sample 95% confidence interval for the difference between true average Copyright 2013 Cengage
=+d. Estimate the difference between the true average measures in a way that conveys information about reliability and precision.
=+c. Predict the adjusted distribution volume of a single healthy individual by calculating a 95% prediction interval. How does this interval’s width compare to the width of the interval calculated in part (b)?
=+b. Calculate an interval for which you can be 95% confident that at least 95% of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval.
=+a. Is it plausible that the population distributions from which these samples were selected are normal?
=+a decrease in benzodiazepine receptor binding in the frontal cortex. The paper “Decreased Benzodiazepine Receptor Binding in Prefrontal Cortex in Combat-Related Posttraumatic Stress Disorder”(Amer. J. of Psychiatry, 2000: 1120–1126) described the first study of benzodiazepine receptor
=+76. Anxiety disorders and symptoms can often be effectively treated with benzodiazepine medications.It is known that animals exposed to stress exhibit
=+d. Calculate a lower prediction bound with a prediction level of 95% for the ultimate tensile strength of the next specimen selected.
=+c. Use a statistical software package to investigate the plausibility of a normal population distribution.
=+b. Is any assumption about the tensile strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.
=+a. Obtain a lower confidence bound for population mean strength. Does the validity of the bound require any assumptions about the population distribution? Explain.
=+75. Exercise 4 of Chapter 1 presented a sample of n =153 observations on ultimate tensile strength.
=+74. In Example 1.8 (Chapter 1), a histogram was fit to the energy consumption data (in BTUs) from a sample of 90 homes. Using this data, experiment with different values of until you find a value that gives a kernel density estimate that approximates the shape of the histogram of this data
=+73. A kernel function is fit to the data in a sample of size n. Later, a researcher realizes that the largest observation in the sample was actually a typographical error and, because the original lab data no longer exists, this data point is removed from the sample, leaving a sample of n21
=+72. Refer to the data in Exercise 42 of Section 7.4.
=+b. Will values of that are greater than dy(3s) lead to choppier- or smoother-looking kernel density estimates?
=+ 5 dy(3s) lead to a kernel density graph with a choppy appearance or a smooth appearance? Why?
=+a. If s denotes the sample standard deviation of the measurements in a sample of size n, would
=+71. Suppose the smallest distance d between any two successive measurements in an ordered set of data(i.e., measurements sorted from smallest to largest)is 3 units.
=+c. Which of the plots in parts (a) and (b) appears to fit the data better?
=+b. Repeat part (a) using a smoothing parameter of 5 .3.
=+a. Use a smoothing parameter of 5 .5 to create a kernel density plot for this data.
=+70. Refer to the data in Exercise 46 of Section 7.4.
=+69. Suppose someone suggests using a smoothing parameter of 5 2 to create a kernel density graph.Do you expect the graph to provide a useful picture of the data? Why?
=+b. Five randomly chosen specimens are weighed, yielding the following data: (3.10, 3.12), (3.52, 3.45), (4.22, 4.30), (2.98, 3.06), and (5.43, 5.38).Use the result in part (a) to find an estimate of 2.
=+2, where(x1, y1), (x2, y2),…,(xn, yn) denote the n pairs of scale measurements. Hint: The sample variance of two measurements z1 and z2 equals(z1 2 z2)2 y2.
=+a. For a random sample of n specimens, show that the maximum likelihood estimator of 2 is given by (1y4n) ^(xi 2 yi)
=+68. A specimen is weighed twice on the same scale. Let x and y denote the two measurements. Suppose x and y are independent of one another and are assumed to follow normal distributions with the same mean (the true weight of the specimen) and the same variance 2.
=+has been modeled by a shifted exponential distribution. For a random sample of ten headway times—3.11, .64, 2.55, 2.20, 5.44, 3.42, 10.39, 8.93, 17.82, and 1.30—use the results from part(a) to find estimates of and .
=+b. In traffic flow research, time headway is defined to be the elapsed time between the moment that one car finishes passing a fixed point and the instant that the next car begins to pass that point. The random variable x = time headway
=+When 5 0, this probability density function reduces to the probability density function of the exponential distribution.a. Obtain maximum likelihood estimators of both and .
=+67. A random sample x1, x2, x3,…, xn is selected from a shifted exponential distribution whose probability density function is given by f(x) 5 ee 2(x2 ) for x $ 0 otherwise
=+b. Use the result in part (a) with the data from Exercise 65(b) to estimate P(x , 400).
=+a. Find a maximum likelihood estimator of the probability that x is less than 400. That is, find the MLE for P(x , 400).
=+66. Refer to Exercise 65. Suppose the strength x of another randomly selected spot weld is measured.
=+b. A random sample of ten spot-weld strengths yields the following data (in psi): 392, 376, 401, 367, 389, 362, 409, 415, 358, 375. Use the result in part (a) to find an estimate of the 95th percentile of the distribution of all weld strengths.
=+a. Find the maximum likelihood estimator of the strength that is exceeded by 5% of the population of welds. That is, find a maximum likelihood estimator for the 95th percentile of the normal distribution based on a random sample of size n. Hint: Determine the relationship between the 95th
=+65. The shear strength x of a random sample of spot welds is measured. Shear strengths (in psi) are assumed to follow a normal distribution.
=+b. A random sample of ten workers yielded the following data on x: .92, .79, .90, .65, .86, .47, .73,.97, .94, .77. Use this data to obtain an estimate of .
=+a. Derive the maximum likelihood estimator of for a random sample of size n.
=+64. Let x denote the proportion of an allotted time frame that a randomly selected worker spends performing a manufacturing task. Suppose the probability density function of x is f(x) 5 e( 1 1)x for 0 # x # 1 0 otherwise where the value of must be larger than 1.
=+b. Is the estimator in part (a) unbiased?c. What is the MLE of (1 2 )5, the probability that none of the next five assemblies tested is defective?
=+a. In terms of x, what is the maximum likelihood estimator of ?
=+63. A random sample of n electronic assemblies is selected from a large shipment, and each assembly is tested on an automatic test station. The number x of assemblies that do not perform correctly is determined. Let denote the proportion of assemblies in the entire shipment that are
=+b. Use the procedure outlined in this exercise to generate a 95% bootstrap interval for the average dye-layer density.c. Compare your result in part (b) to the 95%confidence interval found in Exercise 14(a).
=+a. Under what conditions will this procedure provide a reliable interval estimate?
=+generate a random sample of 69 observations from a normal distribution whose mean and standard deviation are 1.028 and .163, respectively. If necessary, after obtaining the sample, the data are adjusted so that their sample mean and standard deviation coincide exactly with 1.028 and .163. A 95%
=+62. In Exercise 14 (Section 7.2), the sample mean and standard deviation of the dye-layer density of aerial photographs of 69 forest trees were found to be 1.028 and .163, respectively. Because the raw data is not available, a researcher suggests using a computer to
=+b. Compare your result in part (a) to the 95% confidence interval found in Exercise 46(a).
=+a. Use the bootstrap method to find a 95% bootstrap interval for the mean of the population from which this data was obtained.
=+61. Refer to Exercise 46 of Section 7.4.
=+b. Compare your result in part (a) to the 95% confidence interval found in Exercise 42(c).
=+a. Use the bootstrap method to find a 95% bootstrap interval for the mean of the population from which the data of Exercise 42 was obtained.
=+60. Refer to Exercise 42 of Section 7.4.
=+b. For the sample of 45 individuals who had taken olanzapine, the article reported (7.38, 9.69) as a 95% CI for true average weight gain (kg). What is a 99% CI?
=+. This requires setting the derivative of L( ) equal to 0 and solving for . However, to simplify the calculations, we first take the natural logarithm of L( ) 5 3(1 2 )7:ln(L( )) 5 ln3 3(1 2 )7 4 5 3 ln( ) 1 7 ln(1 2 )and then take the derivative1:d d ln(L( )) 5 3 2 7 1 2 Example 7.15
=+. We now ask, For what value of is the observed sample most likely to have occurred? That is, we want to find the value of that maximizes the probability 3(1 2 )7
=+ represents the likelihood of our sample result occurring, and it is abbreviated as L( ) 5 3(1 2 )7
=+scheme introduced in Section 5.6, we can write the data in this sample as x1 5 1, x2 5 0, x3 5 1, x4 5 0, . . . , x10 5 1, where a “0” indicates that the component functioned correctly and a “1” indicates that it did not work correctly.Since this data comes from a random sample, we can
=+a. The sample of 41 individuals who had taken aripiprazole had a mean change in total cholesterol (mg/dL) of 3.75, and the estimated standard error sd y1n was 3.878. Calculate a confidence interval with confidence level approximately 95% for the true average increase in In a random sample of
=+Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially
=+59. Antipsychotic drugs are widely prescribed for conditions such as schizophrenia and bipolar disease. The article “Cardiometabolic Risk of Second-Generation Antipsychotic Medications During First-Time Use in Children and Adolescents” (J. of the Amer. Med. Assoc., 2009: 1765–1773)
=+Does it appear that the true average pressure is different in a dental setting than in a medical setting?
=+58. Dentists make many people nervous (even more so than statisticians!). To assess any effect of such nervousness on blood pressure, the systolic blood pressure of each of 60 subjects was measured both in a dental setting and in a medical setting (“The Effect of the Dental Setting on Blood
=+c. The authors asserted that “pitchers have greater difference in side-to-side anteroposterior translation of their shoulders compared with position players.” Do you agree? Explain.Pos Dom Tr Pos ND Tr Pit Dom Tr Pit ND Tr 1 30.31 32.54 27.63 24.33 2 44.86 40.95 30.57 26.36 3 22.09 23.48
=+b. Repeat part (a) for position players.
=+a. Estimate the true average difference in translation between dominant and nondominant arms for pitchers in a way that conveys information about reliability and precision. Interpret the resulting estimate.
=+kindly supplied the following data (for 19 position players and 17 pitchers) on anteroposterior translation (mm), a measure of the extent of anterior and posterior motion, for both dominant nondominant arms.
=+57. The paper “Quantitative Assessment of Glenohumeral Translation in Baseball Players” (Amer. J. of Sports Med., 2004: 1711–1715) considered various aspects of shoulder motion for a sample of pitchers and another sample of position players [glenohumeral refers to the articulation
=+b. Estimate the difference between true average TBBMC for the two periods of concrete in a way that conveys information about precision and reliability. Does it appear plausible that the true average TBBMCs for the two periods are identical? Why or why not?
=+a. Construct a comparative boxplot of TBBMC for the lactation and postweaning periods and comment on any interesting features.
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