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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

=+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
=+56. Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk production. The paper “Bone Mass Is Recovered from Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes” (Amer. J. of Clinical Nutr., 2004: 1322–1326) gave the following
=+b. If a seventh site were to be randomly selected among locations bearing service signs, between what values would you predict the difference in number of crashes to lie?
=+a. Calculate a confidence interval for the population mean difference in the number of crashes per year before and after the sign changes were made. Provide an interpretation for this interval.
=+ In one investigation, the authors selected six sites along Virginia interstate highways where service signs are posted. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of
=+55. Along any major freeway we often encounter service (or logo) signs that give information on attractions, camping, lodging, food, and gas services in advance of the off-ramp that leads to such services. These signs typically do not provide information on distances. Researchers in Virginia,
=+c. Estimate the difference between the two >s using the two-sample t interval discussed in this section, and compare it to the interval of part (b).
=+15.1. A sample of the second brand gave output values 12.1, 13.6, 11.9, and 11.2 (“Multiple Comparisons of Means Using Simultaneous Confidence Intervals,” J. of Quality Technology, 1989: 232–241). Use the pooled t formula from part (a) to estimate the difference between true average
=+b. A sample of ultrasonic humidifiers of one particular brand was selected for which the observations on maximum output of moisture (oz) in a controlled chamber were 14.0, 14.3, 12.2, and
=+ has a t distribution with n1 1 n2 22 df.a. Use the t variable above to obtain a pooled t confidence interval formula for 1 2 2.
=+ (n1 1 n2 2 2 is the sum of the df’s contributed by the two samples). It can then be shown that the standardized variable t 5 (x1 2 x2) 2 (1 2 2)spA 1n1 11 n2
=+54. Suppose not only that the two population or treatment response distributions are normal but also that they have equal variances. Let 2 denote the common variance. This variance can be estimated by a “pooled” (i.e., combined) sample variance as follows:s 2p 5 a n1 2 1 n1 1 n2 2 2 bs 21 1
=+Estimate for difference: 8.800 95% CI for difference: (6.498, 11.102)Calculate a two-sided 95% confidence interval for the difference in population means and confirm the lower and upper endpoints reported by Minitab. Based on the interval, what conclusion you can draw about the two methods? Why?
=+output for comparing a laboratory method to a new relatively quick and inexpensive field method (from the article “Evaluation of a New Field measurement Method for Arsenic in Drinking Water Samples,”J. of Envir. Engr., 2008: 382–388).Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 3
=+53. Arsenic is a known carcinogen and poison. The standard laboratory procedures for measuring arsenic concentration (g/L) in water are expensive. Consider the accompanying summary data and Minitab Section 7.5 Exercises Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied,
=+b. Calculate a 99% CI for the difference between true average stance duration for the elderly and the younger individuals. Does your interval suggest that true average stance duration is larger among elderly individuals than among younger individuals?
=+a. Calculate and interpret a 99% CI for true average stance duration among elderly individuals.
=+52. The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. The article “Evidence of Mechanical Load Redistribution at the Knee Joint in the Elderly when Ascending Stairs and Ramps” (Annals of Biomed. Engr., 2008: 467–476) presented the
=+b. Calculate a 95% confidence interval for the difference between true average degree of polymerization for the middle range and that for the high range. Does the interval suggest that 1 and 2 may in fact be different? Explain your reasoning.
=+a. Construct a comparative boxplot for the two samples, and comment on any interesting features.
=+51. Refer to Exercise 42 in Section 7.4. The cited article also gave the following observations on degree of polymerization for specimens having viscosity times concentration in a higher range:429 430 430 431 436 437 440 441 445 446 447
=+b. Calculate an estimate for the difference between true average AN IAT and true average control IAT; do so in a way that conveys information about the reliability and precision of the estimation. What does your estimate suggest about true average AN IAT relative to true average control IAT?
=+a. Calculate an estimate for true average IAT under the described AN protocol; do so in a way that conveys information about the reliability and precision of the estimation.
=+ Assume that both samples were selected from normal distributions.
=+imagery to determine various tissue characteristics for both an AN sample of individuals who had undergone acute weight restoration and maintained their weight for a year and a comparable (at the outset of the study) control sample. Here is summary data on intermuscular adipose tissue (IAT, in
=+50. Anorexia nervosa (AN) is a psychiatric condition leading to substantial weight loss among women fearful of becoming overweight. The article“Adipose Tissue Distribution After Weight Restoration and Weight Maintenance in Women with Anorexia Nervosa” (Amer. J. of Clinical Nutr.,
=+.39, respectively. Calculate a confidence interval for the difference between true average firmness for zero-day apples and true average firmness for 20-day apples using a confidence level of 95%, and interpret the interval.
=+20 golden apples with a shelf life of zero days, resulting in a sample mean of 8.74 and a sample standard deviation of .66, and another sample of 20 apples with a shelf life of 20 days, with a sample mean and sample standard deviation of 4.96 and
=+49. The firmness of a piece of fruit is an important indicator of fruit ripeness. The Magness–Taylor firmness (N) was determined for one sample of
=+What is the confidence level for each of the following three confidence intervals for the mean of a normal population distribution? Which of the three intervals would you recommend be used, and why?a. (x 2 .687sy121, x 1 1.725sy121)b. (x 2 .860sy121, x 1 1.325sy121)c. (x 2 1.064sy121, x 1
=+48. A more extensive tabulation of t critical values than what appears in this book shows that for the t distribution with 20 df, the areas to the right of the values .687,.860, and 1.064 are .25, .20, and .15, respectively.
=+b. Calculate an interval for which you can have a high degree of confidence that at least 95%of all pieces of laminate result in amounts of warpage that are between the two limits of the interval.
=+in a sample mean warpage of .0635 and a sample standard deviation of .0065.a. Calculate a prediction for the amount of warpage of a single piece of laminate in a way that provides information about precision and reliability.
=+47. A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected and the amount of warpage (in.) under particular conditions was determined for each piece, resulting
=+c. Calculate an interval that includes at least 99%of the cadences in the population distribution using a confidence level of 95%.
=+a. Calculate and interpret a 95% confidence interval for population mean cadence.b. Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population.
=+ A normal quantile plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:Variable N Mean Median TrMean StDev SEMean cadence 20 0.9255 0.9300 0.9261 0.0809 0.0181 Variable Min Max
=+46. A study of the ability of individuals to walk in a straight line (“Can We Really Walk Straight?” Amer.J. of Physical Anthro., 1992: 19–27) reported the accompanying data on cadence (strides per second)for a sample of n 5 20 randomly selected healthy men:.95 .85 .92 .95 .93 .86 1.00 .92
=+d. Calculate an upper prediction bound with level 95% for the maximum pressure of a single observation. How does the prediction compare to the estimate calculated in part (b)?
=+c. Calculate an upper confidence bound with confidence level 95% for the population mean of maximum pressure.
=+ Calculate a two-sided 95% confidence interval for the population mean of maximum pressure and confirm the lower and upper endpoints reported by SAS.
=+b. SAS reports the following summary information for this data:The MEANS Procedure Analysis Variable:pressure Lower 95% Upper 95%CL for Mean CL for Mean Mean Std Error 36.1892782 39.0373884 37.6133333 0.6639612
=+45. The article “Concrete Pressure on Formwork”(Mag. of Concrete Res., 2009: 407–417) gave the following observations on maximum concrete pressure (kN/m2):33.2 41.8 37.3 40.2 36.7 39.1 36.2 41.8 36.0 35.2 36.7 38.9 35.8 35.2 40.1
=+c. Predict the modulus of elasticity for a single mixture in a way that conveys information about precision and reliability. How does the prediction compare to the estimate calculated in part (b)?
=+b. Estimate true average modulus of elasticity for these mixtures in a way that conveys information about precision and reliability.
=+27.0 25.5 28.5 34.0 31.0 34.5 32.5a. Is it plausible that this sample was selected from a normal population distribution?
=+ One important mechanical property of concrete is its modulus of elasticity (in GPa), which is the material’s tendency to be deformed elastically when subjected to an applied force. A higher modulus of elasticity indicates a stiffer material. As reported in the article, the following are
=+44. A new concrete structure that experiences cracking within the first seven days after setting is often said to have experienced “early-age cracking.”This is usually a precursor to later-age cracking and other problems that lead to an overall weakening of the structure. According to the
=+b. Predict the mileage for a single Porsche Boxster in a way that conveys information about precision and reliability. How does the prediction compare to the estimate calculated in part (a)?
=+a. Estimate true average mileage in a way that conveys information about precision and reliability.
=+ A normal quantile plot supports the assumption that mileage is at least approximately normally distributed. The R software reports the following summary statistics for this data:> summary(odometer, digits=6)Min 1st Qu Median Mean 1445.0 33928.8 68730.5 66221.1 3rd Qu Max 91932.8 137652.0>
=+43. Haven’t you always wanted to own a Porsche? We investigated the Boxster (their cheapest model)and performed an online search at www.cars.com on December 30, 2012. Asking prices were well beyond our meager professorial salaries, so instead we focused on odometer readings (mileage). Here are
=+c. Calculate a two-sided 95% confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree of polymerization? What about 450?
=+b. Is it plausible that the given sample observations were selected from a normal distribution?
=+a. Construct a boxplot of the data and comment on any interesting features.
=+42. The article “Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers” (IEEE Electrical Insul. Mag., 1996: 28–34)contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain
=+b. Calculate a two-sided 95% confidence interval for the true average work of adhesion for UHPC adhered to steel. Does the interval suggest that 107 is a plausible value for the true average work of adhesion for UHPC adhered to steel? What about 110?
=+a. Is it plausible that the given sample observations were selected from a normal distribution?
=+41. Ultra high performance concrete (UHPC) is a relatively new construction material that offers strong adhesive properties with other materials.The authors of “Adhesive Power of Ultra High Performance Concrete from a Thermodynamic Point of View” (J. of Materials in Civil Engr., 2012:
=+20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607. Calculate and interpret a confidence interval at the 99% confidence level for the true average number of cycles to break.(Note: The article presented the results of hypothesis
=+40. According to the article “Fatigue Testing of Condoms” (Polymer Testing, 2009: 567–571), “tests currently used for condoms are surrogates for the challenges they face in use,” including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant
=+39. Determine the t critical value for a lower or an upper confidence bound for each of the situations described in Exercise 38.
=+38. Determine the t critical value for a two-sided confidence interval in each of the following situations:a. Confidence level 5 95%, df 5 10b. Confidence level 5 95%, df 5 15c. Confidence level 5 99%, df 5 15d. Confidence level 5 99%, n 5 5e. Confidence level 5 98%, df 5 24f. Confidence level 5
=+37. Determine the t critical value that will capture the desired t curve area in each of the following cases:a. Central area 5 .95, df 5 10b. Central area 5 .95, df 5 20c. Central area = .99, df 5 20d. Central area 5 .99, df 5 50e. Upper-tail area 5 .01, df 5 25f. Lower-tail area 5 .025, df 5 5
=+36. An investigator wishes to estimate the difference between population mean lifetimes of two different brands of batteries under specified conditions. If the population standard deviations are both roughly 2 hr and equal sample sizes are to be selected, what value of the common sample size n
=+steel and that for the commercial steel using a lower 95% confidence bound. Does your estimate demonstrate conclusively that this difference exceeds 5? Explain your reasoning.
=+35. An experiment was performed to compare the fracture toughness of high-purity Ni-maraging steel with commercial-purity steel of the same type. For 32 high-purity specimens, the sample mean toughness and sample standard deviation of toughness were 65.6 and 1.4, respectively, whereas for 32
=+ Calculate a confidence interval at the 99%level to estimate the true mean GPA difference between full-time and part-time faculty. Does it appear that true average course GPA for part-time faculty differs from that for faculty teaching full-time?Explain your reasoning.
=+34. Is there any systematic tendency for part-time college faculty to hold their students to different standards than full-time faculty do? The article“Are There Instructional Differences Between Full-Time and Part-Time Faculty?” (College Teaching, 2009: 23–26) reported that for a sample of
=+ Estimate the difference between true average densities for the two types of wood in a way that conveys information about reliability and precision.
=+33. Relative density was determined for one sample of second-growth Douglas fir 2 3 4s with a low percentage of juvenile wood and another sample with a moderate percentage of juvenile wood, resulting in the following data (“Bending Strength and Stiffness of Second-Growth Douglas Fir Dimension
=+32. Use the accompanying data to estimate with a 95%confidence interval the difference between true average compressive strength (N/mm2) for 7-day-old concrete specimens and true average strength for 28-dayold specimens (“A Study of Twenty-Five-Year-Old Pulverized Fuel Ash Concrete Used in
=+31. A manufacturer of exercise equipment is interested in estimating the proportion of all purchasers of one of its products who still own the product two years after purchase. What sample size is required to estimate this proportion to within .05 with a confidence level of 90%?
=+30. A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it is considered defective and must be discarded. A very large shipment of handles is received. The proportion of defective handles, , is of interest. How many handles
=+100 questionnaires that included a prepaid cash amount of $5 were returned. Calculate a 95% confidence interval for the ratio of the proportion of questionnaires returned when such a cash incentive is included to the proportion returned in the absence of any incentive. Does the interval suggest
=+b. The article cited in Exercise 27 stated that in addition to 75 of 110 questionnaires without an incentive to respond being returned, 78 of
=+ln(1 y2) and 2(n1 2 u)y(un1) 1 (n2 2 v)yvn2), respectively.a. Use these facts to obtain a large-sample twosided 95% CI for ln(1 y2) and a CI for 1 y2 itself.
=+successes occur three times as frequently in population 1 as they do in population 2. Alternatively, if the ’s refer to success proportions for two different treatments, then a ratio of 3 implies that the first treatment is three times as likely to result in a success as is the second
=+29. Let 1 and 2 denote the proportions of successes in two different populations. Rather than estimate the difference 1 2 2 as described in Exercise 27, an investigator will often wish to estimate the ratio of the two ’s. If, for example, 1 y2 5 3, then
=+reported that 6.8% of a sample of 503 nonsmoking obese women (body mass index . 29) gave birth to children of low birth weight. Calculate a 95% lower confidence bound for the difference between the population proportion of normalweight nonsmoking women and the population proportion of obese
=+J. of Public Health, 1997: 591–596) reported on a random sample of 487 nonsmoking women of normal weight (body mass index between 19.8 and 26.0) who had given birth at a large metropolitan medical center. It was determined that 7.2% of these births resulted in children of low birth weight
=+28. The article “The Effects of Cigarette Smoking and Gestational Weight Change on Birth Outcomes in Obese and Normal-Weight Women” (Amer.
=+c. Recent research has shown that “coverage probability” and small-sample behavior are improved by adding one success and one failure to each sample and then using the formula you obtained in part (a). Do this for the data of part (b).
=+Quarterly, 1996: 542–562). Calculate a twosided 95% CI for the difference between the true response proportions under these circumstances. Does the interval suggest that, in fact, the values of 1 and 2 are different? Explain your reasoning.
=+b. Is the response rate for questionnaires affected by including some sort of incentive to respond along with the questionnaire? In one experiment, 110 questionnaires with no incentive resulted in 75 being returned, whereas 98 questionnaires that included a chance to win a lottery yielded 66
=+a. Use the foregoing facts to obtain a large-sample two-sided 95% confidence interval formula for estimating 1 2 2.
=+estimated standard deviation of this statistic results from replacing each π under the square root by the corresponding p.
=+samples of size n1 and n2, respectively, are independently selected from the two populations, and let p1 and p2 denote the resulting sample proportions of successes. If the sample sizes are sufficiently large(apply the rule of thumb appropriate for a single proportion to each sample separately),
=+27. Let 1 and 2 denote the proportion of successes in population 1 and population 2, respectively. An investigator sometimes wishes to calculate a confidence interval for the difference 1 2 2 between these two population proportions. Suppose random
=+of a 95% confidence level: Add 2 to both the number of successes and the number of failures and then use the traditional formula. Do this for the data described in this exercise, and compare the resulting interval to the one you calculated in part (a).
=+confidence level can deviate dramatically from the nominal one chosen by the investigator(e.g., the actual level may be quite different from the 95% level selected). An article by two statisticians (Agresti, A., and B. A. Coull,“Approximate Is Better Than ‘Exact’ for Interval Estimation of
=+b. The traditional CI for discussed in Section 7.3 is based on the sample proportion p having approximately a normal sampling distribution, so the confidence level is only approximate rather than exact. Recent research has shown that under certain circumstances, its actual
=+a. Calculate and interpret a 95% confidence interval for the proportion of all such trees that would die if the treatment were applied at the tested level.
=+26. Researchers have developed a chemical treatment that retards the growth of trees of a certain type whose branches pose a safety threat to power lines. However, an overly severe application of the treatment can cause trees to die. In an experiment involving one particular treatment level
=+b. Interpret the 95% confidence level used in part (a).
=+a. Calculate a lower confidence bound at the 95%confidence level for the true proportion of such hips that develop squeaking.
=+patients the increased durability has been counterbalanced by an increased incidence of squeaking.The May 11, 2008 issue of The New York Times reported that in one study of 143 individuals who received ceramic hips between 2003 and 2005, 10 developed squeaking problems.
=+25. The technology underlying hip replacements has changed as these operations have become more popular (more than 250,000 in the United States in 2008). Starting in 2003, highly durable ceramic hips were marketed. Unfortunately, for too many
=+process, rebate amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate an upper confidence bound at the 95% confidence level for the true proportion of such consumers who never apply for a rebate. Based on this
=+24. In a sample of 1000 randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, 250 said they never did so (“Rebates: Get What You Deserve,”Consumer Reports, May 2009: 7). Reasons cited for their behavior included too many steps in the

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