New Semester
Started
Get
50% OFF
Study Help!
--h --m --s
Claim Now
Question Answers
Textbooks
Find textbooks, questions and answers
Oops, something went wrong!
Change your search query and then try again
S
Books
FREE
Study Help
Expert Questions
Accounting
General Management
Mathematics
Finance
Organizational Behaviour
Law
Physics
Operating System
Management Leadership
Sociology
Programming
Marketing
Database
Computer Network
Economics
Textbooks Solutions
Accounting
Managerial Accounting
Management Leadership
Cost Accounting
Statistics
Business Law
Corporate Finance
Finance
Economics
Auditing
Tutors
Online Tutors
Find a Tutor
Hire a Tutor
Become a Tutor
AI Tutor
AI Study Planner
NEW
Sell Books
Search
Search
Sign In
Register
study help
business
applied statistics and probability for engineers
Applied Statistics For Engineers And Scientists 3rd Edition Jay L. Devore, Nicholas R. Farnum, Jimmy A. Doi - Solutions
=+select the factor level having the largest sample mean. This strategy has been called the “pick the winner” approach in the literature on experimental design. Explain what is wrong with this approach and why it does not take the place of an ANOVA test.
=+5. As a simple method of determining which of k factor levels maximizes the average value of a certain response variable, inexperienced researchers sometimes calculate the k sample means and then simply
=+item, is not considered to be a valid form of replication. Instead, several different items should each be measured once. What is the danger in using repeated measurements of the same item instead of truly replicating an experimental result? What do you expect the effect on the F statistic to
=+4. Repeated measurements in an ANOVA study are supposed to indicate what would happen if another researcher tried to repeat your study. In particular, simply measuring the same sampled item several times, which gives repeated measurements of that
=+the techniques given in this section, explain what effect you think such data would have on the results of an ANOVA test comparing samples of the three alloys. Do you think an ANOVA test based on accurate measurements of the same samples of alloys will lead to a different conclusion?
=+3. In a one-way ANOVA test for comparing the mean strengths (in kilograms) of three different alloys, suppose that the measuring instrument used is out of calibration, causing it to give readings that are consistently 2.5 kilograms higher than the true measured strength. Using the general
=+correspondingly, more plastic parts at each concentration level? Conversely, what scenario would lead you to use a larger number of concentration levels and, therefore, fewer plastic parts per concentration level? Include the two sources of variation in an ANOVA experiment in your answers.
=+of electroplated plastic parts. Describe in general terms how you would allocate the samples. Specifically, what information would make you want to use fewer levels of chemical concentration and,
=+2. Suppose you have a fixed budget to allocate to the samples used in a study of the effect of the factor“chemical concentration” on the plating thickness
=+c. Suppose the ANOVA test does not reveal any significant differences in strength between the three types of beams. If the builder must use one of the three types, which type should be chosen?
=+b. Suppose an ANOVA test indicates that beams of types A and B are not significantly different in strength from one another, but that both types are significantly stronger than beams of type C. If the builder’s objective is to use as strong a beam Unless otherwise noted, all content on this
=+a. What hypotheses would you test in such a study?Describe, in words, the parameters that appear in the hypotheses.
=+whether there is a difference between the average strengths of the three types of wood, a random sample of ten beams of each type is selected and their strengths are measured.
=+1. Three types of wood (denoted A, B, and C) are being considered for use in a building project. Each type of wood differs in cost, so the builder is interested in keeping costs down as well as in selecting wood that will be strong enough. To determine
=+the fact that for any two independent random variables y1 and y2 and numerical constants a1 and a2, V(a1y1 1 a2y2) 5 a2 1V(y1) 1 a2 2V(y2).Nitrite: 9000 20,000 10,000 20,000 21,000 3000 4000 Thermal: 49,000 23,000 20,000 100,000 114,000 35,000 30,000
=+is more than four times that of the nitride treatment. Hint: Consider the parameter 5 4 1 2 2 with corresponding estimator n 5 4x1 2 x2. This estimator is unbiased and normally distributed provided that the two population distributions are normal, and its variance can be determined from
=+Core and Pin Life” (Die Casting Engineer, March/April 1999: 88) reported on an experiment in which one group of core pins was coated using the traditional nitride process and a second group was coated using a new thermal diffusion process. Use the accompanying data to decide at significance
=+90. One way to reduce the equipment problems that occur during die casting is to apply a thin coating to the core pins. The paper “Tool Treatment Extends
=+1, 2, or 3 winning their regional tournaments are.45883, .18813, and .11032, respectively. Assess the fit of this model.
=+percentiles (percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the range of probabilities is the same as for the model in part (a)). The resulting probabilities of seeds
=+b. A more sophisticated model has Pij 5 .5 1.2813625(zi 2 zj) where the z’s are measures of relative strengths related to standard normal
=+a. One model postulated Pij 5 .5 1 (i 2 j) with 5 1y32 (from which P16.1 5 , P16.2 5 2 , etc.).Based on this, P(seed #1 wins) 5 .27477, P(seed#2 wins) 5 .20834, and P(seed #3 wins) 5.15429. Does this model appear to provide a good fit to the data?
=+against the team ranked j. Once the Pij’s are available, it is possible to compute the probability that any particular seed wins its regional tournament(a complicated calculation because the number of outcomes in the sample space is quite large).The paper “Probability Models for the NCAA
=+team won 10 times, the third-ranked team was 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3. Let Pij denote the probability that the team ranked i in its region is victorious in its game
=+teams in each region are then ranked (seeded)from 1 to 16. During the 12-year period from 1991 to 2002, the top-ranked team won its regional tournament 22 times, the second-ranked
=+89. The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16
=+Be sure to check the validity of any assumptions on which your chosen inferential method is based.
=+ Does the true mean difference between slide retrieval time and digital retrieval time appear to exceed 10 sec?
=+image from a library of slides and while retrieving the same image from a computer database with a Web front end.Subject: 1 2 3 4 5 6 7 Slide: 30 35 40 25 20 30 35 Digital: 25 16 15 15 10 20 7 Difference: 5 19 25 10 10 10 28 Subject: 8 9 10 11 12 13 Slide: 62 40 51 25 42 33 Digital: 16 15 13 11
=+88. Adding computerized medical images to a database promises to provide great resources for physicians. However, there are other methods of obtaining such information, so the issue of efficiency of access needs to be investigated. The article “The Comparative Effectiveness of Conventional
=+ Does the data suggest that true average maximum lean angle for older females is more than 10 degrees smaller than it is for younger females? State and test the relevant hypotheses at significance level .10.
=+a subject is able to lean and still recover in one step—was determined for both a sample of younger females (21–29 years) and a sample of older females(67–81 years). The following observations are consistent with summary data given in the article:YF: 29 34 33 27 28 32 31 34 32 27 OF: 18 15
=+87. As the population ages, there is increasing concern about accident-related injuries to the elderly. The article “Age and Gender Differences in Single-Step Recovery from a Forward Fall” (J. of Gerontology, 1999: M444–M50) reported on an experiment in which the maximum lean angle—the
=+State and test the appropriate hypotheses using 5 .05. Do you think your conclusion can be attributed to a single sport being an anomaly?
=+Do the four sports appear to be identical with respect to the proportion of games won by the late-game leader?
=+determined, and it was noted whether the leader actually ended up winning the game. The leader was defined as the team ahead after three quarters in basketball and football, two periods in hockey, and seven innings in baseball. The results follow:Sport Leader wins Leader loses Basketball 150 39
=+the last quarter, whereas baseball games are “over”by the seventh inning. They also considered football and hockey. Data was collected for a sample of games of each type, selected from all games played during the 1990 season for baseball and football and during the 1990–1991 season for the
=+86. The authors of the article “Predicting Professional Sports Game Outcomes from Intermediate Game Scores” (Chance, 1992: 18–22) used statistical analysis to determine whether there was any merit to the idea that basketball games are not settled until
=+weeks each in random order to 14 patients, resulting in the following total severity index scores (“Double Blind Evaluation of Deanol in Tardive Dyskinesia,”J. of the Amer. Med. Assoc., 1978: 1997–1998):Patient: 1 2 3 4 5 6 7 Deanol: 12.4 6.8 12.6 13.2 12.4 7.6 12.1 Placebo: 9.2 10.2 12.2
=+85. Tardive dyskinesia refers to a syndrome comprising a variety of abnormal involuntary movements assumed to follow long-term use of antipsychotic drugs. An experiment carried out to investigate the effect of the drug deanol also used a placebo treatment, something that resembled deanol in
=+ The authors of the cited article stated that strength for heated specimens appeared to be slightly higher on average than for the control specimens. Is the difference statistically significant? State and test the relevant hypotheses using 5 .05.
=+from the article “Heat Treatment of Cotton: Effect on Endotoxin Content, Fiber and Yarn Properties, and Processability” (Textile Research J., 1996: 727–738):Twist multiple: 1.054 1.141 1.2451.370 .481 Control strength: .45 .60 .61 .73 .69 Heated strength: .51 .59 .63 .73 .74
=+84. Long-term exposure of textile workers to cotton dust released during processing can result in substantial health problems so textile researchers have been investigating methods that will result in reduced risks while preserving important fabric properties. The accompanying data on roving
=+b. Is there strong evidence for concluding that true average force in a wet medium at the lower temperature exceeds that at the higher temperature by more than 50 N?
=+a. Is there strong evidence for concluding that true average force in a dry medium at the higher temperature exceeds that at the lower temperature by more than 100 N?
=+83. The article cited in Exercise 78 of Chapter 7 gave additional data on breaking force (N):Temp Medium n x s 22° Dry 6 170.60 39.08 37° Dry 6 325.73 34.97 22° Wet 6 366.36 34.82 37° Wet 6 306.09 41.97
=+two log concentration distributions, are not equal, would 1 and 2, the means of the concentration distributions, be equal if 1* 5 2*?Explain your reasoning.
=+a. Use the pooled t test (based on assuming normality and equal standard deviations), described in Exercise 37, to decide at significance level.05 whether the two concentration distribution means are equal.b. If 1* and 2* , the standard deviations of the
=+* be the true average log concentration at the first site and define 2* analogously for the second site.
=+data refers to bromine concentration in needles taken from a site near an oil-fired steam plant and from a relatively clean site. The summary values are means and standard deviations of the log-transformed observations.Site Sample Size Mean log concentration Standard Deviation of log
=+82. The article “Pine Needles as Sensors of Atmospheric Pollution” (Environ. Monitoring, 1982: 273–286)reported on the use of neutron-activity analysis to determine pollutant concentration in pine needles.According to the article’s authors, “These observations strongly indicated that
=+b. Is there substantial evidence for concluding that true average strength for males exceeds that for females by more than 25 N? State and test the relevant hypotheses.
=+a. A test carried out to see whether true average strengths for the two genders were different resulted in t 5 2.51 and P-value 5 .019. Does the appropriate test procedure described in this chapter yield this value of t and the stated P-value?
=+81. Information about hand posture and forces generated by the fingers during manipulation of various daily objects is needed for designing high-tech hand prosthetic devices. The article “Grip Posture and Forces During Holding Cylindrical Objects with Circular Grips” (Ergonomics, 1996:
=+Suppose the sample standard deviations are 14.8 and 12.5, respectively (consistent with the sample ranges given in the article). The authors commented that the thicker brand B condom displayed a smaller mean tear length than the thinner brand A condom.Is this difference in fact statistically
=+80. The article “Two Parameters Limiting the Sensitivity of Laboratory Tests of Condoms as Viral Barriers”(J. of Testing and Eval., 1996: 279–286) reported that, in brand A condoms, among 16 tears produced by a puncturing needle, the sample mean tear length was 74.0 m, whereas for the 14
=+steel specimens was 6.43 and the sample mean for aluminum specimens was 7.09. Suppose that the sample standard deviations were 1.08 and 1.19, respectively. Do you agree with the article’s authors that the difference in headability ratings is significant at the 5% level?
=+Heading Quality” (Wire J. Intl., Oct. 1996: 66–72)described the result of a headability impact test applied to 30 specimens of aluminum killed steel and 30 specimens of silicon killed steel. The sample mean headability rating number for the
=+79. Headability is the ability of a cylindrical piece of material to be shaped into the head of a bolt, screw, or other cold-formed part without cracking. The article “New Methods for Assessing Cold
=+b. This poll was conducted December 19–22, just days after a mass shooting at an elementary school in Connecticut. Discuss what effects this event may have had on the poll’s outcome.
=+a. Does this provide strong evidence for concluding that more than 50% of the population of American adults was in favor of making laws covering the sale of firearms more strict? Conduct an appropriate test of hypotheses using a.01 significance level. (Hint: Read the first paragraph of the
=+78. Some of the deadliest mass shootings in U.S. history occurred in 2012. These events led to many calls for stricter national gun control. On December 27, 2012, the Gallup organization reported that roughly 600 of 1038 American adults surveyed said they would be in favor of strengthening laws
=+Head-on Collisions,” Accident Analysis and Prevention, 1995: 143–150). Does this data suggest that less than one-third of all such accidents result in no injuries? State and test the relevant hypotheses using a significance level of .05.
=+ Seat belts help prevent injuries in vehicle accidents, but they don’t offer complete protection in extreme situations. A sample of 319 front-seat occupants involved in head-on collisions in a certain region resulted in 95 who sustained no injuries (“Influencing Factors on the Injury
=+H0 is true and both n 0 . 5 and n(1 2 0) . 5, the sampling distribution of p is approximately normal with mean value 0 and standard deviation 2 0(1 2 0)yn. This implies that a “large-sample”test statistic is z 5 (p 2 0)y2 0(1 2 0)yn (i.e., we standardize p assuming that H0 is true); the
=+77. Let denote the proportion of “successes” in some population. Consider selecting a random sample of size n, and let p denote the sample proportion of successes (number of successes in the sample divided by n). Suppose we wish to test H0: 5 0. When
=+b. Suppose that the investigator who performed the experiment described in part (a) had wished to test H0: 5 .70 versus Ha: , .70. Can this test be carried out using the chi-squared table in this book? Why or why not?
=+.50°C. The softening points of ten different specimens were determined, yielding a sample standard deviation of .58°C. Assume that the distribution from which the observations were selected is normal. Does the data contradict the uniformity specification? State and test the appropriate
=+a. To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most
=+obtain a test statistic. If the alternative hypothesis is Ha: . 0, the P-value is the area under the n 2 1 df chi-squared curve to the right of the calculated X2(an upper-tailed test).
=+a chi- squared distribution with n 2 1 df. This can be used as a basis for testing H0: 5 0, as follows:Replace 2 in X2 by its hypothesized value 2 0 to
=+76. When the population distribution is normal, it can be shown that the variable X2 5 (n 2 1)s 2y 2 has
=+antitoxin necessary was found to be 1.89 mg, and the sample standard deviation was .42. Previous research had indicated that the true average neutralizing amount was 1.75 mg/g of toxin. Does the new data contradict the value suggested by prior research?State and test the relevant hypotheses using
=+75. In an investigation of the toxin produced by a certain poisonous snake, a researcher prepared 26 different vials, each containing 1 g of the toxin, and then determined the amount of antitoxin necessary to neutralize the toxin. The sample average amount of
=+background value for this concentration was 20. The results of various statistical tests described in the article were predicated on assuming normality.Does the data provide strong evidence for concluding that the true average concentration in the sampled region exceeds the stated background
=+74. Contamination of mine soils in China is a serious environmental problem. The article “Heavy Metal Contamination in Soils and Phytoaccumulation in a Manganese Mine Wasteland, South China” (Air, Soil, and Water Res., 2008: 31–41) reported that, for a sample of 3 soil specimens from a
=+b. Suppose it had previously been believed that population mean consumption was at most 200 mg.Does the given data contradict prior belief?
=+a. Does it appear plausible that the population distribution of daily caffeine consumption is normal?Is it necessary to assume a normal population distribution to test hypotheses about population mean consumption? Explain your reasoning.
=+73. The following summary data on daily caffeine consumption for a sample of adult women appeared in the article “Caffeine Knowledge, Attitudes, and Consumption in Adult Women” (J. of Nutrition Educ., 1992: 179–184): n 5 47, x 5 215 mg, s 5 235 mg, range of data: 5–1176.
=+b. Suppose researchers wanted to investigate whether the true average mass crystallinity exceeds 40%. Carry out a test of appropriate hypotheses using a significance level of .05.
=+a. Is it plausible that the mass crystallinity for this type of polymer is normally distributed?
=+72. The article cited in Exercise 25 of Section 8.2 gave the following data on mass crystallinity (in %) for 12 samples of the PHB polymer:42.97 38.81 38.83 41.03 41.25 36.99 49.57 41.77 34.50 44.77 36.92 40.48
=+c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.
=+b. Carry out a test of the appropriate hypotheses using a significance level of .05. Would your conclusion change if a significance level of .01 had been used?
=+a. Check the validity of any assumptions necessary for testing the appropriate hypotheses.
=+are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): .53, .65, .46, .50, .37. Does it appear that the true average amount left
=+71. Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article “Shake, Rattle, and Squeeze:How Much Is Left in That Container?” (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various
=+cth largest difference, where the value of c depends on the desired confidence level. In the case n1 5 n2 5 5, c 5 4 results in a confidence level of 94.4%, which is as close to 95% as can be obtained. Determine this CI for the strength data in Exercise 69
=+70. The confidence interval associated with Wilcoxon’s rank-sum test has the following general form. First, subtract each observation in the first sample from every observation in the second sample to obtain a set of n1n2 differences. Then the confidence interval extends from the cth smallest
=+ Use the Wilcoxon rank-sum test to decide whether true average bond strengths differ for the two adhesives. Hint: For these sample sizes, when H0 is true, P(w $c) 5 .048 for c 5 36, .028 for c 5 37, and .008 for c 5 39. Furthermore, when H0 is true, the distribution of w is symmetric about
=+69. In an experiment to compare the bond strength of two different adhesives, each adhesive was used in five bondings of two surfaces, and the force necessary to separate the two surfaces was determined for each bonding, resulting in the following data:Adhesive 1: 229 286 245 299 250 Adhesive 2:
=+8, when H0 is true, P(w $c) 5 .054 for c 5 58 and is .010 for c 5 63.Pine: .98 1.40 1.33 1.52 .73 1.20 Oak: 1.72 .67 1.55 1.56 1.42 1.23 1.77 .48
=+68. The article “A Study of Wood Stove Particulate Emission” (J. of the Air Pollution Control Fed., 1979: 724–728) reported the following data on burn time (hr) for specimens of oak and pine.Use Wilcoxon’s test at a significance level of .05 to decide whether true average burn time for
=+Alpha = 0.01 Sigma = 0.8 Sample Difference Size Power 0.5 15 0.3311 0.8 15 0.7967 Alpha = 0.01 Sigma = 0.8 Sample Target Actual Difference Size Power Power 0.5 42 0.9000 0.9047 0.8 19 0.9000 0.9147
=+who is familiar with the elements of hypothesis testing but not with type II error probabilities:Testing mean = null (versus not = null)Calculating power for mean = null + difference Alpha = 0.01 Sigma = 1 Sample Difference Size Power 0.5 15 0.1944 0.8 15 0.5619
=+67. A sample of 15 radon detectors of a particular type is to be selected, and each will be exposed to 100 pCi/L of radon. The resulting data will be used to test whether the population mean reading is in fact 100.Suppose that the reading x has a normal distribution within the population. Write a
=+The alloy will not be used unless there is strong evidence that the criterion has been met. Assuming a normal distribution and a test with 5 .01, what is the probability that a type II error will be committed and the alloy not used when in fact 5 72 and 5 5? What is this probability when 5
=+66. The Charpy V-notch impact test is to be applied to a sample of 20 specimens of a certain alloy to determine transverse lateral expansion at 110°F.To be suitable for a particular application, true average expansion should be less than 75 mils.
=+b. At a significance level of .01, rejecting H0 is appropriate if the P-value # .01, equivalent to z $2.33, that is, x $ 100 1 (2.33)(15)y1n. Determine the value of the type II error probability when 5 101 for each of the sample sizes given in part (a). Is a large sample size likely to
=+a. Determine the P-value of the test for each of the following values of n when x 5 101 (which suggests that if there is a departure from H0, it is of little practical significance): i. 100, ii. 400, iii. 1600, iv. 2500.
Showing 1400 - 1500
of 5106
First
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Last
Step by Step Answers
Get 50% OFF
Claim Now
