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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

=+b. Do the differences in rising effort that the researchers expected seem to be confirmed by the data?
=+a. Using a significance level of 5 .05, can you conclude that there is a difference in the average effort required to rise from each type of stool?
=+people could exhibit large differences in effort, even from the same type of stool, a sample of nine people was selected and each was tested on all four stools:Subject 1 2 3 4 5 6 7 8 9 Type of stool A 12 10 7 7 8 9 8 7 9 B 15 14 14 11 11 11 12 11 13 C 12 13 13 10 8 11 12 8 10 D 10 12 9 9 7 10 11
=+38. In the article “The Effects of a Pneumatic Stool and a One-Legged Stool on Lower Limb Joint Load and Muscular Activity During Sitting and Rising”(Ergonomics, 1993: 519–535), the following data is given on the effort (measured on the Borg scale)required by a subject to arise from sitting
=+b. Can you conclude that the different operators have differing effects on product acceptability rate? (Use 5 .05).
=+a. Using the three operators as blocks, can you conclude that there is a difference among the percent of acceptable products due to lathes?(Use 5 .05)
=+of a full workday shift. For each shift, the researchers recorded the percentage of acceptable products manufactured by the operator. The data from the experiment is given here:Lathe Brand 1 2 3 Operator 1 86 86 88 2 85 86 91 3 82 83 85
=+37. The article “A Software-Based Resource Selection Process in Competitive Network Environment Using ANOVA (A Case Study)” (Intl. J. of Comp.Appl., 2012: 17–21) reported on a study in which three types of lathes were compared. Each of three operators used each of the lathes for the
=+b. Based on the ANOVA results, was the use of houses as a blocking factor warranted in this study?
=+a. Using 5 .05, test the hypothesis that there are no differences between the average values reported by the three assessors.
=+A denote the factor “assessors” and B denote the blocking factor “houses.” An ANOVA calculation revealed that SSA 5 11.7, SSB 5 113.5, and SSE 5 25.6.
=+36. A certain county uses three assessors to determine the values of residential properties. To see whether the three assessors differ in their assessments, five houses are selected and each assessor is asked to determine the market value of each house. Let
=+b. Using 5 .01, can you conclude that there are differences in power consumption between the levels of the blocking factor “humidity”? Does this result support the experimenters’ use of humidity as a blocking factor
=+a. Using 5 .01, can you conclude that there is a difference between the power consumptions of the five brands?
=+from moderate to heavy humidity. For each brand, a sample of four humidifiers was randomly assigned, one each, to the four humidity levels. The resulting annual power consumption (in kilowatt-hours) is given in the following table:Humidity level Brand 1 2 3 4 1 685 792 838 875 2 722 806 893 953 3
=+35. A consumer protection organization wants to compare the annual power consumption of five different brands of dehumidifiers. Because power consumption depends on the prevailing humidity level, each brand was tested at four different humidity levels, ranging
=+a randomized block design. Using the characteristics mentioned, describe how the researchers should go about creating the blocks for such an experiment.
=+34. A pharmaceutical company wants to begin testing a drug designed to reduce blood pressure. The company wants to test the drug by measuring the blood pressures of two samples of people, those who take the drug for a prescribed period of time and those who do not take this drug (or any other
=+33. Using a significance level of 5 .05, apply Tukey’s method to the data of Exercise 12. Is there a pulse current that seems to be the best choice to yield maximum average toughness?
=+b. Using 5 .05, apply Tukey’s method to determine which if any of the loading points differ from the others.
=+a. Draw an effects plot for the data.
=+32. In Exercise 16, samples of three different loading points were tested to determine whether there were differences among their average fracture loads.
=+gave the following sample means: 14.18, 17.94, 18.00, 18.00, 25.74, 27.67. Apply Tukey’s method using 5 .05 to identify significant differences, and describe your findings (use MSE 5 13.929).
=+31. Exercise 11 described an experiment in which 26 resistivity observations were made on each of six different concrete mixtures. The article cited there
=+c. Suppose that asphalt that is not aged is taken to be a control group. Use Dunnett’s method with 5 .05 to decide whether one or both of the aged asphalt groups differ from the control group.
=+b. Use Tukey’s method with 5 .05 to determine which age categories differ from each other.
=+30. Refer to the data from Exercise 19.a. Construct an effects plot for this data.
=+29. Repeat Exercise 28 for the case where the sample means are 462.0, 502.8, 427.5, 469.3, 532.1 (i.e., the second and third sample means have been changed from their original values in Exercise 27).
=+to see which population averages can be considered different from one another ( 5 .05). Use the method of placing bars under those means that are not statistically different from one another. Write a short sentence summarizing your conclusions. Assume the MSE remains the same as in Exercise 27.
=+28. In Exercise 27, suppose that the third sample mean is 427.5 (instead of 437.5). Use Tukey’s procedure
=+/gal) for the five brands were:462.0, 512.8, 437.5, 469.3, 532.1. The MSE was 272.8 and the computed F statistic for the ANOVA test was found to be significant at 5 .05. Use Tukey’s test(at 5 .05) to investigate the pairwise differences between the coverage areas of the five brands of
=+27. An experiment to compare the wall coverage area of five different brands of yellow interior latex paint used 4 gallons of each brand. The sample means of the coverage areas (in ft2
=+b. Use Tukey’s multiple comparisons procedure to determine which groups differ from one another with respect to CS activity.
=+a. Create an effects plot of the data.
=+26. Refer to the data from Exercise 18.
=+25. Explain why creating an effects plot does not take the place of performing an ANOVA test.
=+24. Check the validity of the two fundamental ANOVA assumptions for the data in Exercise 21 by following the steps stated in Exercise 14.
=+b. Based on your conclusions in part (a), what general statement can you make about the effect of calibration problems in measuring the response variable of a single-factor ANOVA test?
=+a. Using the formulas for SSTr, SSE, SST, MSE, MSTr, and F, describe the exact effect the calibration problem in Exercise 3 will have on the entries in the ANOVA table.
=+23. In Exercise 3, a measuring instrument that was out of calibration was used to measure strengths(in kg) of three different alloys. Use the formulas in Section 9.2 to give a more specific answer to the question posed in Exercise 3. That is,
=+Using a significance level of 5%, can it be concluded that there is a difference among the true mean temperature measurements for the three structure thicknesses?
=+experiment that compared thickness of the cooling structure (mm) to the corresponding cutting temperature (K) is given here:Thickness Temperature 0.5 425.60 426.95 424.30 1.0 415.38 415.04 418.71 1.5 416.91 418.84 418.63
=+ Within the dry cutting device an interchangeable cooling structure is placed near the cutting tip.The authors of “Design and Analysis of an Internally Cooled Smart Cutting Tool for Dry Cutting” (J. of Engr. Manuf., 2012: 585–591) investigated how various physical characteristics of the
=+hazards, and higher production costs. An alternative and novel process known as dry cutting uses no cooling liquids and has shown great promise for the machining industry to produce components in an economical and ecologically desirable manner.
=+22. Friction in machining processes generates high cutting temperatures that ultimately lead to wear and thermal damage of cutting tools. Fluid is traditionally used to reduce cutting temperature, but this can lead to environmental pollution, health
=+b. Would you say the test results in part (a) are favorable or unfavorable for the practice of using wooden pegs in timber connections?
=+a. Conduct an ANOVA test to determine whether the mean bearing strength of the pegs is affected by the orientation of the pegs in the joint connection (use 5 .05).
=+stress measurement (in MPa) was recorded for each peg:Peg orientation Sample number 0° 45° 90°1 17.7 22.0 19.3 2 17.4 18.7 20.8 3 17.1 20.5 27.5 4 17.3 19.5 19.6 5 16.8 17.4 19.3 6 22.4 22.0 22.3 7 22.3 19.4 22.9 8 20.4 18.3 19.6
=+Engr., 1997: 326–332). To determine whether the bearing strength of a peg is affected by the direction with which forces are applied to the peg, three different peg orientations were used in the study: 0°, 45°, and 90°. The pegs were randomly assigned to one of the three orientations, and a
=+21. Pegged mortise and tenon joints have been used to build wooden structures for centuries. Since the mid-1960s, there has been renewed interest in this method of timber connection because of its inherent strength compared with other methods of connection. In a recent study of the bearing
=+ Using a significance level of 5%, conduct an ANOVA test to determine whether there is a difference in the mean bending strengths among the three types of wood.
=+15% moisture content. The results of the experiment are given here.Bending strength (in MPa)Species Douglas-Fir: 65 46 52 39 41 44 Hem-Fir: 45 48 32 30 47 50 Spruce Pine- Fir: 42 38 30 28 39 40
=+particular stress applied, the board dimensions were kept the same in each of the three wood species. Wood samples were selected from randomly selected sawmills, and, according to ASTM Standard D 4761, each sample was conditioned by kiln and air drying to achieve approximately a
=+20. To assess the reliability of timber structures and related building design codes, many researchers have studied strength factors of structural lumber. In one such study (“Size Effects in Visually Graded Softwood Structural Lumber,” J. of Materials in Civil Engr., 1995: 19–29), three
=+ Can you conclude (using 5 .05) that there is a difference between the means for the three age categories?
=+Another group of samples was not aged. The following table shows the percentage area of the same slice of the chromatograms of these samples(i.e., area of the slice as a percentage of the entire chromatogram):Age of asphalt 0 hours 5 hours 24 hours Mean 3.43 3.18 3.22 Standard deviation .22 .13
=+Civil Engr., 1997: 31–39). To determine whether certain bands or slices of the chromatogram can be used to distinguish different aging conditions in asphalt, samples of grade AC-10 asphalt were sampled from several sources and artificially aged, some samples for 5 hours and others for 24 hours.
=+19. A study was conducted to determine whether certain physical properties of asphalt are related to portions of a gel permeation chromatogram of the asphalt (“Methodology for Defining LMS Portion in Asphalt Chromatogram,” J. of Materials in
=+b. Using 5 .05, can you conclude that true average activity differs for the three groups?
=+a. Construct an ANOVA table for this experiment.
=+18. The accompanying summary data on skeletalmuscle citrate synthase activity (nmol/min/mg)appeared in the article “Impact of Lifelong Sedentary Behavior on Mitochondrial Function of Mice Skeletal Muscle” (J. of Gerontology, 2009: 927–939):Young Old Sedentary Old Active Sample size 10 8
=+b. Based on your conclusions in part (a), what general statement can you make about the effect of changing units of measure in an ANOVA test?
=+a. Use the formulas for SSTr, SSE, SST, MSE, and MSTr to discuss the effect that changing from inches to centimeters has on the ANOVA calculations.
=+inches. At a later date, the experimenter decides that plant heights should have been measured in centimeters instead of inches. After multiplying the data in the original samples by 2.54 (1 in. 5 2.54 cm), the experimenter wants to know what effect this data conversion will have on the
=+17. In an experiment to study the possible effects of four different concentrations of a chemical on heights of newly grown plants, suppose that an ANOVA test is conducted and that plant height is measured in
=+c. Returning to the Minitab output, note that the number reported under P corresponds to the P-value. Is the P-value exactly zero? What does it mean when Minitab reports 0.000?
=+b. At a significance level of .01, can you conclude there is a difference among true average fracture loads for the three loading point locations?
=+a. Use your calculator to confirm Minitab’s computations.
=+the following fracture load data (kN):Distance Fracture Load 42 2.62 2.99 3.39 2.86 36 3.47 3.85 3.77 3.63 31.2 4.78 4.41 4.91 5.06 Here is the corresponding Minitab ANOVA table:One-way ANOVA: Fracture versus Distance Source DF SS MS F P Dist. 2 6.7653 3.3826 48.58 0.000 Error 9 0.6267 0.0696
=+ The test was performed by applying asymmetric three-point bending loads on PMMA specimens and varied the location of one of the three loading points to determine its effect on fracture load.In one experiment, three loading point locations based on different distances (mm) from the center of
=+materials. The authors applied this new fracture test to the brittle polymer polymethylmethacrylate(PMMA), more popularly known as Plexiglas, which is widely used in commercial products.
=+16. According to “Evaluating Fracture Behavior of Brittle Polymeric Materials Using an IASCB Specimen”(J. of Engr. Manuf., 2013: 133–140), researchers have recently proposed an improved test for the investigation of fracture toughness of brittle polymeric
=+15. It is common practice in many countries to destroy(shred) refrigerators at the end of their useful lives.In this process, material from insulating foam may be released into the atmosphere. The article “Release of Fluorocarbons from Insulation Foam in Home Appliances During Shredding” (J.
=+b. The assumption of equal population variances is plausible if the ratio of the largest sample variance to the smallest sample variance is not much more than 4. Is it plausible that the population variances are approximately equal?
=+a. Create a single normal probability (quantile)plot based on the deviations of the sample data from the sample mean for each of the three samples. Does the assumption of normality appear to hold?
=+ Consider conducting an ANOVA test to see if there are any differences in the true mean ATFW caused by the different coolant pressures. The validity of an ANOVA test depends on the extent to which the two fundamental ANOVA assumptions (normal populations; equal population variances) are
=+ In one experiment, the researchers machined six specimens of Inconel 718 at each of three different HPJA coolant pressure levels (.6, 10, and 30 MPa)and recorded the corresponding average tool flank wear (ATFW), a combination of abrasive and depth of cut notch wear:Pressure.6: 145.00 158.14
=+14. In “Investigation on Machining Performance of Inconel 718 Under High Pressure Cooling Conditions” (J. of Mech. Engr., 2012: 683–690), researchers varied selected high-pressure jet-assisted(HPJA) machining parameters for the nickel-based alloy Inconel 718 and investigated their effect
=+b. Using a significance level of .01, can you conclude that there is a difference between the mean shear bond strength of the five groups?
=+a. State the hypotheses of interest in this experiment.
=+13. The article “Influence of Contamination and Cleaning on Bond Strength to Modified Zirconia”(Dental Materials, 2009: 1541–1550) reported on an experiment in which 50 zirconium-oxide disks were divided into 5 groups of 10 each. Then a different contamination/cleaning protocol was used
=+Pulse Current: 100 100 100 120 120 120 140 140 140 Toughness: 39 47 44 52 56 53 40 46 42 Use 5 .05 to conduct the test for whether there are any differences in the true average weld toughness that may be attributable to the different pulse currents.
=+Waste Mgmt. Assoc., 2012: 1978–1988) researched the impact of different PCGTAW process parameters on mechanical properties of the welds of a particular SDSS. One investigation focused on seeing how pulse current (A) of the PCGTAW affects the toughness (J) of the SDSS welds. Here are
=+has shown that the pulsed current gas tungsten arc welding (PCGTAW) process offers superior SDSS welds compared to other methods. The authors of“Optimization of Experimental Conditions of the Pulsed Current GTAW Parameters for Mechanical Properties of SDSS UNS S32760 Welds Based on the Taguchi
=+12. Super duplex stainless steels (SDSS) are ironbased alloys that offer an excellent combination of toughness and mechanical strength. Such alloys are useful for many applications in the chemical and petrochemical industries. Recent research
=+b. State the null and alternative hypotheses of interest in this experiment.c. Use 5 .05 to carry out the hypothesis test in part (b).
=+a. Fill in the missing entries in the ANOVA table.
=+11. An experiment was carried out to compare electrical resistivity for six different low-permeability concrete bridge deck mixtures. There were 26 measurements on concrete cylinders for each mixture;these were obtained 28 days after casting. The entries in the accompanying ANOVA table are
=+10. Five brands of raw materials are tested for their effect on a process yield. Random samples of size 10 are used for each of the materials. Complete the following ANOVA table for this experiment:Source of variation df SS MS F Brand 15.32 Error .64 Total variation
=+in this exercise are df1 5 3, df2 5 20. Using 5 .05, conduct the test to determine whether you can conclude that there are differences between 1, 2, 3, and 4
=+9. In a test of the hypothesis H0: 1 5 2 5 3 5 4, samples of size 6 were selected from each of four populations, and an F statistic value of 4.12 was calculated (using the methods in the next section). The appropriate degrees of freedom for the F distribution
=+b. Analysis of the data using statistical software yielded a P-value of P 5 .029. Using 5 .01, what conclusion would you draw regarding the test in part (a)?
=+a. Samples of five type-A nozzles, six type-B nozzles, seven type-C nozzles, and six type-D nozzles were tested. ANOVA calculations yielded an F value of 3.68 with df1 5 3 and df2 5 20.State and test the relevant hypotheses using 5 .01.
=+8. An experiment was carried out to compare flow rates for four different types of nozzles.
=+7. Based on your answers to Exercise 6(a)–(d), what effect does interchanging df1 and df2 have on the critical F value (for a fixed upper-tail area)?
=+g. The probability P(4.74 # F # 7.87) for df1 5 10 and df2 5 5
=+f. The probability P(F # 6.16) for df1 5 6 and df2 5 4
=+d. The F critical value based on df1 5 8 and df2 5 5 that captures an upper-tail area of .01e. The 95th percentile of an F distribution with df1 5 3 and df2 5 20
=+b. The F critical value based on df1 5 8 and df2 5 5 that captures an upper-tail area of .05c. The F critical value based on df1 5 5 and df2 5 8 that captures an upper-tail area of .01
=+6. Use the table of F distribution critical values (Appendix Table VIII) to finda. The F critical value based on df1 5 5 and df2 5 8 that captures an upper-tail area of .05

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