Solution Manual For Managing, Controlling, and Improving Quality, 1st Edition

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CHAPTER1Introduction to QualityLearning ObjectivesAfter completing this chapter you should be able to:1.Define and discuss quality and quality improvement2.Discuss the different dimensions of quality3.Discuss the evolution of modern qualityimprovement methods4.Discuss the role that variability and statistical methods play in controlling and improving quality5.Explain the links between quality and productivity and between quality and cost6.Discuss product liabilityImportant Terms and ConceptsAcceptance-samplingAppraisal costsCritical-to-quality (CTQ)Dimensions of qualityFitness for useInternal and external failurecostsNonconforming product orservicePrevention costsProduct liabilityQuality assuranceQualitycharacteristicsQuality control andimprovementQuality engineeringQuality of conformanceQuality of designQuality planningSpecificationsVariabilityExercises1.1.Why is it difficult to define quality?Even the American Society for Quality describes “quality” as a subjective term for which each person orsector has its own definition.Given a large set of customers considering purchasing the same product orservice, it is likely that each customerevaluates itin terms of a completely different set of desirablecharacteristics.As a result, it is extremely difficult to come up with a single definition of quality that couldmeet the expectations of allcustomers for all products or services.(For further details refer to page 2)1.7.What are the three primary technical tools used for quality control and improvement?The three primary statistical technical tools include:statistical process control,design of experiments,andacceptance-sampling.1.10.Discuss the statement“Quality is the responsibility of the quality department.”The responsibility for quality spans the entire organization. Quality improvement must be a total,company-wide activity, and that every organizational unitmustactively participate.Because qualityimprovement activities are so broad, successful efforts require, as an initial step, senior management

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commitment. This commitment involves emphasis on the importance of quality, identification of therespective quality responsibilities of the various organizational units, and explicit accountability for qualityimprovement of all managers and employees in the company.Nevertheless, the quality department takescare of the quality planning and analyses in order to make sure all quality efforts are being successfullyimplemented throughout the organization.(For further details refer to pages 26-27)1.12.Explain why it is necessary to consider variability around the mean or nominal dimension as a measure ofquality.The primaryobjectiveof quality engineering efforts is thesystematic reduction of variabilityin the keyquality characteristics of the product. The introduction of statistical process control willhelp tostabilizeprocessesand reduce theirvariability.However, it is not satisfactory just to meet requirements-Ffurtherreduction of variabilityaround the mean or nominal dimensionoften alsoleads to better productperformance and enhanced competitive position.(For further details refer to page17).

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CHAPTER2Managing aspects of QualityLearning ObjectivesAfter completing this chapter you should be able to:1.Describe the quality management philosophies of W. Edwards Deming, Joseph M. Juran, and Armand V.Feigenbaum2.Discuss totalquality management, six-sigma, the Malcolm Baldrige National Quality Award, and qualitysystems and standards3.Understand the importance of selecting good projects for improvement activities4.Explain the five steps of DMAIC5.Know when and when not to use DMAICImportant Terms and ConceptsAnalyze stepControl stepDefine stepDesign for Six-Sigma (DFSS)DMAICFailure modes and effects analysis(FMEA)Improve stepKey process input variables(KPIV)Key process output variables(KPOV)Measure stepProject charterSIPOC diagramSix-SigmaTollgateExercises2.12.Explain the importance of tollgates in the DMAIC process.At a tollgate, a project team presents its work to managers and “owners” of the process. In a six-sigmaorganization, the tollgate participants also would include the project champion, master black belts, andother black belts not working directly on the project. Tollgates are where the project is reviewed to ensurethat it is on track and they provide a continuing opportunity to evaluate whether the team can successfullycomplete the project on schedule. Tollgates also present an opportunity to provide guidance regarding theuse of specific technical tools and other information about the problem. Organization problems and otherbarriers to success—and strategies for dealingwith them—also often are identified during tollgate reviews.Tollgates are critical to the overall problem-solving process; It is important that these reviews be conductedvery soon after the team completes each step.2.18.Suppose that your business is operating at the four-and-a-half-sigma quality level. If projectshave an average improvement rate of 50% annually, how many years will it take to achieve six-sigma quality?Assumingthea1.5shiftin the meanthat is customary with six-sigma applications[𝑋′~𝑁(𝜇′=1.50,𝜎=1)],the percentage within the−4.5𝜎and4.5𝜎limits is:

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𝑃(−4.5≤𝑋′≤4.5)=𝑃(𝑋′≤4.5)−𝑃(𝑋′≤−4.5)=0.9987Then, the𝑝𝑝𝑚defective is:(1−0.9987)∗106≈1,350.Using the equation in Example 2.1:3.4=1,350∗(1−0.5)𝑥𝑥=𝑙𝑛(3.41,350)𝑙𝑛(1−0.5)⁄𝑥=8.6332`It will takethe businessnearly 8 years and 7 monthsto achieve6𝜎quality.2.19.Explain why it is important to separate sources of variability intospecial or assignable causes and commonor chance causes.Common or chance causes are due to the inherent variability in the system and cannot generally becontrolled. Special or assignable causes can be discovered and removed, thus reducing the variability inthe overall system. It is important to distinguish between the two types of variability, because the strategyto reduce variability depends on the source. Chance cause variability can only be removed by changing thesystem, while assignable cause variability can be addressed by finding and eliminating the assignablecauses.2.21.Suppose that during the analyze phase an obvious solution is discovered. Should that solution beimmediately implemented and the remaining steps of DMAIC abandoned? Discuss your answer.The answer is generally NO. The advantage of completing the rest of the DMAIC process is that thesolution will be documented, tested, and it’s applicability to other parts of the business will be evaluated.An immediate implementation of an “obvious” solution may not lead to an appropriate control plan.Completing the rest of DMAICprocesscan also lead to further refinements and improvements to thesolution.Also the transition plan developed in the control phaseincludes a validation check several monthsafter project completion, if the DMAIC processis notcompleted;there is a danger of the originalresultsnot being sustained.

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CHAPTER3Tools and Techniques for Quality Control and ImprovementLearning ObjectivesAftercompleting this chapter you should be able to:1.Understand chance and assignable causes of variability in a process2.Explain the statistical basis of the Shewhart control chart3.Understand the basic process improvement tools of SPC: the histogram or stem-and-leaf plot, the checksheet, the Pareto chart, the cause-and-effect diagram, the defect concentration diagram, the scatter diagram,and the control chart4.Explain how sensitizing rules and pattern recognition are used inconjunction with control chartsImportant Terms and ConceptsAction limitsAssignable causes of variationCause-and-effect diagramChance causes of variationCheck sheetControl chartControl limitsDefect concentration diagramDesigned experimentsFlow charts,operations processcharts, and value stream mappingFactorial experimentIn-control processMagnificent sevenOut-of-control-action plan(OCAP)Out-of-control processPareto chartPatterns on control chartsScatter diagramShewhart control chartsStatistical control of a processStatistical process control (SPC)Three-sigma control limitsExercises3.6.Consider the control charts shown here. Does the pattern appear random?No, the last four samples are located at a distance greater than 1from the center line.

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3.7.Consider the control charts shown here. Does the pattern appear random?Yes, the pattern is random.3.8.Consider the control charts shown here. Does the pattern appear random?Yes, the pattern is random.

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CHAPTER4Statistical Inference about Product and Process QualityLearning ObjectivesAfter completing this chapter you should be able to:1.Construct and interpret visual data displays, including the stem-and-leaf plot, the histogram, and the boxplot2.Compute and interpret the sample mean, the sample variance, the sample standard deviation, and thesample range3.Explain the concepts of a random variable and a probability distribution4.Understand and interpret the mean, variance, and standard deviation of a probability distribution5.Determine probabilities from probability distributions6.Construct and interpret confidence intervals on a single mean and on the difference in two means7.Construct and interpret confidence intervals on a single variance and the ratio of two variances8.Construct and interpret confidence intervals on a single proportion and on the difference in two proportions9.Understand how the analysis of variance (ANOVA) is used to compare more than two samplesImportant Terms and ConceptsAlternative hypothesisAnalysis of variance (ANOVA)Approximations to probabilitydistributionsBinomial distributionBox plotCentral limit theoremChecking assumptions forstatisticalinference proceduresChi-square distributionConfidence intervalConfidence intervals on means,known variance(s)Confidence intervals on means,unknown variance(s)Confidence intervals onproportionsConfidence intervals on thevariance of a normal distributionConfidence intervals on thevariancesof two normal distributionsContinuous distributionControl limit theoremCritical region for a test statisticDescriptive statisticsDiscrete distributionExponential distributionF-distributionHistogramHypothesis testingInterquartile rangeMean of a distributionMedianNegative binomialdistributionNormal distributionNormal probability plotParameters of a distributionPercentilePoisson distributionPopulationPower of a statistical testProbability distributionProbability plottingP-valueQuartileRandom sampleRandom variableResidual analysisRun chartSampling distributionStandard deviationStandard normal distributionStatisticStatisticsStem-and-leaf displayt-distributionTest statisticTests of hypotheses on means,known variance(s)Tests of hypotheses on means,unknown variance(s)Tests of hypotheses onproportionsTests of hypotheses on thevarianceof a normal distributionTests of hypotheses on thevariancesof two normal distributionsTime series plotType I errorType IIerrorUniform distributionVariance of a distribution

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Exercises4.1.Thefill amount of liquid detergent bottles is being analyzed. Twelve bottles, randomly selected from theprocess, are measured, and the results are as follows (in fluid ounces):16.05, 16.03, 16.02, 16.04, 16.05,16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07.a.Calculate the sample average.∑𝑥𝑖𝑛𝑖=1𝑛=16.05+⋯+16.0712=16.0292𝑓𝑙𝑢𝑖𝑑𝑜𝑢𝑛𝑐𝑒𝑠b.Calculate the sample standard deviation.√∑(𝑥𝑖−𝑥̅)2𝑛𝑖=1𝑛−1=√(16.05−16.0255)2+⋯+(16.07−16.0255)211=0.0202𝑓𝑙𝑢𝑖𝑑𝑜𝑢𝑛𝑐𝑒𝑠4.3.Waiting times for customers in an airline reservation system are (in seconds) 953,955, 948,951,957,949,954, 950, 959.a.Calculate the sample average.∑𝑥𝑖𝑛𝑖=1𝑛=953+⋯+9599=952.8889𝑠𝑒𝑐𝑜𝑛𝑑𝑠b.Calculate the sample standard deviation.√∑(𝑥𝑖−𝑥̅)2𝑛𝑖=1𝑛−1=√(96−121.25)2+⋯+(156−121.25)27=3.7231𝑠𝑒𝑐𝑜𝑛𝑑𝑠4.5.The time to complete an order (in seconds) is as follows:96, 102, 104, 108, 126, 128, 150, and 156.a.Calculate the sample average.∑𝑥𝑖𝑛𝑖=1𝑛=96+⋯+1568=121.25𝑠𝑒𝑐𝑜𝑛𝑑𝑠

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b.Calculate the sample standard deviation.√∑(𝑥𝑖−𝑥̅)2𝑛𝑖=1𝑛−1=√(96−121.25)2+⋯+(156−121.25)27=22.6258𝑠𝑒𝑐𝑜𝑛𝑑𝑠4.9.Construct and interpret a normal probability plot ofthe volumes in Exercise 4.1.Eitheritsquestion4.8orits exercise 4.3. If its Question 4.9, the data is different, the whole solution changes.Data:16.05, 16.03, 16.02, 16.04, 16.05, 16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07MTB > Graph >Probability Plot > SingleIt can be assumed that the data follows a normal distribution.4.11.Construct a normal probability plot of the chemical process yield data in Exercise 4.7. Does theassumption that process yield iswell modeled by a normal distribution seem reasonable?13.314.314.915.215.814.2161414.516.113.715.213.716.914.914.415.313.115.215.915.114.913.613.715.315.514.516.513.415.215.313.814.312.615.314.814.114.414.315.614.814.615.615.114.815.215.614.515.214.315.81714.314.616.112.814.515.413.314.914.316.413.916.114.615.214.114.816.414.215.216.614.116.815.41416.915.714.415.6970965960955950945940999590807060504030201051C1PercentMean952.9StDev3.723N9AD0.166P-Value0.908Probability Plot of C1Normal - 95% CIFormatted:Font: Bold

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MTB > Graph > Probability Plot > SingleIt can be assumed that the data follows a normal distribution.4.13.Consider the chemical process yield data in Exercise 4.12. Calculate the sample averageand standarddeviation.94.187.394.192.484.685.493.284.192.190.683.686.690.690.196.489.185.491.791.495.288.288.889.787.588.286.186.486.487.684.286.194.38585.185.185.195.193.284.98489.690.59086.787.393.79095.692.48389.687.790.188.387.395.390.390.694.384.186.694.193.189.497.383.791.297.894.688.696.882.986.193.196.384.194.487.390.486.494.782.696.186.489.187.691.183.19884.5a.Calculate the sample average.∑𝑥𝑖𝑛𝑖=1𝑛=94.1+⋯+84.590=89.475619181716151413121199.99995908070605040302010510.1C3PercentMean14.90StDev0.9804N80AD0.249P-Value0.740Probability Plot of C3Normal - 95% CI

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b.Calculate the samplestandard deviation.√∑(𝑥𝑖−𝑥̅)2𝑛𝑖=1𝑛−1=√(94.1−89.4756)2+⋯+(90−89.4756)289=4.15784.17.Suppose that two fair dice are tossed and the sum of the dice is observed. Determine the probabilitydistribution of𝑥, the sum of the dice.Let𝑥=𝑑1+𝑑2represent the sum of two dice. First, letus determine the possible values for𝑥, and then,letusdetermine how likely it istoobtain each value.For example,𝑥=3can occur when𝑑1=1and𝑑2=2or when𝑑1=2and𝑑2=1. Hence, there are two possible scenarios that lead to𝑥=3.Similarly, let us list all possible scenarios for𝑥=𝑑1+𝑑2:Scenario𝑑1𝑑2𝑥Scenario𝑑1𝑑2𝑥111219415212320426313421437414522448515623459616724461072132551682242652792352753810246285491125729551012268305611133143161714325326281533633639163473464101735835651118369366612From the previous table, we can conclude𝑥can assume integer values between 2 and 12. Now, let usdefine the likelihood of each scenario by counting how many times each scenario occurs and dividing it bythe total number of possible scenarios (36).

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𝑥Number of Scenarios𝑃(𝑑1+𝑑2=𝑥)211/36322/36433/36544/36655/36766/36855/36944/361033/361122/361211/364.18.Find the mean and standard deviation of𝑥in Exercise 4.17.𝑥𝑝(𝑥)21/3632/3643/3654/3665/3676/3685/3694/36103/36112/36121/36Mean:𝐸(𝑥)=∑𝑥∗𝑝(𝑥)=2∗136+⋯+12∗136⁄⁄=7𝑛𝑖=1Standard Deviation:√𝑉(𝑥)=∑𝑥2∗𝑝(𝑥)−[𝐸(𝑥)]2=22∗136+⋯+122∗136⁄⁄−72=5.8333𝑛𝑖=14.19.Thetensile strength of a metal part isnormally distributed with mean 40 lb and standard deviation 5 lb.If50,000 parts are produced, how many would you expect to fail to meet a minimum specification limit of 35lb tensile strength? How many would have a tensile strength in excess of 48 lb?𝑋~𝑁(𝜇=40,𝜎=5)Probability that a part fails to meet 35 lb specification(look upprobability in a normal probability table):𝑃(𝑋<35)=𝑃(𝑍<35−405)=𝑃(𝑍<−1)=0.1587

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Expected number of parts that fail to meet 35 lb:𝑛𝑃(𝑋<35)=50,000∗0.1587≈7,9337935𝑝𝑎𝑟𝑡𝑠4.23.A light bulb has a normally distributed light output with mean 5,000 end foot-candles and standarddeviation of 50 end foot-candles. Find a lower specification limit such that only 0.5% of the bulbswill notexceed this limit.𝑋~𝑁(𝜇=5000,𝜎=50)Lower specification limit (x):𝑃(𝑋<𝑥)=𝑃(𝑍<𝑥−500050)=𝑃(𝑍<𝑧)=0.005Look upinverse value of𝑧in probability table:𝑧=𝑥−500050=−2.5758𝑥=5000+50𝑧=5000−2.5758∗50=4,871.20854.25.The tensile strength of fiber used in manufacturing cloth is of interest to the purchaser. Previousexperience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiberspecimensis selected, and the average tensile strength is found to be 127 psi.a.Build a 95% lower confidence interval in the mean tensile strength.𝜇≥𝑥̅−𝑧0.05∗𝜎√𝑛⁄=127−1.6449∗2√8=125.8369⁄psib.What can you conclude?For95% of thetimeCIsconstructed, the mean tensile strength is expected tobe equal to orexceed125.8369psi.4.27.The service life of a battery used in a cardiac pacemaker is assumed o benormally distributed. A randomsample of 10 batteries is subjected to an accelerated life test by running the, continuously at an elevatedtemperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2,28.4, 25.0, 27.8, 27.3, and 25.7. The manufacturer wants to be certain that the mean battery life exceeds 25hours in accelerated life testing.a.Construct a 90% two-sided confidence interval on mean life in the accelerated test.𝑥̅−𝑡𝛼2,𝑛−1⁄𝑠√𝑛≤𝜇≤⁄𝑥̅+𝑡𝛼2,𝑛−1⁄𝑠√𝑛⁄26−𝑡0.12,10−1⁄∗1.6248√10≤𝜇≤⁄26+𝑡0.12,10−1⁄∗1.6248√10⁄26−1.8331∗1.6248√10≤𝜇≤⁄26+1.8331∗1.6248√10⁄25.0581≤𝜇≤26.9419

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b.Construct a normal probability plot of the battery life data. What conclusions can you draw?MTB > Graph > Probability Plot > SingleNormality assumption is reasonable. Also, the manufacturer canbecertain that the mean exceeds 25 hours,95% of theCIsconstructed will not containa mean of25 hrs.time,given the lower limit for the confidenceinterval exceeds 25 hrs.NOTE:Conclusions on theconfidence interval are highly dependent on thenumber of significant digits usedthroughout the calculations.4.29.A company has just purchased a new billboard near the freeway. Sales for the past 10 days have been 483,532, 444, 510, 467, 461, 450, 444, 540, and 499. Build a 95% two-sided confidenceinterval on sales. Isthere any evidence that the billboard has increased sales from its previous average of 475 per day?𝑥̅−𝑡𝛼2,𝑛−1⁄𝑠√𝑛≤𝜇≤⁄𝑥̅+𝑡𝛼2,𝑛−1⁄𝑠√𝑛⁄483−𝑡0.052,10−1⁄∗35.7553√10≤𝜇≤⁄483+𝑡0.052,10−1⁄∗35.7553√10⁄483−2.2622∗35.7553√10≤𝜇≤⁄483+2.2622∗35.7553√10⁄457.4217≤𝜇≤508.5783Since 475 is within the confidence interval, there is no evidencethat the billboard has increased sales fromits previous average of 475 per day.NOTE: Confidence intervals endpoints are highly dependent on the number of significant digits used throughout thecalculations.4.31.A company is evaluating the quality of aluminum rods received in a recent shipment. Diameters ofaluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001in. A random sample of 25 rods has an average diameter of 0.5046 in. Test whether or not the mean roddiameter is 0.5025 using a two-sided confidence interval.𝑥̅−𝑧0.052⁄∗𝜎√𝑛⁄≤𝜇≤𝑥̅+𝑧0.052⁄∗𝜎√𝑛⁄0.5046−1.96∗0.0001√25⁄≤𝜇≤0.5046−1.96∗0.0001√25⁄0.504561≤𝜇≤0.50463932302826242220999590807060504030201051C1PercentMean26StDev1.625N10AD0.114P-Value0.986Probability Plot of C1Normal - 95% CI

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Since0.5025 is not within the confidence interval, we do not have evidence to state the mean rod diameteris 0.5025 in.4.33.Last month, a large national bank’s average payment time was 33 days with a standard deviation of 4 days.This month, the average payment time was 33.5 days with a standard deviation of 4 days. They had 1,000customers both months.a.Build a 95% confidence interval on the difference between this month’s and last month’s payment times.Is there evidence that payment times are increasing?Using a one-sided confidence interval:𝜇𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝜇𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≥𝑥̅𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝑥̅𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠−𝜎0.05√𝜎𝐶𝑢𝑟𝑟𝑒𝑛𝑡2𝑛𝐶𝑢𝑟𝑟𝑒𝑛𝑡+𝜎𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠2𝑛𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠𝜇𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝜇𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≥33.5−33−1.6449√421000+421000=0.2058Sincethe lower limit for the difference is positive, there is evidence to state that the payment times areincreasing.Usinga two-sided confidence interval:𝑥̅𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝑥̅𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠−𝜎0.052⁄√𝜎𝐶𝑢𝑟𝑟𝑒𝑛𝑡2𝑛𝐶𝑢𝑟𝑟𝑒𝑛𝑡+𝜎𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠2𝑛𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≤𝜇𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝜇𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≤𝑥̅𝐶𝑢𝑟𝑟𝑒𝑛𝑡+𝑥̅𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠+𝜎0.052⁄√𝜎𝐶𝑢𝑟𝑟𝑒𝑛𝑡2𝑛𝐶𝑢𝑟𝑟𝑒𝑛𝑡+𝜎𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠2𝑛𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠33.5−33−1.96√421000+421000≤𝜇𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝜇𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≤33.5−33+1.96√421000+4210000.149≤𝜇𝐶𝑢𝑟𝑟𝑒𝑛𝑡−𝜇𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠≤0.851We still arrive at the same conclusion that payment times have increased given the entire confidenceinterval is greater than zero.4.35.A supplierreceived results on the hardness of metal from two different hardening process (1) salt-waterquenching and (2) oil quenching. The results are shown in Table 4E.4.

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Salt QuenchOil Quench145152150150153147148155141140152146146158154152139151148143a.Construct a 95% confidence interval on the difference in mean hardness.Using a two-sided confidence intervalassuming unequal variances:=(𝑠12𝑛1+𝑠22𝑛2)2(𝑠12𝑛1)2𝑛1−1+(𝑠22𝑛2)2𝑛2−1=(24.711110+29.822210)2(24.711110)210−1+(29.822210)210−1=17.3992⇒17𝑥̅𝑆𝑄−𝑥̅𝑂𝑄−𝑡0.052,⁄√𝑠𝑆𝑄2𝑛𝑆𝑄+𝑠𝑂𝑄2𝑛𝑂𝑄≤𝜇𝑆𝑄−𝜇𝑂𝑄≤𝑥̅𝑆𝑄−𝑥̅𝑂𝑄+𝑡0.052,⁄√𝑠𝑆𝑄2𝑛𝑆𝑄+𝑠𝑂𝑄2𝑛𝑂𝑄147.6−149.4−2.1098√24.711110+29.822210≤𝜇𝑆𝑄−𝜇𝑂𝑄≤33.5−33+2.1098√24.711110+29.822210−6.7269≤𝜇𝑆𝑄−𝜇𝑂𝑄≤3.1269b.Construct a 95% confidence interval on the ratio of variances.𝑠12𝑠22𝐹1−𝛼2,𝑛2−1,𝑛1−1⁄≤𝜎𝑆𝑄2𝜎𝑂𝑄2⁄≤𝑠12𝑠22𝐹𝛼2,𝑛2−1,𝑛1−1⁄24.711129.8222𝐹1−0.052,10−1,10−1⁄≤𝜎𝑆𝑄2𝜎𝑂𝑄2⁄≤24.711129.8222𝐹0.052,10−1,10−1⁄24.711129.8222∗0.2484≤𝜎𝑆𝑄2𝜎𝑂𝑄2⁄≤24.711129.8222∗4.02600.2058≤𝜎𝑆𝑄2𝜎𝑂𝑄2⁄≤3.3360c.Does this assumption of normality seem appropriate for this data?

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MTB > Graph > Probability Plot > SingleThe normality assumption for both samples looks reasonable.d.What can youconclude?For the confidence interval based on the mean difference,zero is includedwithin the interval. For theconfidence interval based on the ratio of the variances, one is includedwithin the interval.Hence, there is nosignificancesignificantdifference inthe samplemeans or variances.4.37.A random sample of 500 connecting rod pins contains 65 nonconforming units. Estimate the processfraction nonconforming. Construct a 90% upper confidence interval on the true process fractionnonconforming. Is it possible that the true fraction nonconforming defective is 10%?Process fraction nonconforming estimate:𝑝̂=65500=0.1390% upper confidence interval:𝑝≤𝑝̅+𝑧0.1√𝑝̅(1−𝑝̅)𝑛𝑝≤0.13+1.2816√0.13∗(1−0.13)500𝑝≤0.1493For90% of thetimeCIs constructed, the true defective fraction is less than14.93%(evidence suggestspmight belarger than 0.10)4.41.A new purification unit is installed in a chemical process. Before and after installation data was collectedregarding the percentage of impurity:Before(1):Sample mean = 9.85, Sample variance= 6.79, Number of samples = 10After(2):Sample mean = 8.08, Sample variance = 6.18, Number of samples = 8170160150140130999590807060504030201051170160150140130Salt QuenchPercentOil QuenchMean147.6StDev4.971N10AD0.218P-Value0.779Salt QuenchMean149.4StDev5.461N10AD0.169P-Value0.906Oil QuenchProbability Plot of Salt Quench, Oil QuenchNormal - 95% CI

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a.Can you conclude that the two variances are equal using a two-sided 95% confidence interval?𝑠12𝑠22𝐹1−𝛼2,𝑛2−1,𝑛1−1⁄≤𝜎12𝜎22⁄≤𝑠12𝑠22𝐹𝛼2,𝑛2−1,𝑛1−1⁄6.796.18𝐹1−0.052,8−1,10−1⁄≤𝜎12𝜎22⁄≤6.796.18𝐹0.052,8−1,10−1⁄24.711129.8222∗0.2073≤𝜎12𝜎22⁄≤24.711129.8222∗4.61130.2278≤𝜎12𝜎22⁄≤4.6113Since one is included in the interval,we canconcludethatthe variancesare equal.b.Can you conclude that the new purification device has reduced the mean percentage of impurity using atwo-sided 95% confidence interval?Using a two-sided confidence interval assuming equal variances:=(𝑠12𝑛1+𝑠22𝑛2)2(𝑠12𝑛1)2𝑛1−1+(𝑠22𝑛2)2𝑛2−1=(6.7910+6.188)2(6.7910)210−1+(6.188)28−1=15.4373⇒15𝑥̅1−𝑥̅2−𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛2≤𝜇1−𝜇2≤𝑥̅1−𝑥̅2+𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛29.85−8.08−2.1315√6.7910+6.188≤𝜇1−𝜇2≤9.85−8.08+2.1315√6.7910+6.188−0.7980≤𝜇1−𝜇2≤4.3380Since zero is within the interval, we cannotconcludethatthe new devicehasreduced the mean percentageof impurity.

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4.43.The diameter of a metal rod is measured by 12 inspectors, eachusing both a micrometer caliper and avernier caliper. The results are shown in Table 4E.5. Is there a difference between the mean measurementsproduced by the two types of caliper? Use alpha = 0.01.InspectorMicrometer CaliperVernier Caliper10.150.15120.1510.1530.1510.15140.1520.1550.1510.15160.150.15170.1510.15380.1530.15590.1520.154100.1510.151110.1510.15120.1510.152=(𝑠12𝑛1+𝑠22𝑛2)2(𝑠12𝑛1)2𝑛1−1+(𝑠22𝑛2)2𝑛2−1=(6.97∗10−712+2.63∗10−612)2(6.97∗10−712)212−1+(2.63∗10−612)212−1=16.4498⇒16𝑥̅1−𝑥̅2−𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛2≤𝜇1−𝜇2≤𝑥̅1−𝑥̅2+𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛20.1512−0.1516−2.9208√6.97∗10−712+2.63∗10−612≤𝜇1−𝜇2≤0.1512−0.1516+2.9208√6.97∗10−712+2.63∗10−612−0.0020≤𝜇1−𝜇2≤0.0011Sincezero iswithin the confidence interval, no evidence suggests the mean measurements produced by thetwo types of calipers are different.

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4.45.Rehab, Inc. is evaluating patient success results for its Northbrook and Southbrook locations. Eachsuccessfully treated 10 patients within the last year following elbow surgery. Total recovery times and %range of motion achieved are listed in Table 4E.6.Northbrook(1)Southbrook(2)Recovery Time% ROM AchievedRecovery Time% ROM Achieved148.810.69135.250.98188.720.89174.990.47186.770.65144.150.85152.720.73161.810.71197.80.79151.350.94162.780.81149.690.56192.180.64136.170.57200.170.88146.250.2181.320.67162.880.84193.030.74183.950.62a.Build a 95%two-sided confidence interval for the differences of the average of the recovery times. Is thereevidence to support that the facilities are different?=(𝑠12𝑛1+𝑠22𝑛2)2(𝑠12𝑛1)2𝑛1−1+(𝑠22𝑛2)2𝑛2−1=(353.012210+258.375110)2(353.012210)210−1+(258.375110)210−1=17.5788⇒17𝑥̅1−𝑥̅2−𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛2≤𝜇1−𝜇2≤𝑥̅1−𝑥̅2+𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛2180.43−154.649−2.1098√353.012210+258.375110≤𝜇1−𝜇2≤180.43−154.649+2.1098√353.012210+258.3751109.2841≤𝜇1−𝜇2≤42.2779The confidence interval is positive; thus,there is evidence thatrecovery times are larger at Northbrook.b.Build a 95% two-sided confidence interval for the ratio of the variance of recovery times. Is there evidenceto support that the facilities are different?𝑠12𝑠22𝐹1−𝛼2,𝑛2−1,𝑛1−1⁄≤𝜎12𝜎22⁄≤𝑠12𝑠22𝐹𝛼2,𝑛2−1,𝑛1−1⁄353.0122258.3751𝐹1−0.052,10−1,10−1⁄≤𝜎12𝜎22⁄≤353.0122258.3751𝐹0.052,10−1,10−1⁄353.0122258.3751∗0.2484≤𝜎12𝜎22⁄≤353.0122258.3751∗4.0260

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0.3394≤𝜎12𝜎22⁄≤5.5006Since one is included in the interval,we canconclude the variancesare equal.c.Build a 95% two-sided confidence interval for the differences on the percentage of motion restored. Isthere evidence to support that the facilities are different?𝑝̅1−𝑝̅2−𝑧0.052⁄√𝑝̅1(1−𝑝̅1)𝑛1+𝑝̅2(1−𝑝̅2)𝑛2≤𝑝1−𝑝2≤𝑝̅1−𝑝̅2+𝑧0.052⁄√𝑝̅1(1−𝑝̅1)𝑛1+𝑝̅2(1−𝑝̅2)𝑛20.7490−0.674−1.96√0.7490(1−0.7490)10+0.674(1−0.674)10≤𝑝1−𝑝2≤0.0909−0.2397+1.96√0.7490(1−0.7490)10+0.674(1−0.674)10−0.3208≤𝑝1−𝑝2≤0.4708Sincezero is within the interval, there is not enough evidence to conclude thefacilities are different interms of the percentage of motion restored.4.47.An article in the ACI Materials Journal (Vol. 84, 1987, pp.213-216) describes several experimentsinvestigating the rodding of concrete to remove entrapped air. A 3-in diameter cylinder was used, and thenumber of times this rod was used is the design variable. The resulting compressive strength of theconcrete specimen is the response. The data are shown in Table 4E.8.a.Is there any difference in compressive strength due to rodding level? Answer the question by usinganalysis of variance with alpha = 0.05.Rodding levelCompressive strength101530101530101440151610151650151500201560201730201530251500251490251510

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MTB > Stat > ANOVA > One-wayOne-way ANOVA: Compressive strength versus Rodding levelSourceDFSSMSFPRodding level_132863395441.870.214Error8409335117Total1169567S= 71.53R-Sq = 41.16%R-Sq(adj) = 19.09%Individual 95% CIs For Mean Based onPooled StDevLevelNMeanStDev----+---------+---------+---------+-----1031500.052.0(-----------*----------)1531586.777.7(-----------*-----------)2031606.7107.9(-----------*-----------)2531500.010.0(-----------*----------)----+---------+---------+---------+-----1440152016001680Pooled StDev = 71.5Sincep-value = 0.214 > 0.05, there is no evidence that suggests rodding level influences compressivestrength.NOTE: Let us now assumeonly data on levels 10 and 15 isavailable, and that we will use a 95% two-sidedconfidence interval to evaluate whether there are any differences in terms of compressive strength betweenthe two rodding levels.=(𝑠12𝑛1+𝑠22𝑛2)2(𝑠12𝑛1)2𝑛1−1+(𝑠22𝑛2)2𝑛2−1=(27003+6033.33333)2(27003)23−1+(6033.33333)23−1=3.4914⇒3𝑥̅1−𝑥̅2−𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛2≤𝜇1−𝜇2≤𝑥̅1−𝑥̅2+𝑡0.052,⁄√𝑠12𝑛1+𝑠22𝑛21500−1586.6667−3.1825√27003+6033.33333≤𝜇1−𝜇2≤1500−1586.6667+3.1825√27003+6033.33333−258.3748≤𝜇1−𝜇2≤85.0415Since zero is within the confidenceinterval, we cannot concludethere is a difference in the compressivestrength associated with the two different rodding levels.b.Construct box plots of compressive strength by rodding level. Provide a practical interpretation of theplots.

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MTB > Graph > Boxplot > One Y > With GroupsThere does not seem to be a significant difference in terms of the mean compressive strengthfor rodding levels10,15, and2520 and between, 10 and 25.Nevertheless, roddingRoddinglevel 25 shows remarkably smallervariabilitycompared to other levels.On the other hand, the boxplotsalso suggests there a significantdifference in terms of the mean compressive strength and variance for rodding levels 20 and 25.c.Construct a normal probability plot of the residuals from this experiment. Does the assumption of anormal distribution for compressive strength seem reasonable?MTB > Graph > Probability Plot > SingleNormality assumption appears reasonable.2520151017501700165016001550150014501400Rodding level_1Compressive strength_1Boxplot of Compressive strength_1150100500-50-100-150999590807060504030201051ResidualPercentNormal Probability Plot(response is Compressive strength_1)

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CHAPTER5Control Charts for VariablesLearning ObjectivesAfter completing this chapter you should be able to:1.Understand chance and assignable causes of variability in a process2.Explain the statistical basis of the Shewhart control chart, including choice of sample size, control limits,and sampling interval3.Explain the rational subgroup concept4.Explain how sensitizing rules and pattern recognition are used in conjunction with control charts5.Know how to design variables control charts6.Know how to set up and use andRcontrol charts7.Know how to set up and use andScontrol charts8.Know how to set up and usecontrol charts for individual measurements9.Understand the importance of the normality assumption for individuals control charts and know how tocheck this assumption10.Set up and use CUSUM control charts for monitoring the process mean11.Set up and use EWMA control charts for monitoring the process mean12.Understand the difference between process capability and process potential13.Calculate and properly interpret process capability ratios14.Understand the role of the normal distribution in interpreting most process capability ratiosImportant Terms and ConceptsAssignable causes of variationControl chartControl limitsCUSUM control chartEWMA control chartIn-control processIndividuals control chartOne-sided process-capabilityratiosOut-of-control processOut-of-control-action plan(OCAP)Pareto chartPCRCpPCRCpkPCRCpmProcess capabilityRcontrol chartRational subgroupss control chartSampling frequency for controlchartsShewhart control chartsSignal resistance of a control chartSpecification limitsStatistical control of a processStatistical process control (SPC)Three-sigma control limitsVariable sample size on controlchartsVariables control chartsxcontrol chart

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Exercises5.14.Consider thecontrol chart shown here. Does the pattern appear random?No.The last four runs appear to plot at a distance of one-sigma or beyond from the center line.5.15.Consider the control chart shown here. Does the pattern appear random?Yes, the pattern appears random.5.16.Consider the control chart shown here. Does the pattern appear random?Yes, the pattern appears random.

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5.17.Apply the Western Electric rules tothe control chart in Exercise 5.17. Are any of the criteria fordeclaringthe process out of control satisfied?Check:1.Any point outside the 3-sigma control limits? NO.2.2 of 3 beyond 2 sigma of centerline? NO.3.4 of 5 at 1 sigma or beyond of centerline? YES. Points 17, 18, 19, and 20 are outside the lower 1-sigma area.4.8 consecutive points on one side of centerline? NO.The process can be declared out-of-controlbecausethe last four runs appear to plot at a distance of one-sigma or beyond from the center line.

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5.19.Consider the time-varying processbehavior shown below. Match each of these several patterns ofprocess performance to the corresponding andRcharts shown in Figures (a) to (e) below.

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BehaviorControl Chart(a)(2)(b)(4)(c)(5)(d)(1)(e)(3)5.21.The net weight of a soft drink is to be monitored by𝑥̅and R control charts using a sample size of𝑛=5.Data for 20 preliminary samples are shown in Table 5E.2.Sample Number𝑥1𝑥2𝑥3𝑥4𝑥5115.816.316.216.116.6216.315.915.916.216.4316.116.216.516.416.3416.316.215.916.416.2516.116.116.416.516616.115.816.716.616.4716.116.316.516.116.5816.216.116.216.116.3916.316.216.416.316.51016.616.316.416.116.51116.216.415.916.316.41215.916.616.716.216.51316.416.116.616.416.11416.516.316.216.316.41516.416.116.316.216.2161616.216.316.316.21716.416.216.416.316.2181616.216.416.516.11916.41616.316.416.42016.416.416.51615.8

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a.Set up𝑋̅and R control charts using these data. Does the process exhibit statistical control?MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R𝑥𝑋̅̅̅̅chart:𝑈𝐶𝐿=16.5420,𝐶𝐿=16.268,𝐿𝐶𝐿=15.9940𝑅𝑐ℎ𝑎𝑟𝑡:𝑈𝐶𝐿=1.004,𝐶𝐿=0.475,𝐿𝐶𝐿=0The process appears to exhibit statistical control.b.Estimate the process mean and standard deviation.𝜇̂=𝑥̿=16.268,𝜎̂=𝑅̅𝑑2⁄=0.4752.326⁄=0.2042c.Does fillweight seem to follow a normal distribution?MTB > Stat > Basic Statistics > Normality TestNormality assumption is reasonable.19171513119753116.616.416.216.0SampleSample M ean__X=16.268U C L=16.5420LC L=15.99401917151311975311.000.750.500.250.00SampleSample Range_R=0.475U C L=1.004LC L=0Xbar-R Chart of Net Weight17.0016.7516.5016.2516.0015.7515.5099.99995908070605040302010510.1Net WeightPercentMean16.27StDev0.2014N100AD1.257P-Value<0.005Probability Plot of Net WeightNormal - 95% CI

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