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1、版本版本: 1.00日期日期: May 20036 6西格玛绿带培训教材西格玛绿带培训教材ONEONE-6-4-20246标准偏差标准偏差标准偏差标准偏差1西格玛绿带培训教材课件DAY1 第一天(定义阶段):- 6西格玛及精简优化与COQ(質量成本）的关系- COQ的脑力风暴- First Pass Yield Exercise II初始直通率的練習 IPO（輸入輸出流程）and flow diagram（IPO和流程圖）Flow analysis of dropping cards onto targetRepeating the exercise重復練習Results and discus
2、sions結論和檢討- 西格玛培训中的某些质量改进工具- 脑力风暴技术- 第一天結束 wrap up通過以上的教學引導學員對品質成本的認識運用六西格瑪就是有效的降低品質成本課程安排2西格玛绿带培训教材课件DAY2 第二天(Variance Reduction降低变差的理解):- The power of Plato chart and the 80/20 rules柏拉圖表的功能和80/20的規則- Construction of a Plato Chart using computer- flow diagram and its associated symbols流程圖和其制作符號含義-
3、Two 实例of flow diagram (using a common scenario)兩個流程圖的實例（使用通用的情節）- Barriers that hinder 6西格玛implementation阻礙開展執行六西格瑪的因素-Break- What is FMEA 什么是FMEA- Example of FMEA關于FMEA的實例- Group exercise on FMEA of barriers to 6西格玛implementation- FMEA presentations關于FMEA的介紹-Lunch- Concept of precision and accuracy
4、對准確和准確的理解（Cp和Ca）- How does it link to the mean and 标准偏差（如何將平均值和標准偏差聯系起來）- Precision and accuracy example (i.e. Selection of fund manager准確和精確的實際例子-Catapult exercise I彈弓拋物發射器的思維練習一-Break- Introducing concept of variance reduction (i.e. PF/CE/CNX/FMEA/SOP)介紹降低變差的觀念-Variance reduction 脑力风暴exercise for
5、Catapult用彈弓發射器進行降低變差的腦力風暴練習-Catapult exercise II彈弓拋物發射器思維練習二- Computation of Catapult exercise result after variance reduction評估計算彈弓發射器游戲中的數據來了解降低變差的含義- Discussion of variance contributors討論降低變差的意義-第二天wrap up 在以上學習中通過彈弓發射器游戲的了解在游戲中掌握了解在六西格瑪中(Variance Reduction）降低变差重要性3西格玛绿带培训教材课件DAY 3 (Measure Phase
6、測量階段):- Recap of statistical terminology全新的統計學朮語- Histogram and a normal data對直方圖和常態數據的理解- Construction of histogram對直方圖的解釋- Transformation of data數據的轉換- Calculate Cp, Cpk from non-normal data計算非正態數據的Cp, Cpk -Break- - The importance of good measurement正確的測量方法的重要性- Direct 和indirect measurement (i.e.
7、Introduction to scatter diagram)- Risk of wrong interpretation錯誤解釋的風險- Under and variance concept in measurement system在變異范圍內的測量系統的觀念- Introduction to Gauge repeatability and reproducibility (GR&R)介紹GR&R-Lunch- GR&R example 對GR&R的計算的例子 - Calculation of measurement variance測量變異的計算- Rules of thumb in
8、GR&R閱讀GR&R 規則手冊- Calculation of GR&R using computer使用電腦計算GR&R -Break- Interpretation graphical of GR&R解釋GR&R的繪制- Balls circumference measurement exercise測量園球周長的游戲- Result and discussion on measurement exercise以上測量結果和方法的練習- The ANOVA(analysis of variance) method of GR&R對GR&R的方差計算方法- Day 3 wrap up 以上的
9、培訓使學員開始接觸品質分析工具4西格玛绿带培训教材课件DAY 4 (Measure + Analyze Phase測量分析階段): - Introduction to GR&R analysis on attribute data介紹GR&R 的數據- Example of attribute data GR&R （GR&R 的實際例子）- Attribute GR&R exercise （GR&R的練習）- Result and discussion on exercise(練習計算和討論）-Break- Computing attribute data GR&R using the com
10、puter使用電腦計算GR&R 的數據- what variable data is better than attribute data為什么變差數據比品質數據好- Converting attribute data to variable data將品質數據轉化成變差數據- Example of attribute data conversion (i.e. Wu Fans project on reducing bubble defect)數據運算的實際例子- Introduction to probability theory介紹概率原理-Lunch- Probability appr
11、oach (classical, relative frequency)概率統計的步驟（古典方式相關頻率）- Probability rules概率規則- Probabilities under statistical independence (Marginal, Joint, Conditional)- Exercise練習-Break- Probability under conditions of statistical dependence概率條件下的統計學原理- Exercise練習- Introduction to probability distributions介紹概率分配-
12、 Day 4 wrap up 本天是學員掌握基本的統計原理5西格玛绿带培训教材课件DAY 5 (Analyze Phase分析階段): - What is a binomial distribution ?什么是二次項分配- Conditions for the use of the Bernoulli trials (流程)- Graphical illustration of a binomial distribution二項次分配繪制的說明- Measures of central tendency and dispersion for binomial distribution 二項次
13、分配的集中趨勢和離散趨勢的測量- Probability calculation with binomial distribution用二項次分配計算概率- Binomial distribution case studies二項次分配的案例學習- Characteristics of the Poisson distribution泊松分配的特點- Probability calculation using Poisson distribution用泊松分配的概率計算- Poisson distribution as an approximation of binomial distribu
14、tion 泊松分配是一個比較接近二次項分配的-Break- Poisson distribution case studies泊松分配案例的學習- Introduction to normal distribution介紹正態分配- Characteristics of normal distribution典型的正態分配- Areas under the normal curve正態曲線內部的區域面積的理解- The use of normal probability distribution table使用概率分配表- Probability calculation using norma
15、l distribution利用正態分配計算概率- Normal distribution case studies正態分配案例的學習- Using computer to calculate probability of different distribution 利用電腦計算不同性質的概率分配-Lunch-6西格玛绿带培训教材课件DAY 5 (Analyze Phase分析階段): - Random sampling: Basis of statistical inference隨意抽樣統計推理的基礎- Introduction to sampling distributions介紹取樣
16、分類- Concept of standard error and sampling from normal population-Probability of the sample mean樣品平均數的概率- Central limit theorem中心極限定理- Exercise to demonstrate central limit theorem中心極限定律的練習- Result and discussion結論和檢討-Break- Confined interval and population mean estimation信賴區間和 Continuous data集中趨勢的數
17、據- Discrete data 離散趨勢的數據- Determining sample size決定樣本大小- Continuous data集中趨勢的數據- Discrete data 離散趨勢的數據- Finite and infinite population and the associated impact to the confidence interval- Exercise on confidence interval信賴區間的練習- Day 5 和week 1 wrap up課程安排7西格玛绿带培训教材课件Asia 6Asia 6西格玛西格玛西格玛西格玛VisionVisi
18、on亞洲六西格瑪進行的遠景亞洲六西格瑪進行的遠景亞洲六西格瑪進行的遠景亞洲六西格瑪進行的遠景 Our Mission 我們的任務To drive operational excellence through the deploy of 6西格玛initiatives in optimal-electronics Asia sites.開展六西格瑪是通往卓越的亞洲最佳電子行業的道路GOAL目标目标 1) To consolidate Asia effort in 流程改进through 6西格玛initiative.2) To share best practices across Perkin
19、elmer Asia Sites.3) To entrench 6西格玛culture in optimal electronics Asia sites, 和make it the way of life for our people.8西格玛绿带培训教材课件Course Course 目标目标目标目标Explain 和motivate others in the 6西格玛哲学Underst和characterize the 6西格玛组织支持结构Describe 和explain the role of 6西格玛黑带, champion 和绿带.Certification 流程in Asia
20、 6西格玛deploymentHow does 6西格玛drive 产品ivity 质量 deployment Perkinelmer 6西格玛implementation strategyUnderst和the importance of motivating others to inculcate 6西格玛culture9西格玛绿带培训教材课件What is 6SIGMA?What is 6SIGMA?什么是六西格瑪什么是六西格瑪什么是六西格瑪什么是六西格瑪Vision 和哲学和哲学:Develop BETTER, FASTER, 和LOWER COST 产品 和services - ag
21、gressively attack the cost of 质量 (COQ), leading to higher customer satisfaction 和retentionPrimary Focus:Eliminates variation in all business 流程es to reduce costs 和increase customer satisfaction Definition:A 方法 to identify 和minimize variation in a 流程, resulting in 3.44 PPM defectiveA Vision,A 哲学哲学一個方
22、法一個方法,一個改进工具一個改进工具工具工具一個基准一個基准一個標准一個標准What is 6SIGMAWhat is 6SIGMA10西格玛绿带培训教材课件Ways to create business, operation 和technology breakthrough in meeting goals associated with doing things better, faster , lower cost in all 产品和services6西格玛develop people towards 世界级别优秀culture in company for competitive a
23、dvantagesMost powerful breakthrough management 工具Drastically improve bottom lineMinimize waste 和increase resources while increasing customer satisfaction 6西格玛can result in increasing market share, reduce operation costs 和profit growth.为什么要使用为什么要使用为什么要使用为什么要使用6 6西格玛西格玛西格玛西格玛? ?11西格玛绿带培训教材课件1970sJapan
24、ese firm took over Motorola TV factory in USA 和reduced 缺陷by 95%!1981Motorola CEO Bob Galvin challenged company to a 10x 质量改进in 5 years1985Paper on “Defect 和Correct vs. Error Free Assembly” by Smith from Motorola1988 Won Malcolm Baldrige award for Total Customer Satisfaction1988 6西格玛Research Institut
25、e formed: Implementation strategy-Guidelines-Advanced 工具s6 6西格玛的历史西格玛的历史西格玛的历史西格玛的历史12西格玛绿带培训教材课件Continuous 改进改进改进Time6西格玛西格玛Breakthrough 方方法法Normal Continuous 改改进进Breakthrough 改进改进改进TimeIncremental 改进改进resulting from a consistent series of many, small 改进改进activities Quantum Leap 改进改进driven from a o
26、ne-time, 10 times 改进改进 使用六西格瑪的巨大突破使用六西格瑪的巨大突破使用六西格瑪的巨大突破使用六西格瑪的巨大突破13西格玛绿带培训教材课件通过通过6西格玛获利的实例西格玛获利的实例交付世界级别优秀质量的产品降低业务活动中的缺陷缺陷和cycle time主打产品和流程知识提供系统的工具和技术Create opportunity for business growth 和increase profitabilityCreating a learning 组织by building teamwork , sharing of success ideas, best practi
27、ce Improve communication 和teamwork14西格玛绿带培训教材课件流程改进流程改进产品和服务改进产品和服务改进投资商关系投资商关系设计方法设计方法供应商改进供应商改进培训和招聘培训和招聘作出决定的流程作出决定的流程6 6西格玛对公司产生的价值西格玛对公司产生的价值西格玛对公司产生的价值西格玛对公司产生的价值Management Leadership in 6Management Leadership in 6西格玛西格玛西格玛西格玛Communicating 6西格玛西格玛visionTake ownership to be a change agent提供支持和提
28、供支持和resourcesRemove barriers to success Visible Visible 和和和和Top Down Commitment in 6Top Down Commitment in 6西格玛西格玛西格玛西格玛DeploymentDeployment Establish a 6Establish a 6西格玛西格玛西格玛西格玛Leadership teamsLeadership teams15西格玛绿带培训教材课件绿带绿带- A part-time person who undertakes projects of lesser scope than a 黑带 p
29、roject. 黑带黑带 - A full-time person who leads critical projects.Master 黑带黑带 - Mentor to Black/绿带s. Champion盟主盟主 - Sponsor of a project.Roles in 6 Sigma DeployRoles in 6 Sigma Deploy開展六六西格瑪的角色開展六六西格瑪的角色開展六六西格瑪的角色開展六六西格瑪的角色16西格玛绿带培训教材课件開展六西格瑪的方法和流程開展六西格瑪的方法和流程 ApplyReviewTrain培训is Conducted by Master 黑带
30、sPlanProject Selection 流程is Managed by ChampionsProjects企化MEASURE (M)IMPROVE (I)ANALYZE (A)CONTROL (C)Dollars SavedPROBLEMDEFINE (D)17西格玛绿带培训教材课件Linking to the Goal Deploy Matrix目標開展矩陣圖目標開展矩陣圖18西格玛绿带培训教材课件Management Roles in 6Management Roles in 6西格玛西格玛西格玛西格玛Identifying key business areas where brea
31、kthrough is neededIdentifying area with the highest saving Identify the right 6西格玛candidates in accordance to selection criteria提供支持of resources to train 和equip people to attain stretch goals Set measurement matrix 和track progressRecognition 和reward successCreate opportunity 和channel to propagate su
32、ccess stories to generate culture change 19西格玛绿带培训教材课件6 6西格玛西格玛西格玛西格玛Leadership TeamLeadership TeamTrained in 6西格玛Develop a 6西格玛Deployment PlanWork with management to identify right 6西格玛candidatesDevelop a Focused Schedule 培训Serve as mentors for candidatesCertify 6西格玛CandidatesIdentify MentorsMonito
33、r project selection 和progressWork with management sponsors on reward system 和propagating successDevelop a 6西格玛Network to enhance communicationReview 和Improve 6西格玛流程20西格玛绿带培训教材课件黑带黑带黑带黑带 Roles Roles Assist Department to identifying operating 和business issue in this area :Set Stretch Goals, Measure, I
34、mprove, Document, TransferLead 和managed 6西格玛ProjectsAct as Breakthrough Strategy expertsLead 和direct team in project executionStimulate Champion thinkingReport progress to appropriate leadership levelsInfluence without direct authorityDetermine the most effective 工具to applyBe a Change AgentMotivate
35、others to set 和accomplish stretch goals with 6西格玛philosophies 和methodologies培训of 绿带s21西格玛绿带培训教材课件绿带绿带绿带绿带Roles Roles Focus on Projects that tie directly to their daily workTo help deploy the success of 6西格玛Lead 改进projectsGathering 和analyzing dataExecuting experimentsCapture 和Sustain 6西格玛Gain提供leader
36、ship in areas of uses 6西格玛methodologies22西格玛绿带培训教材课件Over View of Leaders Roles Over View of Leaders Roles 23西格玛绿带培训教材课件6西格玛西格玛PROJECT MASTER STRATEGYD DEFINE A ANALYZE MMEASUREI IMPROVEC CONTROL1. What 流程es are you responsible for ? Who is the owner of these 流程es ? Who are the team members ? How wel
37、l does the team work together ? 2. Which 流程es have the highest priority for 改进? How did you come to this conclusion ? Where is the data that supports this conclusion ? 3. How is the 流程performed ?4. What are your 流程performance measures ? 为什么? How accurate 和precise is your measurement system ?5. What
38、are the customer driven specifications for all your performance measures ? How good or bad is the current performance ? Show me the data. What are the 改进goals for the 流程?6. What are all the sources of variability in the 流程? Show me what they are.7. Which sources of variability do you control ? How d
39、o you control them 和is it documented ?8. Are any sources of variability 供应商-dependent ? If so, what are they, whos the 供应商和whats being done ?9. What are key variables that affect the average 和variation of the measures of performance ? How do you know this ? Show me the data.10. What are the 关系hips b
40、etween the key variables 和流程output ? Do any key variables interact ? How do you know for sure ? Show me the data.11. What setting for the key variables will optimize the measures of performance ? How do you know this ? Show me the data.12. For the optimal setting of the key variables, what kind of v
41、ariability exists in the performance measures ? How do you know ? Show me the data.13. How much 改进has the 流程shown in the last 3 months ? How do you know this ? Show me the data.14. How much time and/or money have your efforts saved or generated for the company ? How did you document all of your effo
42、rts ? Show me the data.24西格玛绿带培训教材课件Characterize 和和OptimizeDMAICworkwork6s缺陷缺陷are easy to see, but are expensive to fixSustain The GainDetermine business measurement needing 改进改进Identify 产品产品 and/or 流程流程es that impact the targeted business measurementCategorize 缺陷缺陷generated by 产品产品 & 流程流程esDetermin
43、e largest opportunity for reducing defects, 和和estimate potential impact on businessGain business approval for 改进改进projectABCDEDefineCharacterize the response Y, look at the raw dataIs it Bimodal? Skewed? Other Clues?Is the problem with mean or sigma?Identify gap between current 和和desired stateDevelo
44、p cause 和和effect diagram to list potential XsUse graphical analysis, multi-vari, Anova 和和basic statistical 工具工具to screen the likely families of variability1234AnalyzeSelect likely Xs for experimentation; set levels on XsUse 设计设计of experiments to find the critical few Xs 和和equation Y=F(X), SY=G(X)Mov
45、e the distributionShrink the spreadConfirm the results12345ImproveMistake proof the 产品产品or 流程流程Tolerance the 产品产品or 流程流程, Y=F( X)Measure the final capability for YPlace appropriate controls on the critical XsDocument the effort 和和results12345ControlProject ObjectiveDevelop focused problem statement
46、with project teamIdentify the response variable(s) Ys 和和how to measure themAssess the specification(s) for defectIs one in place? Is it the right one?Analyze measurement system capability & accuracy1234MeasurePractical ProblemStatistical ProblemStatistical SolutionPractical SolutionStatistical Contr
47、ol25西格玛绿带培训教材课件Stage of 6Stage of 6西格玛西格玛西格玛西格玛Implementation Implementation RecognizeRecognizeKey business issues from the executive viewpointBaseline major 流程流程: FPY, cycle time, ppm, dpuEvaluate customer perceptionsDetermine the cost of 质量质量 (COQ) Customer requirement Increase market share , prof
48、its margin , expansion , stock price , etc VisionVisionSet customer goalsSet goals for stock price , market share , sales, profiles ,etcSet 6西格玛西格玛goals for dpu , FPY, cycle time ,COQ ,etcSet goal that tied to performanceDefine “世界级别优秀世界级别优秀culture“ in PerkinElmer DeployDeployCommunicate vision , pl
49、an 和和needTrain all level manager to be champion, 黑带黑带s , 绿带绿带sTrain operator 和和supervisors提供提供resources ,software 和和hardwareApply 方法方法 (DMAIC) Schedule weekly mentor 和和expert meetings26西格玛绿带培训教材课件Monitor Monitor Quarterly briefing to executives 和和financeSchedule monthly meetings to share ideas , gro
50、up consulting , 脑力风脑力风暴暴, progress reports, etc.Identify 和和removed barriers Stay focused 和和committed提供提供d consulting on 6西格西格玛工具玛工具through masters For each project : define , measure , analysis , improve , control 和和review Evaluate Evaluate Evaluate goals 和和timelinesEvaluate business impactEvaluate
51、problematic projectsEvaluate executive level 和和mentor or sponsor supportFinance link the gain to bottom lineRefocus 和和re-commitSustain Sustain Develop control planCommunicate success - motivate all personnel through executive briefings videos 和和organizational newslettersReward success as plannedMake
52、 appraisals reflect performanceSelect new projectsStage of 6Stage of 6西格玛西格玛西格玛西格玛Implementation (cont)Implementation (cont)27西格玛绿带培训教材课件Goals of the Certification Goals of the Certification 流程流程流程流程: :Develop associates having the following attributesi. Understanding of 6西格玛工具西格玛工具sii. Appropriate
53、applications of the 工具工具siii. Demonstrate results using the 工具工具sHave consistence certification criteria 和和judgment6 6 Sigma Certification Sigma Certification 流程流程流程流程28西格玛绿带培训教材课件Definition黑带Candidate who are full-time working on 6西格玛projects. Responsible to : Lead the team in effective utilization
54、 of 6西格玛方法 Select, teach 和use the most effective 工具s Oversee data collection 和analysis Complete 2 projects a year绿带 Candidate who is doing 6西格玛projects on a part time basis.Responsible to :Assist the 黑带 Complete 1 project a year29西格玛绿带培训教材课件流程流程to 6西格玛西格玛GB/BB certificationPhase 1; Project / Candida
55、te SelectionCandidate SelectionTo fill in ; Characteristics of 6西格玛candidate Score-sheet 和Summary Leadership Values Score-sheet 6西格玛Candidates Leadership Values Summary 6西格玛CandidatesProject SelectionTo fill in ; 6西格玛Project Selection Summary Ease of Implementation Assessment ROI Impact AssessmentCa
56、ndidates must score greater than 2.5 pts for each to qualifyCandidates must score at least 0.9 pts in the in the 6西西格玛格玛Project Selection Summary to qualify6西格玛西格玛Project SummaryMatching right project to the right people30西格玛绿带培训教材课件Phase 2; 6西格玛培训和application of 6西格玛工具sCollect Baseline Data on Proj
57、ect yield COQ Cost Cycle Time Inventory levelTo fill in ; 6西格玛R0 Project Review Sheet 6西格玛COQ Template 6西格玛Project Progress Report To get all relevant parties approval signatoriesAttend 6西格西格玛玛GB/BB Academic 培训培训Application of 6西格玛工西格玛工具具to Project(s)PF/CE/CNX/SOPs FMEA MSA PARETO PROBABILITY DISTRI
58、BUTION ANOVA DOE STATISTICAL INTERVAL SPCCreate Certification Template Follow DMAIC Presentation Summary Rev A , to complete each phase of the project流程流程to 6西格玛西格玛GB/BB certification31西格玛绿带培训教材课件Phase 3; Certification of CandidateComplete Closure Technical ReportPresentation of 6西格玛西格玛Certification
59、 Project(s)Site Assessment of Project(s)Issue Plaque /Certificate Achievement to candidate Update Candidate LOR Career Pro 培训和Recognition Finance to verify project savings Champions 和MBB to assess candidate understanding 和application of 6西格玛工具s Candidate Certification Evaluation Form流程流程to 6西格玛西格玛GB
60、/BB certification Finance to verify project savings Champions 和MBB to assess candidate understanding 和application of 6西格玛工具s Candidate Certification Evaluation Form32西格玛绿带培训教材课件Requirements for Certification Completion of 6西格玛培训西格玛培训course Successful project completion (goal achievement 和和documentat
61、ion) Demonstration on the understanding of 6西格玛工具西格玛工具s Effective 和和successful completion of steps to “hold the gain” Completion of each project within 1 yr 绿带绿带Certification; completion of 2 projects with minimum saving of US$25,000 per project 黑带黑带 Certification; completion of 2 projects with mini
62、mum saving of US$100,000 per project 交付交付ables for Certification Demonstration of six-sigma though 流程流程 Completion of 6西格玛西格玛R0 Project Review Sheet Completion of 6西格玛西格玛 Project Progress Report Completion of 6西格玛西格玛COQ Template Completion of DMAIC Presentation Summary Completion of Closure Technica
63、l Report33西格玛绿带培训教材课件Certification 结构结构 Candidate Certification Board Site Leader Champions Master 黑带 Certification Board Review 流程Candidate to distribute the Closure Report Summary to the board at least one week before the review date (may be waived at the discretion of the site leader)MBB to act a
64、s the chair of the boardCandidate to present project details using using thought 流程map with emphasis on how each 工具was appliedQuestions, clarifications 和review by the board34西格玛绿带培训教材课件Certification 结构结构(cont)Board members to rate candidate using Certification Evaluation FormPass if ;1.Candidate Tec
65、hnical Assessment is greater than 20 pts for Greenbelt 和greater than 40 pts for Blackbelt2.流程Variation 和Measurable Results Evaluation is at least 4 pts for each categoryInform candidate of the outcome 35西格玛绿带培训教材课件Sample FormsCharacteristics of 6西格玛西格玛CandidatesLeadership Values Score-sheet36西格玛绿带培训
66、教材课件Sample Forms; Leadership Values Summary37西格玛绿带培训教材课件Sample Form; Closure Technical Report38西格玛绿带培训教材课件Sample Form; Candidate Certification Evaluation Form39西格玛绿带培训教材课件Definition of a Definition of a 流程流程流程流程A 流程is an activity which utilize inputs from external source 和transform them into desired
67、 output(s).Example:Manufacturing 流程es(i.e. wire bonding, injection molding, glass sawing)Financial 流程es(i.e. double book keeping, 产品costing)HR 流程es(i.e. recruiting, ranking 和appraisal, 培训)Daily activity(i.e. parking a car, buying lunch, brushing your teeth)40西格玛绿带培训教材课件IPO (Input-IPO (Input-流程流程流程流程
68、-Output) Diagram -Output) Diagram PeoplePeopleMaterialMaterialEquipmentEquipmentPoliciesPoliciesProceduresProceduresMethodsMethodsEnvironmentEnvironmentPerformPerforma servicea serviceProduceProducea a 产品产品产品产品CompleteCompletea taska taskutilizing external utilizing external utilizing external INPUT
69、S INPUTS INPUTS to achieve to achieve to achieve the desired the desired the desired OUTPUT(S)OUTPUT(S)OUTPUT(S)INPUTSINPUTSINPUTS 流程流程流程流程流程流程 OUTPUTSOUTPUTSOUTPUTSA visual representation of a 流程which lists input variables 和output characteristics41西格玛绿带培训教材课件PeoplePeople Kids KidsMaterialMaterial W
70、ater Water Detergent DetergentEquipmentEquipment Cleaning Cloth Cleaning Cloth Pail Pail Water hose Water hoseEnvironmentEnvironment raining raining sunny sunnyCleanliness Cleanliness of car of car surfacesurfaceWashing a CarWashing a CarWashing a CarINPUTSINPUTSINPUTS 流程流程流程流程流程流程 OUTPUTSOUTPUTSOUT
71、PUTSWhat is the purpose of IPOWhat is the purpose of IPOA high level interpretation of a 流程, which enable ease of understanding, through outlining the 关系关系hips between input variables 和output response(s).42西格玛绿带培训教材课件What is a distribution ?It is a pattern form by the collection of data, grouping th
72、e outcome horizontally (x-axis), 和indicates the observed frequency of the outcome vertically (y-axis).In statistic, this generate a theoretical pattern whereby information of entire population can be obtained from observing limited samples.A Normal Distribution2527293133353739414345Response Measurem
73、entFrequency43西格玛绿带培训教材课件Characteristic of a normal distribution- It has a single peak 和a bell shape curve- It is a distribution for continuous data- The mean (average) is is at the center of the curve- The variance describe the spreading of dataWage Distribution of Engineering Fresh Graduate2.252.3
74、02.352.402.452.502.552.602.652.702.75Response X ( $K )Frequency ( No. of person )Mean, m m44西格玛绿带培训教材课件为什么为什么is distribution important to a 流程流程The output of a 流程can most of the time associates with a statistical distribution, giving opportunity for engineer to analyze the data statistically, hence
75、arriving conclusion with a statistical confidence 和at a lower cost.Distributions differs in locationDistributions differs in spreadDistributions differs in shapeNormal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 - 99.9999999998% - 99.73% - 95.45% -68.27%- 3 variation is called natural tolerance 45西格玛
76、绿带培训教材课件流程流程Capability Study流程流程capability potential, CpBased on the assumptions that :Cp= 流程capability potentialCpk= 流程capability indexIt is a measurement of the capability of a 流程, by indexing the 流程natural tolerance with respect to the device specification (i.e. customer tolerance)1.流程is normalNo
77、rmal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 Lower Spec LimitLSLUpper Spec LimitUSLSpecification Center2.It is a 2-sided specification3.流程mean is centered to the device specificationSpread in specificationNatural toleranceCP =USL - LSL6 8 6 = 1.3346西格玛绿带培训教材课件流程流程Capability Index, Cpk1. Based on
78、the assumption that the 流程is normal2. An index that compare the 流程center with specification centerUSL - Y3 Y - LSL3 Cpk = min ,Normal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 Lower Spec LimitLSLUpper Spec LimitUSLSpecification CenterTherefore when ,Cpk Cp ; then 流程is not centeredCpk = Cp ; then 流程
79、is centered47西格玛绿带培训教材课件Arbitrary sample data of starch content (gram) in 1gram anti-biotic pillWhat is the mean starch content ?What is the variance ?What is the 标准偏差标准偏差 ?What is the 流程流程capability potential, Cp ?What is the 流程流程capability index, Cpk ?What can you do to improve the 流程流程?The above
80、activity is a sigma 流程流程.What can you conclude from the below 流程流程?Given:Nominal content = 0.5 gramTolerance = 0.1 gram48西格玛绿带培训教材课件关系关系hip between 流程流程Capability 和和DPPMCp = Cpk = 1Cp = Cpk = 1.33Cp = Cpk = 1.67Cp = Cpk = 2Cp =1Cpk = 0.5Cp =1.33Cpk = 0.83Cp =1.67Cpk = 1.17Cp =2Cpk = 1.5Long Term流程流程
81、CapabilityShort Term流程流程CapabilityFor a 6 Sigma 流程流程Cp = 2.0Cpk = 1.5DPM = 3.4 49西格玛绿带培训教材课件First Pass Yield Exercise I Break into team of 9 persons. Each team consist of 1 供应商, 1 supervisor, 3 operators, 3 inspectors 和1 customer. The objective is to 交付20 complete 产品 to the customer The task of 供应商i
82、s to supply card (i.e. raw material) to the first operator. The task of the operator is to drop the card with edge facing down onto the target area, 和count the number of drop throughout the exercise. The task of the inspector is to verify that the card has drop within the target area If card fall wi
83、thin the target:Pass the card to the next operator (or to the customer for the last inspector)Record down the number of good units that are passed to the next 流程 If card does not fall within the target:Pick up the card 和h和it back to the operator The task of the customer is to receive the 20 cards, 和
84、record the time required (i.e. lead time) At the end of the exercise, the 供应商will count the number of raw material issue to the 流程流程流程1流程流程2流程流程3供应商CustomerInspectorsOperators50西格玛绿带培训教材课件Computation of 产品产品cost51西格玛绿带培训教材课件What is Cost of 质量质量 (COQ)It is the total cost incur from the result of poor
85、 质量 of materials, 产品, 流程es, 和the cost of not doing things right at the first time.What are 某些某些of the COQ that you can see at your site ? Scrap, Rework, Cycle Time Warranty, Liability, Service Audits, Inspections, Evaluations 培训without ROI, Consultation Lost of sales, Lost of market share52西格玛绿带培训教材
86、课件How does 6 Sigma 和Lean related to COQ ?2 ways the entire business 流程can be improved 和reduce COQ : To lean up the 流程chain 和reduce chances for error To improve the sigma capability of the 流程LeanThrough 6 Sigma53西格玛绿带培训教材课件What are the overall yield for your 流程流程?Double click on table to find outGoin
87、g back to the dropping cards exerciseHow can we reduce the 产品产品cost ?What is the main problem that associate to the cost factor ?Knowing the problem, what is the root cause to the problem ?Can changing the way we do thing help to improve the situation ?54西格玛绿带培训教材课件First Pass Yield Exercise II Break
88、 into team of 9 persons. Each team consist of 1 供应商, 1 supervisor, 3 operators, 3 inspectors 和1 customer. The objective is to 交付20 complete 产品 to the customer The task of 供应商is to supply card (i.e. raw material) to the first operator. The task of the operator is to drop the card with the surface fac
89、ing down onto the target area, 和count the number of drop throughout the exercise. The task of the inspector is to verify that the card has drop within the target area If card fall within the target:Pass the card to the next operator (or to the customer for the last inspector)Record down the number o
90、f good units that are passed to the next 流程 If card does not fall within the target:Pick up the card 和h和it back to the operator The task of the customer is to receive the 20 cards, 和record the time required (i.e. lead time) At the end of the exercise, the 供应商will count the number of raw material iss
91、ue to the 流程流程流程1流程流程2流程流程3供应商CustomerInspectorsOperators55西格玛绿带培训教材课件What is the 产品产品cost after the change of 流程流程method56西格玛绿带培训教材课件某些质量改进工具某些质量改进工具in 6西格玛培训西格玛培训Pareto ChartGraphical representation that describe categories in descending occurrence order 80% of a problem is contributed by 20% of t
92、he categoriesSeparate the important few factors from the many trivial factorsExtensively use to define areas to focus on for detail analysisInputInputDecisionDecision流程流程流程流程StepStepOutputOutputScrapScrapNONOYesYes流程流程Flow DiagramGraphical representation of the various 流程activities in sequence.A ext
93、ended vision from IPO diagram to 提供more details about the 流程Use in 流程mapping to identify area where problem surface out.Suggest data collection points for a 流程57西格玛绿带培训教材课件Cause 和和Effect Diagram (Fishbone diagram)DistanceDistanceManManMaterialsMaterialsMeasurementMeasurementMethodsMethodsEnvironment
94、EnvironmentMachinesMachinesDistanceDistanceManManMaterialsMaterialsMeasurementMeasurementMethodsMethodsEnvironmentEnvironmentMachinesMachines提供a 结构d approach during a brain storming session for root cause analysis.Identifies components to be 1) ignored (Noise), 2) to held Constant, 和3) for optimizat
95、ion through DOE (eXperimentation)Failure Mode 和和Effect Analysis (FMEA)Identifies potential failure mode, evaluate its effect 和analyze the cause(s).RPN in FMEA help to focus on developing 改进plan for critical causes.Measurement Capability AnalysisEvaluate the variance in the measurement system.In orde
96、r to achieve 改进, you must first be able to measure with substantial resolution.58西格玛绿带培训教材课件Regression Modeling 和和DOEDevelop model that describe how the inputs of a 流程influence its output.Allow optimization of 流程和cost through the modelInformation about the strength of the model is availableHo = Ho =
97、 m m m m1 = 1 = m m m m2 2H1 = H1 = m m m m1 1 m m m m2 2Hypothesis Testing提供information about whether there is a significant difference between two activities.Allow statistical justification rather than the traditional “Gut” feeling justification.Control ChartDistinguish special causes in the 流程All
98、ow 流程monitoring over timeHelps in controlling the 流程through early trend detection59西格玛绿带培训教材课件脑力风暴脑力风暴It is the generation of ideas in a group situation, based on the principle of suspending judgement. A principle which is highly 产品产品ive in both individual or group effortEffective 脑力风暴脑力风暴Well defin
99、ed 和和clearly stated problemAssign a recorder to write down all the ideas as they occurMaintain manageable group size (i.e 5 to 8 persons)Enforce following guide lines: Suspend judgement Every idea is accepted 和和recorded Encourage people to build on the ideas of others Encourage wild or odd ideas60西格
100、玛绿带培训教材课件某些脑力风暴某些脑力风暴PracticesRandom method:Everyone in the group will throw in ideas randomly as soon as the ideas are generated- Possibility that idea contributions are dominated by few persons- Good ideas may be suspress through feeling inferior Cycle method:Each one in the group will come out wi
101、th one idea, 和和the cycle continues after the the last person in the group has given his contribution- Possibility of putting someone in a uncomfortable position- Idea generation is less dynamic, 和people might feel bored easilySilence method:Everyone in the group will be given 某些某些times to write down
102、 all their ideas on papers 和和submit to a designated recorder, who will then consolidate all the ideas 和和record on the board for everyone to view- Lack of real time interaction- Might hinder the building of ideas from contributions generated by others61西格玛绿带培训教材课件62西格玛绿带培训教材课件为什么为什么为什么为什么do I use it?
103、do I use it? To focus our efforts on the problems that offer the greatest To focus our efforts on the problems that offer the greatest potential for potential for 改进改进改进改进What is it?What is it? A bar chart for non-numerical categories that displays A bar chart for non-numerical categories that displ
104、ays attribute data from highest to lowest frequency (based on attribute data from highest to lowest frequency (based on “Pareto Principle” which states that “Pareto Principle” which states that 80%80% of the effects are of the effects are due to due to 20%20% of the causes) of the causes)What does i
105、t do?What does it do? Identifies the causes that will have the greatest impact on Identifies the causes that will have the greatest impact on solving the problemsolving the problemIntroduction to Pareto ChartIntroduction to Pareto Chart63西格玛绿带培训教材课件Example of Pareto ChartExample of Pareto Chart64西格玛
106、绿带培训教材课件65西格玛绿带培训教材课件Pareto Analysis66西格玛绿带培训教材课件67西格玛绿带培训教材课件 In particular week of the M&M Factory 产品ion have a major 缺陷of Broken M&M 和Rejected Color Now present which is the group of M&M you should look into during 产品ion if the following data is given to you .GroupCost of Manufacturing Broken M&M
107、 Rejected Colorper piece(K)(k)Brown $ 0.27.4 3.1 Red$ 0.52.1 1.1Yellow $ 0.31.3 2.1Blue$ 0.36.2 7.4Orange $ 0.28.2 1.3Green$ 0.22.3 6.2 In particular week of the M&M Factory 产品ion have a major 缺陷of Broken M&M 和Rejected Color Now present which is the group of M&M you should look into during 产品ion if
108、the following data is given to you .GroupCost of Manufacturing Broken M&M Rejected Colorper piece(K)(k)Brown $ 0.27.4 3.1 Red$ 0.52.1 1.1Yellow $ 0.31.3 2.1Blue$ 0.36.2 7.4Orange $ 0.28.2 1.3Green$ 0.22.3 6.2作出决定的using Pareto Chart68西格玛绿带培训教材课件69西格玛绿带培训教材课件70西格玛绿带培训教材课件71西格玛绿带培训教材课件流程Flow Diagram -S
109、ymbolsStartStart流程流程ConnectorConnectorDecisionDecisionFlow/transitionStart/EndStart/EndServices / add value72西格玛绿带培训教材课件流程流程流程流程Flow Diagram - Making my cup of coffee Flow Diagram - Making my cup of coffee StartSip my coffeeNo goodgood73西格玛绿带培训教材课件Inspect the Inspect the tubestubesExternal dirty def
110、ectExternal dirty defectfound that cannot be found that cannot be removed ?removed ?Submit good tubeSubmit good tubeto OQA for buy offto OQA for buy offExternal dirty defectExternal dirty defectfound that cannot be found that cannot be removed ?removed ?PackingPackingTake out the Take out the defect
111、 & scrapdefect & scrapFINAL INSPECTION FINAL INSPECTION 流程流程流程流程FLOW CHARTFLOW CHARTRELATED TO THE EXTERNAL STAINRELATED TO THE EXTERNAL STAINYESYESNONOYESYESNONOAQL AQL Return back theReturn back thelot to lot to 产品产品产品产品ionionFRONTLINEFRONTLINE流程流程流程流程( SGP )( SGP )ADS assemblyADS assemblyRaw Glas
112、sRaw GlassAnode Sealing Anode Sealing Glass washing Glass washing Cathode Sealing Cathode Sealing Glass sawingGlass sawingCathode sealing Cathode sealing AssemblyAssemblyPre CVD washingPre CVD washingCVD coatingCVD coatingSandblastingSandblastingWashing Washing Tinning Tinning Thermal shockThermal s
113、hockWashing Washing FluorescentFluorescentFinal Final inspectioninspectionTestingTestingEndEndBACK- LINEBACK- LINE流程流程流程流程( BTM )( BTM )流程流程流程流程mapping in term of extrenal stain will be mapping in term of extrenal stain will be performed to identify origin sources of the performed to identify origin
114、 sources of the defectdefectNon Value Non Value added added activityactivityCost of Poor Cost of Poor 质量质量质量质量Cost of Poor Cost of Poor 质量质量质量质量流程流程流程流程Flow Example IFlow Example I74西格玛绿带培训教材课件EndFail OnceFail twiceMechanical Dimension(FAIR Result)Light Transmission ( 40mV)Insulation Resistance( 10
115、Gohm)Start Pass Pass Fail流程流程流程流程Flow Example IIFlow Example II75西格玛绿带培训教材课件Exercise -: Putting them togetherDevelop 流程flow for your project Construct a 流程flow chart for your projects Mark ( *) next to the 流程or step which is value added 和产品ive Identify the number of decision box Indicate the time re
116、quire in every step including transport Construct a table to summary the number of 产品ive 和non 产品ive 流程 , decision 和flow Present you Flow chart 和propose changes76西格玛绿带培训教材课件FMEA ( Failure Mode Effect Analysis )FMEA , is a attempt to identify possible failure modes 和和effects or consequences that failu
117、re modes may have on performance ,提供提供a mean of prevention FMEA 提供提供s an excellent basis for classification of characteristicsIt is a fundamental 工具工具in creating a reliability system Do you have an FMEA in Your 流程流程?When you have a Failure do you know what causes it ? How does FMEA help in Daily wor
118、k How does FMEA help in Daily work Improves the 质量质量, reliability, 和和safety of 产品产品Increases customer satisfactionReduces 产品产品development timing 和和costDocuments 和和tracks actions taken to reduce risk 77西格玛绿带培训教材课件Types of FMEA Types of FMEA System FMEA: Used to analyze systems 和subsystems in the earl
119、y concept 和设计stages. Focuses on potential failure modes associated with the functions of a system caused by design.设计设计FMEA: Used to analyze 产品 before they are released to 产品ion.流程流程FMEA: Used to analyze manufacturing 和assembly 流程es.Defect FMEA : Used to analyze 和prioritize 缺陷to prevent reoccurrence
120、 (in 产品 和流程es)l$305,000 Increased Annual Revenuel3220 Minutes Less Maintenance Downtime/ Machine/Monthl452,000 More Piece Increased Annual 产品ionl28 Less Maintenance Calls/Machine/Month l15 Less Sensor Failures Per Yearl$7,200 Annual Reduction in Maintenance 和Replacement Costs (Source from GE , Motor
121、ola ,Bombardier , ABB ,IBM Allied Signal)FMEA Benefits FMEA Benefits 实例实例实例实例From 6From 6西格玛西格玛西格玛西格玛Company Company 78西格玛绿带培训教材课件 FMEA : FMEA : 流程流程流程流程/ / 产品产品产品产品FMEAFMEA79西格玛绿带培训教材课件SeveritySeverity is an assessment of the seriousness of effectSeverity is applies to the effect onlyThe following
122、is the evaluation Criteria80西格玛绿带培训教材课件OccurrenceThis means how Frequently the failure mode is projected to occur as a result of a specific causeIt can be thought of as the probability of occurrence of the failure modeUse a scale of 1(very low) to 5( very high) Frequency81西格玛绿带培训教材课件Detection ( DET
123、) Assume the failure has occurred; then asses the probability that the current control will detect the failure mode before the 产品产品is shipped outUse a scale of 1(very low) to 5( Nil) Detection 82西格玛绿带培训教材课件Possible Controls item :Fixture fool- proofingSPCinspectionTestinggauge R & RPreventive mainte
124、nanceOperator 培训scheduleStandard operating procedurehumidity / pressure / temperature Controlfinite element analysiscalibration etcRisk Priority Number (RPN)RPN = Severity x Occurrence x DetectionNote : In themselves , absolute RPN numbers have no significant . They are use for “ Paretorising ” or r
125、anking failure causes , to decide which require corrective actions83西格玛绿带培训教材课件Building an FMEA - Step 2Enter all of the ways that each 流程流程or sub-assembly can failBuilding an FMEA - Step 3What happens when this failure occurs?Building an FMEA - Step 1Enter All of your 流程流程es or sub-assemblies in th
126、e first column84西格玛绿带培训教材课件Building an FMEA - Step 4What causes this failure?Building an FMEA - Step 5What do you now do to prevent this failure?Building an FMEA - Step 6Assign how severe the failure mode is (1 not severe, 5 catastrophic), how like it is to occur (1 seldom, 5 frequently) 和how likely
127、 you will notice the failure mode before it occurs (1 likely, 5 not likely)85西格玛绿带培训教材课件Sample 流程流程FMEA : Filling Petrol into Car86西格玛绿带培训教材课件87西格玛绿带培训教材课件What is Statistics ?Inferential StatisticsStatistics: The science of collecting, describing & interpreting data The 流程流程of collecting, presenting
128、 和和describing sample data, using graphical 工具和工具和numbersDescriptive StatisticsThe 流程流程of interpreting the sample data to draw conclusions about the population from which the sample was taken Mean=3.2Std dev=2.5T- T-testtestF-F-testtestHypothesis testingLinear RegressionChi2-Chi2-testtest88西格玛绿带培训教材课
129、件Population vs SamplePopulation vs SamplePopulation: The set of all possible objects, individuals or events of interestmean =m m 标准偏差= or sIf we could measure the mileage per gallon for all the cars in the parking lot the mean 和标准偏差和标准偏差 would be, m m =17.3 mpg s= 5.4 mpg89西格玛绿带培训教材课件Population vs S
130、ampleWhat are 某些某些common 实例实例of where we use a sample to make inferences about the population?Instead of taking data for all the cars we take a sample of cars from the parking lot 和和measure the mean 和标准偏差和标准偏差 for their mileage per gallonx = 19.3 mpg s =2.0 mpg From these statistics we can make infe
131、rences regarding the PopulationSample: A subset of the populationmean = x 标准偏差标准偏差= s90西格玛绿带培训教材课件x971.5”x1x2x3x4x5x6x7x856.5”61.5”62.5”64.0”64.5”64.5”65.5”67.5”Mean HeightMean Center Tendency 和和Measure of LocationThe Mean is obtained by adding all the data 和和dividing by thenumber of observations, t
132、he symbol x is used to represent the meanx =56.5 + 61.5 + 62.5 + 64.0 + 64.5 + 64.5 + 65.5 + 67.5 + 71.59= 64.2”x1 + x2 + x3 + xnx =# of observationsSum of Data=nThe mean for this data sample is: 91西格玛绿带培训教材课件To find the Median, order the data from shortest to tallestThe Median is the the measuremen
133、t in the middle, 50% of the data points are above the median 和和50% below Note : If there is an even number of data points then the take the average of the two data points nearest the middle56.5”71.5”61.5”62.5”64.0”64.5”64.5”65.5”67.5”In this case the median = 64.5”Median HeightMedian Center Tendency
134、 和和Measure of Location92西格玛绿带培训教材课件98,000236,00040,00036,00032,00028,000Mean : 46,000 Total Payroll = $1,058,000# on payroll : 23Average Annual Pay :1058000/23 = $ 46,000How Do We Measure Location93西格玛绿带培训教材课件94西格玛绿带培训教材课件95西格玛绿带培训教材课件96西格玛绿带培训教材课件For sniper 2 help him to reduce variation in during
135、firing Concept of Precision 和和AccuracyHow would you help sniper 1 和和2 ?For sniper 1, we adjust the rifle aiming tip to center on the target Precision 和和Accuracy97西格玛绿带培训教材课件Linking Precision 和和Accuracy to 流程流程Capability & ControlAccurateYesNoPreciseYesNoAccurate 和和preciseNot accurate but preciseNot
136、accurate 和和not preciseAccurate but not preciseUCLLCLUCLLCLUCLLCLUCLLCLLSLUSLLSLUSLLSLUSLLSLUSLLSL = Lower Specification LimitUSL = Upper Specification LimitLCL = Lower Control LimitUCL = Upper Control LimitOut Liers98西格玛绿带培训教材课件95.45 %69.12 %99.73 %93.32 %99.9999998 %99.9996549 %Short Term YieldLong
137、 Term YieldLinking Precision 和和Accuracy to 流程流程Capability & ControlHow can we categorise problems found in a 流程?1.Precision problems2.Accuracy problems3.Out-lier problems99西格玛绿带培训教材课件Statapult Exercise 1Separate into teamsEach person takes one turn at being a shooter, recorder 和和measurerEnsure the S
138、tatapult is preset to:Pull back Angle - 177Hook Position - 4Cup Position - 1 Stop Position 3Pin Position - 2Each person will take 3 shotsPractice shots are not allowed 和和there is a 15 seconds time limits between successive shotsRecord the distances in the table 提供提供dCompute the Range, Max, 和和Min of
139、the distanceCupPositionCupPositionPinPositionPinPositionPullBackAnglePullBackAngleHookPositionHookPositionStopPositionStopPosition100西格玛绿带培训教材课件Statapult DataRange = Max - Min = _101西格玛绿带培训教材课件Plot the Data onto Run Chart102西格玛绿带培训教材课件Exercise : Understanding of Data Variance Reduction Project - How
140、 to Hit our Target A) Create a IPO A) Create a IPO 和流程和流程和流程和流程Flow Chart on Firing the catapultFlow Chart on Firing the catapultB) Perform B) Perform 脑力风暴和脑力风暴和脑力风暴和脑力风暴和draw a Cause draw a Cause 和和和和Effect Diagram Effect Diagram C) Discuss C) Discuss 和和和和label on the cause label on the cause 和和和和e
141、ffect diagram effect diagram C- controllable , N- noise , X - Experimental factor C- controllable , N- noise , X - Experimental factor D) Combine with the FMEA for catapult firingD) Combine with the FMEA for catapult firingE) Write down the standard operating procedureE) Write down the standard oper
142、ating procedure103西格玛绿带培训教材课件Cause 和和Effect Diagram -Step 1ManMachineMethodMaterialEnviromentHow to Shoot My Catapult AccuratelyCause 和和Effect Diagram- Step 2 Discuss with your team 和和identified : which are the factors that are controllable , noise 和和experiment Label them next to each factor on the
143、cause 和和effect diagram 104西格玛绿带培训教材课件Standard Operating ProcedureFactorsDocument NoWork instructionAccurate Firing of Catapult HSP-WI-00011) Tape the Statapult on the Floor 2) Have one person to apply pressure on the Base3) Check the Rubber B和和position 4) One person Pull the Firing Arm Backward to t
144、he required angle5) Place the ball on the cup Gently 和和check orientation6) Let Everyone know that we are ready to fire7) Release The Firing ArmsCause 和和Effect Diagram- Step 3PF / CE / CNX / SOP = Variance Reduction105西格玛绿带培训教材课件Statapult Data 2Range = Max - Min = _106西格玛绿带培训教材课件Plot the Data onto Ru
145、n Chart 2107西格玛绿带培训教材课件108西格玛绿带培训教材课件Statistical Terminology Use in 6西格玛西格玛Important points to note:In order to apply Cp, Cpk, 和Sigma studies in a 流程, the 流程must be normally distributed.109西格玛绿带培训教材课件Impact On 质量质量 Measurement With Changes in Variation110西格玛绿带培训教材课件Impact On 质量质量 Measurement With Ch
146、anges in Variation111西格玛绿带培训教材课件Impact On 质量质量 Measurement With Changes in Variation112西格玛绿带培训教材课件113西格玛绿带培训教材课件114西格玛绿带培训教材课件Exercise : Applying what you have learn Using Excel 和和key in your stratapult data before 和和after 改进改进 Find Cpk , Cp 和和dpm for each set of dataWhat is a Histogram ?A histogram
147、 is a graphical representation of the frequency distribution of data. It is presented as a series of rectangles, each proportional in width to the range of values within a class.Histogram can give a graphical indication on whether the data is normally distributed115西格玛绿带培训教材课件Constructing a Histogra
148、mIn a Muffin factory each weight of the muffin is recorded on a weighing machine In a Muffin factory each weight of the muffin is recorded on a weighing machine the following is the data.the following is the data.Use the following rule to draw a histogram Number of data n = 55Range of data : R = Xma
149、x - X min = 40 15 = 25Number of Class : k = I Square Root n I = 7.1416 = 8 (round up)Class Width w = R/k = 25/8 = 4 (round up)116西格玛绿带培训教材课件Lg Taken From Each DataNon Normal DataCpk = .6579Cpk = .1942Transformation of Non normal data 1 117西格玛绿带培训教材课件Transformation of Non normal data example R3 (Resi
150、stance 0.1 GOhm)Insulation resistance data is Insulation resistance data is not normally distributednot normally distributedTransformed R3 (Log (Resistance) 0.1)118西格玛绿带培训教材课件119西格玛绿带培训教材课件120西格玛绿带培训教材课件121西格玛绿带培训教材课件222122西格玛绿带培训教材课件123西格玛绿带培训教材课件124西格玛绿带培训教材课件125西格玛绿带培训教材课件Essential Features of a
151、Measurement System Variability should be small compared to that due to the 流程和流程和the specification limits, or to the total 产品产品variation.As rule of thumb for GR&R:126西格玛绿带培训教材课件Gage R&R Studies of Continuous DataGageGage: Measurement system, : Measurement system, R R: Repeatability, & : Repeatabilit
152、y, & R R: Reproducibility: ReproducibilityShort Method: Quick 和和meaningful way of determining if Gage is acceptable Only five samples 和和two appraisers are needed Can not separate Repeatability 和和ReproducibilityLong Method: Need at least 10 parts, 3 samples 和和multiple tests Can separate Repeatability
153、 和和Reproducibility Can not separate Operator* Part interactionsAnalysis of Variance (ANOVA) Method: Can separate Repeatability 和和Reproducibility Can separate Operator* Part interactions 127西格玛绿带培训教材课件128西格玛绿带培训教材课件129西格玛绿带培训教材课件GR&R Long MethodExampleUsing GR&R matrix of : 10 Parts, 3 Operators, 2 t
154、rialsS2 measurement = S2 repeatability + S2 reproducibility = (0.047)2 + (0.011)2 = 0.02209 + 0.000121 = 0.00233S measurement = 0.048S reproducibility = R(Xbar) / d2 = (4.04 - 4.03) / 1.69 = 0.011Average Rbar = (0.06 + 0.05 + 0.05) / 3 = 0.05333S repeatability = Average Rbar / d2 = 0.05333 / 1.13 =
155、0.047130西格玛绿带培训教材课件GR&R Long Method (cont)Given specification = 3.8 +/- 0.5GR&R (Tolerance) = (6 x 0.048) / (1) = 0.288GR&R (Total) = (0.048) / 0.241 = 0.200131西格玛绿带培训教材课件Graphical Interpretation of GR&RLCL = Avg m m (Rbar x A2)UCL = Avg m m + (Rbar x A2)LCLUCL132西格玛绿带培训教材课件Very good repeatability f
156、or operator 3Poor reproducibility because of operator 1, repeatability for operator 1 is also bad.Graphical Interpretation of GR&R (cont)Assuming operator is not a trained inspector, the expected result will be:133西格玛绿带培训教材课件Conducting GR&R Using Long Method1.Determine GR&R matrix.2.Number the parts
157、, but do not let the operators / inspectors know that you are conducting a gauge study.3.Calibrate the gauge.4.Let the operators / inspectors measure the parts in random order.5.When repeating trials, re-shuffle the parts before letting the operators / inspectors conduct the measurements.6.Use opera
158、tor / inspectors from different shift if possible7.Tally the result 和compute GR&R8.Interpret the result graphically if necessary134西格玛绿带培训教材课件If Repeatability is the problem Replace, repair or adjust the equipment. Revise the standard operating procedure.If Reproducibility is the problem Examine the
159、 difference between operators 和和determine if it is due to insufficient 培训培训, or inspector not following the SOP.If P/TOL 0.30 Check if the specification limits is reasonable or attainableIf P/TOT 0.30 Check the items that were part of the measurement system study 和和see if they are representing at le
160、ast 80% of the actual total 流程流程variability. Check If the measurement system equipment is the best condition 和和is performing up to specifications. (or do we have no choice but to use it)If P/TOL or P/TOT close to 0.3 but if the 流程流程is operating at high capability Cpk 2, then the measurement system i
161、s most probably not the problem Improving Poor Measurement System135西格玛绿带培训教材课件Discuss in group on how to measure the diameter of the catapult ball 和和determine the 流程流程steps.In your group measure the circumference of the group of Catapult Ball 提提供供d to you.Each group is to measure the 10 Balls three
162、 times, n = 3.Collect the data from the other two groups.Once the data is available, combine the result 和和conduct the GR&R study using the long method.Discuss the repeatability 和和reproducibility of the operator. Use graphical analysis if appropriate.Gage Repeatability 和和Reproducibility- Exercise136西
163、格玛绿带培训教材课件Gage Repeatability 和和Reproducibility- Exercise137西格玛绿带培训教材课件GR&R for Attribute Data为什么attribute data ?When the output is qualitative 和not quantitative(i.e. color of a solder bump, black spot on flashcube, print 质量) When it is too costly to measure for continuous output(i.e. analyzing the w
164、avelength of the solder bump)When it is too time consuming to measure for continuous output(i.e. measure the light transmission of the flashtube bodies)When continuous measurement is destructive to the part(i.e. wire pull test, impact test for steel rod)为什么is attribute GR&R important ?It highlights
165、the subjective element in the measurement systemIndicate if the measurement system is biasIt is the basis for establishing accurate 和precise visual inspection138西格玛绿带培训教材课件Terminology in Attribute GR&RGauge effectiveness (E):It is the ability of the inspection system to distinguish between defective
166、 和non-defective parts. Number of correctly identified parts Number of correctly identified partsE = -E = - Total number of parts inspected Total number of parts inspectedProbability of False Rejects, P(FR):It is the probability of rejecting a part when its actual performance is acceptable. Number of
167、 times good parts rejected by the inspectorsNumber of times good parts rejected by the inspectorsP(FR) = -P(FR) = - Total number of time good parts are inspected Total number of time good parts are inspectedProbability of False Acceptance, P(FA):It is the probability of accepting a part when its act
168、ual performance is unacceptable. Number of times defective parts accepted by the inspectorsNumber of times defective parts accepted by the inspectorsP(FA) = -P(FA) = - Total number of time defective parts are inspected Total number of time defective parts are inspected139西格玛绿带培训教材课件Terminology in At
169、tribute GR&R (cont)Gauge Bias (B):It is the measure of the inspection system tendency in classifying a part defective or non-defective. Probability of false reject, P(FR) Probability of false reject, P(FR)B = -B = - Probability of false acceptance, P(FA) Probability of false acceptance, P(FA)Guideli
170、nes for Attribute GR&R:140西格玛绿带培训教材课件ABC company is a food 流程流程ing company that manufactures coffee beans for the up market cafeteria. One of the important 流程流程is the roasting of the coffee bean in order to bring out the aroma of the coffee. The inspection system implemented in this 流程流程is a “smell
171、gauging” method, whereby every batch of roasted coffee beans is determine to be pass or fail by 3 trained inspectors in this gauging 流程流程. The supervisor of the bean roasting 流程流程suspected that a large number of rejects in the line was due to the measurement error, 和和wanted to conduct an experiment
172、to analyze the accuracy of the existing measurement system that consists of the 3 trained inspectors.He deliberately mix 25 marked batches of both good 和和bad roasted bean into the line without the knowing of the inspectors. He then repeatedly put back the same 25 coffee bean batches into the line su
173、ch that each inspectors tested the same 25 coffee bean batches for a total of 3 times unknowingly. At the end of the experiment, the supervisor consolidated all the finding on the 25 batches by each inspector, 和和compared these findings with the actual performance of these 25 coffee bean batches (i.e
174、. the classification of good or bad batches is determined by the supervisor himself).With reference to the data collected by the supervisor (i.e. presented in the next page), What can you conclude about the measurement system of this coffee bean case study ?Attribute GR&R Example141西格玛绿带培训教材课件Attrib
175、ute GR&R Example (Cont)R = RejectA = Accept142西格玛绿带培训教材课件Attribute GR&R Example (Cont)Step 1: With reference to the actual performance, assign 1 point for every correct inspection response.143西格玛绿带培训教材课件Attribute GR&R Example (Cont)Step 2: Find the total reject 和accept opportunitiesWith reference to
176、 actual performance, multiply the total parts, total R, 和total A by the number of trials 和grouped by each inspector.Sum the total opportunities horizontally for all the inspectorsStep 3: Find the actual false rejects 和false accepts in the experiments Correct response is when the assign point for eac
177、h trial is equal to 1 (i.e. test = actual performance). False reject is when the assign point for each trial is 0 和the trial result is reject, R. False accept is when the assign point for each trial is 0 和the trial result is accept, A.144西格玛绿带培训教材课件Step 4: Find the total probabilities from the GR&R
178、equations given earlierAttribute GR&R Example (Cont)Conclusions:1. All 3 inspectors tend to bias towards rejecting good parts.2. Inspector 2 has acceptable gauging skill as far as the supervisor is concerned.3. Inspector 3 has the highest bias towards rejecting good parts.4. Over all effectiveness i
179、s acceptable, but however is bias towards false reject. 145西格玛绿带培训教材课件Variable Data vs Attribute Data in GR&RVariable Data -Data that is continuous 和can be quantify by the measurement system (i.e. Temperature, Voltage, Length)Attribute Data -Data which cannot be quantified, 和exist in one of the 2 st
180、ates, either “accept” or “reject”. (i.e. Taste, Visual inspection) 146西格玛绿带培训教材课件Effective Attribute GR&R- Give concise instructions- Reduce number of criteria in one single test- Identify prominent features- 提供visual comparison when possible- Adequate 培训和familiarization- Use appropriate inspection
181、gauge (i.e. Electronic indicator)- Convert attribute data to variable whenever possible Gauging the clarity of glass tube visually versus measuring the intensity of light passing through the tube. Gauging for fever by h和versus measuring the body temperature. Gauging the 质量 of the red wine versus mea
182、suring the viscosity 和PH level of the wine.147西格玛绿带培训教材课件MSA Design:- No. of Operators: 3- No. of Replicates: 3- No. of Parts: 20Interpretation:- Probability of False Acceptance is too high- Caused: poor sockets cap condition- Lack of audio indicator- Gage is unacceptableExample of improving attribu
183、te GR&RBefore 改进改进Interpretation:- Use ceramic sockets cap with pins guide- Include buzzer as reject indicator- Gage is now acceptableAfter 改进改进Measurement system includes a test socket where detector can be plug in, 和和the green LED will go off if the measured resistance is less than 100 M .148西格玛绿带
184、培训教材课件Attribute GR&R ExerciseBreak into teams of 5 with the following roles:One InstructorOne InstructorOne SupervisorOne SupervisorThree InspectorsThree InspectorsEach team will be assigned with 25 M&Ms of different 质量. The instructor will determine the criteria for a good M&M (i.e. Clear M sign, n
185、o chip-off, color, etc). The instructor will pass down these requirements to the supervisor who will then pass these information to the inspectors. The supervisor will distribute the M&Ms to the inspectors as randomly as possible, 和ask them to rate each M&M as “good” or “bad”. There will only be one
186、 trial for each inspector, 和the inspectors are to submit their rating sheets to the supervisor at the end of the inspection. The inspectors are to assess the M&Ms individually without any communication among each others.149西格玛绿带培训教材课件Probability Theory概率理論概率理論What is probability ?Probability is the
187、chance which something will happen, 和are expressed as fractions or as decimals. It is a value between 0 和1 (i.e. 0% to 100%). What is an event ?An event is one or more of the possible outcomes of doing something. (i.e. tossing of a coil will give rise to a Head event or a Tail event.What is an exper
188、iment ?It is the activity that produces events.EventPeopleMaterialsEquipmentEnvironmentProbability will determine the chance of an event. In 6 sigma 方法, we want to control our inputs so as to increase the probability of a favorable event. Conversely, we will also make decision on what needs to be do
189、ne (or change) based on the probability 知识 attained from experiments.概率理論150西格玛绿带培训教材课件Probability in Everyday Lives Experiments表現手法表現手法Decisions結論結論Weather Forecast Opting between indoor 和outdoor activities 天氣預報Gambling Whether or not to ask for another card in a blackjack game賭博Sales Forecast Inve
190、ntories planning銷售額預測 詳細計划Examination questions Which chapters of the books to revise for the examination猜測考試題目 書中哪一章會被考試Statistics shows that 9 out of 10 patients die in this operation. Since you are my 10th patient, you dont have to worry at all, as your chance is perfect.It is really comforting t
191、o know about this.這是一句安慰人的話吧日常生活的概率實例日常生活的概率實例151西格玛绿带培训教材课件Probability Approach1. Classical approach古典定義古典定義2. Relative frequency approach頻率定義頻率定義Classical approach defines that the probability for an event to occur as: Number of outcomes where the event occursProbability of an event = - Total numb
192、er of possible outcomes 1Probability (head in one toss) = - = 0.5 (1 + 1) 1Probability (2 in a dice roll) = - = 0.167 (1+1+1+1+1+1)Classical approach is based on logical reasoning before any experiment takes place概率定義152西格玛绿带培训教材课件Relative frequency approach頻率定義頻率定義It defines probability as the prop
193、ortion of times which an event occurs in the long run when the conditions are stable.Example:From historical data, it was found out that 75,000 out of every 100,000 children who are over-weight after the age of 10, remain to be over-weight when reaching adulthood.With this information, it is estimat
194、ed that the probability of over-weight children after age of 10 to become a over-weight adult is:75,000 - = 0.75 (i.e. 75%) 100,000 Relative frequency probability in getting “head” from tossing a unbiased coin.153西格玛绿带培训教材课件Probability Rules“Mutually exclusive events”Events are said to be mutually e
195、xclusive when only one event can take place at a time. (i.e. tossing of a coin)Conversely, if more than one output can occur at the same time, the event are not mutually exclusive. (i.e. Probability of drawing an ace or a heart from a deck of cards)Most people who makes use of probabilities are conc
196、erned with two conditions:1. Marginal or unconditioned probability - the case where only one event or another will take place (i.e. mutually exclusive event)2. Conditioned probability - the case where two or more events will both occur概率的規則154西格玛绿带培训教材课件Probability Rules (Cont)“Marginal probability”
197、example:25 members of a class drew lots to determine which student would get a free trip to the much sought after National Day Festival in Thailand. The probability for any student of that class to win the draw can be calculated as:P(Winning) = 1/25 = 0.04P(Winning) = 1/25 = 0.04In this case, a stud
198、ent chance is one in 25, as we are certain that only one student can win at a time (i.e. events are mutually exclusive).Probability of something happening(i.e. Total area = P(any event) = 1)Probability of an event(i.e. P(Winning) of 1 student)P(1 event) = 0.04P(1 event) = 0.04155西格玛绿带培训教材课件Probabili
199、ty Rules (Cont)“Addition rule for marginal probability”Additional rule in marginal probability is used when we are interested in the probability that one thing or another will occur, 和is denote as:P(A or B) = P(A) + P(B)example:The management of a company will select one out of the 4 equally capable
200、 employees to be promoted to the supervisor position. The employees consist of John, Sally, Bill, 和Karen. What is the probability that Sally will be the candidate ?P(Sally) = = 0.25P(Sally) = = 0.25What is the probability that Sally or Bill will be the candidate ?P(P(SallySally or or BillBill) ) = P
201、(= P(SallySally) + P() + P(BillBill) ) = + = + = 0.5= 0.5BillSallyJohnKaren156西格玛绿带培训教材课件Probability Rules (Cont)“Addition rule for conditioned probability”Conditioned probability is used when we are interested in the probability that one thing or another will occur, when the events are not mutually
202、 exclusive (i.e. when events can occur together):AceHeartP(Ace)P(Heart)P(Ace 和和Heart)P(Ace or Heart) = P(Ace) + P(Heart) - P(Ace 和和Heart) = 4/52 + 13/52 1/52 = 16/52 = 0.3077Example:What is the probability of getting an Ace or a heart in a deck of 52 cards ?157西格玛绿带培训教材课件Probability under Statistica
203、l Independence“Statistical independence”Under statistical independence, the occurrence of one event has no effect on the probability of the occurrence of any other event.“Joint probabilities under statistical independence”The probability of two or more independent events occurring together, or in su
204、ccession is denoted as: P(AB) = P(A) x P(B)where,P(A)= Marginal probability of event A occurringP(B)= Marginal probability of event B occurringP(AB)= Joint probability of event A 和B occurring together or in succession.Example: What is the probability of getting 2 heads on two successive tosses of a
205、fair coin ?P(H1 和和H2)= P(H1) x P(H2) = 1/2 x 1/2= 1/4= 0.25独立性与独立事件的概率158西格玛绿带培训教材课件“The probability tree”A graphical approach that describe the probabilities of event under the condition of statistical independence.How should the probability tree of tossing a fair coin look like ?Toss 1Total : 1.0
206、Toss 21.00 Toss 31.00 .5 .5P(H) = 0.5P(T) = 0.5.125.125.125.125.125.125.125.125P(H) = 0.5P(H) = 0.5P(H) = 0.5P(H) = 0.5P(T) = 0.5P(T) = 0.5P(T) = 0.5P(T) = 0.5 .25 .25 .25 .25P(H) = 0.5P(H) = 0.5P(T) = 0.5P(T) = 0.5159西格玛绿带培训教材课件“Transferring the probability tree to outcomes table”1 TossPossible Out
207、comesProbability3 TossesPossible OutcomesProbability2 TossesPossible OutcomesProbabilityH1T10.50.5H1, H2H1, T20.250.25T1, H2T1, T20.250.25H1, H2, H3H1, H2, T30.1250.125H1, T2, H3H1, T2, T30.1250.125T1, H2, H3T1, H2, T30.1250.125T1, T2, H3T1, T2, T30.1250.1251.01.001.00160西格玛绿带培训教材课件“Conditional prob
208、abilities under statistical independence”The probability that a second event B will occur if a first event A has already happened. As mentioned earlier, under statistical independence, the probability of one event will not affect the probability of the subsequent event:Therefore, P(B|A) = P(B)In the
209、 tossing of a fair coin, what is the probability that the second toss is head ?Since the tossing of a coin is a statistical independence experiment, the result of the first toss will not influence the out come of the second toss, therefore the probability of the second toss to be head is still : = 1
210、 out of 2= 0.5161西格玛绿带培训教材课件Probability under Statistical Dependence“Statistical Dependence”Statistical dependence exists when the probability of 某些event is dependent upon or affected by the occurrence of 某些other event.“Statistical dependence example”Assuming there is one bag containing 10 balls as
211、follows:4 x Red Striped Box3 x Green Dotted Box2 x Red Dotted Box1 x Green Striped Box162西格玛绿带培训教材课件“Marginal probabilities under statistical dependence”Under statistical dependence, conditional probabilities are computed by summing up the probabilities of all joint events in which the simple event
212、occurs.P(Green) = P(Green Dotted) + P(Green Striped) = 0.3 + 0.1 = 0.4 P(Dotted) = P(Green Dotted) + P(Red Dotted) = 0.3 + 0.2 = 0.5P(Striped) = P(Red Striped) + P(Green Striped) = 0.4 + 0.1 = 0.5163西格玛绿带培训教材课件ExampleSuppose someone draws a green box from the bag. What is the probability that it is
213、dotted ?The question can be rephrase as : What is the conditional probability that a drawn box is dotted, given that the box is green in color ?Symbolically = P(Symbolically = P(DottedDotted | | GreenGreen) ) = = P(P(DottedDotted 和和和和GreenGreen) / P() / P(GreenGreen) ); ; for statistical dependence
214、event P(P(DottedDotted 和和和和GreenGreen) = 3/10) = 3/10P(P(GreenGreen) = 4/10) = 4/10 = 0.3 / 0.4 = 0.3 / 0.4 = 0.75 = 0.75164西格玛绿带培训教材课件Solving the problem with a tableWhat is the probability that a drawn box is striped, given that the box is red ?Symbolically = P(P(StripedStriped | | RedRed) )Dotted
215、StripedGreen0.30.10.4Red0.20.40.60.50.5What is the probability that the drawn box is red ?What is the probability of drawing a red striped box ?What is the conditional probability of the drawn box to be striped when drawn red ?0.40.40.60.6 = P(P(StripedStriped 和和和和RedRed) / P() / P(RedRed) ) = 0.4 /
216、 0.6 = 0.667 = 0.4 / 0.6 = 0.667165西格玛绿带培训教材课件“Joint probabilities under statistical dependence”We know that the formula for conditional probability under statistical dependence is :P(P(A A| |B B) = P() = P(A A 和和和和B B) / P() / P(B B) )If we shift P(A 和B) to the left 和P(A|B) to the right, we will ha
217、ve:P(P(A A 和和和和B B) = P() = P(A A| |B B) x P() x P(B B) )Joint probability of events A 和和B happening together or in successionProbability of event A given that event B has happenedProbability that event B will happened166西格玛绿带培训教材课件With reference to the scenario of 10 boxes in a bag, what is the pro
218、bability to draw a green dotted box ?P(P(DottedDotted 和和和和GreenGreen) = P() = P(DottedDotted| |GreenGreen) x P() x P(GreenGreen) ) = 0.75 x 0.4 = 0.75 x 0.4 = 0.3 = 0.3DottedStripedGreen0.30.10.4Red0.20.40.60.50.5P(G 和和D)P(R 和和S)P(R 和和D)P(G 和和S)P(D)P(S)P(G)P(R)167西格玛绿带培训教材课件Posterior Probability (Ba
219、yes Theorem)Posterior probability defines that certain probabilities were Posterior probability defines that certain probabilities were altered after the people involved obtained additional altered after the people involved obtained additional information. This new probabilities are known as revised
220、, information. This new probabilities are known as revised, or posterior probabilities.or posterior probabilities.A Chef has formulated a new chicken recipe 和based on his experience with formulating a steak recipe, the probability of selling a recipe well is to increase the amount of garlic powder i
221、n the ingredients. However, after introducing the new recipe for one week, he found that the new chicken recipe does not sell as well as the old one, he must therefore revise his prior probabilities 和use other ingredients in the recipe. Calculating posterior probabilitiesAssuming that we have equal
222、numbers of 2 types of “biased” dice in a bowl. One type of the biased die will roll out a value of Five 40% of the time, 和the other type will roll out the value Five, 70% of the time. If one dice is drawn 和roll to the value Five, what is the probability that it is a type 1 dice ?168西格玛绿带培训教材课件Type 1
223、Type 20.40.70.50.51.01.00.4 x 0.5 = 0.200.7 x 0.5 = 0.35P(Five) = 0.55P(Five) = 0.55ElementaryEventProbability(Elementary Event)Probability(Five|Elementary Event)Probability(Five, Elementary Event)Calculating posterior probabilities (Cont)P(P(A A 和和和和B B) = P() = P(A A| |B B) x P() x P(B B) )To find
224、 the probability that the dice we have drawn is type 1, we should adopt the formula for conditional probability under statistical dependence:P(P(B B | | A A) = P() = P(BABA) / P() / P(A A) )P(P(Type 1 Type 1 | | FiveFive) = P() = P(Type 1, FiveType 1, Five) / P() / P(FiveFive) ) = 0.20 / 0.55 = 0.20
225、 / 0.55 = 0.364 = 0.364Therefore the probability of drawing a type 1 dice is 0.364169西格玛绿带培训教材课件Calculating posterior probabilities (Cont)As such, what is the probability of drawing a type 1 dice before the dice was rolled ?0.5What is the revised probability of drawing a type 1 dice after the dice w
226、as rolled 和having a value of Five ?0.364If we are to roll the same dice again 和achieve a value of Five, what is the probability that it is a type 1 dice ?P(P(Type 1 Type 1 | | 2 Fives2 Fives) = P() = P(Type 1, 2 FivesType 1, 2 Fives) / P() / P(2 Fives2 Fives) )ElemEventProbability(Elem Event)Probabi
227、lity(2 Fives|Elem Event)Probability(2 Fives, Elem Event)Type 1Type 20.4 x 0.4 = 0.160.7 x 0.7 = 0.490.50.51.01.00.16 x 0.5 = 0.080.49 x 0.5 = 0.245P(2 Fives) = 0.325P(2 Fives) = 0.325 = (0.08 / 0.325) = (0.08 / 0.325) = 0.246 = 0.246170西格玛绿带培训教材课件Further example of posterior probabilities更多的實際例子更多的實
228、際例子Kenneth is a manager of a automation 设计company 和has previously 8 engineering staffs reporting to him. Based on his assessments on the 8 engineers, he is only satisfied with the performance of 6 of his engineers. He further concluded from past 5 years observation that if a project was assigned to
229、those engineers who had met his work expectation, 85% of the time the project will be completed before dateline. On the other hand, the 2 engineers whose performance is unsatisfactory, has only 35% of the time completed their projects before dateline.One of the 2 “unsatisfactory” engineer had resign
230、ed from the company 8 months ago, 和his replacement, Joe had since then join the company for a period of 6 months, 和it is now time for Kenneth to review Joes probation status, which he would like to do this through probability study.During the pass 6 months in the company, Joe had successfully comple
231、ted 3 projects before the given dateline. Based on this information, what is the chance that Joes actual ability is up to Kenneth expectation ?171西格玛绿带培训教材课件Further example of posterior probabilities (cont)Event = Employing the correct or incorrect personStrike = Success in closing the project befor
232、e datelineEventP(Event)P(3 Strikes|Event)P(3 Strikes, Event)P(P(Correct Correct | | 3 Strikes3 Strikes) = P() = P(Correct, 3 StrikesCorrect, 3 Strikes) / P() / P(3 Strikes3 Strikes) ) = (0.4606 / 0.4713) = (0.4606 / 0.4713) = 0.9773 = 0.9773If Joe has completed 3 successive projects before dateline,
233、 the posterior probability that he is the correct candidate for the job is 0.9773 (or 97.73%).CorrectIncorrect0.853 = 0.61410.353 = 0.04296/8 = 0.752/8 = 0.251.01.00.6141 x 0.75 = 0.46060.0429 x 0.25 = 0.0107P(3 Strikes) = 0.4713P(3 Strikes) = 0.4713172西格玛绿带培训教材课件Further example of posterior probabi
234、lities (cont)If Joe does not meet his dateline target for the 4th project, does it mean that Kenneth has make the wrong judgement, 和what is the probability associated ? EventP(Event)P(SSSN|Event)P(SSSN, Event)CorrectIncorrect0.853 x 0.15 = 0.09210.353 x 0.65 = 0.02796/8 = 0.752/8 = 0.251.01.00.0921
235、x 0.75 = 0.06910.0279 x 0.25 = 0.0070P(SSSN) = 0.0761P(SSSN) = 0.0761P(P(Correct Correct | | SSSNSSSN) = P() = P(Correct, SSSNCorrect, SSSN) / P() / P(SSSNSSSN) ) = (0.0691 / 0.0761) = (0.0691 / 0.0761) = 0.9080 = 0.9080If Joe does not meet the dateline for his 4th projects, then Kenneth will be 90.
236、8% sure that Joe is the correct candidate for the job.173西格玛绿带培训教材课件StartAre events mutually exclusive ?P(A or B) = P(A) + P(B)P(A or B) =P(A) + P(B) P(AB)Are events statistically independent ?Marginal probability of event A is P(A)Marginal probability of event A is sum of all joint events where A o
237、ccursJoint probability of 2 events is P(AB) = P(A) x P(B)Joint probability of 2 events is P(AB) = P(A|B) x P(B)Conditional probability is P(A|B) = P(A)Conditional probability is P(A|B) = P(AB)/P(B)Determined posterior probabilitiesEnd174西格玛绿带培训教材课件Introduction to Probability DistributionWhat is prob
238、ability distribution?It is a theoretical frequency distribution that describe how outcome are expected to vary, 和are useful models in making inferences under conditions of uncertainty.实例实例(tossing of a fair coin)If we toss a fair coin twice, the possible outcomes are:1st Toss2nd TossTotal no. of tai
239、lsP(outcomes)TTHHTHTH21100.5 x 0.5 = 0.250.5 x 0.5 = 0.250.5 x 0.5 = 0.250.5 x 0.5 = 0.25No. of TailsTossesP(Tail)012P(H,H)P(T,H) + P(H,T)P(T,T)0.250.500.25No. of tailsProbability0120.250.5175西格玛绿带培训教材课件Example of statistical inference using probability distributionMartin is one of the authorized ma
240、gazine distributors for the monthly business journal “Financial Times”. The distribution contract with the magazine publisher has stated clearly that no back issues are allowed in the store in order to prevent readers from obtaining / misled by back-dated market information, causing creditability 和l
241、iability impact to the publisher. Due to the popularity of the magazine, all distributor are allowed to place only one order every month, 和the publisher will not refund any excess order placed by the distributor if the magazine was not sold out for any particular month.Each magazine cost US$2, 和the
242、distributor earn US$3 from each copy sold. In order to maximize his profit by reducing loss, Martin decide to use probability study to help him make his decision on the number of magazine to order each month. He collected his sale data over a period of 1 year as below:Monthly Sales100001100012000130
243、00No. of Months2442P(Monthly Sales)0.167 (i.e. 2/12)0.333 (i.e. 4/12)0.333 (i.e. 4/12)0.167 (i.e. 2/12)176西格玛绿带培训教材课件Example of statistical inference using probability distribution (cont)There are 2 types of possible losses; cost of obsolescence (i.e. excess order that has to be throw away), 和cost o
244、f loss opportunity (i.e. cost of losing sales).Cost of obsolescence = (Monthly Order - Monthly Sale) x US$2 x P(M_Sale)where negative value equal to no lossCost of loss opportunity = (Monthly Sale - Monthly Order) x US$3 x P(M_Sale)where negative value equal to no lossTotal Losses = Obsolescence Cos
245、t + Opportunity Cost177西格玛绿带培训教材课件Monthly Sales10000110001200013000ObsolescenceLosses0000P(M_Sales)0.1670.3330.3330.167Monthly Order from Publisher = 10000Total Losses4500OpportunityLosses099919981503Monthly Sales10000110001200013000ObsolescenceLosses334000P(M_Sales)0.1670.3330.3330.167Monthly Order
246、 from Publisher = 11000Total Losses2335OpportunityLosses009991002Monthly Sales10000110001200013000ObsolescenceLosses66866600P(M_Sales)0.1670.3330.3330.167Monthly Order from Publisher = 12000Total Losses1885OpportunityLosses000501Monthly Sales10000110001200013000ObsolescenceLosses100213326660P(M_Sale
247、s)0.1670.3330.3330.167Monthly Order from Publisher = 13000Total Losses3000OpportunityLosses0000Example of statistical inference using probability distribution (cont)Therefore in order to maximize his profit, Martin will need to order 12000 magazine per month.178西格玛绿带培训教材课件Types of Probability Distri
248、butionsWhat is the difference between frequency 和和probability distributions ?Frequency distribution is a listing of observed frequencies for all the outcomes that occurred when an experiment was done,while probability distribution is a listing of the probabilities for all the possible outcomes if th
249、e same experiment is to be carried out.Discrete probability distributionsA distribution of probabilities which allowed to take on a limited number of values for its outcomes. (i.e. the rolling of a dice) Binomial distribution Poisson distributionContinuous probability distributionsA distribution of
250、probabilities which allowed to take on a any number of values within a given range of outcomes. (i.e. the rolling of a dice) Normal distribution Student T distribution179西格玛绿带培训教材课件Binomial DistributionsWhat is the binomial distributions ?A probability distribution of discrete random variable, resul
251、ting from an experiment known as Bernoulli 流程.Conditions for the use of binomial distributions (i.e. Bernoulli trial)?Each experimental trial has only two possible outcomes: success or failure, head or tail.Probability of the outcome of any trial remains fixed over time (i.e. statistical independenc
252、e). 和the outcome of one event does not affect the outcome of the subsequent event.Characteristic of binomial distributionp = Probability of successq = Probability of failurer = Number of success desiredn = Number of undertaken trials180西格玛绿带培训教材课件Probability mass function for binomial distribution n
253、!Probability of r successes in n trials = - x pr x qn-r r! (n - r)!Example of using binomial distributionPaul is a manufacturing manager of a disk drive company who has five 流程engineers reporting to him. He is facing a situation where his engineers are often late for work, 和after 某些observation, he h
254、as determined that there is a 0.4 chance that any one engineer will be late. If at any one time, the 产品ion will required 3 流程engineers to run smoothly, what is the probability that his 产品ion will run smoothly at present situation ?Probability of success (late), p = 0.4Probability of failure (early),
255、 q = 0.6Number of trial (engineers), n = 5Number of success (late), r = 2 or less181西格玛绿带培训教材课件 n!Probability of r late arrival out of n student = - x pr x qn-r r! (n - r)!Example of using binomial distribution (cont)For r = 0, 5!P(0) = - (0.40) (0.65) 0!(5 - 4)! = 0.07776For r = 1, 5!P(1) = - (0.41
256、) (0.65-1) 1!(5 - 1)! = 0.2592For r = 2, 5!P(2) = - (0.42) (0.65-2) 2!(5 - 2)! = 0.3456For r = 3, 5!P(0) = - (0.43) (0.65-3) 3!(5 - 3)! = 0.2304For r = 4, 5!P(1) = - (0.44) (0.65-4) 4!(5 - 4)! = 0.0768For r = 5, 5!P(2) = - (0.45) (0.65-5) 5!(5 - 5)! = 0.01024Total probability = 0.0776 + 0.2592 + 0.3
257、456 + 0.2304 + 0.0768 + 0.01024 = 1Total probability = 0.0776 + 0.2592 + 0.3456 + 0.2304 + 0.0768 + 0.01024 = 1182西格玛绿带培训教材课件Example of using binomial distribution (cont)No. of engineers lateProbability0120.10.23450.30.4Since Paul required at least 3 engineers at one time for the 产品产品ion to be smoot
258、h, he can only allow 2 or less engineers to be late at any one day.r = 2P( r = 2 ) = P(0) + P(1) + P(2)P( r 1)= 1 - 0.04042= 0.9596Therefore there is 95.96% chance that earth will be hit more than one time by the solar wind for any given month.187西格玛绿带培训教材课件Conditions of using Poisson as an estimati
259、on of binomial:- The number of trials, n is large- The probability of success, p is smallPoisson distribution as an estimation of binomial distributionThe rule of thumb is:A Poisson is a good approximation of the binomial when the number of trial, n is equal to or greater than 20, 和the probability o
260、f success p, is equal to or less than 0.05.Under this condition;Poisson Mean, substitute by binomial mean so that; (np)X X e-npP(X) = - X!188西格玛绿带培训教材课件189西格玛绿带培训教材课件Normal Distributions常態分配常態分配What is the normal distributions ?A distribution that is used to describe 流程es with continuous random vari
261、able output, 和it comes very close to the actual observed frequency distributions of many phenomena.Conditions of a Poisson distribution ?Important features of a normal distribution are:1.The distribution has a single peak defined by a bell shaped curve.2.The mean of the normally distributed populati
262、on lies at the center of its normal curve3.Because it is symmetric, the median 和mode of the distribution are also at the center of the distribution. Hence mean = median = mode.MeanModeMedian190西格玛绿带培训教材课件What describes a normal distributions ?A population that is normally distributed can be describe
263、d by 2 parameters. Its mean (i.e. the center of the location), 和its 标准偏差 (i.e. the dispersion of the distribution).Yellow curve has very small 标标准偏差准偏差. (i.e. = 0.1)Red curve has larger 标准偏差标准偏差. (i.e. = 1)Green curve has very large 标标准偏差准偏差. (i.e. = 10)Yellow curve has small mean. (i.e. m m = 10)Gr
264、een curve has larger mean. (i.e. m m = 20)191西格玛绿带培训教材课件Area under a normal curveRegardless of what is the value of mean (m) 和标准偏差 () for a normal distribution, the total area under the curve is 1. (i.e. total probability is 1 or 100%)Normal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 - 99.9999998% -
265、 99.73% - 95.45% -68.27%- 3 variation is called natural tolerance 1. Approximately 68.27% of all values in a normally distributed population lie within one 标准偏差 from the mean.2. Approximately 95.45% of all values in a normally distributed population lie within two 标准偏差s from the mean.3. Approximatel
266、y 99.73% of all values in a normally distributed population lie within three 标准偏差s from the mean.192西格玛绿带培训教材课件Using the standard normal probability distribution tableAs the probability distribution of a normal curve is associated with its mean 和标准偏差, we are able to determine the area under the curv
267、e by normalizing the distance a random variable is away from its mean, with its 标准偏差.This is denoted as: X - m mZ = - Where;X= value of the random variable which we are concernedm m= mean of the distribution of this random variable = 标准偏差标准偏差 of this distributionz= number of 标准偏差标准偏差s from X to its
268、mean of this distribution193西格玛绿带培训教材课件Using the standard normal probability distribution table (cont)Assuming, Mean, m = 5标准偏差, = 0.1What is the probability associate with the area under the normal curve if random variable X = 5.05.Z=(X m)m) / =(5.05 5) / 0.1=0.5m m = 55.050.5 From the table:Area u
269、nder the curve when Z = 0.5 is: 0.6915194西格玛绿带培训教材课件Example of using normal distributionA self-administered 培训program aiming at upgrading the statistic skills of the line operators has been initiated in ABC company for 12 months. As this is a self-administered program, different candidates will take
270、 different times to complete the 培训. From the 12 months records, the average length of time spent to complete the program is 400 hours with a 标准偏差 of 80 hours.What is the probability that a candidate selected at random will require more than 400 hours to complete the 培训?m m = 400 hoursTotal area und
271、er normal curve = 1hence, P(400) = 0.550% chance that a random selected candidate to take more than 400 hours to complete the 培训培训.195西格玛绿带培训教材课件What is the probability that a candidate selected at random will require between 400 和550 hours to complete the 培训?Example of using normal distribution (co
272、nt)From the standardized normal distribution table,Z = 1.875 = 0.9696Area between 400 to 500 = 0.9696 0.5 = 0.4696As such 46.96% chance that a random candidate will take between 400 to 550 hours to complete the 培训培训.Z=(X m)m) / =(550 400) / 80=1.875Area when Z = 1.8750.5196西格玛绿带培训教材课件What is the pro
273、bability that a candidate selected at random will takes more than 650 hours to complete the 培训?Example of using normal distribution (cont)From the standardized normal distribution table,Z = 3.125 = 0.99911Area 650 = 1 0.99911 = 0.00089As such there is only 0.089% chance that a random candidate will
274、take more than 650 hours to complete the 培训培训.Z=(X m)m) / =(650 400) / 80=3.125Area when Z = 3.125197西格玛绿带培训教材课件What is the probability that a candidate selected at random will takes between 360 hours 和510 hours to complete the 培训?Example of using normal distribution (cont)01002003004005006007008009
275、00Z1Z2Z1 =(X m)m) / =(360 400) / 80=-0.5Z2 =(X m)m) / =(510 400) / 80=1.375From the standardized normal distribution table,Area under the curve when Z1 = -0.5 = 0.3085Area under the curve when Z2 = 1.375 = 0.9154 Area between Z1 和和Z2 = 0.9154 0.3085 = 0.6069As such 60.69% chance that a random candid
276、ate will take between 360 hours 和和510 hours to complete the 培训培训.198西格玛绿带培训教材课件Normal distribution as an binomial distribution approximationNormal distribution can sometimes be used as an approximation for binomial distribution, giving the conditions:a) np 5b) nq 5ExampleWhat is the probability of g
277、etting 5,6,7, or 8 tails in 10 tosses of fair coin ?p = 0.5,q = 0.5,n = 10,r = 5, 6, 7, or 8From binomial probability distributions table,P(5,6,7,8) = P(5) + P(6) + P(7) + P(8) = 0.2461 + 0.2051 + 0.1172 + 0.0439 = 0.6123199西格玛绿带培训教材课件Example (cont)Substitute the mean 和标准偏差 of binomial distribution
278、into a normal distribution, we have:Mean, m m = np = 5标准偏差, = npq = 1.581Probability0123456789104.5 to 8.5To improve the accuracy of the approximation, subtract 和add a continuity correction factor of 0.5 to the range 5 to 8 respectively. The area under this range is what we are interested in.200西格玛绿
279、带培训教材课件Example (cont)At x = 4.5,z = (x - m m) / = (4.5 - 5) / 1.581 = -0.32At x = 8.5,z = (x - m m) / = (8.5 - 5) / 1.581 = 2.21From standardised normal distribution table,at z = -0.32P(4.5) = 1 - 0.6255 = 0.3745at z = 2.21P(8.5) = 0.9864Hence, P(4.5 x 8.5) = 0.9864 - 0.3745 = 0.6119 (approx equal t
280、o 0.6123 by binomial distribution)201西格玛绿带培训教材课件Introduction to Sampling取樣的介紹取樣的介紹What is population in statistic ?A population in statistic refers to all items that have been chosen for study.What is a sample in statistic ?A sample in statistic refers to a portion chosen from a population, by which
281、 the data obtain can be used to infer on the actual performance of the population PopulationSample 2Sample 6Sample 8Sample 1Sample 3Sample 7Sample 4Sample 5202西格玛绿带培训教材课件Symbols for population 和和sampleFor population,Population mean = mPopulation size= NPopulation 标准偏差= For sample,Sample mean = X Sam
282、ple size= nSample 标准偏差 = SSampling distribution - a distribution of sample meansIf you take 10 samples out of the same populations, you will most likely end up with 10 different sample means 和sample 标准偏差s. A Sampling distribution describes the probability of all possible means of the samples taken f
283、rom the same population.Population mean, m = 150Sampling distributionmean, XSample 2,mean X2Sample 1,mean X1Sample 3,mean X3Sample n,mean XnCollection of sample means203西格玛绿带培训教材课件m m = 150Population distribution with = 25When sample size increases, the standard error (or the std deviation of sampli
284、ng distribution) will get smaller.Sample distribution with std error x much less than 25 (when n = 30)Sample distribution with std error x 25 (when n = 5)Sampling distribution (cont)取樣分配取樣分配Like all normal curve, the sampling distribution can be described by its mean, x 和its 标准偏差 x (which is also kn
285、own as standard error of the mean). As such, the standard error measures the extent to which we expect the means from different samples to vary due to chance error in sampling 流程.Standard error = Population 标准偏差 / square root of sample size x = / n 204西格玛绿带培训教材课件Central Limit Theorem中心線規則中心線規則1.The
286、mean x of the sampling distribution will approximately equal to the population mean regardless of the sample size. The larger the sample size, the closer the sample mean is towards the population mean.2. The sampling distribution of the mean will approach normality regardless of the actual populatio
287、n distribution.3.It assures us that the sampling distribution of the mean approaches normal as the sample size increases.It allow us to use sample statistics to make inferences about population parameters without knowing anything about the actual population distribution, other than what we can obtai
288、n from the sample.m = 150Population distributionx = 150Sampling distribution(n = 5)x = 150Sampling distribution(n = 20)205西格玛绿带培训教材课件Example of sampling distributionThe population distribution of annual income of engineers is skewed negatively. This distribution has a mean of $48,000, 和a 标准偏差 of $50
289、00. If we draw a sample of 100 engineers, what is the probability that their average annual income is $48700 和more.m = $48KPopulation distribution = $5000x = $48KSampling distribution(n = 100)x = ?$48.7KSince population mean is equal to sampling distribution mean (i.e. central limit theorem), henceX
290、 = m m = 48000Sampling distribution mean = Standard error = x = / n = 5000 / 100 = 5000 / 10= 500206西格玛绿带培训教材课件Example of sampling distribution (cont)x = $48KSampling distribution(n = 100)x = 500$48.7KTherefore,mean = 48000sigma = 500X = 50000Z = (48700 - 48000) / 500 = 700 / 500 = 1.4From the stand
291、ardized normal distribution table, P(X $48700) = 0.9192Therefore;P(X $48700) = 1 - 0.9192 = 0.0808Thus, we have determined that it has only 8.08% chance for the average annual income of 100 engineers to be more than $48700.207西格玛绿带培训教材课件Central Limit Theorem ExerciseBreak into 4 groups as below:Grou
292、p 1: The population group.Group 2 to 4: The sample sub-groupThe population group will have 3 of their members throwing a single dices 60 times each. A total of 180 throws will be recorded 和this data will be the population data.Each sample sub-group will have 3 of their members throwing 5 dices at on
293、e time, 和collect the sum 和average value of the particular throw. Each member is to conduct 20 throws 和obtain the sample mean of each throw. At the end of the exercise, a total of 180 sample means will be collected from the 3 sub-groups.From the arrived data, plot the histogram 和comment on the distri
294、bution of both the population 和the samplings.208西格玛绿带培训教材课件The finite population multiplierPreviously we say that:This equation however applies only when the population is infinite or relatively large when compare to the sample size. In the case when the population is finite or relatively small when
295、 compare to the sample size, standard error is calculated as:Finite Population MultiplierFinite population multiplier with respect to population 和和sample sizeRule of thumbsThe finite population multiplier need only be included if population size to sample size ratio is less than 25.209西格玛绿带培训教材课件Con
296、fidence Interval“Point estimates”A point estimate is a single number that is used to estimate an unknown population parameter.“实例of point estimates”Sample mean, x as the estimator of the population mean, m m.Sample 标准偏差, S as the estimator of the population 标准偏差, .210西格玛绿带培训教材课件What does 95% confide
297、nce intervals means ?It defines 95% of the time, the average value of a random sampling will fall within a value range which is +/- 1.96 standard error from the sample mean.为什么1.96 standard error ?Referring to the standardized normal distribution table, when z = 1.96, the associated probability is 0
298、.975 (or 97.5%) as below:But this is a 2 tails interval (i.e. +/- 1.96), hence we need to minus off another 2.5% from the other end, giving a total coverage of 95%.2.5% uncovered at one end+1.962.5% uncovered at another end-1.96211西格玛绿带培训教材课件Equation for computing confidence intervalsFor continuous
299、data:集中的數據For discrete data:離散的數據212西格玛绿带培训教材课件Confidence interval for continuous data集中的信賴區間的數據A large disk drive manufacturerneeds an estimate of themean life it can expect from the magnetic media by reciprocally switching its digital state at 1MHz frequency. The development team has determined pr
300、eviously that the variance of the population life is 36 months, and had conducted a reliability testing for 100 samples, collecting data of its useful life as below:有一個大型的軟驅制造廠必須評估一個From the above data, what is the 95% confidence interval for the useful mean life of the magnetic media in a disk driv
301、e ? What does this mean ?213西格玛绿带培训教材课件Confidence interval for continuous data (cont)Applying the confidence intervals equation:Upper confidence limit = 27.75 + 1.96 (0.6) = 28.926Lower confidence limit = 27.75 1.96 (0.6) = 26.574From the sample data given,Sample size, n = 100Mean=(Sum of useful lif
302、e) / 100=2775 / 100=27.75 monthsVariance, 2=36 months标准偏差, =6 monthsStandard error, x= 6 / 10=0.6As such, there is 95% confidence level that the average useful life of the magnetic media to fall between 26.574 和28.926 months.214西格玛绿带培训教材课件Confidence interval for discrete dataBreak into 4 or 5 teams,
303、 combined all the M&Ms in one team 和calculate the confidence intervals for each color type, using below table.Combine all the data from the 4 teams, what are the changes to the confidence intervals ?215西格玛绿带培训教材课件Determine the sample size for confidence intervals (continuous data)GivenSample mean =
304、21S = 6 What is the sample size required, such that there is a 95% confidence level that the average value will fall within +/- 1.176 from its mean ?Ans:Applying the equation,n = (1.96 x 6) / 1.1762 = (10) 2 = 100The sample size must be 100.216西格玛绿带培训教材课件Determine the sample size for confidence inte
305、rvals (discrete data)Givenp = 0.4 (Proportion agree)q = 0.6 (Proportion not agree) What is the sample size required, such that there is a 99% confidence level that the proportion agree will fall within +/- 0.146 from p ?Ans:Applying the equation,n = (2.58) 2 (0.4)(0. 6) / (0.146)2 = 74.95 = 75The sample size must be 75.217西格玛绿带培训教材课件
