六西格玛绿带培训.ppt

1、Revision: 1.00 Date: June 2001,6西格玛绿带培训Materials TWO,第二天: Tests of Hypotheses Week 1 recap of Statistics Terminology Introduction to Student T distribution Example in using Student T distribution Summary of formula for Confidence Limits Introduction to Hypothesis Testing The elements of Hypothesis T

2、esting -Break- Large sample Test of Hypothesis about a population mean p-Values, the observed significance levels Small sample Test of Hypothesis about a population mean Measuring the power of hypothesis testing Calculating Type II Error probabilities Hypothesis Exercise I -Lunch- Hypothesis Exercis

3、e I Presentation Comparing 2 population Means: Independent Sampling Comparing 2 population Means: Paired Difference Experiments Comparing 2 population Proportions: F-Test -Break- Hypothesis Testing Exercise II (paper clip) Hypothesis Testing Presentation 第一天wrap up,第二天: Analysis of variance 和simple

4、linear regression Chi-square : A test of independence Chi-square : Inferences about a population variance Chi-square exercise ANOVA - Analysis of variance ANOVA Analysis of variance case study -Break- Testing the fittness of a probability distribution Chi-square: a goodness of fit test The Kolmogoro

5、v-Smirnov Test Goodness of fit exercise using dice Result 和discussion on exercise -Lunch- Probabilistic 关系hip of a regression model Fitting model with least square approach Assumptions 和variance estimator Making inference about the slope Coefficient of Correlation 和Determination Example of simple li

6、near regression Simple linear regression exercise (using statapult) -Break- Simple linear regression exercise (cont) Presentation of results 第二天wrap up,Day 3: Multiple regression 和model building Introduction to multiple regression model Building a model Fitting the model with least squares approach

7、Assumptions for model Usefulness of a model Analysis of variance Using the model for estimation 和prediction Pitfalls in prediction model -Break- Multiple regression exercise (statapult) Presentation for multiple regression exercise -Lunch- - Qualitative data 和dummy variables Models with 2 or more qu

8、antitative independent variables Testing the model Models with one qualitative independent variable Comparing slopes 和response curve -Break- Model building example Stepwise regression an approach to screen out factors Day 3 wrap up,Day 4: 设计of Experiment Overview of Experimental Design What is a des

9、igned experiment Objective of experimental 设计和its capability in identifying the effect of factors One factor at a time (OFAT) versus 设计of experiment (DOE) for modelling Orthogonality 和its importance to DOE H和calculation for building simple linear model Type 和uses of DOE, (i.e. linear screening, line

10、ar modelling, 和non-linear modelling) OFAT versus DOE 和its impact in a screening experiment Types of screening DOEs -Break- Points to note when conducting DOE Screening DOE exercise using statapult Interpretating the screening DOEs result -Lunch- Modelling DOE (Full factoria with interactions) Interp

11、reting interaction of factors Pareto of factors significance Graphical interpretation of DOE results 某些rules of thumb in DOE 实例of Modelling DOE 和its analysis -Break- Modelling DOE exercise with statapult Target practice 和confirmation run Day 4 wrap up,Day 5: Statistical 流程Control What is Statistical

12、 流程Control Control chart the voice of the 流程 流程control versus 流程capability Types of control chart available 和its application Observing trends for control chart Out of Control reaction Introduction to Xbar R Chart Xbar R Chart example Assignable 和Chance causes in SPC Rule of thumb for SPC run test -B

13、reak- Xbar R Chart exercise (using Dice) Introduction to Xbar S Chart Implementing Xbar S Chart 为什么Xbar S Chart ? Introduction to Individual Moving Range Chart Implementing Individual Moving Range Chart 为什么Xbar S Chart ? -Lunch- Choosing the sub-group Choosing the correct sample size Sampling freque

14、ncy Introduction to control charts for attribute data np Charts, p Charts, c Charts, u Charts -Break- Attribute control chart exercise (paper clip) Out of control not necessarily is bad Day 5 wrap up,Recap of Statistical Terminology,Area under a Normal Distribution,流程capability potential, Cp Based o

15、n the assumptions that :,流程is normal,It is a 2-sided specification,流程mean is centered to the device specification,Spread in specification,Natural tolerance,流程Capability Index, Cpk,Based on the assumption that the 流程is normal 和in control 2. An index that compare the 流程center with specification center

16、,Therefore when , Cpk Cp ; then 流程is not centered,Cpk = Cp ; then 流程is centered,The 流程of collecting, presenting 和describing sample data, using graphical 工具和numbers. Pareto Chart Population mean Histogram Population 标准偏差,Descriptive Statistics,Probability Theory,Probability is the chance for an event

17、 to occur. Statistical dependence / independence Posterior probability Relative frequency Make decision through probability distributions (i.e. Binomial, Poisson, Normal),Inferential Statistics,The 流程of interpreting the sample data to draw conclusions about the population from which the sample was t

18、aken. Confidence Interval (Determine confidence level for a sampling mean to fluctuate) T-Test 和F-Test (Determine if the underlying populations is significantly different in terms of the means 和variations) Chi-Square Test of Independence (Test if the sample proportions are significantly different) C

19、orrelation 和Regression (Determine if 关系hip between variables exists, 和generate model equation to predict the outcome of a single output variable),Central Limit Theorem,某些take aways for sample size 和sampling distribution,Percentiles of the t Distribution,Whereby, df = Degree of freedom = n (sample si

20、ze) 1 Shaded area = one-tailed probability of occurence a = 1 Shaded area Applicable when: Sample size 30 标准偏差 is unknown Population distribution is at least approximately normally distributed,Percentiles of the Normal Distribution / Z Distribution,Whereby, Shaded area = one-tailed probability of oc

21、curence a = 1 Shaded area,Student t Distrbution example,FDA requires pharmaceutical companies to perform extensive tests on all new drugs before they can be marketed to the public. The first phase of testing will be on animals, while the second phase will be on human on a limited basis. PWD is a pha

22、rmaceutical company currently in the second phase of testing on a new antibiotic project. The chemists are interested to know the effect of the new antibiotic on the human blood pressure, 和they are only allowed to test on 6 patients. The result of the increase in blood pressure of the 6 tested patie

23、nts are as below: ( 1.7 , 3.0 , 0.8 , 3.4 , 2.7 , 2.1 ) Construct a 95% confidence interval for the average increase in blood pressure for patients taking the new antibiotic, using both normal 和t distributions.,Student t Distrbution example (cont),Using normal or z distribution,Using student t distr

24、ibution,Although the confidence level is the same, using t distribution will result in a larger interval value, because: 标准偏差, S for small sample size is probably not accurate 标准偏差, S for small sample size is probably too optimistic Wider interval is therefore necessary to achieve the required confi

25、dence level,Summary of formula for confidence limit,6 Sigma 流程和1.5 Sigma Shift in Mean,Statistically, a 流程that is 6 Sigma with respect to its specifications is:,But Motorola defines 6 Sigma with a scenario of 1.5 Sigma shift in mean,DPM = 3.4 Cp = 2 Cpk = 1.5,某些Explanations on 1.5 Sigma Mean Shift,M

26、otorla has conducted a lot of experiments, 和found that in long term, the 流程mean will shift within 1.5 sigma if the 流程is under control. 1.5 sigma mean shift in a 3 Sigma 流程control plan will be translated to approximately 14% of the time a data point will be out of control, 和this is deem acceptable in

27、 statistical 流程control (SPC) practices.,Our Explanation,Most frequently used sample size for SPC in industry is 3 to 5 units per sampling. Take the middle value of 4 as an average sample size used in the sampling. Assuming the 流程is of 6 sigma capability, is in control, 和is normally distributed. Unde

28、r the confidence interval for sampling distribution, we expect the average value of the samples to fluctuate within 3 standard errors (i.e. natural tolerance), giving confidence interval of:,Introduction to Hypothesis Testing ?,What is hypothesis testing in statistic ? A hypothesis is “a tentative a

29、ssumption made in order to draw out or test its logical or empirical consequences.” A statistical hypothesis is a statement about the value of one of the characteristics for one or more populations. The purpose of the hypothesis is to establish a basis, so that one can gather evidence to either disp

30、rove the statement or accept it as true.,Example of statistical hypothesis The average commute time using Highway 92 is shorter than using France Avenue. This 流程change will not cause any effect on the downstream 流程es. The variation of Vendor Bs parts are 40% wider than those of Vendor A.,Elements of

31、 Hypothesis Testing,Possible outcomes for hypothesis testing on two tested populations:,为什么Hypothesis Testing ?,Many problems require a decision to accept or reject a statement about a parameter. That statement is a Hypothesis. It represents the translation of a practical question into a statistical

32、 question. Statistical testing 提供s an objective solution, with known risks, to questions which are traditionally answered subjectively. It is a stepping stone to 设计of Experiment, DOE.,Hypothesis Testing Descriptions,Hypothesis Testing answers the practical question: “Is there a real difference betwe

33、en A 和B ?” In hypothesis testing, relatively small samples are used to answer questions about population parameters. There is always a chance that a sample that is not representative of the population being selected 和results in drawing a wrong conclusion.,Elements of Hypothesis Testing (cont),The Nu

34、ll Hypothesis Statement generally assumed to be true unless sufficient evidence is found to be contrary Often assumed to be the status quo, or the preferred outcome. However, it sometimes represents a state you strongly want to disprove. Designated as H0 In hypothesis testing, we always bias toward

35、null hypothesis,The Alternative Hypothesis (or Research Hypothesis) Statement that will be accepted only if data 提供convincing evidence of its truth (i.e. by rejecting the null hypothesis). Instead of comparing two populations, it can also be based on a specific engineering difference in a characteri

36、stic value that one desires to detect (i.e. instead of asking is m1 = m2, we ask is m1 450). Designated as H1,Elements of Hypothesis Testing (cont),Example if we want to test whether a population mean is equal to 500, we would translate it to: Null Hypothesis, H0 : mp = 500 和consider alternate hypot

37、hesis as: Alternate Hypothesis, H1 : mp 500 ; (2 tails test),Remember confidence interval, at 95% confidence level states that: 95% of the time the mean value will fluctuate within the confidence interval (limit) 5% chance that the mean is natural fluctuation, but we think it is not alpha (a) probab

38、ility,Type II Error Accepting a null hypothesis (H0), when it is false. Probability of this error equals b,Type I Error Rejecting the null hypothesis (H0), when it is true. Probability of this error equals a,Use the std error observed from the sample to set confidence limit on 500 (mH0). The assumpt

39、ion is mH0 has the same variance as mp.,Elements of Hypothesis Testing (cont),Other possible alternate hypothesis are: Alternate Hypothesis, H1 : mp 500 ; (1 tail test) Alternate Hypothesis, H1 : mp 500 ; (1 tail test),Taking example for alternate hypothesis, H1 : mp 500 For 95% confidence level, a

40、= 0.05. Since H1 is one tail test, reject area does not need to be divided by 2.,某些hypothesis testings that are applicable to engineers: The impact on response measurement with new 和old 流程parameters. Comparison of a new vendors parts (which are slightly more expensive) to the present vendor, when va

41、riation is a major issue. Is the yield on Tester ECTZ21 the same as the yield on Tester ECTZ33 ?,流程Situations Comparison of one population from a single 流程to a desirable standard Comparison of two populations from two different 流程es or Single sided: comparison considers a difference only if it is gr

42、eater or only if it is less, but not both. Two sided: comparison considers any difference of ine质量 important,Inferences based on a single sample,“Large sample test of hypothesis about a population mean”,Example: An automotive manufacturer wants to evaluate if their new throttle 设计on all the latest c

43、ar model is able to give an adequate response time, resulting in an predictable pick-up of the vehicle speed when the fuel pedal is being depressed. Based on finite element modelling, the 设计team committed that the throttle response time is 1.2 msec, 和this is the recommended value that will give the

44、driver the best control over the vehicle acceleration. The test engineer of this project has tested on 100 vehicles with the new throttle 设计和obtain an average throttle response time of 1.05 msec with a 标准偏差 S of 0.5 msec. Based on 99% confidence level, can he concluded that the new throttle 设计will g

45、ive an average response time of 1.2 msec ?,“Large sample test of hypothesis about a population mean” (cont),From standard normal distribution table, The Z value corresponding to 0.005 tail area is 2.58.,a = 0.01 (2 tails), since 2 tails test, therefore tail area = a/2 = 0.005 ;,“Large sample test of

46、 hypothesis about a population mean” (cont),What does 99% confidence level means in the above example ? It defines the limits whereby 99% of the average sampling value should fall within, given the desirable (hypothesised) mean as mH0. Any value fall outside this confidence limit indicates the sampl

47、e mean is significantly different from mH0. In other words, we will only conclude the alternate hypothesis H1 (that the means are different) if we are more than 99% sure.,The Observed Significance level, p-value,p-value is the probability for concluding the null hypothesis H0 that both population me

48、ans are equal with the observed sample data. Hence 1 pvalue will be the confidence level we have on the alternate hypothesis.,Using the throttle question as an example: We know that the mean response time is 3 standard error away from 1.2msec (mH0), therefore Z = 3. Since this is a 2 tails test, p-v

49、alue = P(Z 3) = 2 P(Z 3) From standard normal distribution table, P(Z = 3) = 0.9987 P(Z 3) = 1 0.9987 = 0.0013 p-value = 2 P(Z 3) = 0.0026,In statistical term, it means there is only 0.0026 probability that the average throttle response time to be 1.02msec if the actual population mean is 1.2msec as

50、 suggested by finite element analysis.,“Small sample test of hypothesis about a population mean”,Example: Amy is the Personnel Officer of a multi-national company who is in charge of recruiting a large number of employees for an overseas assignment. As these overseas assignments are very crucial for

51、 the company success in meeting their business plan, an aptitude test was formulated to test the 质量 of all potential candidates head-hunted by the 招聘Agency. The management wants to know the effectiveness of the 招聘Agency, as it was believed that the average test score for all the identified candidate

52、s should be equal or more than 90 in order to reduce the risk of assigning the wrong candidates for the task. When Amy reviews the tests result of a particular batch of 20 candidates, she finds that the mean score is 84 和the 标准偏差 is 11. As this is a very critical 招聘project, Amy wants to be more rese

53、rve with her analysis, 和decided to be more bias towards proving that the population mean is lesser than 90. As a result, a confidence level of 90% will be used in her analysis.,“Small sample test of hypothesis about a population mean” (cont),From student t distribution table, The t value with 19 df

54、corresponding to 0.1 tail area is = 1.3277.,“Large sample test of hypothesis about a population proportion”,A method currently used by doctors to screen for possible stomach ulcer fails to detect the ulcer in 20% of the patients who actually have the disease. Suppose a new method has been developed that researchers hope will detect stomach ulcer more accurately. This new method was used to screen a random sample of 140 patients known to have stomach ulce