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    6西格玛绿带培训教材2

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    • 1. Revision : 1.00 Date : June 20016西格玛绿带培训Materials TWO-6-4-20246标准偏差
    • 2. 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 Testing -----------------------------------------------------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 Exercise 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
    • 3. 第二天: Analysis of variance 和simple 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 Kolmogorov-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 linear regression Simple linear regression exercise (using statapult) ------------------------------------------------------Break------------------------------------------------------------------- Simple linear regression exercise (con’t) Presentation of results 第二天wrap up
    • 4. Day 3: Multiple regression 和model building Introduction to multiple regression model Building a model Fitting the model with least squares approach 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 quantitative 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
    • 5. Day 4: 设计of Experiment Overview of Experimental Design What is a designed 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, linear 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 DOE’s result ---------------------------------------------------Lunch---------------------------------------------------------------------- Modelling DOE (Full factoria with interactions) Interpreting 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
    • 6. Day 5: Statistical 流程Control What is Statistical 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 -------------------------------------------------------------Break------------------------------------------------------------ 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 frequency 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
    • 7. Recap of Statistical TerminologyDistributions differs in locationDistributions differs in spreadDistributions differs in shapeNormal Distribution-6-5-4-3-2-10123456 ------------------------------ 99.9999998% ------------------------------------------- 99.73% -------------------- 95.45% ---------68.27%--± 3 variation is called natural tolerance Area under a Normal Distribution
    • 8. 流程capability potential, Cp Based on the assumptions that : 流程is normalNormal Distribution-6-5-4-3-2-10123456Lower Spec Limit LSLUpper Spec Limit USLSpecification CenterIt is a 2-sided specification流程mean is centered to the device specificationSpread in specificationNatural toleranceCP =USL - LSL686= 1.33
    • 9. 流程Capability Index, CpkBased on the assumption that the 流程is normal 和in control 2. An index that compare the 流程center with specification centerNormal Distribution-6-5-4-3-2-10123456Lower Spec Limit LSLUpper Spec Limit USLSpecification CenterTherefore when , Cpk < Cp ; then 流程is not centered Cpk = Cp ; then 流程is centeredUSL - Y3Y - LSL3Cpk = min ,
    • 10. 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 StatisticsEstimates for Descriptive StatisticsThe 流程of estimating the population parameters from sample(s) that was taken from the population. Sample mean, X  Population mean, m Sample 标准偏差, S  Population 标准偏差,  (when sample size, n > 20) Estimated 标准偏差, R/d2  Population 标准偏差,  (when sample size, n  20)
    • 11. Probability TheoryProbability is the chance for an event to occur. Statistical dependence / independence Posterior probability Relative frequency Make decision through probability distributions (i.e. Binomial, Poisson, Normal)Central Limit TheoremRegardless the actual distribution of the population, the distribution of the mean for sub-groups of sample from that distribution, will be normally distributed with sample mean approximately equal to the population mean. Set confidence interval for sample based on normal distribution. A basis to compare samples using normal distribution, hence making statistical comparison of the actual populations. It does not implies that the population is always normally distributed. (Cp, Cpk must always based on the assumption that 流程is normal)
    • 12. Inferential StatisticsThe 流程of interpreting the sample data to draw conclusions about the population from which the sample was taken. 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) Correlation 和Regression (Determine if 关系hip between variables exists, 和generate model equation to predict the outcome of a single output variable)
    • 13. Central Limit TheoremThe 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 population distribution. 3. It assures us that the sampling distribution of the mean approaches normal as the sample size increases.m = 150Population distributionx = 150Sampling distribution (n = 5)x = 150Sampling distribution (n = 20)x = 150Sampling distribution (n = 30)m = 150Population distributionx = 150Sampling distribution (n = 5)
    • 14. 某些take aways for sample size 和sampling distribution For large sample size (i.e. n  30), the sampling distribution of x will approach normality regardless the actual distribution of the sampled population. For small sample size (i.e. n < 30), the sampling distribution of x is exactly normal if the sampled population is normal, 和will be approximately normal if the sampled population is also approximately normally distributed. The point estimate of population 标准偏差 using S equation may 提供a poor estimation if the sample size is small.Introduction to Student t Distrbution Discovered in 1908 by W.S. Gosset from Guinness Brewery in Ireland. To compensate for 标准偏差 dependence on small sample size. Contain two random quantities (x 和S), whereas normal distribution contains only one random quantity (x only) As sample size increases, the t distribution will become closer to that of standard normal distribution (or z distribution).
    • 15. Percentiles of the t DistributionWhereby, df = Degree of freedom = n (sample size) – 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 distributedt ( a, u )aArea under the curve
    • 16. Percentiles of the Normal Distribution / Z DistributionZaArea under the curveWhereby, Shaded area = one-tailed probability of occurence a = 1 – Shaded area
    • 17. Student t Distrbution exampleFDA 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 pharmaceutical 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 patients 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.
    • 18. Student t Distrbution example (con’t)Using normal or z distributionUsing student t distributionAlthough 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 confidence level
    • 19. Summary of formula for confidence limit
    • 20. 6 Sigma 流程和1.5 Sigma Shift in MeanStatistically, a 流程that is 6 Sigma with respect to its specifications is:Normal Distribution-6-5-4-3-2-10123456 ------------------------------ 99.9999999998% ----------------------------LSLUSLDPM = 0.002 Cp = 2 Cpk = 2But Motorola defines 6 Sigma with a scenario of 1.5 Sigma shift in meanDPM = 3.4 Cp = 2 Cpk = 1.51.5
    • 21. 某些Explanations on 1.5 Sigma Mean Shift Motorla 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 statistical 流程control (SPC) practices.Normal Distribution-3-2-10123------------------ 99.74% -----------------LCLUCLDistribution with 1.5 Sigma Shift-3-2-10123----------------- 86.64% ----------------LCLUCLOut of control data points
    • 22. 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. Under 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:
    • 23. Introduction to Hypothesis Testing ?What is hypothesis testing in statistic ? A hypothesis is “a tentative assumption 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 disprove 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 B’s parts are 40% wider than those of Vendor A.
    • 24. Elements of Hypothesis TestingPossible outcomes for hypothesis testing on two tested populations:No Significant DifferenceSignificant Difference in VariationSignificant Difference in MeanSignificant Difference in both Mean 和Variationm1 <> m2 1 = 2m1 <> m2 1 <> 2m1 = m2 1 <> 2m1 = m2 1 = 2
    • 25. 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 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 between 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.
    • 26. Elements of Hypothesis Testing (con’t)The Null 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 null hypothesisThe 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 characteristic value that one desires to detect (i.e. instead of asking is m1 = m2, we ask is m1 > 450). Designated as H1
    • 27. Elements of Hypothesis Testing (con’t)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 hypothesis 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) probability --- Confidence limit ---mH0 = 5000.025 of area0.025 of area(a/2) reject area(a/2) reject area1.96 std error1.96 std errorType II Error Accepting a null hypothesis (H0), when it is false. Probability of this error equals bType I Error Rejecting the null hypothesis (H0), when it is true. Probability of this error equals aIf mp is within confidence limit, accept the null hypothesis H0.If mp is in reject area, reject the null hypothesis H0.Use the std error observed from the sample to set confidence limit on 500 (mH0). The assumption is mH0 has the same variance as mp.
    • 28. Elements of Hypothesis Testing (con’t)Other possible alternate hypothesis are: Alternate Hypothesis, H1 : mp > 500 ; (1 tail test) Alternate Hypothesis, H1 : mp < 500 ; (1 tail test) 1.645 std errorAcceptance area mH0 = 5000.05 of area(a) reject areaTaking example for alternate hypothesis, H1 : mp > 500 For 95% confidence level, a = 0.05. Since H1 is one tail test, reject area does not need to be divided by 2.From standard normal distribution table:Z-value of 1.645 will give 0.95 area, leaving a to be 0.05.Therefore if mp is more than 500 by 1.645 std error, it will be in the reject area, 和we will reject the null hypothesis H0, concluding on alternate hypothesis H1 that mp is > 500.
    • 29. 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 variation 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 greater or only if it is less, but not both. Two sided: comparison considers any difference of ine质量 important
    • 30. 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 car 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 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 give an average response time of 1.2 msec ?
    • 31. “Large sample test of hypothesis about a population mean” (con’t)Solution: Since the sample size is relatively large (i.e. >30), we should use z statistic. m  X = 1.05 msec ; s  S = 0.5 msec ; n = 100 ; Null hypothesis H0 : m = mH0 (1.2 msec) Alternate hypothesis H1 : m <> mH0 (1.2 msec) ----- Acceptance Area -----mH0 = 1.20.005 of area0.005 of area(a/2) Reject Area(a/2) Reject Area2.58 std error2.58 std errorFrom 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 ;How many std error is X away from 1.2 msec ?X = 1.02Therefore X is –3 std errors away from 1.2 msec.
    • 32. ----- Acceptance Area -----mH0 = 1.20.005 of area0.005 of area(a/2) Reject Area(a/2) Reject Area2.58 std error2.58 std errorX = 1.02“Large sample test of hypothesis about a population mean” (con’t)Baed on 99% confidence level, since X is at the negative reject area, we will reject the null hypothesis 和conclude on the alternate hypothesis that the average response time is significantly different than 1.2 msec The average response time appears to be lower than 1.2 msec.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 sample 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.
    • 33. The Observed Significance level, p-value p-value is the probability for concluding the null hypothesis H0 that both population means are equal with the observed sample data. Hence 1 – pvalue will be the confidence level we have on the alternate hypothesis.-- Acceptance Area --mH0 = 1.22.58 std error2.58 std errorX = 1.023 std errorP/2Acceptance Area mH0 = 1.22.58 std errorX = 1.023 std errorPFor 2 tails testFor 1 tail testP/2-Z+Z3 std error-ZUsing 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-value = P(Z < -3, or 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 suggested by finite element analysis.
    • 34. “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 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 candidates 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 reserve 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.
    • 35. “Small sample test of hypothesis about a population mean” (con’t)Solution: Since the sample size is relatively small (i.e. < 30), we should use t statistic. m  X = 84 ; s  S = 11 ; n = 20 ; a = 0.1 (1 tail) Degree of freedom, df = 19 Null hypothesis H0 : m = mH0 (90) Alternate hypothesis H1 : m < mH0 (90)mH0 = 900.1 of area(a) Reject Area1.3277 std errorX = 84From student ‘t’ distribution table, The t value with 19 df corresponding to 0.1 tail area is = 1.3277.How many std error is X lesser than 90 ?Therefore X is –2.439 std errors away from 90 和in the reject area. Amy will reject the null hypothesis 和conclude that the average score is less than 90 marks.2.439 std errorAcceptance Area
    • 36. “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 ulcer. Of these, the new method failed to detect ulcer in 12 of the patients. Using 95% confidence level, does this sample 提供evidence that the failure rate of the new method differs from the one currently in use ?Solution: Let the probability of success in missed detection as p therefore; H0 : p = 0.2 (i.e. pH0) H1 : p <> 0.2 Sample size, n = 140 (i.e. use standard normal z as the test statistic) Compute the standard error for null hypothesis (i.e. when p = 0.2)
    • 37. Test if pH0  3 std error will give reasonable value (i.e. between 0 to 1) pH0  3 std error  0.2  3(0.034) = (0.166, 0.234)“Large sample test of hypothesis about a population proportion” (con’t)Calculate the number of standard errors between the sampled 和hypothesised value:-- Acceptance Area --pH0 = 0.20.025 of area0.025 of area(a/2) Reject Area(a/2) Reject Area1.96 std error1.96 std errorp = 0.0863.36 std errorConclusions: Since p = 0.086 is in reject area, we reject null hypothesis 和conclude that the new screen method is significantly different than the old screen method with 95% confidence level. With 3.36 std error from pH0 (0.2), the p-value is calculated to be 0.00078, hence there is a 99.922% confidence in the alternate hypothesis. It appears that the new screen method will give lesser miss detection for stomach ulcer.
    • 38. Power of a Hypothesis Testing95% Confidence Interval (Acceptance Area)mH0 (a/2) Reject Area(a/2) Reject AreaType I Error Rejecting the null hypothesis (H0), when it is true. Probability of this error equals a. Type II Error Accepting a null hypothesis (H0), when it is false. Probability of this error equals b. In hypothesis testing, we are always bias towards H0. Therefore a 95% confidence limit will only tell us if we are more than 95% sure that the two population means are different. However the true states of the population means can be different even if we are less than 95% sure. In other words, if there is no significant difference between the 2 means, it does not indicate that they are equal, it could be that they are not far enough apart. Illustrate in the above distribution, assuming m1 has the same variance as mH0 和they are different, the area under m1 curve that is fall within 95% confidence limit of mH0 will be b (probability for type II error).m1 Control by aControl by b
    • 39. Power of a Hypothesis Testing (con’t)As such there is a trade off between a 和b. As a decreases, b increases 和vice versa.mH0 (a/2) Reject Area(a/2) Reject Aream1 75% (a = 0.25)95% (a = 0.05)Confidence Interval b for m1 when a = 0.05b for m1 when a = 0.25
    • 40. Power of a Hypothesis Testing (con’t)A hospital uses large quantities of packaged doses of a particular drug. The individual dose of this drug is 100 cc. The action of the drug is such that the body will harmlessly pass off excessive doses. On the other hand, insufficient doses (i.e. 99.6 cc 和below) do not produce the desired medical effect, 和they interfere with patient treatment. The hospital has purchased its requirements of this drug from the same manufacturer for a number of years 和knows that the population 标准偏差 is 2 cc. The hospital inspects 50 doses of this drug at random from a very large shippment 和finds the mean of these doses to be 99.75 cc. With 90% confidence level, how can the hospital conclude wether the dosages in this shipment are too small ?mH0 = 100 cc (hypothesised value of population mean)  = 2 (known population 标准偏差) X = 99.75 (sample mean) n = 50 (sample size) H0 : m = 100 (null hypothesis that mean dosage from shippment is 100 cc) H1 : m < 100 (alternate hypothesis that mean dosage from shippment is < 100 cc) a = 0.1 (probability of type I error)
    • 41. Power of a Hypothesis Testing (con’t)90% Confidence Interval 100 (a) Reject Area1.28 std error99.64 (i.e. 100 – 1.28 std error)From standard normal distribution table, the z-value that associate with 0.9 probability is 1.28.Since X = 99.75 falls within the acceptable area, there is not enough evidence to reject null hypothesis 和conclude that the sample mean is not significantly lower than 100 cc.99.75The hypothesis test indicates that sample mean of 99.75 is not significantly different than 100, 和therefore there is not enough evidence to say that the underlying population mean is not equal to 100. However it does indicate that they are equal. It signifies that there is a 10% chance to reflect the population mean to be not equal to 100 if the population mean is actually 100. If the actual population mean is 99.75 as the sample mean, what is the probability that the above hypothesis test (of 90% confidence) to mistakenly reflect that the population mean is equal to 100 (i.e. b error) ?
    • 42. Power of a Hypothesis Testing (con’t)90% Confidence Interval for mH0 = 100100 99.6499.75Area inside the accept area will be probability to mistakenly accept 99.75 as equal to 100 (i.e. b error)Area in the reject area will be probability of rejecting 99.75 as equal to 100 (i.e. 1-b)Setting 90% confidence level (i.e. a = 0.1) for the hypothesis testing, if the actual population mean is 99.75: 34.83% chance to reject H0 (correct conclusion), 65.17% chance to accept H0 (incorrect conclusion)In hypothesis testing we want to achieve small a 和b (or big 1-a 和1-b).
    • 43. Power of a Hypothesis Testing (con’t)The Power Curve – A graphical presentation of the possible population means against the probabilities of rejecting H0 when H1 is true (i.e. 1-b), after fixing a at a certain value.90% Confidence Interval for mH0 = 10099.6499.61(1-b) = 0.543890% Confidence Interval for mH0 = 10099.6499.42(1-b) = 0.782390% Confidence Interval for mH0 = 10099.6499.75(1-b) = 0.3483The power curve shows that: The probability of rejecting H0 when H1 is true increases as the actual population mean deviates from the hypothesised mean of 100. If the actual population mean is 99.28, there will be 90% confidence in both 1-a 和1-b. (i.e. a 和b error will be 10%)
    • 44. Power of a Hypothesis Testing (con’t)Central Limit Theorem – The sample standard error will 降低with a factor that is square root of the sample size. When sample size increases, the standard error decreases.90% Confidence Interval for mH0 = 10099.7399.61(1-b) = 0.664390% Confidence Interval for mH0 = 10099.7399.42(1-b) = 0.863590% Confidence Interval for mH0 = 10099.7399.75(1-b) = 0.4718The power curve shows that: As sample size increase, 1-b reaches 90% with a closer value to 100 (i.e. 99.46 instead of 99.28).
    • 45. 某些take aways in Hypothesis Testing:In Hypothesis Testing : a is fixed 和b is controlled by sample size. When sample size increases, b error decreases. a  b ; when the actual population mean equal Hypothesised mean – 2 x Z(a) x Std Error (i.e. for n = 50, when population mean = 100 – [2 x 1.28 x 0.2829] = 99.28) You might want to verify b error if the hypothesis test fail to reject null hypothesis, H0. In the example of drug dosage, even though the hypothesis testing shows that the average should be equal to 100cc, we should reject the batch because the b risk is too high.
    • 46. Comparing 2 Population Means with Large SamplesA dietitian has developed a diet that is low in fats, carbohydrates, 和cholesterol. Although the diet was initially intended to be used by people with heart disease, the dietitian wishes to examine the effect this diet has on the weights of obese people. 2 random samples of 100 obese people each are selected, 和one group of 100 is placed on the low-fat diet. The other 100 are placed on a diet that contains approximately the same quantity of food but is not as low in fats, carbohydrates, 和cholesterol. For each person, the amount of weight loss or gained in a 3 weeks period is recorded. The data is presented below, how can we interpret the result with a 95% confidence.
    • 47. Let m1 = mean weight losses for people taking low-fat diet Let m2 = mean weight losses for people taking other diet Form 95% confidence on the difference in mean for both diet types using X1 和X2 as estimators: SolutionAssuming the 2 samples are independent, the 标准偏差 of the difference between the sample means (or the standard error of the statistics) is:Assumptions The 2 samples are randomly selected in an independent manner from the 2 populations. The sample size n1 和n2 are large enough so that both sample means have near normal sampling distributions. S1 和S2 提供good estimate for s1 和s2, hence the 2 population variances does not need to be equal. (i.e. n  30)
    • 48. H0 : (m1 – m2) = 0 (i.e. m1 = m2) H1 : (m1 – m2) <> 0 (i.e. 2 tails test, m1 <> m2) For 2 tails test, Z value for 95% confidence level is Z(a/2) = 1.96 std error Test Statistics: Solution (con’t)Since Z = 3.05 is in the rejection area, we will reject null hypothesis H0, 和conclude on alternate hypothesis H1 that both population means are significantly different. From the data, the low fat diet is more effective to reduce weight than the other diet. 0 +1.96 std err-1.96 std err Z = 3.05
    • 49. Comparing 2 Population Proportions with Large SamplesJoan is a HR officer who wants to compare the preference of employee towards the new implemented employee stock purchase plan in Singapore to those in Canada. Such a comparison would help her to determine if there is a need to either divide or concentrate the campaign efforts in these two sites. Joan randomly chose 1000 employees from each site, 和interview each to learn about the employees’ likelihood in signing up for the stock purchase plan. Her objective is to use this sample information to make inference about the difference (p1 – p2) between the proportion p1 of all employee in Singapore who are likely to sign up 和p2 of all employee in Canada who are likely to sign up. The two samples represent independent binomial experiments, 和the number of employee in each site who indicate that they will enrol in the stock purchase plan is as below: Singapore : X1 = 546 ; Sample size, n = 1000 Canada : X2 = 475 ; Sample size, n = 1000 Based on the above data, should Joan assume that employee in Canada is less willing to enrol in the plan 和concentrate her promotional campaign in that site ?
    • 50. SolutionLet p1 = 546/1000 = 0.546 Let p2 = 475/1000 = 0.475Properties of Sampling Distribution of (p1 – p2) (P1 – P2) is an unbiased estimator of the population proportions. The best estimate of the overall proportion of success if the population are hypothesised to be equal: The 标准偏差 of the sampling distribution (or standard error of the distribution) is: If the sample sizes n1 和n2 are large, the sampling distribution is approximately normal. The two samples selected are independent random samples(When both proportions are equal)(When both proportions are not equal)
    • 51. 95% Confidence Interval Since test statistic of 3.176 is in the rejection area, Joan shall reject H0 和conclude on H1 that proportions from both sites are significantly different. Based on the 95% confidence interval, we infer that there are between 2.7% 和11.5% more employee likely to enrol to the plan in Singapore site than in Canada site. She should therefore concentrate her campaign on Canada to balance up the 2 sites. 0 +1.96 std err-1.96 std err Z = 3.176H0 : (p1 – p2) = 0 (i.e. p1 = p2) H1 : (p1 – p2) <> 0 (i.e. 2 tails test, p1 <> p2) For 2 tails test, Z value for 95% confidence level is Z(a/2) = 1.96 std error Test Statistics: Solution (con’t)
    • 52. Cup PositionPin PositionPull Back AngleHook PositionStop PositionHypothesis ExerciseSeparate into 4 teams Ensure the Statapult is preset to: Pull back Angle - 177° Hook Position – 4 Cup Position - 1 Stop Position – 3 Pin Position - 2 Each team will evaluate on the differences in the mean distances with respect to a) different operators b) different ball types c) different statpults d) different pull back angle by 3 degrees A total of 100 shots for each set-up are required using the SOP outlined in week 1. Compute statistically using hypothesis testing to comment on the mean distance of the 2 statpults. What is the a error in your analysis. If null hypothesis is accepted, what is the b risk that you have to bear ? Present your result in the class
    • 53. Comparing 2 Population Means with Small Samples (T-Test)Suppose you wish to compare a new method of teaching reading to “slow learners” to the current standard method. You decide to base this comparison on the results of a reading test given at the end of a learning period of 6 months. Of a random sample of 20 slow learners, 8 are taught by qualified instructors under similar conditions for a 6 months period. The results of the reading test at the end of this period are as below: With 95% confidence level, is there any difference between the 2 methods ? What are 某些of the assumptions in order for the estimate to be valid ? New Method, X1Standard Method, X280 80 79 81 76 66 79 76 79 62 70 68 73 76 86 73 72 68 75 66
    • 54. Since the sample is small (i.e. <30), we cannot use the previous method as the point estimate for 标准偏差 will not be accurate. When sample size is less than 30, use t-test with below assumptions: Both sampled populations have relative frequency distributions that are approximately normal. (i.e. to ensure effectiveness of central limit theorem) The population variance are equal The samples are randomly 和independently selected from the populations. SolutionSince we assume that both population variances are equal, we can use the data from both samples to construct a pooled sample estimator of variance S2p for use in confidence interval.Small sample confidence interval for (m1 – m2) is:
    • 55. Solution (con’t)H0 : (m1 – m2) = 0 (i.e. m1 = m2) H1 : (m1 – m2) <> 0 (i.e. 2 tails) For 2 tails test of 18 degree of freedom, t value for 95% confidence level is t(a/2) = 2.101 std error Since t = 1.807 is in the acceptance area, we will not reject null hypothesis H0, 和conclude that there is not enough evidence that both methods will result in different means. Need to further verify if population variances are equal, 和b error is assume to be not a concern. 0 +2.101 std err-2.101 std err t = 1.807
    • 56. Comparing 2 Population Means (Paired Difference Experiment)The previous example is based on independent sampling. Suppose it is possible to measure the slow learners’ “reading IQ” before they are subjected to a teaching method, then eight pairs of slow learners with similar reading IQs are found, 和one member of each pair is randomly assigned to the standard teaching method while the other is assigned to the new method. The resulting data are given in below table. From the data, is it true that the new method will give better test result as compared to the standard method ?Pair1 2 3 4 5 6 7 8Standard Method72 68 76 68 84 68 61 76New Method77 74 82 73 87 69 66 80
    • 57. Paired Difference Experiment SolutionH0: m1 – m2 = 0 where m1 is mean test result for new method H1: m1 – m2 > 0 m2 is mean test result for standard method If we employ the t statistic for independent sample, we have: 0 +1.76 std err t = 1.26From the t-distribution table, the t value associate with 14 degree of freedom 和0.95 is 1.76 (i.e. one tail test). Since 1.26 is lesser than 1.76 和in the acceptance area, there is therefore not enough evidence to suggest that the new teaching method can give better test result than the standard method. If we examine the data carefully, we will find that this result is difficult to accept, as the test score of the new method is larger than the corresponding test score of the standard teaching method for every single pair of slow learner ! Showing that m1 should infact greater than m2 from the look of it. Infact the b risk is around 69% if the true difference in mean is 4.375 (i.e. X1 – X2).
    • 58. Paired Difference Experiment Solution (con’t)The t-test is inappropriate for this example because the assumption of independent sample is invalid. The pairs of test score have been randomly selected, 和once we have chosen the sample for the new method, we have not independently chosen the sample for the standard method (i.e. The test score fluctuate dependently as shown in the graph). We should use paired difference experiment for dependent sampling like this: Pair1 2 3 4 5 6 7 8Standard Method72 68 76 68 84 68 61 76New Method77 74 82 73 87 69 66 80Difference5 6 6 5 3 1 5 4
    • 59. Paired Difference Experiment Solution (con’t)H0: mD = 0 (i.e. m1 = m2) H1: mD > 0 (i.e. m1 = m2) XD = Sample mean difference = 4.375 SD = Sample 标准偏差 of differences = 1.685 ND = Number of pair (differences) = 8 0 +1.76 std err t = 7.34Since Test statistic, t = 7.34 is in the reject area, there is therefore enough evidence to reject H0 和conclude on H1that there the new teaching method will give better mean score as compare to the standard method.
    • 60. 某些possible paired difference experiment Suppose the government wants to estimate the difference in mean starting salaries for men 和women graduates in the country. If the samples selected are independent, the starting salaries may vary because of their different majors, different academic achievement, etc. To eliminate this source of variability, the government could match male 和female job seeker according to their majors 和academic achievement, 和use paired difference experiment to analyse. If you wish to estimate the difference in mean absorption rate into bloodstream for two pain relief drug. When select the sample independently, the absorption rate may vary due to age, weight, sex, blood pressure etc. In this case, since there is so many possible variation, it might be difficult to pair the experiment using 2 persons as a single pair. However, we can evaluate the time for the drug to take effect by experimenting on the same person to minimise other variations, 和used a single person as a pair experiment for 2 different drugs.
    • 61. Paired Difference Experiment SummaryFor large sample (i.e. n  30)For small sample (i.e. n < 30)Pair difference experiment is useful when you can assure input dependency, 和neligible input variation apart from the experiment. What is the hypothesis result for the earlier statapult exercise if you analyse the data using pair difference method ? Is there any difference in the result interpretation ?
    • 62. Comparing Two Population Variances (F-Test)Frequently we need to compare the varainces of 2 populations. Example in t-test we assume that the population variance is equal as there is not enough data for estimation. How can we be sure then that the variances are equal ? Therefore there is a need to develop such technique to analyse variance.The associated equation in excel worksheet to obtain the F value is: “ = FINV(a, n1, n2) “ Example the F value for n1 = 2, n2 = 5, 和a area of 0.99 is: “ = FINV(0.01, 2, 5) “ n = Degree of freedomF(a ; v1, v2)a
    • 63. F-Test ExampleAn experimenter wants to compare the metabolic rates of white mice subjected to different drugs. The weights of the mice may affect their methabolic rates, 和thus it is hopefully that the obtained mice are homogeneous with respect to weight. Five hundred mice will be needed to complete the study. Currently 18 mice from 供应商A 和13 mice from 供应商B are available for comparison, 和the result is as below:供应商A 供应商B nA = 18 nB = 13 XbarA = 4.21 ounces X barB = 4.18 ounces S2A = 0.019 S2B = 0.049 Df, nA = 17 Df, nB = 12Do these data 提供enough evidence to indicate a difference in the variability of weights obtained from the two 供应商s ? Use a = 0.1 for the analysis.
    • 64. F-Test Example (con’t)Let S2A = Population variance of weights of white mice from 供应商A Let S2B = Population variance of weights of white mice from 供应商B H0: S2A/S2B = 1 H1: S2A/S2B <> 1 F Statistic = Larger sample variance / Smaller sample variance = S2B/S2A (Hence 供应商B will be sample 1 和供应商A will be sample 2) = 0.049 / 0.019 = 2.58 Using the excel formula, input “= FINV(0.05, 12, 17)” into excel spreadsheet, we have: F value = 2.38 (i.e. 2 tails test, hence divide a by 2)
    • 65. F-Test Example (con’t)Since F statistic of 2.58 is greater than F value of 2.38 和in the rejection area, the data had therefore 提供sufficient evidence to indicate that the population variances differ.From the sample variance data, it appears that 供应商A tend to be more homogeneous than 供应商B.Confidence Interval for s21 / s22Where FL,a/2 is based on n1 = (n1 –1) 和n2 = (n2 –1). Where FU,a/2 is based on n1 = (n2 –1) 和n2 = (n1 –1). It was estimated that the variance in the weights of mice from 供应商A could be as small as 0.15 or as large as 0.923 times the weight variance of mice from 供应商B.F(0.05 ; 12, 17)a/2F = 2.38Fstatistic = 2.58
    • 66. F-Test SummaryF statistic = Larger sample variance / Smaller sample variance Rejection region: Fstatistic > Fa (one tail test) or Fstatistic > Fa/2 (two tails test) Assumptions: Both sampled populations are normally distributed The samples are random 和independent When Fstatistic < F we accept the null hypothesis that both variances are equal (or not significantly different), based on the assumption that b risk is not a concern.
    • 67. Hypothesis Exercise IIBreak into groups. Take 25 paper clips from each 供应商(i.e. A 和B). Bent the paper clip until they break, associate the total numebr of bents with the ductility of the paper clip (i.e. Use only one operator in each team). Discuss in your team how should you approach the problem with both descriptive 和inferential statistic (i.e. remember the assumptions in hypothesis testing, how can we give assurance). Perform appropriate statistical tests to 支持your analysis 和give explanation. Compare the performance between 供应商和between batches (i.e. obtain the data from another group). Perform all calculation manually, 和present your result with calculation 和numerous graphs. (i.e. Use a = 0.05)
    • 68. Chi-Square - Testing For IndependenceStep 1 The individual data point is term as the Observed Frequency, fo. H0 = Cause of rejection 和work week are independent H1 = Cause of rejection are depending on work week  = 0.05Company : Natrella (1663) Rejects of metal castings were classified by cause of rejection for three different weeks. Does the distribution of rejects differ from week to week ?
    • 69. Chi-Square - Testing For Independence (con’t)Step 2 Computed the Expected Frequency, fe in each cell . fo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fe
    • 70. Chi-Square - Testing For Independence (con’t)Step 3 Computed the Chi-square statistic, c2. fo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo fefo feStep 4 Computed the degrees of freedom. Degree of freedom = (Number of row –1) (Number of column –1) = (2)(6) = 12
    • 71. Chi-Square - Testing For Independence (con’t)Step 5 Obtain the Chi-square,c2 value for the a level of significanceFor the casting example, The Chi-square value with a = 0.05 和12 degrees of freedom is 21.026 Alternatively you can compute using Excel spredsheet with the equation: “=CHIINV(a , Degrees of freedom)” “=CHIINV(0.05,12)” c2(a ; Df)a area
    • 72. Chi-Square - Testing For Independence (con’t)Step 6 Interpret the hypothesis result 和obtain the p value21c2 valuep valuec2 statistic45.6Since the c2 statistic is in the rejection area, we will reject the null hypothesis 和conclude on the alternate hypothesis that the cause of reject (i.e. column) depends on the work week (i.e. row). The associated P value can be obtained using Excel function: “= CHIDIST(c2 statistic, Degrees of freedom)” “= CHIDIST(45.6, 12)” = 0.00000813 (i.e. only 0.000813% that the 2 tested factors are independent)
    • 73. Chi-Square - Testing For Independence (con’t)Summary H0: The two classifications are independent H1: The two classifications are dependent Test Statistic:Rejection criteria: c2 statistic > c2a, where c2 statistic has (r-1)(c-1) degrees of freedomAssumptions: The observed counts are a random sample from the population of interest. The sample size, n will be large enough so that for every cell, the expected frequency will be equal to 5 or more.
    • 74. Independence Test MatrixGoodBrown stainMarginalcotton glove376011154875Gloves type4180.609756694.3902439palm fit glove415119943503730.390244619.6097561Marginals791113149225Expected Values in Redp Value4.9025E-139Chi Statistic630.0386347df1Chi-Square - Testing For Independence case studyProject objective is to prevent external stain defect on photo flash tube.Using the Chi-square test for independence, the 流程engineer is able to quantity if the corrective action is effective or not. By determining the independency between the yield 和actions classifications, an indepent scenario would mean the corrective action is not effective 和vice versa.
    • 75. Chi-Square - Inferences About a Population Variance F-Test make inferences about two population variances (assuming normality) In F-Test, confidence interval is an ration index between s2 A 和s2 B. Chi-square test make inference about a single population variance (assuming normality) In Chi-square test, confidence interval for the variance of the activity of concerned is expressed as value.A 流程engineer who is in-charge of back-end testing has conducted a new evaluation, 和found that he is able to reduce the variation in terms of electrical response by 50% (i.e. 标准偏差 from 0.232 to 0.116). Before he report this result to his boss he wants to make sure that he is 95% confidence in his analysis, how should he approach this problem ?Step 1 - From his experiment gather the descriptive statistical data
    • 76. Step 2 - Plot histogram to examine distribution normalityChi-Square - Inferences About a Population VarianceStep 3 - Identify outliers which is likely due to special causes (i.e. outside natural tolerance) Sample mean = 3.84 Sample Std Deviation = 0.153 Natural Tolerance = Mean +/- 3 Std Deviation = 3.38 to 4.30 Therefore remove data that are <3.38 和>4.3 is likely due to special causes.At this moment we accept that what looks like a normal distribution must be at least approximately normal. A more detail method to test for normality will be discussed in later section.-6-5-4-3-2-10123456<3.38>4.30OutliersOutliersNatural Tolerance
    • 77. Chi-Square - Inferences About a Population VarianceStep 4 - Remove outlier from the data, 和re-compute descriptive statisticStep 5 - Compute confidence intervals for the population variance based on assumption that the samples are taken from approximately normal population=“CHIINV(0.025, 519)” = 584.02c2 valuec2 Uc2 L=“CHIINV(0.975, 519)” = 457.770.025 of area0.025 of areaTherefore the 流程engineer can report to his boss that he is 95% confidence that the variance of the new 流程is between 0.012 to 0.0153 (or Std deviation between 0.110 to 0.124).
    • 78. Chi-square ExerciseBreak into 4 Groups Each group will be given a packet of M&M “melt in your h和not in your mouth” chocolate candy. In your groups count the number of M&Ms in their Color. Combine the Data 和conduct a Chi-Square test to see if the distribution of color are independent with respect to different M&M packages.M&M ColourTeam 4Team 3Team 2Team 1BrownRedBlueYellowGreenOrangeRemember the paper clip ductility test previously. Each team is to compute the confidence interval for the 标准偏差 of the paper clip ductility of the 2 供应商s. Also indicate if the ductility is depending on different team (i.e. operators). Present your result on the 2 case studies to the class.
    • 79. ANOVA – Analysis of VarianceWhat is ANOVA ? A statistical approach using F distribution to test for the significance of the differences among more than 2 sample means. Completely randomised design A 设计for which independent random samples of experimental units are selected for each treatment. Objective of completely randomised design The objective is usually to compare the treatment means. Example for population means of p treatment, the null hypothesis, H0 states that all treatment means are equal, 和alternate hypothesis, H1 states that at least two of the treatment means differ. H0: m1 = m2 = .… = mp H1: At least two of the p treatment means differ
    • 80. ANOVA – Analysis of Variance ExampleThe International football assocaiation wants to compare the mean distances associated with four different brands of foot balls when kick by a striker. A completely randomised 设计is employed, with the kicking simulated by a pneumatic activated swing that will strike a consistent 100N force at a fixed contact angle to the balls. The experiment was conducted with a random sample of 10 balls of each br和hitted at a random sequence. The distance of each trial was recorded 和the data was organised as below:Br和A251.2 245.1 248.0 251.1 260.5 250.0 253.9 244.6 254.6 248.8Br和B263.2 262.9 265.0 254.5 264.3 257.0 262.8 264.4 260.6 255.9Br和C269.7 263.2 277.5 267.4 270.5 265.5 270.7 272.9 275.6 266.5Br和D251.6 248.6 249.4 242.0 246.5 251.3 261.8 249.0 247.1 245.9Mean250.78261.06269.95249.32Variance22.421814.947120.258327.0729
    • 81. ANOVA – Analysis of Variance ExampleStep 2: Compute the Sum of Squares for Treatments (SST), which is the measurement of variation between treatment means.Step 1: Compute the gr和average for all the samples taken. Gr和average X = (250.8 + 261.1 + 270 + 249.3) / 4 = 257.8Step 3: Compute the Sum of Squares for Errors (SSE), which is the measurement of variation within treatment. H0: m1 = m2 = .… = mp H1: At least two of the p treatment means differ
    • 82. Step 4: Compute the Mean Square for Treatment (MST), 和Mean Square for Error (MSE).ANOVA – Analysis of Variance ExampleStep 5: Compute the test statistic F, reject null hypothesis, H0 when Fstatistic > Fa. Degree of freedom, n1 = 3 (i.e. p – 1) ; 和n2 = 36 (i.e. n – p).The associated p-value can be computed by “=FDIST(Fstatistic, n1, n2)”. Therefore p-value = FDIST(43.99, 3, 36) = 3.97 E-12.aF = 2.243Fstatistic = 43.99p-value
    • 83. ANOVA – Analysis of Variance ExampleStep 6: Compile the summary table 和draw conclusion.Since Fstatistic > Fa, we will reject the null hypothesis H0, 和conclude that at least two of the treatment (brand) means are significantly different. The p-value shows that there is only 0.000000000397% chance that all the means are equal (i.e. very strong evidence).Assumptions The sample are selected randomly 和independently from the respective populations. All treatment population probability distributions are normal. All treatment population variances are equal.Rejecting the null hypothesis testing only indicate that more than 2 treatment means differ, but which are the two ? Total possible pair of combination = C = p(p –1) / 2 ; p = number of treatments = (4)(3) / 2 = 6 pairs of means that can be compared
    • 84. ANOVA – Analysis of Variance Examplep = Number of treatments df = Degree of freedom for error
    • 85. ANOVA – Analysis of Variance ExampleFor ANOVA, we use studentized range instead of t-distribution to compute confidence interval, therefore critical value of studentized range at a = 0.05, p = 4, 和dferror = 36 is  3.8.Compute the table of confidence interavals for differences of the 6 possible combination pairs:Not significantly different
    • 86. ANOVA – Analysis of Variance Example (optional)Suppose that the International Football Association wants to compare the mean distances assoicated with the 4 brands of balls when kick by a striker, but wishes to employ human striker instead of the pneumatic swing. There will be a total of 10 balls for each br和to be tested 和there are two ways that the experiment can be conducted ; completely randomised 和randomised block design, which is a better 设计in this case ?Br和A 和striker 1 to 10Br和B 和striker 11 to 20Br和C 和striker 21 to 30Br和D 和striker 31 to 40Completely Randomised DesignRandomised Block DesignBr和A Ball 1Br和B Ball 1Br和C Ball 1Br和D Ball 1Striker 1Br和A Ball 2Br和B Ball 2Br和C Ball 2Br和D Ball 2Striker 2Br和A Ball 10Br和B Ball 10Br和C Ball 10Br和D Ball 10Striker 10Until
    • 87. ANOVA – Analysis of Variance Example (optional)Using randomised block 设计as an example; 2 Sets of hypothesis [i.e. treatment (brands) 和block (strikers)] H0: m1 = m2 = m3 = m4 H1: At least two of the brands differ in mean H0: m1 = m2 = m3 = m4 = m5 = m6 = m7 = m8 = m9 = m10 H1: At least two of the strikers differ in meanTotal sum of squares, SS(Total)Sum of squares for treatment, SSTSum of squares for error, SSESum of squares for treatment, SSTSum of squares for error, SSESum of squares for blocks, SSBdf = p - 1df = n - pdf = n - 1df = p - 1df = b - 1df = n – b – p + 1Completely Randomised DesignRandomised Block Design
    • 88. ANOVA – Analysis of Variance Example (optional)Gr和average = (227.1 + 233.2 + 245.3 + 220.7) / 4 = 231.6Compute the sum of squares for treatments:Distance Data for Randomised Block 设计
    • 89. Compute the sum of squares for blocks:ANOVA – Analysis of Variance Example (optional)Compute the sum of squares for total 和sum of squares for error :Taking a = 0.05 (95% confidence level), compute the critical F value using Excel spread sheet for both treatment (i.e. brand) 和block (i.e. striker):
    • 90. ANOVA – Analysis of Variance Example (optional)Compute a ANOVA summary table for analysis:Both Fstatistic for treatment 和block exceed the respective critical F values, hence we will reject the 2 sets of null hypothesis 和conclude that at least 2 brands differ in mean distance as well as at least 2 strikers differ in mean distance. The mean distance is dependent on the striker, hence a randomised block 设计is better in this case than a completely randomised 设计(i.e. prevent striker variation in the analysis)Assumptions The probability distributions of observations corresponding to all the block-treatment combinations are normal. The variances of all the block-treatment combinations are equalANOVA method though enabled multiple means comparison, it is based on F-distribution having the assumption that all the population means are normal with equal variance. Unlike hypothesis 和t-test which utilise central limit theorem, ANOVA method is therefore more sensitive to deviation from normality. Nevertheless it serves an important role in DOE.
    • 91. ANOVA – Analysis of Variance Case StudyA Chemical Vapour Deposition equipment designer wants to evaluate 2 new brands of high vacuum pump which is to be used in their latest deposition system because of possible cost benefit from the new 供应商s. To qualify any new pump 供应商, the designer need to ensure that the 质量 in terms of pumping down time to 6 torr need to be at least the same if not better than the br和of vacuuum pump currently used in the 产品ion. Assuming that the new pump 供应商only 提供15 sets of pump for the evaluation purpose, 和the result is given as below. Disscuss in teams 和give a presentation on your team analysis 和recommendation using the following questions as guidelines.Questions Use randomised block 设计to analyse the data. What can we tell from the analysis ? Is there an impact from different CVD system ? How about a comparison with Chi-square method ? With 95% confidence, how do the different brands differ. What are the underlying assumptions ? Compare the result with other (appropriate) hypothesis testing method. What are the assumptions ? can we lower the risk of these assumptions ? If none of the brands are significantly different in mean, how should we deal with it ?
    • 92. Word of Caution About Hypothesis Testing Ensure that the experiment carried out satisfied the underlying assumptions of whichever statistical method was employed in the analysis. Be careful about falsely concluding null hypothesis, H0 due to beta risk (i.e. Tyoe II Error). Analyse beta risk (if possible) 和see if it is acceptable. Increase sample size if needed to. Be careful about falsely concluding alternate hypothesis, H1 due to alpha risk (i.e. Tyoe I Error). Increase confidence level (or a level of significance) if needed to. Alternatively look at the p-value (if possible), 和see if the alternate hypothesis is reasonable to you. In dependency test, a dependent conclusion does not necessary imply a unique dependent 关系hip between the classifications. Analyse your result with cost-effect factor in mind. Re-confirm result if possible (unless analysis is supported by strong theoretical reasoning and/or too expensive to duplicate).
    • 93. Distribution Analysis – Chi-Square: Goodness of Fit TestSupposing that ABC company requires that all their potential future leaders to be interviewed by three different directors, so that the company can obtain a consensus evaluation of each candidate. Each director will give the candidate either a positive or negative rating as below data of 100 candidates: Positive rating out of the 3 interviews No. of candidates 0 18 1 47 2 24 3 11 If the HR Director for ABC company thinks that the interview 流程can be approximated by a binomial distribution of 40% chance of any candidate receiving a positive rating on any one interview. How can he verify this with a 90% confidence level ?What is the Chi-Square: Goodness of Fit Test used for ? It is used to to decide if the underlying probability distribution of a population is what we expected it to be (I.e. poisson, binomial or normal), by fitting the actual frequency with the expected frequency predicted by the distribution.
    • 94. Distribution Analysis – Chi-Square: Goodness of Fit TestH0: A binomial distribution of p = 0.4 is a good description H1: A binomial distribution of p = 0.4 is not a good description a = 0.1 (i.e. 90% confidence level)We need to firstly determine if the discrepancies between the observed frequencies 和those we would expect (I.e. binomial distribution of p = 0.4) are actually due to chance. Step 1: Compute the binomial probability in accordance to the scenario We can firstly use Excel spreadsheet to obtain the binomial probability using the equations: “=BINOMDIST (No. of successes, No. of trials, Proportion of success, Cumulative)”
    • 95. Distribution Analysis – Chi-Square: Goodness of Fit TestStep 2: Compute the expected frequencyStep 3: Compute the associated Chi-square statisticStep 4: Determine the degree of freedom 和obtain the critical c2 value at a = 0.1 Degree of freedom = Total observation – 1 – other estimation(s). = 4 – 1 – 0 = 3 Therefore critical F value from excel = “=CHIINV(0.1,3)” = 6.2514. Since c2statistic < c2 和in the acceptance region, we will accept the null hypothesis that the binomial distribution of p = 0.4 is a good description of the population distribution (i.e. 16.8% chance if interpret from p-value). Is 16.8% good enough ?
    • 96. Distribution Analysis – Kolmogorov-Smirnov TestWhat is the Kolmogorov-Smirnov Test used for ? It is a non-parametric method for testing if there is significant difference between an observed frequency distribution 和a theoretical frequency distribution. It is like chi-square goodness of fit test, except it is more simple in term of appliaction.K-S Goodness of fit Table
    • 97. Distribution Analysis – Kolmogorov-Smirnov TestAn airline food-服务adminstrator, has examined past records from 200 randomly selected cross-country flights to determine the frequency with which low-sodium meals were requested. The number of flights in which 0, 1, 2, 3 or 4 or more low-sodium meals were requested was 25, 45, 67, 43, 和20, respectively. At the 0.05 level of significance (95% confidence level), can she conclude that these requests follow a Poisson distribution with mean l = 1 ?H0: A Poisson distribution with l = 1 is a good description of the pattern of usage H1: A Poisson distribution with l = 1 is not a good description of the pattern of usage a = 0.05Step 1: Compute the Cumulative Poisson probability in accordance to the scenario We can firstly use Excel spreadsheet to obtain the Cumulative Poisson Probability using the equations: “=POISSON (No. of events, Mean (l), Cumulative)”
    • 98. Step 2: Compute the K-S statistic To compute the K-S statistic, simply pick out the maximum absolute value of Fe from Fo. K-S statistic = 0.3858 (i.e. at 1 meal request in this example)Distribution Analysis – Kolmogorov-Smirnov TestStep 3: Determine the critical K-S value From the K-S goodness of fit table, at a = 0.05 和sample size n = 200, the critical K-S value is: K-S critical = 1.36 / Sqrt (200) = 0.0962Step 4: Analyse the result If K-S statistic  K-S critical then accept H0 If K-S statistic > K-S critical then reject H0 和conclude on H1 Since 0.3858 > 0.0962, we shall conclude on alternate hypothesis, H1 that Poisson distribution of l = 1 is not a good description of the low sodium meal request pattern.
    • 99. Distribution Analysis – Case StudyBased on 520 data (i.e. n = 520), the electrical response of infra-red detector was found to be having a mean of 3.85 kV/W 和a 标准偏差 of 0.1161 kV/W. The histogram analysis using the standard classification method in week one result in the below diagram:Data do not look too normal. It is believed that the data should be normal, 和the diagram is simply due to too many classes identified (i.e. 23 classes).After reducing the number of classes from 23 to 8 with a class width of 0.1, the histogram looks more toward the expected distribution.
    • 100. Distribution Analysis – Case StudyWith the histogram of 23 classes, both Chi-square 和K-S goodness of fit tests were performed, 和the results using a = 0.05 are as below: Hint: Use excel function “=NORMDIST(x, mean, std dev, cumulative = true)” for probability computation.For Chi-square test: c2 statistic = 978.12 c2 critical = CHIINV(0.05, 517) = 571 Df = n - 1 - estimation = 520 - 1 - 2 (i.e. mean 和std dev) c2 statistic > c2 critical, therefore distribution is not normal.For K-S test: K-S statistic = 0.1382 K-S critical = 1.36/Sqrt(520) = 0.05964 K-S statistic > K-S critical, therefore distribution is not normal.
    • 101. Distribution Analysis – Case StudyWith the new histogram of 8 classes, both Chi-square 和K-S goodness of fit tests were performed, 和the results using a = 0.05 are as below: Hint: Use excel function “=NORMDIST(x, mean, std dev, cumulative = true)” for probability computation. For Chi-square test: c2 statistic = 63.576 c2 critical = CHIINV(0.05, 517) = 571 Df = n - 1 - estimation = 520 - 1 - 2 (i.e. mean 和std dev) c2 statistic  c2 critical, therefore accept null hypothesis H0, 和distribution is considered to be normal. p-value = CHIDIST(63.576, 517) = 0.99999For K-S test: K-S statistic = 0.1356 K-S critical = 1.36/Sqrt(520) = 0.05964 K-S statistic > K-S critical, therefore reject null hypothesis 和conclude that distribution is not normal.
    • 102. Distribution Analysis – Case StudyWith the new histogram of 8 classes, adjust the start point from 3.49 to 3.45, the K-S test result is as below:K-S statistic = 00217 K-S critical = 1.36/Sqrt(520) = 0.05964 K-S statistic  K-S critical, therefore accept null hypothesis 和conclude that distribution is normal with mean = 3.85 和标准偏差 0.1161.Distribution Analysis Take Away K-S test seems to be very sensitive toward classification of data when sample size is large. When performing distribution analysis, use histogram with all possible classifications to make judgement. Couple engineering judgement with statistical 方法 during analysis. Do not always trust the software result without questioning. Perform 2 types of goodness of fit test as confirmation if possible.
    • 103. Distribution Analysis – Exercise Knowing that the chance of rolling a 3 in a unbiased dice is 1 out of 6 (i.e. 0.1667), 和the experiment is statistically independent. Therefore we should be able to associate the experiment with a binomial distribution of p = 0.1667 (i.e. success in getting a value of 3). Break into 4 teams. Use 5 dices in a single rolling of dice. Conduct an experiment that consist of 500 rolls of 5 dices. The number of dices use in each roll (i.e. 5) will therefore be the number of trial, n in a binomial distribution. The 500 repeating rolls is to take care of the relative frequency behaviour of the experiment. The number of success will be the number of 3s in a single roll. Including also in your experiment the average values of the 5 dices in each roll. Count the number of success (i.e. value of 3) in each roll. Sum them up for the total 500 rolls. Conduct a Chi-square 和K-S goodness of fit test to determine if p = 0.1667 is a good description of obtaining a value of 3 in the rolling of unbiased dice. Also test if central limit theorem is applicable for the average values in the 500 rolls.
    • 104. Regression 和Correlation What is regression 和correlation analyses ? It is a statistical method that will show us how to determine both the strength of a 关系hip, 和the asssociated probabilistic model between two variables.Origin of regression technique Originated in 1877 by Sir Francis Galton. Study showed that height of children born to tall parent will tend to move back (or “regress”) toward the mean height of the population. The word “regression” was then designated as the name of the general 流程of predicting one variable (the children height) from another (the parent height).
    • 105. Regression 和Correlation A Deterministic Model A model constructed to hypothesised an exact 关系hip between variables Example: Force = Mass * Acceleration Given a fixed mass, Force can always be determined if accerleration is known No allowance for error in the predictionA Probabilistic Model A model include both a deterministic component 和a random error component Y = Deterministic Component + Random error Unexplained variation will be taking into account by the random error component
    • 106. Regression 和Correlation 为什么do we allow errors in the prediction model ? Might be a more realistic model since variation is usually unavoidable. All models are wrong, but 某些are useful.In high speed space travelling, variation in velocity will be an important factor to consider in a Force model to determine the thrust power 和fuel consumption.
    • 107. Regression 和Correlation A First-Order (Linear) Probabilistic Modelb0 = S$2.40b1 = y increase / x increase = 9.6 / 30 = 0.32E(y) = Taxi fare = b0 + b0x (line of means)Model for taxi farey-axis = S$36912x axis = KM travelled510152025300
    • 108. Regression 和Correlation Kelvin is a neuro specialist who is interested in how alcohol can affect human reaction time. He conducted an experiment with a single volunteer in five different sessions, 和obtain the reaction time with respect to the volunteer amount of alocohol content in percentage. How should he proceed to get a model out of these data points ?Step 1: Hypothesised the deterministic component of the probabilistic model Kelvin believe that this model should be linear, 和therefore the deterministic component is: E(y) = b0 + b1x
    • 109. Step 2: Use sample data to estimate unknown parameters in the model (with least squares approach) If we plot the indivial data points in the form of scattergram, 和fit a linear visual line to pass through the most data points, we will be able to obtain the magnitude of the deviations (i.e. vertical differences between observed 和predicted valuses). If we add the square of all the deviations, we can obtain the sum of squared errors (SSE) = 12 + 02 + 02 + 12 + 02 = 2 (for the visually fitted line) For a linear model there is only one line for which the SSE is a minimum, 和this line is called the regression line. Regression 和Correlation Reaction Time1234Alocohol Content123450Deviation = 1Deviation = 1Deviation = 0Deviation = 0Deviation = 0
    • 110. Regression 和Correlation Mathematically, the slope (b1) 和y-intercept (b1) of the regression line can be achieved by the below equations:The regression line is: E(y) = b0 + b1x E(y) = -0.1 + 0.7x
    • 111. Regression 和Correlation Reaction Time1234Alocohol Content123450E(y) = -0.1 + 0.7xWith the least squares approach, Ypredict = -0.1 + 0.7x is found to be the best fit line with the minimum sum of squared errors SSE of 1.1.
    • 112. Regression 和Correlation Step 3: Specify the probability distribution of the random error (epsilon, e). Four basic assumptions for the probability distribution of the predicted model are: The mean of the probability distribution of is e 0. The variance of the probability distribution of e is constant for all settings of the independent variable x. The probability distribution of e is normal. The value of e associated with any two observed values of y are independent. (i.e. e of yi will not affect e of yj).Reaction Time1234Alocohol Content123450Positive ErrorNegativeErrorError probability distribution of equal variance 和normal
    • 113. Regression 和Correlation To estimate the 标准偏差 of e, calculate S as the estimated standard error of the regression model using:Reaction Time1234Alocohol Content123450E(y) = -0.1 + 0.7xE(yL) = -0.1 + 0.7x – 3SE(yH) = -0.1 + 0.7x + 3SApproximate prediction interval 3S (i.e. 99.73% of all observed dependent variables should lie within this range)
    • 114. Regression 和Correlation Step 4: Assessing the usefulness of the model using hypothesis testing Since b1 is the slope of a linear model, a value of 0 would mean that the line is flat, 和as the independent variable x changes, the dependent variable y will not (i.e. no co-relation between x 和y)H0: b1 = 0 (i.e. the regression model is not useful for y prediction) H1: b1 <> 0 (i.e. the regression model is useful for y prediction) a = 0.05 (I.e. 95% confidence, 2 tails test)Independent variable, xDependent variable, y b1 = 0Sampling Distribution of b1 Based on the 4 assumptions for e, we can estimate:
    • 115. Regression 和Correlation Compute the test statistic (since n  30 use t-test):Critical t value = “=TINV(a, df)” (i.e. Default to be 2 tails) =TINV(0.05, 3) = 3.1824 Since t statistic > t critical, we will reject H0 和conclude on H1 that there is sufficient evidence that b1 is not equal to 0, hence the regression model is useful for y prediction. What if we fail to reject null hypothesis ? In the case if t statistic  t critical, it does not lead directly to accept the null hypothesis as there might be high Type II error risk. A more complex 关系hip might exist between x 和y, requiring other fitting of a model.
    • 116. Regression 和Correlation The Pearson 产品moment coefficient of correlation, r It is a measure of the strength of the linear 关系hip between two variables x 和y, 和computed as:Positive r: y increases as x increasesNegative r: y decreases as x increasesr = 1: Perfect positive 关系hipr = -1: Perfect negative 关系hipr near 0: little or no 关系hip between y 和xr near 0: no linear 关系hip between y 和x
    • 117. Regression 和Correlation The coefficient of determination, r2 It represent the proportion of the total sample variability around y that is explained by the regression line. (In linear regression, it may also be computed as the square of the coefficient of correlation, r)In our example, The Pearson 产品moment coefficient of correlation, r is:Therefore the model has a positive linear 关系hip.The coefficient of determination, r2 is:r2 value of 0.8167 implies that about 81.67% of the sample variation in reaction time (y) is explained by the blood alcohol content (x).
    • 118. Regression 和Correlation Step 5: Using the model for estimation 和prediction After confirming that the model is useful through hypothesis testing 和r2 value, the probabilistic model is now ready to be used for the prediction of dependent variable, y. The most common use of a probabilistic model are 1) to estimate the mean value of y for a specifc value of x, 和2) to predict a new individual y value for a given x.
    • 119. Regression 和Correlation Assuming that Kelvin is only interested in the reaction time of a person whose blood contains 4% of alocohol, what is the confidence 和prediction interval at 95% confidence level ? At xp = 4%, the predicted dependent variable, yp = -0.1 + (0.7)(4) = 2.7The very large confidence 和prediction interval is due to the small sample size used in the simple linear regression experiment. By increasing the number of data points, these intervals limits can be reduced for a more meaningful estimation 和prediction.
    • 120. Regression 和Correlation
    • 121. Regression 和Correlation Exercise ICup PositionPin PositionPull Back AngleHook PositionStop PositionSeparate into 4 teams Ensure the Statapult is preset to: Hook Position – 4 Pin Position - 2 Cup Position - 1 Stop Position – 3 Each team is given the task to find out the realtionship between the ball distance travelled (y) 和the pull angle (x). The pull back angle should have a range from 160o to 180o. A total of minimum 100 shots are required for each experiment set. Your team will decide on the resolution for experiment. Use the 5 steps for conducting a regression experiment to find out the probabilistic model of distance travelled 和pull back angle. Set the pull back angle to 177o, 和conduct another 100 shots. Is the data in accordance to your prediction interval from the model (using a = 0.3, 0.15 和0.05) ? Present your results in the class.
    • 122. Multiple RegressionModel building In regression, we need to always start with hypothesising a suitable probabilistic model. 1) Single quantitative independent variable (i.e. single x variable) The hypothesised deterministic component of such model can be described as: First-order model: Second-order model: Third-order model: E(y) = b0 + b1x E(y) = b0 + b1x + b2x2 E(y) = b0 + b1x + b2x2 + b3x3 b1 = slopeA Linear Modelb2 < 0A Quadratic Modelb2 > 0b3 < 0b3 > 0A Cubic ModelCaution in hypothesising the deterministic component It is not common that a very high order model (i.e. 4th 和above) to be used in regression. This is because we might end up over-fitting the data points, resulting in unnecessary noise (variation) modelling. Always start the hypothesis with the reasonably lowest order.
    • 123. Multiple Regression2) Two or more quantitative independent variables The hypothesised deterministic component of such model can be described as: First-order model: E(y) = b0 + b1x1 + b2x2 Interaction model: E(y) = b0 + b1x1 + b2x2 + b3x1x2 Complete second-order model: E(y) = b0 + b1x1 + b2x2 + b3x1x2 + b4x12 + b5x22 The model will get more 和more complex as the number of independent variables (xs) increase. The computation of b coefficeients for such model is also very complex 和usually taken care in DOE software.
    • 124. Multiple RegressionIn simple linear regression we say that:Before proceeding to the least square approach for multiple regression, we need to underst和the fundatmental in deriving the b coefficients. First we need to refresh on the “The General Power Rule” for differentiation:We need also to refresh how to break up summation notation in equation:
    • 125. Multiple RegressionIn linear model we have:Step 1: Transform into sum of squared error equationStep 2: To obtain the minimum value of sum of squared error, we can use derivatives 和set the derived equations equal to zero. (i.e. Use partial derivatives with respect to all the b coefficients b0 和b1). (i.e. General power rule)(i.e. Divide equation by 2)(i.e. Break up the summation notation)(i.e. Equation 1)(i.e. Partial derivatives with respect to b0)
    • 126. Multiple Regression(i.e. General power rule)(i.e. Break up the summation notation)(i.e. Equation 2)(i.e. Partial derivatives with respect to b1)Step 3: Solve simultaneously for equation 1 和2:(i.e. Equation 3 from 1)(i.e. Divide equation by 2)
    • 127. Multiple RegressionSubsitute equation 3 into 2:From equation 2:Multiply equation by n:Rearrange the equation:
    • 128. Multiple Regression Quadratic Model ExampleMark is working for a Mayor of a small town that is facing water shortages problem. As such the Mayor had assigned him a task to 设计a water saving campaign targetting at the right target audience in order to achieve maximum result. Mark think that the water usage should have a 关系hip to the home size, 和wants to know what this 关系hip is like, so that he can 设计和deploy his campaign more effectively. To begin with, he collected the data between home size 和water usage of a particular month for 10 families in the town, 和the data is given below table: From the scatter diagram Mark feels that this should be a quadratic model instead of a linear model.
    • 129. Multiple Regression Quadratic Model ExampleStep 1: Hypothesised the deterministic component of the probabilistic model Mark believe that this model should be quadratic, 和therefore the hypothesised probabilistic model is: y = b0 + b1x + b2x2 + e (i.e. y = b0 + b1x + b2x2 + e)Step 2: Estimate the unknown parameters b0, b1, 和b2 Similar to linear regression example, firstly take the partial derivative of the equation with respect to b0, 和form equation 1Equation 1:
    • 130. Next take the partial derivative of the equation with respect to b1, 和form equation 2.Multiple Regression Quadratic Model ExampleEquation 2:
    • 131. Multiple Regression Quadratic Model ExampleTake partial derivative of the equation with respect to b2, 和form equation 3.Equation 3:Next is to solve for the 3 derived equations.
    • 132. Multiple Regression Quadratic Model ExampleObtain the repective values for all xi 和yi combinations appear in the 3 derived equations.Create a new equations in the excel table as below example
    • 133. Multiple Regression Quadratic Model ExampleSolve for b1 和b2 using below table:b1 = 2.399 b2 = -0.00045
    • 134. Multiple Regression Quadratic Model ExampleSince b1 = 2.399 和b2 = -0.00045, subsitute these values into equation 1:Hence the equation that minimise the sum of squared error (SSE) is: E(y) = -1216.4 + 2.399x - 0.00045x2 和the probabilistic model is: y = -1216.4 + 2.399x - 0.00045x2 + e Assumations: For any given set of xs, the random error e has a normal probability distribution with mean equal to 0 和equal variance. The random errors are independent.
    • 135. Multiple Regression Quadratic Model ExampleStep 3: Specify the probability distribution of the random error.To estimate the 标准偏差 of random error, calculate S as the estimated standard error of the regression model using:Sum of squared error (SSE) = 15332.7
    • 136. Multiple Regression Quadratic Model ExampleStep 4: Determine the usefulness of the model.In multiple regression, the multiple coefficient of detemination, R2 is: The result implies that 98.19% of the total sample variation of water usage can be explained by using the independent home size in a quardratic model.Total:15332.7846402.1
    • 137. Multiple Regression Quadratic Model ExampleTesting the global usefulness of the Model with ANOVA: H0: b1 = b2 = 0 (i.e. until bk) H1: At least one of the coefficient is non-zero Use a = 0.05n is the sample size 和k is the number of terms in the model. Reject when Fstatistic > Fcritical, with k numerator degrees of freedom 和(n-k-1) denominator degree of freedom. Caution: A rejection in hypothesis leads to the conclusion that the model is useful with the confidence level used in the test, but does not mean that it is the best model.
    • 138. Multiple Regression Quadratic Model ExampleFcritical = “=FINV(0.05, 2, 7)“ = 4.74 Since Fstatistic > Fcritical, we shall reject the null hypothesis 和conclude that at least one of the coefficients (i.e. b1 or b2) is non-zero, 和the global F-test indicate that the model is useful in predicting the water usage.Step 5: Using the model for prediction.The prediction interval for the regression can be estimated by  ta/2 S Using 95% confidence level, the ta/2 value is 2.365. (i.e. df = n-k-1) Therefore the upper prediction level is: y = -1216.4 + 2.399x – 0.00045x2 + 2.365S = -1105.7 + 2.399x – 0.00045x2 Therefore the lower prediction level is: y = -1216.4 + 2.399x – 0.00045x2 - 2.365S = -1327.1 + 2.399x – 0.00045x2
    • 139. Regression - Residual AnalysisWhat is it used for ? Since all the assumptions in regression model concern the random error component, residual analysis can therefore be employed to indicate if the assumptions has been violatedUsing the water usage regression as an example If we plot a linear model for the water usage data, it will end up with a R2 value of 0.8317, which is not too bad for strength indication.
    • 140. Regression - Residual AnalysisStep 1 Calculate 和plot the residuals against each of the independent variables, whereby: Residual = Yi - Ypredict Step 2 Analyse each plot, looking for curvature (i.e. a bowl or mound) shape. As in residual plot of linear model a mound can be seen, indicating need of a quadratic model.Step 3 Examine the plot for outlier based on +/- 2S from 0. See if the outlier can be explained, if it is due to special cause, remove the data from the regression analysis.
    • 141. Regression - Residual AnalysisStep 4 Plot a histogram on the residuals 和see if there is any obvious departure from normality (i.e. can employ goodness of fit test to check). Extreme skewness of the frequency distribution may indicate the need to transform the dependent y variable. (There are not enough data in the water usage example for distribution analysis)Regression 和Correlation Exercise IICup PositionPin PositionPull Back AngleHook PositionStop PositionSeparate into 4 teams Ensure the Statapult is preset to: Hook Position – 4 Pin Position - 2 Cup Position - 1 Pull Back Angle – 180o Each team is given the task to find out the realtionship between the ball distance travelled (y) 和the stop position (x). Each team member is to take 3 shots at each stop position. Perform the appropriate analysis on the data collected 和present in the class.
    • 142. Time Series AnalysisTime series analysis is used to detect patterns of change in statistical information over regular intervals of time. An estimate for the future can be projected based on the analysed pattern. It is a quantitative method of forecasting.Variations in Time Series Secular trend - where the value of variable tends to increase or 降低over a long period of time (i.e. increase in the cost of living recorded by Consumer Price Index) .Y axisX axis - Time in yearsSecular Trend
    • 143. Time Series AnalysisCyclical fluctuation - where the value of variable tends to increase 和降低in a cyclical manner over a long period of time (i.e. Business cycle, economic performance) .Y axisX axis - Time in yearsStraight Line TrendCyclical FluctuationSeasonal variation - where the involved pattern changes within a year 和tends to repeat year after year (i.e. Business that involve 产品 selling in season).Y axisX axis - Time in yearsSeasonal variationYear 1Year 2Year 3
    • 144. Time Series AnalysisIrregular variation - where the value of variable changes in a random manner which is unpredictable (i.e. lottery number).Y axisX axis - Time in yearsIrregular variationIn most cases, a time series will contain several of the variation components. Therefore, it is possible to describe the overall variation in a single time series in terms of these four different kinds of variation.Making prediction from trend analysis Fitting a linear trend by least square method to the data points collected over time. Equation for linear trend: Y = a + bX X = Years (or time) Y = Response variable
    • 145. Time Series Analysis – Trend AnalysisTranslating 和coding time As we are using least squares approach, coding the time will eliminate the need to square the number as large as the year, 和it will set the mean year (X bar) to zero, simplying the trend equation.
    • 146. Time Series Analysis – Trend AnalysisTherefore the linear trend equation is: Y(trend) = 139.3 + 7.536 X(coded) If you want to predict the sales in 1990 from the trend equation, first code the year 1990: X(coded) = (1990 – 1985.5) x 2 = 9 Subsitute X(coded) = 9 into trend equation: Y(trend) = 139.3 + (7.356) (9) = $207.124M
    • 147. Time Series Analysis – Trend AnalysisFitting a quadratic trend by least square method to the data points Equation for quadratic trend: Y = a + bX + cX2a, b, 和c can be found by solving these equations simultaneously !
    • 148. Time Series Analysis – Trend AnalysisTherefore Y(predict) = 39.3 + 22.7 X(coded) + 5.07 X(coded)2Time Series Analysis – Cyclical variationCyclical variation is the component of a time series that tends to oscillate above 和below the secular trend line for periods longer than 1 year. The procedure used to identify such variation is termed the residual method.X = 1991 X(coded) = 1991-1987 = 4 Y(trend) = 39.3 + (22.7)(4) + (5.07)(16) = $211.2M (predicted for year 1991)Assuming the trend will continue, what will be the expected sales in year 1991 ?
    • 149. Time Series Analysis – Cyclical variationThe measure of cyclical variation is expressed as a percent of trend: Example: A comic book distributor wants to measure the variations in the sales figure of the: “X-Men” comic which is released monthly, so that he can have 某些information about the variation he is seeing for the past 8 years.
    • 150. Time Series Analysis – Cyclical variationBy plotting the residual plot as a percent of trend line, the trend issue will be eliminated, 和the cyclical component is isolated from the time series. It is to be noted that this method is used only for describing past variations 和not to predict future cyclical variation.Time Series Analysis – Seasonal variationThere are three main reasons that seasonal variation is important: To establish the pattern of past changes, giving opportunity to compare 2 time intervals that would otherwise be too dissimilar (i.e. compare December ‘2000 to December’ 1999, instead of June’ 1999). It is useful to project past pattern into the future, in particular for making short-run decision. Once existing seasonal pattern can be established, its effect can be eliminated from the time series.
    • 151. Time Series Analysis – Seasonal variationIn order to measure seasonal variation, the ratio-to-moving-average method is employed to 提供an index to describe the degree of seasonal variation. The index has a base of 100, 和degree of seasonality is measured by the variations away from this base. Example: The management team for telecommunication equipment wants to forecast their sale revenue for the 3rd quarter of 1990. As the sales of such equipment might have a seasonal impact from the bonus period of consumer, how should the team conduct a time series study to find out the indices that describe the degree of seasonal variation.
    • 152. Time Series Analysis – Seasonal variationStep 1: Add the total of the 4 quarters in the 1st year 和put it the value in the middle of the quarter (i.e. 2.5 quarter of year 1)Move 1 quarter down 和sum the next 4 quarters (i.e. Year 1 quarter 2 to year 2 quarter 1). Put this value in the quarter that is incremental by 1 from the previous starting quarter (i.e. Quarter 3.5 in year 1). Repeat the procedure until the summation has reach the last quarter of the last year.
    • 153. Time Series Analysis – Seasonal variationStep 2: Compute the 4-quarter moving average by dividing step 1 values by 4.
    • 154. Time Series Analysis – Seasonal variationStep 3: Since the moving average value start at year 1 quarter 2.5, we need to bring it down half a quarter to quarter 3 by averaging the value between 2.5 和3.5.
    • 155. Time Series Analysis – Seasonal variationStep 4: Calculate the % of the actual value to the moving average value for each quarter in the time series having a 4-quarter moving average entry.
    • 156. Time Series Analysis – Seasonal variationStep 5: Calculate the modified mean (seasonal indices) by discarding the highest 和lowest values (i.e. shaded cell in the table) for each quarter 和averaging the remaining values for each quarter.Step 6: Since the base for an index is 100, the total indices for 4 quarters should therefore be 400 和not 397.45. We must then adjust the individual quarter index so that the total indices = 400.
    • 157. Time Series Analysis – A combined approachIn this case, the approach is a combination of three stages: Deseasonalising the time series Developing the trend line Finding the cyclical variation around the trend line To continue on the example for seasonal variation, if the management team wants to forecast the revenue for 3rd quarter in 1990. How should they approach this problem ?Step 1: Computing seasonal indices 和find the deseasonalised values for every quarter in the data:Once the seasonal effect has been eliminated, the deseasonalised values that remain reflects only the trend, cyclical, 和irregular component of the time series.
    • 158. Time Series Analysis – A combined approach
    • 159. Time Series Analysis – A combined approachStep 2: Developing the trend line with the deseasonalised values (assuming linear)Step 3: Find the cyclical variationThe residual plot seems random, no suggestion visually that the linear trend is a bad estimate.
    • 160. Time Series Analysis – A combined approachUsing the time series model to forecast sales revenue for 3rd quarter 1990: Third quarter in 1990 will be equal to the 23rd quarter according to our table. The coded X will therefore be: = (2) (23 – 10.5) = 25 Subsituting X(coded) = 25 into the deseasonalised trend equation, we have: Y(trend) = (0.159)(25) + 18.06 = $22.035 M However this is a deseasonalised forecast, we must therefore seasonalised this estimate by multiplying the adjusted third quarter seasonal index expressed in ratio to a base of 100: Y(23rd quarter) = 22.035 x 0.6119 = $13.48 M
    • 161. Time Series Analysis – A combined approachTrend 和Cyclical component
    • 162. Time Series Analysis – QuestionThe number of faculty-owned personal computers at the University of Berkeley has increased drmatically over the last 6 years: Year 1994 1995 1996 1997 1998 1999 No. of PCs 50 110 350 1020 1950 3710 a) Develop a linear estimating model that best describes these data. b) Develop a quadratic estimating model that best described these data c) Estimate the number of PCs that will be in use at the university in 2003 using both equations. d) If there are 8000 faculty members at the university, which equation is the better predictor. 为什么? The following data describe the marketing performance of a regional beer producer: Use the above data to predict the sales for 2nd 和3rd quarter of year 2001.
    • 163. Time Series Analysis – Question3. The data in the above table is the monthly sales of jet-engine to the commercial aeroplane manufacturer. Do a time series analysis of jet-engine sales over the last 10 years. Deseasonalise the sales by month (with a 12 months centered moving average method). Determine the correct type of model to use for a sales prediction, look at the residual plot for a conclusion. Use your result to forecast sales for each month of 2001.
    • 164. Introduction为什么do we need to built reliability into our 流程es ? Reliability has Gain importance in many operation 和产品 in the market Competition in the market 和dem和is the service. Customer wants more reliable services, 流程or 产品. As a consumer. We are concern about buying 产品 that last longer 和are cheaper. This call to have High Reliability to be build in our 产品 和services. Reduce risk of liabilityReliability AnalysisThere are many areas where we need reliabilityCritical Raw Material Power system Consumer 产品Life safety System Measuring Instruments Bio Medical FieldTransportation - Aircraft Defense Medical industry
    • 165. How do we measure reliability of your 流程和system?Reliability Analysis Reliability is measure of the probability that the 产品or system will operate without failure for a given time (time, t) Reliability index is a number from 0 to 1, measuring the probability that the part will fail for a given testing time Reliability R(t) of a component or system at time t is given by R(t) = P( T > t ) Where T = a continuous random variable denoting time to failure. Typically, we will assure that T follows 某些probability distribution (e.g. Normal or exponential) t = 某些specific desirable time ( e.g 450 hours)
    • 166. F(to)R(to)t0tF(t) = P(Tt) R(t0) = 1 - F(t)Probability that the component will survive after t0 timeProbability that the component will fail prior to t0 timeReliability AnalysisF(t) is the probability density function (pdf) for the part to fail before time t. R(t) is the probability density function (pdf) for the part to fail after time t. Two possible failure function for investigate: a) Normal distribution 和b) Exponential distribution Since reliability is a probability ,it is also the area under the pdf curve
    • 167. h(t)Burn In zoneWear out zoneOperating ZonetBath Tub CurveReliability AnalysisAnother Term use in Reliability is the hazard function h(t) = f(t) / R (t) Where f(t) is the Failure function 和R(t) is the reliability function Hazard Function is interpreted as the “instantaneous failure rate”. It is the rate of failure at an instant time.
    • 168. Exponential Failure Model If the random variable T denoting time to failure is exponentially distributed, then the following properties hold: f(t) =  e -t , t  0 F(t) = P(failure before time t) = P(T  t) = 1- e -t R(t) = P( failure after time t) = P(T  t) = e -t E(T) = Meantime between failure = MTBF = 1/   = average number of failure per time interval, which is the only parameter in the exponential distribution. Therefore the Hazard Function h(t) = f(t)/R(t) =  e -t / e -t =  Exponential model is often use to model useful life period or operation zoneReliability Analysis
    • 169. Since MTBF is equal to 1/, then R(t)= P(failure after time t)= e-t R(1/)= P(failure after MTBF)= e-t = e-(1/) = e-1 = 0.368 Therefore, from figure 1, MTBF indicates the time whereby 36.8% of the part tested will survive beyond this time period. This is however based on the assumption that all 缺陷follow the exponential failure behaviour. Reliability Analysis
    • 170. Another important property of the exponential distribution is called the “ Forgetfulness” property.This property says that, no matter how long the item has been working, it is as good as new. We express this property asExponential Failure Model Reliability Analysis
    • 171. Example : Reliability AnalysisSuppose data is collected on 40 identical components 和they are found to have a mean time between failure (MTBF) of 25 hrs . Assuming a constant failure rate. Find the reliability function Since the failure rate is constant, we will use exponential distribution. Also, the MTBF is 25 hrs; we can therefore estimate the parameter  by 1/E(T) = 1/25 Reliability function is app. By R(t) = e -t/25 What is the reliability of the item at 30 hours ? R(30) = e -30/25 = 0.3012 What is the reliability of the item at 70 hours, given that it has already lasted 40 hours ? Using the forgetfulness property of the exponential distribution P(T>70 I T > 40) = P(T > 30) = e -30/25 = 0.3012
    • 172. Confidence Limit of ReliabilityGenerally, we are not concerned with upper confidence limits for reliability , only lower limitsWhere t = time l = estimate average number of failure per time interval = 1/ MTBF n = sample size c2 is a Chi-square value, with = level of significanceReliability AnalysisUsing this 关系hip to calculate a 95% lower confidence limit for the reliability at 30 hours of the component given in the previous example
    • 173. Normal Failure ModelExample Suppose we sample 36 machined parts 和put them on test until they fail. We record their failure times, construct a histogram, perform a 2 goodness of fit 和concluded that time to failure is normally distributed. We also calculate thatt = 50 hours 和S = 10 hours a) What is the estimate reliability of the parts at 40 hours? Using t 和S as estimates of  和 ? we let  = 50 和 = 10 Therefore R(40) = P(T > 40) = 1 - P (T40) = 1-P( (T-  )/   (40-50) /10 ) = 1- P( Z  -1) = 1 - 0.1587 = 0.8413Reliability AnalysisThe parameter  和 can be estimate by computing t 和s from a random sample of failure times
    • 174. Component 1Component 4Component 3Component 2Series System all Components must be operating in order for the system to operateFor N number of system R(t) = R1(t) .R2(t).R3(t)….Rn(t)Parallel SystemComponent 1Component 3Component 2For N number of system R(t) =1-[ (1-R1(t)) .(1-R2(t)).( 1-R3(t))…. (1-Rn(t)) ]Parallel System system fail only if all the individual components fail. Reliability Analysis
    • 175. Reliability AnalysisThe expoential reliability equation mentioned earlier is based on a reliability testing that end only after all parts have failed. However sometime due to time constraint, we might perform a reliability in another way: Censored Data I Test until a fixed time, t0, 和observed the number of items fail. where t0 = Total reliability time r = Number of parts fail after t0 n = Total sample tested Assumption: Exponential failure timesFor 95% lower confidence limit on R(t)
    • 176. Censored Data II Test n items until r failures occur. where r = Number of parts fail (fixed) n = Total sample tested Xr = Maximum time of failure detected Assumption: Exponential failure timesReliability AnalysisFor 95% lower confidence limit on R(t)
    • 177. What is a 设计of ExperimentIt is an Experiment that we run to learn about our 流程 We purposefully make changes to the inputs ( or Factors) in order to observe corresponding changes in the outputs ( or response) The information Gain from properly designed Experiment can be use to improve performance characteristic, to reduce costs 和time associated with 产品development , 设计和产品ion To build a mathematical models which approximate the true 关系hip between the inputs 和the outputs. Turning an Art to Science. The mathematical model will contain information to optimize the 流程, reduce variation ,make response robust to noise
    • 178. Past Statistical Guru 和the the 工具ing they useWe will use all this Tooling to Make a quartum Leap In Improvement
    • 179. Definition Of A 流程for DOEMaterialPoliciesEquipmentPeopleProceduresMethods流程EnvironmentPerform a serviceProduce a 产品Complete a task A blending of inputs to achieve the desire outputsIn order that we can improve the 流程, we must have something to measure 和reflect on Better Rate or critical dimension Faster ( such as cycle time , 流程time development time) Lower cost ( such as COPQ 和流程和产品cost
    • 180. What is a 设计Experiment It is a way to change the inputs factors of a 流程in order to observe corresponding changes in the output ( respond) TemperaturepHConcentrationVoltageElectrical Plating 流程Coating ThicknessUse this column to model S hat ( a predicted 标准偏差)Use this column to model Y hat ( a predicted average)Multiple replicatesA set of condition we use to study the 流程
    • 181. Strategies For Experimentation Screening - Identify the key 流程Input Variable : K > 5 Modeling - To Model The 流程和Generate a Mathematical Model : K < 5 where K is the number of factors 知识 Gained From Experimentation Sensitivity Characterization Optimization Robustness Tolerance 设计For X’sY hat (mean) 和 S hat (std Dev) Model 提供s 知识 on all of these
    • 182. 目标 of An Experimental 设计Obtain the maximum amount of information using a minimum amount of resources Determine which factor ( inputs) shift the average response , which shift the variability 和which have no effect Building empirical models relating the response of interest to the input factors Find factor settings that optimize the response 和minimize the cost Validate ( common ) resultsExperimental 设计will help you identify the following types of factors Factor A affects the average Factor B affects the 标准偏差 Factor C affects the average 和the 标准偏差 Factor D Has not Effects No Shift in Mean or VariationVariation ShiftMean ShiftsMean 和Variation Shifts
    • 183. Typical Way Experiment is conductedOne Factor at a time results look X1%X2%Varying The Parameter X1 Temperature 和set at the highest yieldVarying X2 Time set at the highest yield-90807070Max YieldActual 流程 look like this therefore you might not get the optimize point
    • 184. Ways of collecting Experimental dataOne Factor at a time (cannot be assure of detecting the unique effects of combinations of factors) All combinations ( full factorial)- Too Big Best Guess- Never Conclusive DOE Example of DOE : Kraft Cheese packaging Packaging 流程Response Variable Variable InputsY1: Bond StrengthX2: Paper Source X1: Polymer TemperatureA manufacturer of laminated paper take two rolls of Kraft paper 和extrudes a layer of Polymer in between , simultaneously pressing the three components into sandwich . The customer has complained about the lack of adherence of the polymer to the two paper layers. A Team is set up to maximise the bond strength .After apply PF/CE /CNX /SOP the X factors are this:
    • 185. Factors Low Level High Level X1: Polymer TemperatureºF 580 600 X2: Paper Source Vendor Y Vendor X Quantitative factorsQualitative factorscontinuousdiscreteDefine the DOE : Full Factorial mean all combination
    • 186. They conduct an Experiment 和Collect the DataDo we know which factor is the the one that influence the Packaging process ?Next they analysis the Data 和this was the finding
    • 187. Effect of Temperature 和Vendor on Bond strengthThe mean of the Bond strength is influence by Temperature 和Vendor but the Variation of bond strength was not influence by this two parameter
    • 188. There is interaction (combine ) effects between Factor A & B
    • 189. (本页无文本内容)
    • 190. DOE : Manufacturing of Capsule Wetting Solution Pressing strengthMixing timeFlour VendorCapsule Manufacturing 流程Compression StrengthImpact Strength Dissolve RateTemperature curing time Degassing time
    • 191. Number of Factor for this 流程is = 7 Total Number of combination : 2 7 total run is128 One factor at a time approachIt is too big to do 128 : We can do subsetTo detect a significant difference in 2 标准偏差 , a 17 replicate give 95% confidenceA total 8 combinations tested No math model available to predict the response for the other 120 combination No ability to estimate if variable interact
    • 192. INTERACTION- Family Benz Distance per litre of petrol use with passengerDistance per litre of petrol use with Engine TimingY LoHiYLoHiKm/litreKm/litreA-Number of passengerA -Engine Timing Low Octane (B)High Octane (B) High Octane(B) Low Octane(B) To model this picture AB component is the degree of interactionTo model this picture a simple linear model is all that is require Non Parallel slopeParallel slopes This is Main Effect Analysis
    • 193. (本页无文本内容)
    • 194. Terminology Use in DOEFactors -Controllable ( Independent ) variables that may cause a change in the response ( dependent) variable : Example Temperature ,流程ing Speed Levels - The specific Setting or options for a factor in an experiment Full factorial 设计- Combine all levels of all factors in an experiments Replication - A repetition of run of a setup Response - The output form the experiment Main Effect - The average change in the response due to the change in the levels of a factor
    • 195. What is a Full Factorial Cost of Running an Experiment
    • 196. When we can’t do all Combinations this is best Guess by Dr KnowItallIf Run 1- 8 were the results of a series of “Best Guesses“ 和you lump them together , you could evaluate things like the following: Does Factor P shift the average ? It appear P shifts the average from P1 to P but if you were to realize that M Row are the same : This is call “ Perfectly Confounded or Aliased“ They are BAD because we can’t tell the difference between P 和M Are Setting P 和D Aliased ? No but they are Partially Confounded we can’t evaluated they independently. V 和W are not aliased or confounded .Call orthogonal “ Balance” this is Good Ave 5.5Ave 9.5
    • 197. Illustration of Confound 和Aliased on the previous designP=MDTotal Variability of YPerfectly ConfoundedPartially ConfoundedMDPHow do we obtain this independence of variables? Good 设计make every variable independence
    • 198. (本页无文本内容)
    • 199. Simple Definition of Two level orthogonal DesignThis new column is not longer orthogonal with A 和BThis new column are tested independently 和they are orthogonal Use to model interaction of A 和BHow do we know that the column is orthogonal a) sum of 产品between row = 0Way to transfer form Actual to CODED
    • 200. Introduction to DOE Simple Model Building Example
    • 201. Introduction to DOE Simple Model Building YAPareto Chart Y barBSABAY-+-+-+S-+-+-+8 6 4 2ABABABAB25 20 15 10 5 Pareto Chart SABB25 20 15 10 5 8 6 4 25.62515
    • 202. Building an Empirical Model This empirical model is similar to Taylor Series Approximation which relates to the output Y to the coded inputs A 和B.Now How will the Talyor series for A, B , C look like ?
    • 203. Comparing a Screening 设计Vs OFAT : 7 Factors-2 level12X 4 = 48 runPrevious Example: Manufacturing of Capsule using 设计of OFAT in 7 factor is =136 run
    • 204. Comparing a Screening 设计Vs OFAT : 11 Factors -2 levelOFAT 12 x 17= 204 Total RunNeed to get 95% confidence in standard deviation analysis for each run 12 x4=48 Total Run4 Replication will give 95% confidence in S 和99.9% in Y hat
    • 205. Things to note when conducting a 设计of ExperimentAll the experimental Factors should be considered All Controllable factors must be controlled The 流程must be running at its stable stage Eliminate as many noise as possible Identify all the possible response Ensure that the Measurement system is capable Try to use Continuous Factor Try to use measurable responses Use an appropriate 设计 Prepare the experiment in advance Identify potential mistake during the experiment
    • 206. Exercise : Screening DOEUsing the PF/CE/CNX/SOPs from Week 1 minimize the variability Conduct a L 12 Screening 设计using the following factors: Using Software to 设计the Experiment Perform Data Collection for the Experiment
    • 207. Robust 设计for 产品 和流程es You are a Business manager that want to launch your 产品into M&M Candy all over the World. Two area we are in particular to look into you 产品.It must be able to withst和temperature variation 和Altitude A Few Approach use in the market a) Warning Sign “Keep below 200 degree F” b) Sell together with a Portable cooling unit c) Come out with a 设计that is robust to change in environment RespondsLowHigh20 10 Room Which is the most productive ?How can we do this ?
    • 208. (本页无文本内容)
    • 209. 设计a 产品or 流程- Robust to Noise Factor A has substantial Shift . If we optimize A we can make the 流程RobustSet A to Positive the variation of the hardness will reduce 和robust to Changes in temperature
    • 210. Increase the Time to Market for new TechnologiesRear Wing Angle ( -30, 0 , 30 )Side Wing Angle ( -0 , 5 , 10 )Side Wing Span ( 30 , 50 , 100 )Side Wing Area ( 0 , 5 , 10 )Weight ( 100 ,400, 500 )Develop A New Energy Saver -Two Wing Slider ShuttleEngine Power ( 500 ,600, 700 )Aerodynamic CharacteristicWind Tuner Testing to Model the Aircraft Sliding AbilityWhat are some Area that we can use DOE ?
    • 211. Putting all together Developing into 流程Manual流程Name & Description Team members General 流程Flow Determine the measurement Matrix PF/CE/CNX/SOPs/FMEA ; Poka Yoke Control document for all Controllable item Specification , Cpk 和COPQ for Each Y Measure the capability of X 和Y Look at the Historical Data Designed Experiment Control Charts Trouble Shooting Results Track Yield , Time , Cost 和Capability
    • 212. How is Regression linked to DOEWhich Factor is Significant
    • 213. How do we read the F table ?Not significant Significant at 95% confidenceHow would you write the predicted Equation for Y ?Let try to use the software at this pointCreate a 设计(2 level 3 factor)
    • 214. Output from the RegressionMeasure the degree of independence orthogonality ROT: Tol > 0.5 Indicate that the Factors is Significant if P(2tail) < 0.05 Regression Strength :proportion of values that is accounted by the model F statistics ROT: F <0.05 to have a strong model for prediction Estimate of Std DEV of Residual Standard Error = SQRT(MSE)What is the predicted Equation ? Should we include B into the Predicted Equation?
    • 215. Modeling of Deviation SRule of Thumb : For 2 Level Designwrite down the Predicted equation for S hat.Next we have to reduce the variation by minimize S hat : we can set factor B to high +1. S - hat = 0.669 this will minimize the variation To accomplished the desire objective is to set the Y- hat to the desired value 120 = 115.93+4.64A-0.41B+4.14AB where B= + 1 120 = 115.93 +4.64A - 0.41 ( 1 ) + 4.14 A (1) A= 0.51 Decode the coded value
    • 216. Confirmation Run To test for confirmation we should expect all response to fall within y hat +/- 3 S-hat For this example it should be 120 +/- 3(0.669) or (117.993, 122.007) Compute the Cpk, Cp, yc 和c
    • 217. Summary For 设计of Experiment for Linear ModelA) Define the Problem B) Objective of the Experiment C) Identify the number of factor you have in the system D) Identify the number of Response E) Determine the number of resources for the experiment D) Determine 和select the type of DOE 设计和Analysis Strategies F) Randomize of the Run in the 设计Matrix G) Conducting the Experiment 和Recording the Data H) Analyze the Data 和Draw Conclusion ,Make predictions G) Do Confirmation Run
    • 218. Exercise - Linear Modeling DOEBreak into teams Review the step 流程for conducting DOE Using the cause 和effect diagram Using the Full Factorial Design Consider this Factors in your stratapult Pull Back Angle ( 110 ,177) Tension Pin position ( 1 , 3 ) Rubber B和Hook Position ( 1 , 3 ) Stop angle Position ( 2 , 4 ) Measure the x cm +/- 5 inch Build the model 和Find the Y- hat 和S - hat equation Reduce Variation by minimizing S Use optimizer to estimate setting : Target 150 cm+/- 5 cm Conduct Confirmation Run 20 shot : Calculate CPK
    • 219. Flow Chart In 设计of Experiment StartStatement of problem & objectiveDetermine what to measure & complete PE/CE /CNX SOP/FMEA/POKE YOKEHow Many Levels For Each Factors ?How Many Levels For Each Factors ?Full Factorial K = 2 n reps  9 K = 3 n reps  5 K = 4 n reps  3K  4K = 56  K  116  K  82 5 -1 1/2 Factorial n reps  3 12 Run PLACKTT-BURMAN or TAGUCHI L12 Screening n reps  416 Run FRACTIONAL FACTORIAL n reps  3How Many Factors ?Central Composite or Box-Behnken Box K = 2 n reps  9 (CCD) K = 3 n reps  5 (CCD or BB) K = 4 n reps  3(CCD or BB) K = 5 n reps  3(CCD)Modeling K  5Screening 6 K  74 K  7K 3TAGUCHI L18 Screening (also Include One 2 Level Factor n reps  4FULL Factorial K =2 reps  7 K = 3 reps  3Modeling or Screening ?Quantitative OnlyType of FactorsNot all Quantitative (I.e at least 1 Qualitative)Note 1 : Sample Size (n reps) is for 95% confidence in Predicted S 和99.99% Confidence in Predicted Y: If n reps/2 it will 提供75% confidence in Predicted S 和95% confident in predicted Y For Other 2 Level 设计Refer to the next slide3 levels2 Levels
    • 220. Statistical 流程Control ( SPC)Statistical 流程Control is used to “Listen To The Voice Of The 流程”
    • 221. DAYS COSTED INVENTORY (DCI)UPPER CONTROL LIMITLOWER CONTROL LIMITCENTER LINEControl chart
    • 222. Underst和the theory behind control charts Construct the following Control Charts : Variables: Xbar & R, Xbar & s, Individuals 和Moving Range, 和Batch 流程es Attributes: p, np, c, u, 和Low Frequency Pre-Control Interpret Control Charts Methods for choosing appropriate subgroup sample sizes 和frequency of sampling Methods of implementation 和reaction 提供s a graphical representation of 流程performance 和stability Monitoring output variable 和control Input VariableLesson 目标:
    • 223. 为什么do we use Statistical 流程Control ?It 提供s a graphical representation of 流程performance 和stability Help to monitoring Key 流程Output Variable 和 control Key 流程Input Variable SPC help to detect special Causes 和 minimizing reaction to normal variation SPC is a 工具use for observing variation . It can be use to monitor statistical signals 和improve performance. It can be applied to many area. Equipment performance Document errors Sales per Quarter 缺陷rates Time taken to complete by task SPC should be apply to 流程characteristics (X’s) rather than finished goods (Y’s) . Until the 流程inputs become the focus of our effort, the full power of SPC methods to improve 质量, increase 产品ivity, 和reduce cost cannot be realized.
    • 224. What is SPC?SPC is a statistically based method of: 1. Evaluating the performance (stability) of 流程和产品variables 2. Pointing out the existence of special causes of variation 3. Removing the special causes 和maintain 和monitor normal 流程variation
    • 225. All 流程es have natural variation we call them common causes 和unnatural variation we call assignable special causes We use SPC to monitor 和improve the 流程. While control charts can indicate special causes through Out-of-Control signals. Control charts are the means through which 流程和产品parameters are tracked statistically over time. Control charts has upper 和lower control limits that reflect the natural limits of (random) variability. These limits must not be compared to the customer specification limits. Control limits are based on establishing + 3s mean limits for X or Y. Based on statistical principles, control charts allow for the identification of unnatural (nonrandom) patterns: special causes have changed the 流程. The actions we take to attack the special causes are the key to successful use of SPC.What is SPC ?
    • 226. Walter Shewhart invented the control chart in 1924Walter A. ShewhartShewhart: The Father of Control Charts
    • 227. We want to monitor 和improve the Inputs Input 流程/System Output2. Identify Root Cause3. Implement Corrective Action 4. Verify 和Monitor1. Detect Assignable or Special CauseStatistical 流程ControlThe Goal of SPC
    • 228. Where should we use SPC Critical 流程Parameters Highest RPN’s from FMEA When a mistake proofing device is not feasible Customer requirements 和Critical to 质量 measure Management commitments Initially, the 流程outputs may need to be monitored
    • 229. Select the appropriate variables to control chart High RPN’s In 流程monitoring Critical 流程parameters Select the data collection point Select the type of control charts Attribute (np,p, c, u) Variable (X-bar/R, X-bar/S, Individual) Establish basis for rational subgrouping What constitutes a rational subgroup? Does it represent the population from where the samples are taken? Determine sample size Determine measurement method / criteria How to Implement SPC Charts
    • 230. How to Implement SPC Charts (cont)Step 1 We have to verify gage capability of the measurement system (MSA) Step 2 Collect data 和Set up a run chart to study the 流程variability Step 3 Establish control limits Step 4 Establish a method of periodically verifying their accuracy Step 5 Create forms for charting the data Prepare a Work Instruction for the charting Documentation for sampling Data to be collect 和measurement method Corrective action plan when out of control occur Step 6 Train personnel Operators Supervisors Engineers Step 7 Institutionalize the charting
    • 231. Implementation of SPC System SPC on X’s or Y’s without proper 培训= WALLPAPER. SPC on X’s or Y’s with fully trained operators. Operators will underst和the signals, but management will not empower them to stop for investigation. Operators will learn to ignore or disconnect the warning signals once 产品ion becomes the #1 priority. SPC, with Response Action = Inspection. Short term containment equals auditing or 100% inspection. SPC, with Response Action = Equipment Flag. 流程is stopped or equipment automatically shuts down, so that 缺陷will not move forward. SPC, with Response Action = Countermeasure. 改进 are made so that defect cannot occur again.
    • 232. Advantages SPC is a proven method for improving 流程 It is effective in prevention 缺陷 Help us to prevent unnecessary 流程changes It is a good diagnostic 工具 SPC 提供information about the 流程capability Can be used for both attribute 和variable data type Control chart is able to track the 流程over time Disadvantages People must be trained 和retrained on the 知识. Data must be gathered correctly Must be able to maintain all the information 和charted correctly Charts must be analyzed correctly Reponses to the out of control must be appropriate 和every time Advantages 和Disadvantages of SPC
    • 233. The theory behind a control chart follows the principles of simple confidence intervalsmUCLLCLa/2a/2When the true 流程is running at mean m, the probability that a sample mean falls outside the control limits equals aThe Theory of Control Charts
    • 234. Model for Center line, UCL & LCL for Variables ChartUpper Control LimitLower Control LimitCenter LineUCL = m + k1s Center Line = m LCL = m - k2sBecause we must use estimates from samples: m  x = the mean of the sample means s  s = the 标准偏差 derived from the samples k1, k2 = constants implying unusual observations (often = 3)=Components of a Control Chart
    • 235. The Time Element of Control ChartsSPCTime SequenceUCLLCL
    • 236. One advantage of a control chart involves its ability to track the 流程over timeThis time-based dependence means that certain trends or patterns may indicate that special causes are occurring over time.TimeComponents of a Control Chart
    • 237. Assignable Causes:- Things that are unpredictable (hour-to-hour, day-to-day, week-to-week)- 目标 is to : Detect 和eliminate assignable causesCommon Causes:- In statistical control- Natural or typical 流程variationsTimeTypes of Variation
    • 238. TimeTimeIf only common causes of variability are present, then the 流程output is constant 和predictable over timeIf only special causes of variability are present, the 流程output is neither CONSTANT, nor PREDICTABLEVariation vs SPCPF/CE/CNX/SOP
    • 239. 流程 IN-CONTROL, BUT NOT CAPABLE (out of specs)(Only typical 流程variability, but excessive)流程IN-CONTROL 和CAPABLE(The 流程variability has been reduced 和now meets 流程specs)Lower specification limit (LSL)Upper specification limit (USL)TimeVariation vs Specifications
    • 240. Control Chart TypesChoosing the correct control chart Type of dataIs sample size fixed 和n < 10 ?Constant sample size?constant area of opportunity?Counting 缺陷or defectives?ucp, nppX bar - S IMR ChartData tends to be normally distributed because of central limit theoremAttributesVariablesDefectivesYesNoDefectsNoSub-groupsYesNeed to Construct a control Chart BinomialPoissonYesNoX bar-R Chartn=12< n < 9n = sample size or subgroup size Binomial = if a unit it is either pass or fail , defectives or not Poisson = if a unit may contain several 缺陷
    • 241. The Xbar, R chart is useful for machine dominant 流程es This chart plots averages 和ranges Xbar, R charts enable 流程owners to evaluate changes in central tendency 和stability using subgroupingTypes of Control ChartsX, R
    • 242. Types of Control ChartsX, R
    • 243. Variables Control Charts: the Xbar ChartVariables Control Charts involve continuous variables, We use this chart when we are interested in the central tendency 和variability (dispersion) The X-bar Chart measures the central tendency of the variable over time. It uses the mean, or X-bar, from samples of size n. The center line of the chart is represented by the long-term average of the averages, or X-double bar The control limits bound the +3 sigma, or 99.7% confidence interval on the mean
    • 244. Variables Control Charts: The Range (R) ChartThe Range, or R Chart monitors the variability within subgroup over time The center line of the chart is represented by the long-term average of the ranges, or R-bar The control limits bound the +3 [sigma of the range] confidence interval on the range
    • 245. Xbar & R Control Chart LimitsThe Xbar & R Control Chart limits are: Range X bar A2, D3, D4 和d2 are constants based on statistical confidence intervals. These Xbar & R Control Chart constants (A2, D3, D4 和d2) have been tabulated for various sample sizes (see Appendix)
    • 246. Control Charts 和Specification LimitsControl charts are intended to monitor stability The purpose of the charts is for monitoring where the 流程is now or even if it is unable to meet specs Note that the control limits have absolutely no 关系hip to specification limits The control chart monitors means 和the variability of groups of parts; specifications involve individual values Specification limits should not even be placed on control charts* Capability assessment is to be performed separately--although stability is a prerequisite for declaring the 流程“capable.” *Individuals charts can be an exception
    • 247. Creating the Xbar & R Control ChartsUse two six sided dice to construct an X-bar-R Create at least 20 subgroups Record in the Chart 和determine the control limits Range 和X-bar Next to add on for Sub group 21 to 30 use a ten side dice To make the computation easy use n = 2
    • 248. (本页无文本内容)
    • 249. Creating the Xbar & R Control Charts (cont)If all points show only random variation: Use the calculated control limits for monitoring the next 75 or so subgroups If the 流程is still in control, recompute the control limits using all of the data (then keep them there) If patterns indicate special causes of variation: If an assignable cause is present: If the cause relates to isolated points, remove the point(s) 和recalculate If a broader pattern exists, fix the assignable cause 和take another set of data If no assignable cause is known, the 流程must be considered “Out of (Statistical) Control” If continuing evidence indicates the 流程is stable, conclude that the out of control condition was a false alarm Unless the pattern is due solely to being a batch phenomenon (can be handled separately), the 流程must be considered unstable, 和work must be done to find the root cause of the instability
    • 250. What is “Out of Control” ?A 流程that shows a non-random pattern is considered to be statistically out of control Be careful with the terminology, since the 流程may still be producing good parts. The non-random pattern only means that there is strong statistical evidence that the 流程has changed. Specific rules have been devised to assist in determining non-random patterns Control Chart zones 和normal distribution based probabilities for 3 Sigma X-bar Chart
    • 251. Shewhart Charts: Tests for Special CausesUCLLCLAABCCBUCLLCLAABCCBRule 1. One Point Beyond Zone ARule 2. Nine Points in a Row on One Side of the Center LineRule 3. Six Points in a Row Steadily Increasing or DecreasingRule 4. Fourteen Points in a Row Alternating Up 和DownFrom Journal of 质量 Technology, October, 1984UCLLCLAABCCBUCLLCLAABCCB
    • 252. Shewhart Charts: Tests for Special Causes UCLLCLAABCCBUCLLCLAABCCBRule 5. Two Out of Three Points in a Row in Zone A or BeyondRule 6. Four Out of Five Points in a Row in Zone B 和BeyondRule 7. Fifteen Points in a Row in Zone C (Above 和Below Centerline)Rule 8. Eight Points in a Row on Both Sides of Centerline with None in Zone CFrom Journal of 质量 Technology, October, 1984UCLLCLAABCCBUCLLCLAABCCB
    • 253. 20100ObservationNumberUCLLCL20100ObservationNumberUCLLCLSeasonal Trends: These could be caused by systematic changes like: temperature, operator fatigue, voltage or pressure fluctuations, etc.Mixtures: A clear indication is when most of the points are close to the control limits, 和almost no points near the center line. These are usually due to underestimating the 流程variability or “over-control” , where operators intervene with the 流程frequently trying to respond to inherent variability instead of assignable causes. This also occurs when controlling two 流程es on the same chart.Observing trend for Control Charts
    • 254. 20100ObservationNumberUCLLCL151050ObservationNumberUCLLCLChanges in 流程means: These are usually due to the introduction of new operators, machines, procedures, etc. Can also happen because of 流程改进. Upward/Downward Trends: Continuous trending on the same direction. These are usually caused by component wear among others. Observing Trend for Control Charts
    • 255. 20100ObservationNumberUCLLCL“White Space”: Points tend to cluster around the mean. Usually caused by: - incorrect control limits (overestimating 流程variability) - continuous 改进 paying-off. Recommend limits should be recalculated when this situation is found.Observing Trend for Control Charts
    • 256. Applying the the Rules for Special CausesTypically, only a subset of the 8 tests is applied to control chart interpretation Rules 1 和5 should be generally applied Sensitivity to smaller 流程shifts may benefit from applying Rule 2 和6 Consistent mean shifts may be best detected by Rule 4 和8 Rule 2 will uncover an 改进of the 流程s Rule 3 should be used if the 流程inherently drifts
    • 257. Reaction to Out-of-Control It is very important to investigate every signal of special-cause variation Without attention to the charts, those responsible will begin ignoring the signals It is also important to have a planned approach Step 1 Verify that data is correct 和has been entered correctly Step 2 Determine if a simple explanation of special cause is available (e.g. parts were dropped) Step 3 Verify 流程change by taking a bigger sample (>20).Use proper sample size to detect critical difference. Step 4 If there is a 流程change, assess the impact of the change to determine next courses of action : Line shut down Root cause investigation If you have a very capable 流程(Cpk > 1.5) may benefit from relaxed control limits, such that out-of-control signals correspond more to out-of-spec material
    • 258. 流程Capability Evaluation -Short TermCpk Calculation using X-bar-R chart a) Use to estimate  b) Use to estimate  using this ignore the shift between data set measure within group variability c) Use LSL 和USL to calculate Short term Cpk 流程Capability Evaluation - Long TermPpk Calculation using X-bar-R chart a) Use to estimate  b) Use to estimate  using this include the shift in all data measure total group variability c) Use LSL 和USL to calculate . Long term Ppk USLLSLUSLLSL
    • 259. Partitioning the sources of variabilityExercise For the previous x bar R chart accomplished in class find Long term  short term  shift Total Variability = Within Group Variability + Between Group Variability
    • 260. Summary of 流程Capability Measures for Long Term Vs Short TermShort Term ( ST )Long Term (LT)
    • 261. Shewhart control chart constants n= Subgroup size Constant is from ASTM Manual
    • 262. ExerciseLet fire the catapult , with a fix setting Each person fire 3 shots as a subgroup size of 3 an in x bar- R chart. Use software to draw the x bar-R chart 和evaluate 流程control Assume the specification limits to be +/- 5 inches Calculate Cpk Find Long term  short term  shift
    • 263. Xbar-R chart
    • 264. Individuals Moving Range Chart - waste tank readings Where n is the sub group size D3 , D4 & E2 are Shewhart constants R Bar Is the average of the moving ranges x Bar is the average of individual readings
    • 265. Now Let use the previous catapult data to construct a Individual Moving Average Chart Take the Data with only on the First Row Calculate the Individuals Chart limits 和range limits Exercise
    • 266. Individuals Moving Range Chart
    • 267. X bar 和S charts Where n is the sub group size A3 ,B3 & B4 , C4 are Shewhart constants s Is the 标准偏差 x Double Bar Is the average of sample x barWhen Subgroup n > then 10 标准偏差 is a better estimator that R . Thus we suggest that for N> 10 X bar - S chart is recommended. X bar - S chart is also recommended for n sizes varies form one time increment to the next X .For example the sample size from lot to lot can be 5 , 8 , 4 , 和10
    • 268. Xbar & s Chart
    • 269. Range vs 标准偏差For subgroups of size 2, 3 or 4, there is little difference in accuracy As subgroups exceed 4, the 标准偏差 becomes increasingly more accurate than the range, to the point that the range should not be used for subgroups greater than 10 Strategy: use the std dev approach except when... Manual calculation is required Operators (or others), without understanding of the 标准偏差, will be interpreting the charts The reporting 结构is allergic to Greek lettersHistorical Note: when Shewhart developed these charts in the 1920’s, there was no easy way to calculate the 标准偏差. Thus, the range approach became ingrained in SPC application.
    • 270. Choosing the SubgroupSeveral considerations must be made when choosing the subgroups The subgroup is “rational” The sample adequately reflects the 流程performance during the time period which it represents The sample facilitates a specific identification of probable sources of assignable causes The sample can be taken with a consistent 方法 和accurately measured The subgroup is of sufficient size to detect a change in the 流程that is considered significant The subgroup size must also not be too large, or else signals for 流程shifts considered insignificant will occur The frequency of sampling is often enough to avoid excess 产品ion of unacceptable material
    • 271. Choosing a Correct Sample SizeThe capability of the 流程will often 提供essential information for choosing a sample size Present capability 提供s a ppm value 和a Sigma level (assuming a normal distribution) An undesirable increase in ppm to a given value of ppm* can be translated into a Sigma Level* The difference between Sigma Level 和Sigma Level * represents the undesirable shift in mean--in number of 标准偏差 units. This becomes the “critical shift.” A confidence level can be assigned to the probability of detecting this shift The following chart can be used to determine the best sample size to detect the critical shift with the desired confidence
    • 272. Sample Size: Sensitivity with 1 GroupProbability of Detecting a Shift in Mean in the Next Subgroup for Various Control Chart Subgroups of Size nFrom “One Good Idea” by Lyle Dockendorf, 质量 Progress, October 1992n=20n=15n=10n=8n=5n=4n=3n=2
    • 273. Sample Size: Sensitivity with 1 GroupProbability of Detecting a Shift in Mean in the Next Subgroup for Various Control Chart Subgroups of Size nFrom “One Good Idea” by Lyle Dockendorf, 质量 Progress, October 1992n=20n=15n=10n=8n=5n=4n=3n=21.5sSample Size = 890% confidence of detecting
    • 274. Sample Size: Sensitivity in Next 3 GroupsProbability of Detecting a Shift in Mean in the Next 3 Subgroups Also Using the 2-Points-out-of-3-in-the-Warning-Zone Rule for Various Control Chart Subgroups of Size nFrom “One Good Idea” by Lyle Dockendorf, 质量 Progress, October 1992n=20n=15n=10n=8n=5n=4n=3n=2
    • 275. Frequency of Taking SubgroupsThe frequency with which subgroups are taken should reflect: Historical behavior of 流程 Frequent out-of-control occurrences require more frequent sampling If shift changes may cause a change, then each shift should be sampled The impact to 质量 of running in an out-of-control state The ease of implementing a sampling plan The cost of taking 和measuring samplesOften the frequency is set as a matter of convenience; e.g. once or twice each day or shift. Although this may work in practice, a method of calculating an appropriate frequency has been developed.
    • 276. Frequency CalculationDefine the following: S = Cost of taking 和measuring an entire sample L = Loss per unit (averaged over all units) when operating out of statistical control T = Time lag (in number of units) it takes to react F = Average number of units between out-of-control conditions if they were found 和fixed immediately (i.e. no time lag) p = Probability of discovering an out-of-control condition with one sample n = sampling frequency (in number of units produced) Then: 2p(F+T)S n = L(2-p)
    • 277. Exercise: Subgroup Size 和FrequencySuppose you have a 流程with Cpk=1.33. You want to guard against a 流程mean shift that will produce 3% defective with about 90% confidence. What subgroup size should you use? You know the following about your 流程 It runs 2 shifts/day 和5 days/week Each shift produces about 400 parts A scrapped part represents a loss of $5 Historically, the 流程has gone out of statistical control about once every 2 weeks It takes about 1 hour to react 和fix a 流程problem It costs $.50 to measure a sample unit Given that you use the subgroup size calculated in the first section, what frequency should you sample at?
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