Six sigma統(tǒng)計(jì)概念培訓(xùn)100頁(yè)

1、AbcotExample (cont)At x = 4.5,z = (x - m m) / = (4.5 - 5) / 1.581 = -0.32At x = 8.5,z = (x - m m) / = (8.5 - 5) / 1.581 = 2.21From standardised normal distribution table,at z = -0.32P(4.5) = 1 - 0.6255 = 0.3745at z = 2.21P(8.5) = 0.9864Hence, P(4.5 x 8.5) = 0.9864 - 0.3745 = 0.6119 (approx equal to

2、0.6123 by binomial distribution)AbcotIntroduction to SamplingWhat is population in statistic ?A population in statistic refers to all items that have been chosen for study.What is a sample in statistic ?A sample in statistic refers to a portion chosen from a population, by which the data obtain can

3、be used to infer on the actual performance of the population PopulationSample 2Sample 6Sample 8Sample 1Sample 3Sample 7Sample 4Sample 5AbcotSymbols for population 和和sampleFor population,Population mean = mPopulation size= NPopulation 標(biāo)準(zhǔn)偏差= For sample,Sample mean = X Sample size= nSample 標(biāo)準(zhǔn)偏差 = SSamp

4、ling distribution - a distribution of sample meansIf you take 10 samples out of the same populations, you will most likely end up with 10 different sample means 和sample 標(biāo)準(zhǔn)偏差s. A Sampling distribution describes the probability of all possible means of the samples taken from the same population.Popula

5、tion mean, m = 150Sampling distributionmean, XSample 2,mean X2Sample 1,mean X1Sample 3,mean X3Sample n,mean XnCollection of sample meansAbcotm m = 150Population distribution with = 25When sample size increases, the standard error (or the std deviation of sampling distribution) will get smaller.Sampl

6、e distribution with std error x much less than 25 (when n = 30)Sample distribution with std error x 25 (when n = 5)Sampling distribution (cont)Like all normal curve, the sampling distribution can be described by its mean, x 和its 標(biāo)準(zhǔn)偏差 x (which is also known as standard error of the mean). As such, th

7、e standard error measures the extent to which we expect the means from different samples to vary due to chance error in sampling 流程.Standard error = Population 標(biāo)準(zhǔn)偏差 / square root of sample size x = / n AbcotCentral Limit Theorem1.The mean x of the sampling distribution will approximately equal to th

8、e 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 t

9、he mean approaches normal as the sample size increases.It allow us to use sample statistics to make inferences about population parameters without knowing anything about the actual population distribution, other than what we can obtain from the sample.m = 150Population distributionx = 150Sampling di

10、stribution(n = 5)x = 150Sampling distribution(n = 20)AbcotExample of sampling distributionThe population distribution of annual income of engineers is skewed negatively. This distribution has a mean of $48,000, 和a 標(biāo)準(zhǔn)偏差 of $5000. If we draw a sample of 100 engineers, what is the probability that thei

11、r average annual income is $48700 和more.m = $48KPopulation distribution = $5000 x = $48KSampling distribution(n = 100)x = ?$48.7KSince population mean is equal to sampling distribution mean (i.e. central limit theorem), henceX = m m = 48000Sampling distribution mean = Standard error = x = / n = 5000

12、 / 100 = 5000 / 10= 500AbcotExample of sampling distribution (cont)x = $48KSampling distribution(n = 100)x = 500$48.7KTherefore,mean = 48000sigma = 500X = 50000Z = (48700 - 48000) / 500 = 700 / 500 = 1.4From the standardized normal distribution table, P(X $48700) = 0.9192Therefore;P(X $48700) = 1 -

13、0.9192 = 0.0808Thus, we have determined that it has only 8.08% chance for the average annual income of 100 engineers to be more than $48700.AbcotCentral Limit Theorem ExerciseBreak into 4 groups as below:Group 1: The population group.Group 2 to 4: The sample sub-groupThe population group will have 3

14、 of their members throwing a single dices 60 times each. A total of 180 throws will be recorded 和this data will be the population data.Each sample sub-group will have 3 of their members throwing 5 dices at one time, 和collect the sum 和average value of the particular throw. Each member is to conduct 2

15、0 throws 和obtain the sample mean of each throw. At the end of the exercise, a total of 180 sample means will be collected from the 3 sub-groups.From the arrived data, plot the histogram 和comment on the distribution of both the population 和the samplings.AbcotThe finite population multiplierPreviously

16、 we say that:n x error StandardThis equation however applies only when the population is infinite or relatively large when compare to the sample size. In the case when the population is finite or relatively small when compare to the sample size, standard error is calculated as:1n x NnNerror Standard

17、Finite Population MultiplierFinite population multiplier with respect to population 和和sample sizeRule of thumbsThe finite population multiplier need only be included if population size to sample size ratio is less than 25.AbcotConfidence Interval“Point estimates”A point estimate is a single number t

18、hat is used to estimate an unknown population parameter.“實(shí)例of point estimates”Sample mean, x as the estimator of the population mean, m m.nxxSample 標(biāo)準(zhǔn)偏差, S as the estimator of the population 標(biāo)準(zhǔn)偏差, .1-nx -x S2AbcotWhat does 95% confidence intervals means ?It defines 95% of the time, the average value

19、 of a random sampling will fall within a value range which is +/- 1.96 standard error from the sample mean.為什么1.96 standard error ?Referring to the standardized normal distribution table, when z = 1.96, the associated probability is 0.975 (or 97.5%) as below:But this is a 2 tails interval (i.e. +/-

20、1.96), hence we need to minus off another 2.5% from the other end, giving a total coverage of 95%.2.5% uncovered at one end+1.962.5% uncovered at another end-1.96AbcotEquation for computing confidence intervalsFor continuous data:size Sample n)confidence 99% (for 2.58 )confidence 95% (for 1.96 Zaver

21、age Sample X Error StdError StdZ - x limit confidence LowerError StdZ x limit confidence Upperunknown) is (if Sor nnFor discrete data:size Sample ndata continuous as same Zp)1- (i.e. failure of Proportion qsuccess of Proportion pnpq Error StdError StdZ - p limit confidence LowerError StdZ p limit co

22、nfidence UpperAbcotConfidence interval for continuous dataA large disk drive manufacturer needs an estimate of the mean life it can expect from the magnetic media by reciprocally switching its digital state at 1MHz frequency. The development team has determined previously that the variance of the po

23、pulation life is 36 months, 和had conducted a reliability testing for 100 samples, collecting data of its useful life as below:From the above data, what is the 95% confidence interval for the useful mean life of the magnetic media in a disk drive ? What does this mean ?AbcotConfidence interval for co

24、ntinuous data (cont)Applying the confidence intervals equation:Upper confidence limit = 27.75 + 1.96 (0.6) = 28.926Lower confidence limit = 27.75 1.96 (0.6) = 26.574From the sample data given,Sample size, n = 100Mean=(Sum of useful life) / 100=2775 / 100=27.75 monthsVariance, 2=36 months標(biāo)準(zhǔn)偏差, =6 mon

25、thsStandard error, x= 6 / 10=0.6As such, there is 95% confidence level that the average useful life of the magnetic media to fall between 26.574 和28.926 months.AbcotConfidence interval for discrete dataBreak into 4 or 5 teams, combined all the M&Ms in one team 和calculate the confidence intervals

26、 for each color type, using below table.Combine all the data from the 4 teams, what are the changes to the confidence intervals ?AbcotDetermine the sample size for confidence intervals (continuous data)mean from deviation allowable the to equal X - UCL Wherehave weequation, the grearrangin Therefore

27、X - UCLZZX - UCLZ x - UCL Error) Z(Std Error StdError StdZ - x limit confidence LowerError StdZ x limit confidence Upper2 n X - UCL unknown) is (if Sor nnnnGivenSample mean = 21S = 6What is the sample size required, such that there is a 95% confidence level that the average value will fall within +/

28、- 1.176 from its mean ?Ans:Applying the equation,n = (1.96 x 6) / 1.1762 = (10) 2 = 100The sample size must be 100.AbcotDetermine the sample size for confidence intervals (discrete data)mean from deviation allowable the to equal p) - (UCL Where:have weequation, the grearrangin Thereforep - UCLpqZ n

29、Zp - UCLnpqZp - UCL npqpUCLnpqZ p - UCL Error) Z(Stdnpq Error StdError StdZ - p limit confidence LowerError StdZ p limit confidence Upper222 Given,p = 0.4 (Proportion agree)q = 0.6 (Proportion not agree)What is the sample size required, such that there is a 99% confidence level that the proportion a

30、gree will fall within +/- 0.146 from p ?Ans:Applying the equation,n = (2.58) 2 (0.4)(0. 6) / (0.146)2 = 74.95 = 75The sample size must be 75.AbcotRevision: 1.00Date: June 2001-6-4-20246Abcot第二天: Tests of Hypotheses- Week 1 recap of Statistics Terminology - Introduction to Student T distribution- Exa

31、mple 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 p

32、opulation 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

33、 Proportions: F-Test-Break- Hypothesis Testing Exercise II (paper clip)- Hypothesis Testing Presentation- 第一天wrap upAbcot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 - Analysi

34、s 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 關(guān)系hip of a regression model- Fit

35、ting 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 (cont)- Presentation

36、of results- 第二天wrap upAbcotDay 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

37、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

38、 和response curve-Break- Model building example- Stepwise regression an approach to screen out factors- Day 3 wrap upAbcotDay 4: 設(shè)計(jì)of Experiment- Overview of Experimental Design- What is a designed experiment- Objective of experimental 設(shè)計(jì)和its capability in identifying the effect of factors- One facto

39、r at a time (OFAT) versus 設(shè)計(jì)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 sc

40、reening DOEs-Break- Points to note when conducting DOE- Screening DOE exercise using statapult- Interpretating the screening DOEs result-Lunch- Modelling DOE (Full factoria with interactions)- Interpreting interaction of factors- Pareto of factors significance- Graphical interpretation of DOE result

41、s- 某些rules of thumb in DOE- 實(shí)例of Modelling DOE 和its analysis-Break- Modelling DOE exercise with statapult- Target practice 和confirmation run- Day 4 wrap upAbcotDay 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 cont

42、rol 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- Implementin

43、g Xbar S Chart- 為什么Xbar S Chart ?- Introduction to Individual Moving Range Chart- Implementing Individual Moving Range Chart- 為什么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,

44、 c Charts, u Charts-Break- Attribute control chart exercise (paper clip)- Out of control not necessarily is bad- Day 5 wrap upAbcotRecap of Statistical TerminologyDistributions differs in locationDistributions differs in spreadDistributions differs in shapeNormal Distribution-6 -5 -4 -3 -2 -1 01 2 3

45、 4 5 6 - 99.9999998% - 99.73% - 95.45% -68.27%- 3 variation is called natural tolerance Area under a Normal DistributionAbcot流程流程capability potential, CpBased on the assumptions that :1.流程is normalNormal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 Lower Spec LimitLSLUpper Spec LimitUSLSpecification C

46、enter2.It is a 2-sided specification3.流程mean is centered to the device specificationSpread in specificationNatural toleranceCP =USL - LSL6 8 6 = 1.33Abcot流程流程Capability Index, Cpk1.Based on the assumption that the 流程is normal 和in control2. An index that compare the 流程center with specification center

47、Normal Distribution-6 -5 -4 -3 -2 -1 01 2 3 4 5 6 Lower Spec LimitLSLUpper Spec LimitUSLSpecification CenterTherefore when ,Cpk 20) Estimated 標(biāo)準(zhǔn)偏差標(biāo)準(zhǔn)偏差, R/d2 Population 標(biāo)準(zhǔn)偏差標(biāo)準(zhǔn)偏差, (when sample size, n 20) AbcotProbability TheoryProbability is the chance for an event to occur. Statistical dependence /

48、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 normal

49、ly 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

50、 always normally distributed.(Cp, Cpk must always based on the assumption that 流程流程is normal)AbcotInferential StatisticsThe 流程流程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

51、 to fluctuate) T-Test 和和F-Test(Determine if the underlying populations is significantly different in terms of the means 和和variations) Chi-Square Test of Independence(Test if the sample proportions are significantly different) Correlation 和和Regression(Determine if 關(guān)系關(guān)系hip between variables exists, 和和

52、generate model equation to predict the outcome of a single output variable)AbcotCentral Limit Theorem1.The 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 populati

53、on 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 = 150

54、Sampling distribution(n = 20)x = 150Sampling distribution(n = 30)m = 150Population distributionx = 150Sampling distribution(n = 5)Abcot某些某些take aways for sample size 和和sampling distribution For large sample size (i.e. n 30), the sampling distribution of x will approach normality regardless the actua

55、l 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 標(biāo)準(zhǔn)偏差

56、 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 標(biāo)準(zhǔn)偏差 dependence on small sample size. Contain two random quantities (x 和S), whereas normal distribution co

57、ntains 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).AbcotPercentiles of the t DistributionWhereby,df = Degree of freedom = n (sample size) 1Shaded area = one-tailed probability of occuren

58、cea = 1 Shaded areaApplicable when: Sample size 30 標(biāo)準(zhǔn)偏差 is unknown Population distribution is at least approximately normally distributed0.750.900.950.9750.990.9950.999511.00003.07776.313712.706231.821063.6559636.577620.81651.88562.92004.30276.96459.925031.599830.76491.63772.35343.18244.54075.840812

59、.924440.74071.53322.13182.77653.74694.60418.610150.72671.47592.01502.57063.36494.03216.868560.71761.43981.94322.44693.14273.70745.958770.71111.41491.89462.36462.99793.49955.408180.70641.39681.85952.30602.89653.35545.041490.70271.38301.83312.26222.82143.24984.7809100.69981.37221.81252.22812.76383.169

60、34.5868110.69741.36341.79592.20102.71813.10584.4369120.69551.35621.78232.17882.68103.05454.3178130.69381.35021.77092.16042.65033.01234.2209140.69241.34501.76132.14482.62452.97684.1403150.69121.34061.75312.13152.60252.94674.0728160.69011.33681.74592.11992.58352.92084.0149170.68921.33341.73962.10982.56692