常用质量统计方法ANOVAppt课件.ppt

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1、ANOVA,Hypothesis Testing Methods,2,At the end of this module you will be able to:Explain the principles of multiple parameter testingDescribe the general method for multiple parameter testingPerform an Analysis of Variance for 1, 2 or 3 factors and interpret the results,Module Objectives,3,We never

2、know the true population parameters, but we can use sample statistics to estimate them.One type of analysis is called Analysis of Variance (ANOVA).Allows comparison of two or more process s or s .We can test statistically whether these samples represent a single population, or if the s or s are diff

3、erent.The OUTPUT variable (KPOV) is generally measured on a continuous scale (Yield, Temperature, Volts, % Impurities, etc.)The INPUT variables (KPIVs) are known as FACTORS. In ANOVA, the LEVELS of the FACTORS are treated as categorical in nature even though they may not be.When there is only one fa

4、ctor, the type of analysis used is called “One-Way ANOVA.” For 2 factors, the analysis is called “Two-Way ANOVA. And “n” factors entail “n-Way ANOVA.”,Characteristics About Multiple Parameter Testing,4,ANOVA Application,To study the effect of one or more factors on a response, each factor having two

5、 or more levels. ANOVA can be used todetermine the statistical significance of effectscalculate the components of varianceestimate the contribution to variation by each identified sourceestimate the underlying noise within the process,5,Are the levels independent (different m or s), or do they all c

6、ome from a distribution with the same mean and standard deviation,1,2,3,Ho: mpop1 = mpop2 = mpop3 = . . .Ha: at least two are different,Ho: spop1 = spop2 = spop3 = . . .Ha: at least two are different,The Basic ANOVA Question,6,Step 1: State the Practical Problem Step 2: Do the assumptions for the mo

7、del hold?Response means are independent and normally distributedPopulation variances are equal across all levels of the factorRun a homogeneity of variance analysis-by factor level-first !Step 3: State the Null and Alternate HypothesesStep 4: Construct the ANOVA TableStep 5: Do the assumptions for t

8、he errors hold (residual analysis)?Errors of the model are independent and normally distributedStep 6: Interpret the P-Value (or the F-statistic) for the factor effect P-Value 0.05, then REJECT HoOtherwise, operate as if the null hypothesis is trueStep 7: Calculate epsilon squared for the treatment

9、and error termsStep 8: Translate the statistical conclusion into process terms,General Method,7,Are the means independent and normally distributedRandomize runs during the experimentEnsure adequate sample sizesRun a normality test on the data by levelMinitab:Stat Basic Stats Normality TestPopulation

10、 variances are equal for each factor level (run a homogeneity of variance analysis first)For sHo: pop1 = pop2 = pop3 = pop4 = .Ha: at least two are different,Note: The assumption of equal variances generally holds, especially if your test is BALANCED (same # of observations in each level). If varian

11、ces are not equal, and a transformation does not succeed, then non-parametric tests are required.,Step 2: Do the Assumptions for the Model Hold?,8,Ho: 1 = 2 = 3 = 4 = 5Ha: At least one k is different,The Hypotheses in Graphical Form (One-Way) Level 12 3 4 5,Step 3: State the Hypotheses,9,Where:g = n

12、umber of subgroupsn = number of readings per subgroup,This analysis determines if thedifferences between the averageof the levels is greater than couldreasonably be expected from thevariation that occurs within eachlevel,Total variation,Between groupvariation,Within groupvariation,Exercise: ANOVATab

13、le Worksheet,Step 4: Construct the ANOVA Table,10,Where: g = number of subgroups n = number of readings per subgroup,One-Way Analysis of VarianceAnalysis of Variance for TimeSourceDF SS MS F POperator 3149.549.84.350.016Error 20229.211.5Total 23378.6,Whats important the probabilitythat the Operator

14、variation in meanscould have happened by chance.,Step 4: Construct the ANOVA Table,11,The measure of the overall within-mean variability equals the square root of the mean square error (MSerror), called the Root Mean Square Error, or RMSE.This number can often be compared to previously observed with

15、in-group standard deviations to verify that the results of the ANOVA are consistent with historical observations.An inconsistency usually indicates a source of variability (such as an interaction) that is not being accounted for in the ANOVA model.,RMSE: Root Mean Square Error,12,Step 5:Do the assum

16、ptions for the errors hold (residual analysis) ?Errors of the model are independent and normally distributedRandomize runs during the experimentEnsure adequate sample sizePlot histogram of error termsRun a normality check on error termsPlot error against run order (I-Chart)Plot error against model f

17、itStep 6:Interpret the P-Value (or the F-statistic) for the factor effectP-Value 0.05, then REJECT Ho.Otherwise, operate as if the null hypothesis is true.,ResidualAnalysis,Steps 5 - 6,13,Step 7: Calculate epsilon squared for the treatment and error termsStep 8:Translate the statistical conclusion i

18、nto process terms,Epsilon-Square is a controversial statistic. It provides agood guideline of the practical significance of the effect.Epsilon-Squared is a measure of the amount of variationof the Output that the Input of interest explains. Youshould always inspect the P-Value before taking action.,

19、Steps 7 - 8,14,Twenty-four animals receive one of four diets.The type of diet is the KPIV (factor of interest).Blood coagulation time is the KPOVDuring the experiment, diets were assigned randomly to animals. Blood samples taken and tested in random order. Why ?,File: Anova_Diets.mtw,DIET ADIET BDIE

20、T CDIET D626368566067666263717160596467616568636668646359,Example: Experimental Setup,15,State the practical problem:We, as diet researchers, would like to know if your diet can affect the blood coagulation time in your body.Its good practice to ALWAYS plot the data.Minitab:Graph PlotMinitab:Stat AN

21、OVA Main Effects Plot,Example: Step 1,16,Do the assumptions for the model hold?Response means are independent and normally distributedStat Basic Stats Normality TestPopulation variances are equal across all levels of the factorStat ANOVA Homogeneity of Variance,Example: Step 2,17,State the Null and

22、Alternate HypothesesHo: diet1 = diet2 = diet3 = diet4Ha: at least two diets differ from each otherInterpretation of the null hypothesis: the average bloodcoagulation time of each diet is the same (or) what you eat will NOT affect your blood coagulation time.Interpretation of the alternate hypothesis

23、: at least onediet will affect the average blood coagulation timedifferently than another (or) what type of diet you keepdoes affect blood coagulation time.,Example: Step 3,18,Construct the ANOVA Table (using Minitab):Stat ANOVA One-way .,Hint: Store Residuals & Fits for later use,Example: Step 4,19

24、,One-way Analysis of VarianceAnalysis of Variance for Coag_TimSource DF SS MS F PDiet_Num 3 228.00 76.00 13.57 0.000Error 20 112.00 5.60Total 23 340.00 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -+-+-+-+- 1 4 61.00 1.826 (-*-) 2 6 66.00 2.828 (-*-) 3 6 68.00 1.673 (-*-) 4 8

25、61.00 2.619 (-*-) -+-+-+-+-Pooled StDev = 2.366 59.5 63.0 66.5 70.0,Example: Step 4,20,RowDietFITS1CoagTime RESI1116162 1216160-1316163 2416159-2526663-3626667 1726671 5826664-2926665-11026666 01136868 01236866-21336871 31436867-11536868 01636868 01746156-51846162 1,The Fit is just theMean of each D

26、iet,The Residual is:(Raw Score - Diet Mean),What Are Fits & Residuals ?,21,Do the assumptions for the errors hold?Best way to check is through a “residual analysis”Stat Regression Residual Plots .Determine if residuals are normally distributedAscertain that the histogram of the residuals looks norma

27、lMake sure there are no trends in the residuals (its often best to graph these as a function of the time order in which the data was taken)The residuals should be evenly distributed about their expected (fitted) values,Example: Step 5,22,How normal arethe residuals ?,Histogram - bell curve ?Ignore f

28、or small datasets (30),Individual residuals -trends? Or outliers?,This graph investigateshow the Residuals behave across the experiment. This is probably the mostimportant graph, since itwill signal that somethingoutside the experimentmay be operating.Nonrandom patternsare warnings.,This graph inves

29、tigateswhether the mathematicalmodel fits equally for lowto high values of the Fits,Random about zerowithout trends?,Example: Step 5,23,Analysis of Variance on CoagTimeSource DF SS MS FpDiet 3228.0076.0013.570.000Error 20112.005.60Total 23340.00,Interpret the P-Value (or the F-statistic) for the fac

30、tor effectAssuming the residual assumptions are satisfied:If P-Value 0.05, then REJECT HoOtherwise, operate as if null hypothesis is true,When group sizes are equal,If P is less than 5% thenat least one group meanis different. In this case,we reject the hypothesisthat all the group meansare equal. A

31、t least oneDiet mean is different.An F-test this large couldhappen by chance, but inless than one time out of2000 chances. Thiswould be like getting 11heads in a row from afair coin.,F-test is close to 1.00when group meansare similar. In thiscase, The F-test ismuch greater.,Example: Step 6,24,Calcul

32、ate epsilon squared for the treatment and error termsTranslate the statistical conclusion to process termsWe have found that diet type DOES affect blood coagulation time. In particular if you would like to increase blood coagulation time, our laboratory recommends diets 2 or 3. If the goal is to dec

33、rease blood coagulation, then you should try diets 1 or 4.,67% of the Variation in Coag. Timecan be explained by Diet.,Example: Step 7 - 8,25,A quality control engineer conducted an experiment to investigate the effect of experience on an assembly line in terms of the average time required to comple

34、te an assembly task. If experience is found to be a factor, a training program will be planned for newer employees. The engineer randomly selected four employees from groups who had completed 1, 2, 3, and 4 years of work experience, respectively. He set up the experiment with years of experience as

35、treatments. The resulting data are given in HypoHW.mtw (C8-C12). A) Write the appropriate statistical model for this experiment. B) Test for any significant differences among years of experience for average assembly time ( a = 0.05).,Exercise: One-Way ANOVA Problem,26,Often, there are two variables

36、(factors) that we are interested in. These variables may have multiple levels. We may be interested to know if these levels contribute to additional variability.Furthermore, there may be an interaction between these variables, such that the effects of the different levels are not simply additive.Two

37、-Way ANOVA can provide the statistical analysis for these situations.,Two-Way ANOVA,27,Multi-Vari Example,Is there an operator variability as well as a day-to-day variability?,Day-to-Day vs Operator Variability,28,The simplest experiment that is useful involves 2 factorsOne example of a useful 2-fac

38、tor experiment is a Gage R&R study:Multiple parts measured by multiple operators on a single measurement systemMultiple parts measured by a single operator on multiple measurement systemsA single part measured by multiple operators on multiple measurement systems,2-Way ANOVA Application,29,Analysis

39、of Variance for Time Source DF SS MS F POperator 3 149.46 49.82 6.91 0.006Machine 2 12.25 6.12 0.85 0.452Interaction 6 130.42 21.74 3.02 0.049Error 12 86.50 7.21Total 23 378.63,Based on the probability values, at least one of the operators times is different than the others. It is also likely that a

40、t least one interaction term-or operator/machine combination-is different than the rest.,MiniTabsStat ANOVA Two-way.,Data is in ANOVA Example.xls,2-Way ANOVA Table,30,Machine 2 is low here,Machine 2 is high here,Arnies times are high,*,A Multi-Vari Look at 2-Way ANOVA,31,Fixed Effects Model:The anal

41、yst predetermines the treatment effects (or levels) prior to taking data, in this case, the conclusion cannot be extended to similar treatments that were not consideredRandom Effects Model:The treatments are randomly drawn from a much larger population of treatments, in this situation, the conclusio

42、n reached will be valid for the entire population of factor treatments. The majority of Gage R&R studies will be analyzed assuming that the random effects model is appropriate.,The option of “Two-way ANOVA” in Minitab assumes that the effects are fixed. The Gage Study options assume that they are ra

43、ndom. It is important to know the difference, because the analytical conclusions will be different.,2-Way ANOVA Analysis Considerations,32,Redo the exercise given for 1-Way ANOVA HypoHW.mtw (C8-C12), except this time, along with years of experience as one variable, consider “task” as another variabl

44、e.Is there a difference in the results?,Exercise: 2-Way ANOVA Problem,33,In many situations there are more than two variables (factors) that we are interested in. This happens particularly in designed experiments. These variables may have multiple levels. We may be interested in knowing if any of th

45、ese levels contribute to additional variability.Furthermore, there may be an interaction between some of these variables, such that the effects of the different levels are not simply additive.N-Way ANOVA can provide the statistical analysis for some of these situations.,N-Way ANOVA,34,As the complex

46、ity of the model increases, the options we have for analysis becomes more complex. One of the major considerations is whether the data is “balanced.” A balanced data set means that there are an equal number of observations for every combination of the levels of the factors. In these cases, Minitab p

47、rovides the option:Some unbalanced data sets can be analyzed with Minitabs:,Stat ANOVA Balanced ANOVA,Stat ANOVA General Linear Model,Balanced and Unbalanced ANOVA,35,Example,The engineer conducts a 23 (3 factors at 2 levels) experiment to understand the effects of stencil thickness, squeegee speed

48、and force on solder pad height.Verify if there are any significant main effects or interactions.The data is available in the Multi-Vari.MPW worksheet.,36,MiniTabsStat ANOVA Balanced ANOVA,Note interaction designations,Example,37,MiniTabs Session WindowAnalysis of Variance (Balanced Designs)Source DF

49、 SS MS F PStencil 1 6.984 6.984 2.14 0.156Speed 1 0.088 0.088 0.03 0.871Force 1 0.454 0.454 0.14 0.712Stencil*Speed 1 0.146 0.146 0.04 0.834Stencil*Force 1 0.001 0.001 0.00 0.987Speed*Force 1 0.256 0.256 0.08 0.782Error 25 81.534 3.261Total 31 89.462,Nothing appears to be significant. Can you guess

50、why?(Recall that you had viewed this data set earlier in the Multi-Vari module),N-Way ANOVA: Balanced Design,38,MiniTabsStat Quality Tools Multi-Vari Chart,Example,39,Example,40,Perform an N-Way ANOVA on the variable “Roughness” from the data set we dealt with in the Multi-Vari module: Multi-Vari Ro