[信息与通信]第六章英文-简化版-方差分析参考资料-ch10-DesignofExperimentsandAnalysis.ppt

Chapter 10,Design of Experiments and Analysis of Variance,One-Way ANOVA F-Test,Types of Regression Models,Experimental,Designs,One-Way Anova,Completely Randomized,Randomized Block,Two-Way Anova,Factorial,One-Way ANOVA F-Test,1. Tests the Equality of 2 or More (p) Population Means 2. Variables One Nominal Scaled Independent Variable 2 or More (p) Treatment Levels or Classifications One Interval or Ratio Scaled Dependent Variable 3. Used to Analyze Completely Randomized Experimental Designs,One-Way ANOVA F-Test Assumptions,1. Randomness & Independence of Errors Independent Random Samples are Drawn for each condition 2. Normality Populations (for each condition) are Normally Distributed 3. Homogeneity of Variance Populations (for each condition) have Equal Variances,One-Way ANOVA F-Test Hypotheses,H0: 1 = 2 = 3 = . = p All Population Means are Equal No Treatment Effect Ha: Not All j Are Equal At Least 1 Pop. Mean is Different Treatment Effect NOT 1 2 . p,One-Way ANOVA F-Test Hypotheses,H0: 1 = 2 = 3 = . = p All Population Means are Equal No Treatment Effect Ha: Not All j Are Equal At Least 1 Pop. Mean is Different Treatment Effect NOT 1 2 . p,X,f(X),1,=,2,=,3,X,f(X),1,=,2,3,Why Variances?,Observe one sample from each treatment group Their means may be slightly different How different is enough to conclude population means are different? Depends on variability within each population Higher variance in population higher variance in means Statistical tests are conducted by comparing variability between means to variability within each sample,Two Possible Experiment Outcomes,Same treatment variation Different random variation,A,Cant reject equality of means!,Reject equality of means!,Two More Possible Experiment Outcomes,Same treatment variation Different random variation,A,B,Different treatment variation Same random variation,Cant reject equality of means!,Reject,Reject,1. Compares 2 Types of Variation to Test Equality of Means 2. Comparison Basis Is Ratio of Variances 3. If Treatment Variation Is Significantly Greater Than Random Variation then Means Are Not Equal 4. Variation Measures Are Obtained by Partitioning Total Variation,One-Way ANOVA Basic Idea,One-Way ANOVA Partitions Total Variation,One-Way ANOVA Partitions Total Variation,Total variation,One-Way ANOVA Partitions Total Variation,Variation due to treatment,Total variation,One-Way ANOVA Partitions Total Variation,Variation due to treatment,Variation due to random sampling,Total variation,One-Way ANOVA Partitions Total Variation,Variation due to treatment,Variation due to random sampling,Total variation,Sum of Squares Among Sum of Squares Between Sum of Squares Treatment Among Groups Variation,One-Way ANOVA Partitions Total Variation,Variation due to treatment,Variation due to random sampling,Total variation,Sum of Squares Within Sum of Squares Error (SSE) Within Groups Variation,Sum of Squares Among Sum of Squares Between Sum of Squares Treatment (SST) Among Groups Variation,Total Variation,X,Group 1,Group 2,Group 3,Response, X,Treatment Variation,X,X3,X2,X1,Group 1,Group 2,Group 3,Response, X,Random (Error) Variation,X2,X1,X3,Group 1,Group 2,Group 3,Response, X,SS=SSE+SST,But,Thus, SS=SSE+SST,One-Way ANOVA F-Test Test Statistic,1. Test Statistic F = MST / MSE MST Is Mean Square for Treatment MSE Is Mean Square for Error 2. Degrees of Freedom 1 = p -1 2 = n - p p = # Populations, Groups, or Levels n = Total Sample Size,One-Way ANOVA Summary Table,Source of,Variation,Degrees,of,Freedom,Sum of,Squares,Mean,Square,(Variance),F,Treatment,p - 1,SST,MST =,SST/(p - 1),MST,MSE,Error,n - p,SSE,MSE =,SSE/(n - p),Total,n - 1,SS(Total) =,SST+SSE,The F distribution,Two parameters increasing either one decreases F-alpha (except for v23) I.e., the distribution gets smushed to the left See Section 9.5,F,v1,v2,(,),0,F,One-Way ANOVA F-Test Critical Value,If means are equal, F = MST / MSE 1. Only reject large F!,Always One-Tail!,F,a,p,n,p,(,),1,0,Reject H,0,Do Not,Reject H,0,F, 1984-1994 T/Maker Co.,One-Way ANOVA F-Test Example,As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the .05 level, is there a difference in mean filling times?,Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40,F,0,3.89,One-Way ANOVA F-Test Solution,H0: 1 = 2 = 3 Ha: Not All Equal = .05 1 = 2 2 = 12 Critical Value(s):,Test Statistic: Decision: Conclusion:,Reject at = .05,There Is Evidence Pop. Means Are Different, = .05,F,MST,MSE,23,5820,9211,25,6,.,.,.,Summary Table Solution,From Computer,Source of,Variation,Degrees of,Freedom,Sum of,Squares,Mean,Square,(Variance),F,Treatment,(Machines),3 - 1 = 2,47.1640,23.5820,25.60,Error,15 - 3 = 12,11.0532,.9211,Total,15 - 1 = 14,58.2172,Reminder: Assumptions for Equality of Means Test,Independent random samples from each population All population probabilities are normally distributed All populations have equal variances (Test starts with assumption of equal means as well, but that may be rejected as a result of the test),Exercise 10.26,| Summary of value condition | Mean Std. Dev. Freq. -+- 1 | 30.64 20.035438 50 2 | 26.214286 23.701946 42 3 | 15.12766 15.703249 47 -+- Total | 24.057554 20.878451 139,Exercise 10.26,Analysis of Variance Source SS df MS F Prob F - Between groups 6109.7141 2 3054.85705 7.69 0.0007 Within groups 54045.8255 136 397.395776 - Total 60155.5396 138 435.909707 Bartletts test for equal variances: chi2(2) = 7.1931 Probchi2 = 0.027,One-Way ANOVA F-Test Thinking Challenge,Youre a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods ( =.05)? M1 M2 M3 M4 10 11 13 18 9 16 8 23 5 9 9 25 Use the following table., 1984-1994 T/Maker Co.,Summary Table (Partially Completed),Source of,Variation,Degrees of,Freedom,Sum of,Squares,Mean,Square,(Variance),F,Treatment,(Methods),348,Error,80,Total,F,0,4.07,One-Way ANOVA F-Test Solution*,H0: 1 = 2 = 3 = 4 Ha: Not All Equal = .05 1 = 3 2 = 8 Critical Value(s):,Test Statistic: Decision: Conclusion:,Reject at = .05,There Is Evidence Pop. Means Are Different, = .05,F,MST,MSE,116,10,11,6,.,Summary Table Solution*,Source of,Variation,Degrees of,Freedom,Sum of,Squares,Mean,Square,(Variance),F,Treatment,(Methods),4 - 1 = 3,348,116,11.6,Error,12 - 4 = 8,80,10,Total,12 - 1 = 11,428,10.26: condition 1 vs. 2,Two-sample t test with equal variances - Group | Obs Mean Std. Err. Std. Dev. 95% Conf. Interval -+- 1 | 50 30.64 2.833439 20.03544 24.94599 36.33401 2 | 42 26.21429 3.65729 23.70195 18.82824 33.60033 -+- combined | 92 28.61957 2.270253 21.7755 24.10999 33.12914 -+- diff | 4.425714 4.559216 -4.631963 13.48339 - Degrees of freedom: 90 Ho: mean(1) - mean(2) = diff = 0 Ha: diff 0 t = 0.9707 t = 0.9707 t = 0.9707 P |t| = 0.3343 P t = 0.1671,10.26 condition 2 vs. 3,Two-sample t test with equal variances - Group | Obs Mean Std. Err. Std. Dev. 95% Conf. Interval -+- 2 | 42 26.21429 3.65729 23.70195 18.82824 33.60033 3 | 47 15.12766 2.290554 15.70325 10.51701 19.73831 -+- combined | 89 20.35955 2.176533 20.53337 16.03415 24.68495 -+- diff | 11.08663 4.220767 2.697394 19.47586 - Degrees of freedom: 87 Ho: mean(2) - mean(3) = diff = 0 Ha: diff 0 t = 2.6267 t = 2.6267 t = 2.6267 P |t| = 0.0102 P t = 0.0051,Multiple Comparisons Problem,PAt least one of p intervals fails to contain the true difference = 1 PAll c intervals contain the true differences = 1 (1-alpha)c alpha If comparing many pairs, need greater confidence for any one of them than you would for rejecting equality of any one pair,Multiple Comparisons Procedure,1. Tells Which Population Means Are Significantly Different Example: 1 = 2 3 2. Post Hoc Procedure Done After Rejection of Equal Means in ANOVA Output From Many Statistical computer Programs various versions (Tukey, Bonferroni, etc.),10.26 Multiple Comparisons,(Bonferroni) Row Mean-| Col Mean | 1 2 -+- 2 | -4.42571 | 0.872 | 3 | -15.5123 -11.0866 | 0.001 0.029,Randomized Block Design,Types of Regression Models,Experimental,Designs,One-Way Anova,Completely Randomized,Randomized Block,Two-Way Anova,Factorial,Randomized Block Design,1. Experimental Units (Subjects) Are Assigned Randomly to Blocks Blocks are Assumed Homogeneous 2. One Factor or Independent Variable of Interest 2 or More Treatment Levels or Classifications 3. One Blocking Factor,Randomized Block Design,Randomized Block F-Test,1. Tests the Equality of 2 or More (p) Population Means 2. Variables One Nominal Scaled Independent Variable 2 or More (p) Treatment Levels or Classifications One Nominal Scaled Blocking Variable One Interval or Ratio Scaled Dependent Variable 3. Used with Randomized Block Designs,Randomized Block F-Test Assumptions,1. Normality Probability Distribution of each Block-Treatment combination is Normal 2. Homogeneity of Variance Probability Distributions of all Block-Treatment combinations have Equal Variances,Randomized Block F-Test Hypotheses,H0: 1 = 2 = 3 = . = p All Population Means are Equal No Treatment Effect Ha: Not All j Are Equal At Least 1 Pop. Mean is Different Treatment Effect 1 2 . p Is Wrong,Randomized Block F-Test Hypotheses,H0: 1 = 2 = 3 = . = p All Population Means are Equal No Treatment Effect Ha: Not All j Are Equal At Least 1 Pop. Mean is Different Treatment Effect 1 2 . p Is Wrong,X,f(X),1,=,2,=,3,X,f(X),1,=,2,3,The F Ratio for Randomized Block Designs,SS=SSE+SSB+SST,Randomized Block F-Test Test Statistic,1. Test Statistic F = MST / MSE MST Is Mean Square for Treatment MSE Is Mean Square for Error 2. Degrees of Freedom 1 = p -1 2 = n b p +1 p = # Treatments, b = # Blocks, n = Total Sample Size,Randomized Block F-Test Critical Value,If means are equal, F = MST / MSE 1. Only reject large F!,Always One-Tail!,F,a,p,n,p,(,),1,0,Reject H,0,Do Not,Reject H,0,F, 1984-1994 T/Maker Co.,Randomized Block F-Test Example,You wish to determine which of four brands of tires has the longest tread life. You randomly assign one of each brand (A, B, C, and D) to a tire location on each of 5 cars. At the .05 level, is there a difference in mean tread life?,F,0,3.49,Randomized Block F-Test Solution,H0: 1 = 2 = 3= 4 Ha: Not All Equal = .05 1 = 3 2 = 12 Critical Value(s):,Test Statistic: Decision: Conclusion:,Reject at = .05,There Is Evidence Pop. Means Are Different, = .05,F = 11.9933,Exercise 10.47,What is the purpose of blocking on weeks in this study? c. Are the mean number of walkers different among the prompting conditions? d. Which pairwise means are significantly different? e. What assumptions are required for the analysis in c and d?,Factorial Experiments,Types of Regression Models,Experimental,Designs,One-Way Anova,Completely Randomized,Randomized Block,Two-Way Anova,Factorial,Factorial Design,1. Experimental Units (Subjects) Are Assigned Randomly to Treatments Subjects are Assumed Homogeneous 2. Two or More Factors or Independent Variables Each Has 2 or More Treatments (Levels) 3. Analyzed by Two-Way ANOVA,Factorial Design Example,Treatment,Factor 2 (Training Method),Factor,Levels,Level 1,Level 2,Level 3,Level 1,19 hr.,20 hr.,22 hr.,Factor,1,(High),11 hr.,17 hr.,31 hr.,(Motivation),Level 2,27 hr.,25 hr.,31 hr.,(Low),29 hr.,30 hr.,49 hr.,Advantages of Factorial Designs,1. Saves Time & Effort e.g., Could Use Separate Completely Randomized Designs for Each Variable 2. Controls Confounding Effects by Putting Other Variables into Model 3. Can Explore Interaction Between Variables,Two-Way ANOVA,Types of Regression Models,Experimental,Designs,One-Way Anova,Completely Randomized,Randomized Block,Two-Way Anova,Factorial,Two-Way ANOVA,1. Tests the Equality of 2 or More Population Means When Several Independent Variables Are Used 2. Same Results as Separate One-Way ANOVA on Each Variable But Interaction Can Be Tested 3. Used to Analyze Factorial Designs,Two-Way ANOVA Assumptions,1. Normality Populations are Normally Distributed 2. Homogeneity of Variance Populations have Equal Variances 3. Independence of Errors Independent Random Samples are Drawn,Two-Way ANOVA Data Table,Xijk,Level i Factor A,Level j Factor B,Observation k,Factor,Factor B,A,1,2,.,b,1,X,111,X,121,.,X,1b1,X,112,X,122,.,X,1b2,2,X,211,X,221,.,X,2b1,X,212,X,222,.,X,2b2,:,:,:,:,:,a,X,a11,X,a21,.,X,ab1,X,a12,X,a22,.,X,ab2,Two-Way ANOVA Null Hypotheses,1. No Dif