Fu_ch12_Anova - Chapter 12 Multisample inference: Analysis...

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Unformatted text preview: Chapter 12 Multisample inference: Analysis of Variance Analysis EPI809/Spring 2008 EPI809/Spring 1 Learning Objectives 1.Describe Analysis of Variance (ANOVA) 2.Explain the Rationale of ANOVA 3.Compare Experimental Designs 4.Test the Equality of 2 or More Means Completely Randomized Design Randomized Block Design Factorial Design EPI809/Spring 2008 EPI809/Spring 2 Analysis of Variance A analysis of variance is a technique that partitions the total sum of squares of deviations of the observations about their mean into portions associated with independent variables in the experiment experiment and a portion associated with error and EPI809/Spring 2008 EPI809/Spring 3 Analysis of Variance The ANOVA table was previously The discussed in the context of regression models with quantitative independent variables, in this chapter the focus will be on nominal independent variables (factors) on EPI809/Spring 2008 EPI809/Spring 4 Analysis of Variance A factor refers to a categorical factor quantity under examination in an experiment as a possible cause of variation in the response variable. variation EPI809/Spring 2008 EPI809/Spring 5 Analysis of Variance Levels refer to the categories, measurements, or strata of a factor of interest in the experiment. interest EPI809/Spring 2008 EPI809/Spring 6 Types of Experimental Designs Experimental Designs Completely Randomized Randomized Block One-Way Anova Factorial Two-Way Anova EPI809/Spring 2008 EPI809/Spring 7 Completely Randomized Completely Design Design EPI809/Spring 2008 EPI809/Spring 8 Completely Randomized Design 1. Experimental Units (Subjects) Are Assigned Randomly to Treatments Assigned Subjects are Assumed Homogeneous 2. One Factor or Independent Variable 2 or More Treatment Levels or groups 3. Analyzed by One-Way ANOVA EPI809/Spring 2008 EPI809/Spring 9 One-Way ANOVA F-Test 1. Tests the Equality of 2 or More (p) Population Means Population 1. Variables One Nominal Independent Variable One Continuous Dependent Variable EPI809/Spring 2008 EPI809/Spring 10 One-Way ANOVA F-Test One-Way Assumptions Assumptions 1. Randomness & Independence of Errors 2. Normality Populations (for each condition) are Populations Normally Distributed Normally 3.Homogeneity of Variance Populations (for each condition) have Equal Populations Variances Variances EPI809/Spring 2008 EPI809/Spring 11 One-Way ANOVA F-Test One-Way Hypotheses Hypotheses H0: µ 1 = µ 2 = µ 3 = ... = µp All Population Means All are Equal are No Treatment Effect Ha: Not All µ j Are Equal Not At Least 1 Pop. Mean At is Different is Treatment Effect Ν Ο Τ µ 1 ≠ µ 2 ≠ .EPI809/Spring 2008 .. ≠ EPI809/Spring 12 One-Way ANOVA F-Test One-Way Hypotheses Hypotheses H0: µ 1 = µ 2 = µ 3 = ... = µ p All Population Means All are Equal are No Treatment Effect f(X) µ1 = µ2 = µ3 Ha: Not All µ j Are Equal Not At Least 1 Pop. Mean is At Different Different Treatment Effect Ν Ο Τ µ 1 = µ 2 = ... = µ p Or µ i ≠ µ j for some i, j. X f(X) EPI809/Spring 2008 EPI809/Spring µ1 = µ2 µ3 X 13 One-Way ANOVA Basic Idea Basic 1. Compares 2 Types of Variation to Test Equality of Means Equality 2. If Treatment Variation Is Significantly 2. Greater Than Random Variation then Means Are Not Equal Not 3.Variation Measures Are Obtained by 3.Variation ‘Partitioning’ Total Variation ‘Partitioning’ EPI809/Spring 2008 EPI809/Spring 14 One-Way ANOVA Partitions Total Variation Partitions EPI809/Spring 2008 EPI809/Spring 15 One-Way ANOVA Partitions Total Variation Partitions Total variation EPI809/Spring 2008 EPI809/Spring 16 One-Way ANOVA Partitions Total Variation Partitions Total variation Variation due to treatment EPI809/Spring 2008 EPI809/Spring 17 One-Way ANOVA Partitions Total Variation Partitions Total variation Variation due to treatment Variation due to random sampling EPI809/Spring 2008 EPI809/Spring 18 One-Way ANOVA Partitions Total Variation Partitions Total variation Variation due to treatment Variation due to random sampling Sum of Squares Among Sum of Squares Between Sum of Squares Treatment Among Groups Variation EPI809/Spring 2008 EPI809/Spring 19 One-Way ANOVA Partitions Total Variation Partitions Total variation Variation due to treatment Variation due to random sampling Sum of Squares Among Sum of Squares Between Sum of Squares Treatment Sum (SST) (SST) Among Groups Variation Sum of Squares Within Sum of Squares Error Sum (SSE) (SSE) Within Groups Variation EPI809/Spring 2008 EPI809/Spring 20 Total Variation SS ( Total ) = (Y11 − Y ) + (Y21 − Y ) + + (Yij − Y ) 2 2 2 Response, Y Y Group 1 Group 2 EPI809/Spring 2008 EPI809/Spring Group 3 21 Treatment Variation SST = n1 (Y1 − Y ) + n2 (Y2 − Y ) + + n p (Y p − Y ) 2 2 2 Response, Y Y3 Y Y1 Group 1 Y2 Group 2 EPI809/Spring 2008 EPI809/Spring Group 3 22 Random (Error) Variation SSE = ( Y11 − Y1 ) + ( Y21 − Y1 ) + + (Y pj − Y p ) 2 2 2 Response, Y Y3 Y1 Group 1 Y2 Group 2 EPI809/Spring 2008 EPI809/Spring Group 3 23 One-Way ANOVA F-Test One-Way Test Statistic Test 1. Test Statistic F = MST / MSE MST MSE STT / ( p − 1) = SSE / ( n − p ) • MST Is Mean Square for Treatment • MSE Is Mean Square for Error 2. Degrees of Freedom ν 1 = p -1 ν =n-p 2 • p = # Populations, Groups, or Levels • n = Total Sample Size EPI809/Spring 2008 EPI809/Spring 24 One-Way ANOVA One-Way Summary Table Summary Source of Degrees Sum of Variation of Squares Freedom Treatment p-1 SST Mean F Square (Variance) MST = MST SST/(p - 1) MSE Error n-p SSE MSE = SSE/(n - p) Total n-1 SS(Total) = SST+SSE EPI809/Spring 2008 EPI809/Spring 25 One-Way ANOVA F-Test One-Way Critical Value Critical If means are equal, If F = MST / MSE ≈ 1. MST MSE Only reject large F! Reject H0 Do Not Reject H0 0 α Fa ( p −1, n −p) F Always One-Tail! © 1984-1994 T/Maker Co. EPI809/Spring 2008 EPI809/Spring 26 One-Way ANOVA F-Test One-Way Example Example As a vet epidemiologist you As want to see if 3 food supplements have different mean milk yields. You assign 15 cows, 5 per food supplement. Question: At the .05 level, is .05 there a difference in mean mean yields? yields? Food1 Food2 25.40 23.40 26.31 21.80 24.10 23.50 23.74 22.75 25.10 21.60 EPI809/Spring 2008 EPI809/Spring Food3 20.00 22.20 19.75 20.60 20.40 27 One-Way ANOVA F-Test One-Way Solution Solution H0: µ 1 = µ 2 = µ 3 Ha: Not All Equal Not α = .05 .05 ν 1 = 2 ν 2 = 12 Critical Value(s): Test Statistic: Test MST 23.5820 F= = = 25.6 MSE .9211 α = .05 .05 0 3.89 F Decision: Decision: Reject at α = .05 Reject Conclusion: There Is Evidence Pop. There Means Are Different Means EPI809/Spring 2008 EPI809/Spring 28 Summary Table Solution Source of Degrees of Sum of Variation Freedom Squares Food 3-1=2 Mean F Square (Variance) 47.1640 23.5820 25.60 Error 15 - 3 = 12 11.0532 Total 15 - 1 = 14 58.2172 EPI809/Spring 2008 EPI809/Spring .9211 29 SAS CODES FOR ANOVA Data Anova; input group$ milk @@; cards; food1 25.40 food2 23.40 food1 26.31 food2 21.80 food1 24.10 food2 23.50 food1 23.74 food2 22.75 food1 25.10 food2 21.60 ; run; food3 20.00 food3 food3 22.20 food3 19.75 food3 20.60 food3 20.40 proc anova; /* or PROC GLM */ anova /* class group; model milk=group; run; EPI809/Spring 2008 EPI809/Spring 30 SAS OUTPUT - ANOVA Source DF Model 2 12 Sum of Squares Error Corrected Total Mean Square F Value Pr > F 47.16400000 23.58200000 25.60 <.0001 11.05320000 0.92110000 14 58.21720000 EPI809/Spring 2008 EPI809/Spring 31 Pair-wise comparisons Needed when the overall F test is rejected Needed Can be done without adjustment of type I error if Can other comparisons were planned in advance (least significant difference - LSD method) (least Type I error needs to be adjusted if other Type comparisons were not planned in advance (Bonferroni’s and scheffe’s methods) EPI809/Spring 2008 EPI809/Spring 32 Fisher’s Least Significant Fisher’s Difference (LSD) Test Difference To compare level 1 and level 2 To t = ( y1 − y2 ) 1 1 MSE + n n 1 2 Compare this to tα/2 = Upper-tailed value or - tα//2 Compare 2 lower-tailed from Student’s t-distribution for α/2 and lower /2 (n - p) degrees of freedom (n MSE = Mean square within from ANOVA table MSE n = Number of subjects p = Number of levels Number EPI809/Spring 2008 EPI809/Spring 33 Bonferroni’s method To compare level 1 and level 2 t = ( y1 − y2 ) 1 1 MSE + n n 1 2 Adjust the significance level α by taking the new significance level α* p α =α / 2 * EPI809/Spring 2008 EPI809/Spring 34 SAS CODES FOR multiple SAS comparisons comparisons proc anova; class group; class model milk=group; means group/ lsd bon; run; EPI809/Spring 2008 EPI809/Spring 35 SAS OUTPUT - LSD t Tests (LSD) for milk NOTE: This test controls the Type I comparisonwise error rate, not the experimentwise error rate. Alpha Error Degrees of Freedom Error Mean Square 0.05 12 0.9211 .975,12 Critical Value of t 2.17881 = t Least Significant Difference 1.3225 Means with the same letter are not significantly different. t Grouping A B C Mean N group 24.9300 5 food1 22.6100 5 food2 20.5900 EPI809/Spring 5 food3 EPI809/Spring 2008 36 SAS OUTPUT - Bonferroni Bonferroni (Dunn) t Tests for milk NOTE: This test controls the Type I experimentwise error rate Alpha Error Degrees of Freedom Error Mean Square Critical Value of t EPI809/Spring 2008 EPI809/Spring 0.05 12 0.9211 2.77947=t 37 Randomized Block Design EPI809/Spring 2008 EPI809/Spring 38 Randomized Block Design 1.Experimental Units (Subjects) Are Assigned Experimental Randomly within Blocks Randomly Blocks are Assumed Homogeneous 2.One Factor or Independent Variable of One Interest Interest 2 or More Treatment Levels or Classifications 3. One Blocking Factor EPI809/Spring 2008 EPI809/Spring 39 Randomized Block Design Factor Levels: (Treatments) Experimental Units Block 1 Block 2 Block 3 . . . Block b A, B, C, D Treatments are randomly assigned within blocks A C B C D A D B D B A C . . . . . . . . . . . . D C A B EPI809/Spring 2008 EPI809/Spring 40 Randomized Block F-Test 1.Tests the Equality of 2 or More (p) Population Means Population 2.Variables One Nominal Independent Variable One Nominal Blocking Variable One Continuous Dependent Variable EPI809/Spring 2008 EPI809/Spring 41 Randomized Block F-Test Randomized Assumptions Assumptions 1.Normality Probability Distribution of each BlockTreatment combination is Normal 2.Homogeneity of Variance Probability Distributions of all BlockTreatment combinations have Equal Treatment Variances Variances EPI809/Spring 2008 EPI809/Spring 42 Randomized Block F-Test Randomized Hypotheses Hypotheses H0: µ 1 = µ 2 = µ 3 = ... = µ p All Population Means are All Equal Equal No Treatment Effect Ha: Not All µ j Are Equal Not At Least 1 Pop. Mean is At Different Different Treatment Effect µ 1 ≠ µ 2 ≠ ... ≠ µ p Is wrong wrong EPI809/Spring 2008 EPI809/Spring 43 Randomized Block F-Test Randomized Hypotheses Hypotheses H0: µ 1 = µ 2 = ... = µ p All Population All Means are Equal Means No Treatment Effect Ha: Not All µ j Are Not Equal Equal f(X) µ1 = µ2 = µ3 X f(X) At Least 1 Pop. At Mean is Different Mean Treatment Effect µ 1 ≠ µ 2 ≠ ... ≠ µ p Is EPI809/Spring 2008 EPI809/Spring wrong wrong µ1 = µ2 µ3 X 44 The F Ratio for Randomized Block Designs Block SS=SSE+SSB+SST SST / ( p − 1) MST F= = MSE SSE / ( n − 1 − p + 1 − b + 1) = SST / ( p − 1) SSE / ( n − p − b + 1) Randomized Block F-Test Randomized Test Statistic Test 1. Test Statistic F = MST / MSE MST MSE • MST Is Mean Square for Treatment • MSE Is Mean Square for Error 2. Degrees of Freedom ν 1 = p -1 ν 2 = n – b – p +1 +1 • p = # Treatments, b = # Blocks, n = Total Sample Size Size EPI809/Spring 2008 EPI809/Spring 46 Randomized Block F-Test Randomized Critical Value Critical If means are equal, If F = MST / MSE ≈ 1. MST MSE Only reject large F! Reject H0 Do Not Reject H0 0 α Fa ( p −1, n −p) F Always One-Tail! © 1984-1994 T/Maker Co. EPI809/Spring 2008 EPI809/Spring 47 Randomized Block F-Test Randomized Example Example You wish to determine which of four brands of tires has You 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 .05 mean tread life? tread Tire Location Block Left Front Right Front Left Rear Right Rear Car 1 A: 42,000 C: 58,000 B: 38,000 D: 44,000 Car 2 B: 40,000 D: 48,000 A: 39,000 C: 50,000 Car 3 C: 48,000 D: 39,000 B: 36,000 A: 39,000 Car 4 A: 41,000 B: 38,000 D: 42,000 C: 43,000 Car 5 D: 51,000 A: 44,000 C: 52,000 B: 35,000 EPI809/Spring 2008 EPI809/Spring 48 Randomized Block F-Test Randomized Solution Solution H0: µ 1 = µ 2 = µ 3= µ 4 Ha: Not All Equal Not α = .05 .05 ν 1 = 3 ν 2 = 12 Critical Value(s): F = 11.9933 Decision: Decision: Reject at α = .05 Reject α = .05 .05 0 3.49 Test Statistic: Test Conclusion: There Is Evidence Pop. There Means Are Different Means F EPI809/Spring 2008 EPI809/Spring 49 SAS CODES FOR ANOVA data block; input Block$ trt$ resp @@; cards; Car1 A: 42000 Car1 C: 58000 Car1 Car2 B: 40000 Car2 D: 48000 Car2 Car3 C: 48000 Car3 D: 39000 Car3 Car4 A: 41000 Car4 B: 38000 Car4 Car5 D: 51000 Car5 A: 44000 Car5 ; run; B: 38000 Car1 D: 44000 A: 39000 Car2 C: 50000 B: 36000 Car3 A: 39000 D: 42000 Car4 C: 43000 C: 52000 Car5 B: 35000 proc anova; anova class trt block; model resp=trt block; Means trt /lsd bon; Means run; run EPI809/Spring 2008 EPI809/Spring 50 SAS OUTPUT - ANOVA Dependent Variable: resp Source DF Model Error Corrected Total 7 12 19 R-Square 0.784033 Source trt Block Sum of Squares Mean Square 544550000.0 150000000.0 694550000.0 Coeff Var 8.155788 77792857.1 12500000.0 Root MSE 3535.534 F Value 6.22 Pr > F 0.0030 resp Mean 43350.00 DF Anova SS Mean Square F Value Pr > F 3 4 449750000.0 94800000.0 149916666.7 23700000.0 11.99 1.90 0.0006 0.1759 EPI809/Spring 2008 EPI809/Spring 51 SAS OUTPUT - LSD Means with the same letter are not significantly different. t Grouping Mean N trt A 50200 5 C: B B CB C C 44800 5 D: 41000 5 A: 37400 5 B: EPI809/Spring 2008 EPI809/Spring 52 SAS OUTPUT - Bonferroni Means with the same letter are not significantly different. Bon Grouping Mean N trt A A A 50200 5 C: 44800 5 D: C C C 41000 5 A: 37400 5 B: B B B EPI809/Spring 2008 EPI809/Spring 53 Factorial Experiments EPI809/Spring 2008 EPI809/Spring 54 Factorial Design 1. Experimental Units (Subjects) Are Experimental Assigned Randomly to Treatments Assigned Subjects are Assumed Homogeneous 2. Two or More Factors or Independent Two Factors Variables Variables 3. Each Has 2 or More Treatments (Levels) Analyzed by Two-Way ANOVA EPI809/Spring 2008 EPI809/Spring 55 Advantages Advantages of Factorial Designs of 1.Saves Time & Effort e.g., Could Use Separate Completely e.g., Randomized Designs for Each Variable Randomized 2.Controls Confounding Effects by Putting Controls Other Variables into Model Other 3.Can Explore Interaction Between Variables EPI809/Spring 2008 EPI809/Spring 56 Two-Way ANOVA 1. Tests the Equality of 2 or More Tests Population Means When Several Independent Variables Are Used Independent 1. Same Results as Separate One-Way Same ANOVA on Each Variable ANOVA - But Interaction Can Be Tested EPI809/Spring 2008 EPI809/Spring 57 Two-Way ANOVA Two-Way Assumptions Assumptions 1.Normality Populations are Normally Distributed 2.Homogeneity of Variance Populations have Equal Variances 3.Independence of Errors Independent Random Samples are Drawn EPI809/Spring 2008 EPI809/Spring 58 Two-Way ANOVA Two-Way Data Table Data Factor A 1 Y111 1 Y112 Y211 2 Y212 : : Ya11 a Ya12 Factor B 2 ... Y121 ... Y122 ... Y221 ... Y222 ... : : Ya21 ... Ya22 ... EPI809/Spring 2008 EPI809/Spring b Y1b1 Y1b2 Y2b1 YX 2b2 2b2 : Yab1 Yab2 Observation k Yi j Level i Level j Level Level Factor Factor kB A 59 Two-Way ANOVA Two-Way Null Hypotheses Null 1.No Difference in Means Due to Factor A H0: µ 1. = µ 2. =... = µ a. 2.No Difference in Means Due to Factor B H0: µ .1 = µ .2 =... = µ .b 3.No Interaction of Factors A & B H0: ABij = 0 EPI809/Spring 2008 EPI809/Spring 60 Two-Way ANOVA Two-Way Total Variation Partitioning Total Variation SS(Total) Variation Due to Treatment A Treatment Variation Due to Variation Treatment B Treatment SSA SSB Variation Due to Variation Interaction Interaction Variation Due to Variation Random Sampling Random SS(AB) EPI809/Spring 2008 EPI809/Spring SSE 61 Two-Way ANOVA Two-Way Summary Table Summary Source of Degrees of Sum of Variation Freedom Squares Mean Square F A (Row) a-1 SS(A) MS(A) MS(A) MSE B (Column) b-1 SS(B) MS(B) MS(B) MSE AB (a-1)(b-1) (Interaction) SS(AB) Error n - ab SSE Total n-1 SS(Total) EPI809/Spring 2008 EPI809/Spring MS(AB) MS(AB) MSE MSE Same as Same Other Designs Designs 62 Interaction 1.Occurs When Effects of One Factor Occurs Vary According to Levels of Other Factor Factor 2.When Significant, Interpretation of Main When Effects (A & B) Is Complicated Effects 3.Can Be Detected In Data Table, Pattern of Cell Means in One In Row Differs From Another Row Row In Graph of Cell Means, Lines Cross EPI809/Spring 2008 63 EPI809/Spring Graphs of Interaction Effects of Gender (male or female) & dietary Effects group (sv, lv, nor) on systolic blood pressure group (sv, Interaction Average Response No Interaction male Average Response male female sv lv nor EPI809/Spring 2008 EPI809/Spring female sv lv nor 64 Two-Way ANOVA F-Test Two-Way Example Example Effect of diet (sv-strict vegetarians, lvllactovegetarians, nor-normal) and gender (female, actovegetarians, male) on systolic blood pressure. Question: Test for interaction and main effects at the .05 level. .05 EPI809/Spring 2008 EPI809/Spring 65 SAS CODES FOR ANOVA data factorial; input dietary$ sex$ sbp; cards; sv male 109.9 sv sv male 101.9 sv male 100.9 sv male 119.9 sv male 104.9 sv male 189.9 sv female 102.6 sv female 99 sv female 83 .6 sv female 99.6 sv female 102.6 sv female 112.6 lv male 116.5 lv male 118.5 lv male 119.5 lv male 110.5 lv male 115.5 lv male 105.2 nor male 128.3 nor male 129.3 nor male 126.3 nor male 127.3 nor male 126.3 nor male 125.3 nor female 119.1 nor female 119.2 nor female 115.6 nor female 119.9 nor female 119.8 nor female 119.7 ; run; EPI809/Spring 2008 EPI809/Spring 66 SAS CODES FOR ANOVA proc glm; glm class dietary sex; model sbp=dietary sex dietary*sex; run; proc glm; glm class dietary sex; model sbp=dietary sex; run; EPI809/Spring 2008 EPI809/Spring 67 SAS OUTPUT - ANOVA Dependent Variable: sbp Source Model Error Corrected Total R-Square 0.289916 Sum of DF Squares Mean Square 5 2627.399667 525.479933 24 6435.215000 268.133958 29 9062.614667 Coeff Var 14.08140 Root MSE 16.37480 Source dietary sex dietary*sex DF Type I SS 2 958.870500 1 1400.686992 2 267.842175 Source dietary sex dietary*sex DF 2 1 2 F Value Pr > F 1.96 0.1215 sbp Mean 116.2867 Mean Square 479.435250 1400.686992 133.921087 Type III SS Mean Square 1039.020874 519.510437 877.982292 877.982292 267.842175 133.921087 EPI809/Spring 2008 EPI809/Spring F Value Pr > F 1.79 0.1889 5.22 0.0314 0.50 0.6130 F Value Pr > F 1.94 0.1659 3.27 0.0829 0.50 0.6130 68 Linear Contrast Linear Contrast is a linear combination of the Linear means of populations Purpose: to test relationship among different group Purpose: means means L = ∑cjµ j with ∑c j =0 Example: 4 populations on treatments T1, T2, T3 and T4. Contrast T1 T2 T3 T4 relation to test L1 L2 1 1 0 -1/2 -1 -1/2 0 0 EPI809/Spring 2008 EPI809/Spring μ1 - μ3 = 0 μ1 – μ2/2 – μ3/2 = 0 69 T-test for Linear Contrast (LSD) Construct a t statistic involving k group means. Construct Degrees of freedom of t - test: df = n-k. test: n-k. k To test H0: L = ∑ c j µ j = 0 Construct L t= j =1 s 2 k c 2 j ∑n j =1 j Compare with critical value t1-α/2,, n-k. Reject H0 if |t| ≥ t1-α/2,, n-k. SAS uses contrast statement and performs an F – test df (1, n-k); Or estimate statement and perform a t-test df (n-k). EPI809/Spring 2008 EPI809/Spring 70 T-test for Linear Contrast (Scheffe) Construct multiple contrasts involving k group Construct means. Trying to search for significant contrast means. k To test H0: L = ∑ c j µ j = 0 Construct j =1 Compare with critical value. L t= s 2 k c 2 j ∑n j =1 j a = (k − 1) Fk −1,n − k ,1−α Reject H0 if |t| ≥ a EPI809/Spring 2008 EPI809/Spring 71 SAS Code for contrast testing pr oc gl m ; cl ass t r t bl m odel r esp=t M eans t r t / l cont r ast ' A cont r ast ' A cont r ast ' A cont r ast ' A l sm eans t r t / l sm eans t r t / l sm eans t r t / */ est i m e ‘ A at r un; ock; r t bl ock ; sd bon s c hef - B = 0' t r t - B/ 2 - C/ 2 - B/ 3 - C/ 3 +B- C- D st der r pdi f f st der r pdi f f st der r pdi f f - B' f e; 1 -1 0 0 ; = 0' t r t 1 - . - D/ 3 = 0' t r t = 0' t r t 1 1 ; adj us t =s c hef adj us t =bon; 5 -.5 0 ; 3 -1 -1 -1 ; -1 -1 ; f e; / * Sc hef f e' s t es t */ / * Bonef er oni ' s t es t t r t 1 - 1 0 0 0; EPI809/Spring 2008 EPI809/Spring 72 Regression representation of Anova Regression EPI809/Spring 2008 EPI809/Spring 73 Regression representation of Regression Anova yij = µ i + eij = µ + α i + eij One-way anova: One-way p ∑α i =1 i =0 Two-way anova: Two-way yijk = µij + eijk = µ + α i + β j + γ ij + eijk a ∑α i =1 b i = 0, ∑ β j = 0, j =1 b ∑γ j =1 ij = 0 for all i and a ∑γ i =1 ij = 0 for all j SAS uses a different constraint EPI809/Spring 2008 EPI809/Spring 74 Regression representation of Regression Anova One-way anova: Dummy variables of factor One-way with p levels with y = β 0 + β1 x1 + β 2 x2 + ... + β p −1 x p −1 + e 1 where xi = 0 if level i if otherwise This is the parameterization used by SAS EPI809/Spring 2008 EPI809/Spring 75 Conclusion: should be able to 1. Recognize the applications that uses ANOVA 2. Understand the logic of analysis of variance. 3. Be aware of several different analysis of variance designs and understand when to use each one. each 4. Perform a single factor hypothesis test using 4. analysis of variance manually and with the aid of SAS or any statistical software. SAS EPI809/Spring 2008 EPI809/Spring 76 Conclusion: should be able to 5. Conduct and interpret post-analysis of 5. variance pairwise comparisons procedures. variance 6. Recognize when randomized block 6. analysis of variance is useful and be able to perform the randomized block analysis. perform 7. Perform two factor analysis of variance 7. tests with replications using SAS and interpret the output. interpret EPI809/Spring 2008 EPI809/Spring 77 Key Terms Between-Sample Between-Sample Variation Variation Completely Completely Randomized Design Randomized Experiment-Wide Experiment-Wide Error Rate Error Factor Levels One-Way Analysis of Variance Total Variation Treatment Within-Sample Variation EPI809/Spring 2008 EPI809/Spring 78 ...
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