1. Repeated Measures ANOVA Prof. Wei Zhu Department of Applied Mathematics & Statistics Stony Brook University

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3. The One-way ANOVA we have just learnt can test the equality of several population means. It is an extension of the pooled variance t-test That is:  H 0 (null hypothesis) : µ 1 = µ 2 = µ 3 =…….. = µ k  H a (alternative hypothesis): At least one of means differs from the rest. Assumptions:  Unknown but equal population variances  Normal populations  Independent samples 3

4. The F statistic where x ij = the j th observation in the i th sample. 4

5. The ANOVA table SourceS.Sd.f,M.S.F Between Within The ANOVA table is a tool for displaying the computations for the F test. It is very important when the Between Sample variability is due to two or more factors. We reject the ANOVA null hypothesis of equal means if F > F k-1,N-k, α 5

6. The most distinct disadvantage to the analysis of variance (ANOVA) method is that it requires three assumptions to be made: ☼ The samples must be independent to each other. ☼ All variances from each data group, though unknown, must be equal.The normality assumption. Limitations of the one-way ANOVA: 6

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8. ☼ A repeated measures design is one in which at least one of the factors consists of repeated measurements on the same subjects or experimental units, under different conditions and/or at different time points. 8

9.  A repeated measures design often involves measuring subjects at different points in time – or subjects measured under different experimental conditions  It can be viewed as an extension of the paired-samples t-test (which involved only two related measures)  Thus, the measures—unlike in the “regular” ANOVA—are correlated, that is, the observations are not independent 9

10.  Example: Data collected in a sequence of (often) evenly spaced points in time – this is usually referred to as the ‘longitudinal data’  Example: Different treatments are assigned to the each experimental unit 10

11.  By collecting data from the same participants under repeated conditions, the individual differences can be eliminated or reduced as a source of between group differences.  Also, the sample size is not divided between conditions or groups and thus inferential testing becomes more powerful.  This design also proves to be economical when sample members are difficult to recruit. 11

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13.  As with any ANOVA, repeated measures ANOVA tests the equality of means. However, repeated measures ANOVA is used when all members of a random sample are measured under a number of different conditions or at different time points.  As the sample is exposed to each condition, the measurement of the dependent variable is repeated.  Using a standard ANOVA in this case is not appropriate because it fails to model the correlation between the repeated measures: the data violate the ANOVA assumption of independence. 13

14.  The simplest example of a repeated measures design is a paired samples t-test: Each subject is measured twice, for example, time 1 and time 2, on the same variable; or, each pair of matched participants are assigned to two treatment levels. If we observe participants at more than two time-points, then we need to conduct a repeated measures ANOVA. 14

15.  What we would like to do is to decompose the variability into ： (1) A random ‘subject’ effect (2) A fixed ‘treatment’ or ‘time’ effect  Treating the subject as a random effect has the added advantage that we can extend our conclusion to the population where these subjects were drawn from. 15

16. Y ij = μ j +S i +ε ij μ j = The fixed effect, j=1, ⋯,k S i = The random effect of subject i, i=1, ⋯,n ε ij = The random error independent of S i Furthermore, we set the total number of observations as: N = n*k 16

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19. We will reject the ANOVA null hypothesis if F k-1,(k-)(n-1) > α 19

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21. Assumptions – summary: For a repeated measures design, we start with the same assumptions as a paired samples t-test :  Participants are independent and randomly selected from the population  Normality  Then, very importantly, there are two approaches to repeated measures ANOVA depends on the assumption of the variance-covariance matrix: ☺the univariate approach (where compound symmetry is imposed), and ☺the multivariate approach (where no assumption is imposed) 21

22.  Example: Consider the following experiment: We have four drugs (1,2,3 and 4) that relieve pain. Each subject is given each of the four drugs. The subject’s pain tolerance is then measured. Enough time is allowed to pass between successive drug administrations so that we can be sure there is no residual effect from the previous drug. The null hypothesis is: Mean(1)=Mean(2)=Mean(3)=Mean(4) 22

23. SubjectDrug1Drug2Drug 3Drug 4 159611 271289 311121014 43858 Table: Pain tolerance under 4 different drugs 23

24. In the one-way analysis of variance without a repeated measure, we would have each subject receive only one of the four drugs. In this design, each subjects is measured under each of the drug conditions. This has several important advantages as enumerated before. 24

25.  Each subject acts as his own control. i.e. : drugs effects are calculated by recording deviations between each drug score and the average drug score for each subject.  The normal subject-to-subject variation can thus be removed from the error sum of squares. 25

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28. SAS code using the REPEATED Statement DATA REPEAT; INPUT SUBJ PAIN1-PAIN4; DATALINES; 1 5 9 6 11 2 7 12 8 9 3 11 12 10 14 4 3 8 5 8 ; PROC ANOVA DATA=REPEAT; TITLE 'using repeated statement'; MODEL PAIN1-PAIN4 = / NOUNI; REPEATED DRUG 4 (1 2 3 4) / printe; RUN; 28 ☺ Please see Quiz 3 solutions for detailed explanations of the SAS output.

29. SAS code using the REPEATED Statement Remark 1 : about the data set We need the data set in the form: SUBJ PAIN1 PAIN2 PAIN3 PAIN4 NOTICE that it does not have a DRUG variable 29

30. SAS code using the REPEATED Statement Remark 2 : about the REPEATED Statement The general form: REPEATED factor_name CONTRAST(n); To compute pairwise comparisons N is a number from 1 to k, with k being # levels of repeated factor; To get all pairwise contrasts, we need k-1 repeated statements 30

31. SAS code using the REPEATED Statement Remark 2 : about the REPEATED Statement In our example: PROC ANOVA DATA=REPEAT; TITLE 'using repeated statement'; MODEL PAIN1-PAIN4 = / NOUNI; REPEATED DRUG 4 CONTRAST(1) / SUMMARY; REPEATED DRUG 4 CONTRAST(2) / SUMMARY; REPEATED DRUG 4 CONTRAST(3) / SUMMARY; RUN; Request ANOVA tables for each contrast 31

32. SAS code using the REPEATED Statement Remark 3 : more explanation of the ANOVA procedure PROC ANOVA DATA=REPEAT; TITLE 'using repeated statement'; MODEL PAIN1-PAIN4 = / NOUNI; REPEATED DRUG 4 (1 2 3 4) / printe; RUN; No CLASS: our data set does not have an independent variable NOUNI: not to conduct a separate analysis for each of the four PAIN 4: the repeated factor “DRUG” has four levels; optional (1 2 3 4): the labels we want printed for each level of DRUG printe calls for the Machly’s Sphericity Test for compound symmetry 32

33. SAS code using the REPEATED Statement Remark 4 : Data conversion: the long form & the broad form DATA PAIN; INPUT SUBJ DRUG PAIN; DATALINES; 1 1 5 1 2 9 1 3 6 1 4 11 2 1 7 2 2 12 …… ; DATA PAIN; INPUT SUBJ @; DO DRUG = 1 to 4; INPUT PAIN @; OUTPUT; END; DATALINES; 1 5 9 6 11 2 7 12 8 9 3 11 12 10 14 4 3 8 5 8 ; DATA REPEAT; INPUT SUBJ PAIN1-PAIN4; DATALINES; 1 5 9 6 11 2 7 12 8 9 3 11 12 10 14 4 3 8 5 8 ; The following is the data format, the broad form, that You should master first. 33

34.  Generate data in long form from the broad form DATA PAIN; INPUT SUBJ @; DATALINES; DO DRUG = 1 to 4; INPUT PAIN @; OUTPUT; END; iterative loop To keep reading from the same line of data 1 5 9 6 11 2 7 12 8 9 3 11 12 10 14 4 3 8 5 8 ; The broad form! The long form! 34

35. SAS: The DO loop the general form: Do variable = start TO end BY increment; (SAS Statements) END; initial value ending value Default: 1 35

36. in our example: initial value: 1 ending value: 4 DO DRUG = 1 to 4; INPUT PAIN @; OUTPUT; END; to keep reading from the same line of data return to “DO” SAS: The DO loop 36

37. Repeated measure / Longitudinal data: broad form 37 id time1 time2 time3 time4 1 31 29 15 26 2 24 28 20 32 3 14 20 28 30 4 38 34 30 34 5 25 29 25 29 6 30 28 16 34 Hypothetical data from Twisk, chapter 3, page 26, table 3.4 Jos W. R. Twisk. Applied Longitudinal Data Analysis for Epidemiology: A Practical Guide. Cambridge University Press, 2003.

38. Repeated measure/ Longitudinal data: Long form 38 Hypothetical data from Twisk, chapter 3, page 26, table 3.4 id time score 1 1 31 1 2 29 1 3 15 1 4 26 2 1 24 2 2 28 2 3 20 2 4 32 3 1 14 3 2 20 3 3 28 3 4 30 id time score 4 1 38 4 2 34 4 3 30 4 4 34 5 1 25 5 2 29 5 3 25 5 4 29 6 1 30 6 2 28 6 3 16 6 4 34

39. Another way to convert data from broad form to long form in SAS: 39 data long; set broad; time=1; score=time1; output; time=2; score=time2; output; time=3; score=time3; output; time=4; score=time4; output; run; The long form is used to plot the profile plot in the same way as the usual ANOVA

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41. 18083 28586 38388 48294 58793 68498 Subject Control Treatment PREPOST Factor B: TIME Factor A: GROUP Repeated 41

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44. Total Variance df=N-1 Between subjects Within subjects Treatment df=a-1 Error due to subjects within treatment df=a(n-1) Time df=b-1 Treatment × time df =(a-1)×(b-1) Error or residual df =a×(n-1)×(b-1) a: # of treatment groups b: # of time points n: # of subjects per treatment N=a×b×n: total # of measurements 44

45. Sourced.f.SSMS Factor Aa-1SSAMSA = SSA/(a-1) Factor Bb-1SSBMSB = SSB/(b-1) AB interaction(a-1)(b-1)SSABMSAB = SSAB/(a-1)(b-1) Subjects (within A)a(n-1)SSWAMSWA = SSWA/a(n-1) Errora(n-1)(b-1)SSEMSE = SSE/a(n-1)(b-1) Totalnab-1SST 45

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47. Data prepost; Input subj group $ pretest postest; datalines; 1 c 80 83 2 c 85 86 3 c 83 88 4 t 82 94 5 t 87 93 6 t 84 98 ; run; proc anova data=prepost; title 'Two-way ANOVA with a Repeated Measure on One Factor'; class group; model pretest postest = group/ nouni ; repeated time 2 (0 1); run; 47

48. Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.13216314 26.27 1 4 0.0069 Pillai's Trace 0.86783686 26.27 1 4 0.0069 Hotelling-Lawley Trace 6.56640625 26.27 1 4 0.0069 Roy's Greatest Root 6.56640625 26.27 1 4 0.0069 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no time Effect MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no time*group Effect Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.32611465 8.27 1 4 0.0452 Pillai's Trace 0.67388535 8.27 1 4 0.0452 Hotelling-Lawley Trace 2.06640625 8.27 1 4 0.0452 Roy's Greatest Root 2.06640625 8.27 1 4 0.0452 48

49. Tests of Hypotheses for Between Subjects Effects Source DF Anova SS Mean Square F Value Pr > F group 1 90.75000000 90.75000000 11.84 0.0263 Error 4 30.66666667 7.66666667 Univariate Tests of Hypotheses for Within Subject Effects Source DF Anova SS Mean Square F Value Pr > F time 1 140.0833333 140.0833333 26.27 0.0069 time*group 1 44.0833333 44.0833333 8.27 0.0452 Error(time) 4 21.3333333 5.3333333 Level of -----------pretest----------- -----------postest----------- group N Mean Std Dev Mean Std Dev c 3 82.6666667 2.51661148 85.6666667 2.51661148 t 3 84.3333333 2.51661148 95.0000000 2.64575131 49

50. Another Example 50

51. Two-Factor ANOVA with Repeated Measures on One Factor 51

52. Hypothesis 52

53. ANOVA TABLE SourceDFSSMSF Factor A a-1SSASSA/(a-1) MSA/MSWA ～ F (a-1),n(a-1) Factor B b-1SSBSSB/(b-1) MSB/MSE ～ F (b-1),n(a- 1)(b-1) AB Interaction (a-1)(b-1)SSA B SSAB/(a-1)(b-1) MSAB/MSE ～ F (a-1)(b-1,n(a- 1)(b-1) Subjects within A (n-1)aSSW A SSWA/(n-1)a Error (n-1)a(b-1)SSESSE/((n-1)a(b-1)) Total nab-1SST 53

54. Example Six subjects, equally divided between two calibration method groups (calib), and each with 4 accuracy scores measured over 4 different dial shapes (repeated measures). 54

55. Data dials; Input subj calib shape_1 shape_2 shape_3 shape_4; datalines; 1 1 0 0 5 3 2 1 3 1 5 4 3 1 4 3 6 2 4 2 4 2 7 8 5 2 5 4 6 6 6 2 7 5 8 9 ; run; proc anova data=dials; title 'Two-way ANOVA with a Repeated Measure on One Factor'; class calib; model shape_1 shape_2 shape_3 shape_4 = calib/ nouni ; repeated shape 4; run; 55

56. MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no shape Effect H = Anova SSCP Matrix for shape E = Error SSCP Matrix S=1 M=0.5 N=0 StatisticValueF ValueNum DFDen DFPr > F Wilks' Lambda0.0252957325.69320.0377 Pillai's Trace0.9747042725.69320.0377 Hotelling-Lawley Trace38.5323660725.69320.0377 Roy's Greatest Root38.5323660725.69320.0377 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no shape*calib Effect H = Anova SSCP Matrix for shape*calib E = Error SSCP Matrix S=1 M=0.5 N=0 StatisticValueF ValueNum DFDen DFPr > F Wilks' Lambda0.167507953.31320.2404 Pillai's Trace0.832492053.31320.2404 Hotelling-Lawley Trace4.969866073.31320.2404 Roy's Greatest Root4.969866073.31320.2404 At α=0.05,we reject the null hypothesis and conclude that there is shape Effect At α=0.05,we cannot reject the null hypothesis and conclude that there is no shape*calib Effect Analysis of SAS Output 56

57. SourceDFAnova SSMean SquareF ValuePr > F calib151.04166667 11.890.0261 Error417.166666674.29166667 SourceDFAnova SSMean SquareF ValuePr > F Adj Pr > F G - GH - F shape347.4583333315.8194444412.800.00050.00990.0011 shape*calib37.458333332.486111112.010.16620.21520.1791 Error(shape)1214.833333331.23611111 Univariate Tests of Hypotheses for Within Subject Effects Tests of Hypotheses for Between Subjects Effects 57 At α=0.05,we reject the null hypothesis and conclude there is calibration Effect Greenhouse-Geisser Epsilon 0.4751 Huynh-Feldt Epsilon 0.8483 Since the G-G epsilon is rather small, we choose to use the conservative G-G correction.

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59. Two-Factor ANOVA with Repeated Measures on both Factors 59

60. SourceDFSSMSF Subjectsn-1SSSSSS/I-1MSS/MSE Factor A a-1SSASSA/(a-1) MSA/MSA*S ～ F(a-1),(n-1)(a-1) Factor B b-1SSBSSB/(b-1) MSB/MSB*S ～ F(b-1),(n-1)(b-1) AB Interaction (a-1)(b-1)SSABSSAB/((a- 1)(b-1)) MSAB/MSE ～ F(a-1)(b-1),(n-1)(a- 1)(b-1) A*Subjects (n-1)(a-1)SSA*SSSWA/((n- 1)a) SSA*S/MSE F(a-1)(n-1),(n-1)(a- 1)(b-1) B*Subjects (n-1)(b-1)SSB*SSSWB/((n- 1)b) SSA*S/MSE F(n-1)(b-1),(n-1)(a- 1)(b-1) Error(n-a)(a- 1)(b-1) SSESSE/((n- 1)(a-1)(b- 1)) Total nab-1SST ANOVA TABLE 60

61. Example  Three subjects, each with nine accuracy scores on all combinations of the three different dials and three different periods. Here subject is a random factor and both dial and period are fixed factors. 61

62. SAS Code Data Q2; Input subj Mins1-Mins9; Datalines; 1 45 53 60 40 52 57 28 37 46 2 35 41 50 30 37 47 28 32 41 3 60 65 75 58 54 70 40 47 50 ; ODS RTF STYLE=BarrettsBlue; Proc anova data=Q2; Model Mins1-Mins9=/nouni; Repeated period 3, dial 3/nom; Run; ods rtf close; 62

63. SAS Output SourceDFAnova SSMean SquareF ValuePr > F Adj Pr > F G - GH - F period21072.666667536.33333314.450.01480.05630.0394 Error(period)4148.44444437.111111 Greenhouse-Geisser Epsilon0.5364 Huynh-Feldt Epsilon0.6569 SourceDFAnova SSMean SquareF ValuePr > F Adj Pr > F G - GH - F dial2978.6666667489.333333350.910.00140.01690.0115 Error(dail)438.44444449.6111111 Greenhouse-Geisser Epsilon0.5227 Huynh-Feldt Epsilon0.5952 SourceDFAnova SSMean SquareF ValuePr > F Adj Pr > F G - GH - F period*dial48.666666672.166666670.300.87150.66030.7194 Error(period*dial)858.222222227.27777778 Greenhouse-Geisser Epsilon0.2827 Huynh-Feldt Epsilon0.4006 Univariate Tests of Hypotheses for Within Subject Effects At α=0.05,we cannot reject the null hypothesis and conclude the there is no period*dail Effect At α=0.05,we reject the null hypothesis and conclude that there is dial Effect At α=0.05,we can not reject the null hypothesis and conclude that the period Effect is marginal. 63 Since all epsilons < 0.75, we Should use the G-G correction.

64. Another Example 64

65. Example: A group of subjects is treated in the morning and afternoon of two different days. On one of the days, the subjects receive a strong sleeping aid the night before the experiment is to be conducted; on the other, a placebo. treat controldrug subjectreactionsubjectreaction Time A.m. 165170 272278 390397 P.M155160 264268 380385 65

66. data repeat; input subj react1-react4; datalines; 1 65 70 55 60 2 72 78 64 68 3 90 97 80 85 ; run; SAS Code proc anova data=repeat; model react1-react4= /nouni; repeated time 2, treat 2; run; 66

67. SAS Output 67 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no time Effect H = Anova SSCP Matrix for time E = Error SSCP Matrix S=1 M=-0.5 N=0 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.00662252 300.00 1 2 0.0033 Pillai's Trace 0.99337748 300.00 1 2 0.0033 Hotelling-Lawley Trace 150.00000000 300.00 1 2 0.0033 Roy's Greatest Root 150.00000000 300.00 1 2 0.0033 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no treat Effect H = Anova SSCP Matrix for treat E = Error SSCP Matrix S=1 M=-0.5 N=0 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.00775194 256.00 1 2 0.0039 Pillai's Trace 0.99224806 256.00 1 2 0.0039 Hotelling-Lawley Trace 128.00000000 256.00 1 2 0.0039 Roy's Greatest Root 128.00000000 256.00 1 2 0.0039

68. MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no time*treat Effect H = Anova SSCP Matrix for time*treat E = Error SSCP Matrix S=1 M=-0.5 N=0 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.33333333 4.00 1 2 0.1835 Pillai's Trace 0.66666667 4.00 1 2 0.1835 Hotelling-Lawley Trace 2.00000000 4.00 1 2 0.1835 Roy's Greatest Root 2.00000000 4.00 1 2 0.1835 Thus we conclude that there are significant main effects, but their interaction is not significant.

69. Interpretation  According to the observed p-values, except for the interactions, we can reject the hypotheses that time and treat are not significantly different.  The drug increase reaction time  Reaction time is longer in the morning compared to the afternoon  The interaction of treat and time is not significant 69

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71. Consider a marketing experiment:  Male and female subjects are offered one of three different brands of coffee.  Each brand is tasted twice; once after breakfast, the other time after dinner.  The preference of each brand is measured on a scale from 1 to 10 (1=lowest, 10=highest). 71

72. The experimental design is shown below: Three-Factor Experiment with a Repeated Measure on the last factor Meal: Repeated Measure Factor 72

73. SAS Program: 73

74. OUTPUT(Part 1/4): 74

75. OUTPUT(Part 2/4): 75

76. OUTPUT(Part 3/4): 76

77. OUTPUT(Part 4/4): 77

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80.  A group of high- and low-SES children is selected for the experiment. Their reading comprehension is tested each spring and fall for three consecutive years. A Diagram of the design is shown here: 80

81. 81  Notice that each subject is measured every spring and fall of each year so that the variables SEASON and YEAR are both repeated measures factors.

82. DATA READ; INPUT SUBJ SES $ READ1-READ6; DATALINES; 1 HIGH 61 50 60 55 59 62 2 HIGH 64 55 62 57 63 63 3 HIGH 59 49 58 52 60 58 4 HIGH 63 59 65 64 67 70 5 HIGH 62 51 61 56 60 63 6 LOW 57 42 56 46 54 50 7 LOW 61 47 58 48 59 55 8 LOW 55 40 55 46 57 52 9 LOW 59 44 61 50 63 60 10 LOW 58 44 56 49 55 49 ; run; PROC ANOVA DATA=READ; CLASS SES; MODEL READ1-READ6 = SES / NOUNI; REPEATED YEAR 3, SEASON 2; RUN; 82

83. ☺Since the REPEATED statement is confusing when we have more than one repeated factor, it is important for you to know how to determine the order of the factor names. Look at the REPEATED statement in this example: REPEATED YEAR 3, SEASON 2; This statement instructs the ANOVA procedure to choose the first level of YEAR(1), then loop through two levels of SEASON(SPRING FALL), then return to the next level of YEAR(2), followed by two levels of SEASON, etc. 83

84. 84 SAS Output MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no YEAR Effect H = Anova SSCP Matrix for YEAR E = Error SSCP Matrix S=1 M=0 N=2.5 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.19558215 14.40 2 7 0.0033 Pillai's Trace 0.80441785 14.40 2 7 0.0033 Hotelling-Lawley Trace 4.11294094 14.40 2 7 0.0033 Roy's Greatest Root 4.11294094 14.40 2 7 0.0033 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no YEAR*SES Effect H = Anova SSCP Matrix for YEAR*SES E = Error SSCP Matrix S=1 M=0 N=2.5 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.96176044 0.14 2 7 0.8724 Pillai's Trace 0.03823956 0.14 2 7 0.8724 Hotelling-Lawley Trace 0.03975996 0.14 2 7 0.8724 Roy's Greatest Root 0.03975996 0.14 2 7 0.8724

85. MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no SEASON Effect H = Anova SSCP Matrix for SEASON E = Error SSCP Matrix S=1 M=-0.5 N=3 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.03436198 224.82 1 8 <.0001 Pillai's Trace 0.96563802 224.82 1 8 <.0001 Hotelling-Lawley Trace 28.10192837 224.82 1 8 <.0001 Roy's Greatest Root 28.10192837 224.82 1 8 <.0001 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no SEASON*SES Effect H = Anova SSCP Matrix for SEASON*SES E = Error SSCP Matrix S=1 M=-0.5 N=3 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.17759295 37.05 1 8 0.0003 Pillai's Trace 0.82240705 37.05 1 8 0.0003 Hotelling-Lawley Trace 4.63085399 37.05 1 8 0.0003 Roy's Greatest Root 4.63085399 37.05 1 8 0.0003

86. MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no YEAR*SEASON Effect H = Anova SSCP Matrix for YEAR*SEASON E = Error SSCP Matrix S=1 M=0 N=2.5 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.03714436 90.73 2 7 <.0001 Pillai's Trace 0.96285564 90.73 2 7 <.0001 Hotelling-Lawley Trace 25.92198764 90.73 2 7 <.0001 Roy's Greatest Root 25.92198764 90.73 2 7 <.0001 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no YEAR*SEASON*SES Effect H = Anova SSCP Matrix for YEAR*SEASON*SES E = Error SSCP Matrix S=1 M=0 N=2.5 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.91603774 0.32 2 7 0.7357 Pillai's Trace 0.08396226 0.32 2 7 0.7357 Hotelling-Lawley Trace 0.09165808 0.32 2 7 0.7357 Roy's Greatest Root 0.09165808 0.32 2 7 0.7357

87. The ANOVA Procedure Repeated Measures Analysis of Variance Tests of Hypotheses for Between Subjects Effects Source DF Anova SS Mean Square F Value Pr > F SES 1 680.0666667 680.0666667 13.54 0.0062 Error 8 401.6666667 50.2083333

88. The ANOVA Procedure Repeated Measures Analysis of Variance Univariate Tests of Hypotheses for Within Subject Effects Adj Pr > F Source DF Anova SS Mean Square F Value Pr > F G - G H - F YEAR 2 252.0333333 126.0166667 26.91 <.0001 0.0002 <.0001 YEAR*SES 2 1.0333333 0.5166667 0.11 0.8962 0.8186 0.8700 Error(YEAR) 16 74.9333333 4.6833333 Greenhouse-Geisser Epsilon 0.6757 Huynh-Feldt Epsilon 0.8658 Source DF Anova SS Mean Square F Value Pr > F SEASON 1 680.0666667 680.0666667 224.82 <.0001 SEASON*SES 1 112.0666667 112.0666667 37.05 0.0003 Error(SEASON) 8 24.2000000 3.0250000 Adj Pr > F Source DF Anova SS Mean Square F Value Pr > F G - G H - F YEAR*SEASON 2 265.4333333 132.7166667 112.95 <.0001 <.0001 <.0001 YEAR*SEASON*SES 2 0.4333333 0.2166667 0.18 0.8333 0.7592 0.8168 Error(YEAR*SEASON) 16 18.8000000 1.1750000 Greenhouse-Geisser Epsilon 0.7073 Huynh-Feldt Epsilon 0.9221 In this case, one can use either the G-G or the H-F corrections.

89. We can use other SAS procedures such as:  “Proc GLM”  “Proc Mixed” Alternative SAS Procedures for Repeated Measures ANOVA 89

90. My thanks go to:  My students!  Cody & Smith & their little SAS book where several examples were selected from. Acknowledgement 90