Correlated-Samples ANOVA The Univariate Approach

An ANOVA Factor Can Be Independent Samples –Between Subjects Correlated Samples –Within Subjects, Repeated Measures –Randomized Blocks, Split Plot Matched Pairs if k = 2

The Design DV = cumulative duration of headaches Factor 1 = Weeks Factor 2 = Subjects (crossed with weeks) The first two weeks represent a baseline period. The remaining three weeks are the treatment weeks. The treatment was designed to reduce headaches.

The Data SubjectWk1Wk2Wk3Wk4Wk

Crossed and Nested Factors Subjects is crossed with Weeks here – we have score for each subject at each level of Week. That is, we have a Weeks x Subjects ANOVA. In independent samples ANOVA subjects is nested within the other factor –If I knew the subject ID, I would know which treatment e got.

Order Effects Suppose the within-subjects effect was dose of drug given (0, 5, 10 mg) DV = score on reaction time task. All subject tested first at 0 mg, second at 5 mg, and thirdly at 10 mg Are observed differences due to dose of drug or the effect of order Practice effects and fatigue effects

Complete Counterbalancing There are k! possible orderings of the treatments. Run equal numbers of subjects in each of the possible orderings. Were k = 5, that would be 120 different orderings.

Asymmetrical Transfer We assume that the effect of A preceding B is the same as the effect of B preceding A. Accordingly, complete counterbalancing will cancel out any order effects If there is asymmetrical transfer, it will not.

Incomplete Counterbalancing Each treatment occurs once in each ordinal position. Latin Square A B C D E E A B C D D E A B C C D E A B B C D E A

Power If the correlations between conditions are positive and substantial, power will be greater than with the independent samples designs Even though error df will be reduced Because we are able to remove subject effects from the error term Decreasing the denominator of the F ratio.

Reducing Extraneous Variance Matched pairs, randomized blocks, split- plot. Repeated measures or within-subjects. Variance due to the blocking variable is removed from error variance.

Partitioning the SS The sum of all 5 x 9 = 45 squared scores is 11,060. The correction for the mean, CM, is (596) 2 / 45 = = The total SS is then 11,060 ‑ =

SS weeks From the marginal totals for week we compute the SS for the main effect of Week as: ( ) / 9 ‑ = W j is the sum of scores for the j th week.

SS Subjects From the subject totals, the SS for subjects is: ( ) / 5 ‑ = S is the sum of score for one subject

SS error We have only one score in each of the 5 weeks x 9 subjects = 45 cells. So the traditional within-cells error variance does not exist. The appropriate error term is the Subjects x Weeks Interaction. SS Subjects x Weeks = SS total – SS subjects – SS weeks = ‑ ‑ =

df, MS, F, p The df are computed as usual in a factorial ANOVA ‑‑ (s ‑ 1) = (9 ‑ 1) = 8 for Subjects, (w ‑ 1) = (5 ‑ 1) = 4 for Week, and 8 x 4 = 32 for the interaction. The F(4, 32) for the effect of Week is then (2449.2/4) / (230.4/32) = 612.3/7.2 = 85.04, p <.01.

Assumptions Normality Homogeneity of Variance Sphericity –For each (ij) pair of levels of the Factor –Compute (Y i  Y j ) for each subject –The standard deviation of these difference scores is constant – that is, you get the same SD regardless of which pair of levels you select.

Sphericity Test it with Mauchley’s criterion Correct for violation of sphericity by using a procedure that adjust downwards the df Or by using a procedure that does not assume sphericity.

Mixed Designs You may have one or more correlated ANOVA factors and one or more independent ANOVA factors

Multiple Comparisons You can employ any of the procedures that we earlier applied with independent samples ANOVA. Example: I want to compare the two baseline weeks with the three treatment weeks. The means are ( )/18 = for baseline, ( )/27 = 7.30 for treatment.

t The 7.20 is the MSE from the overall analysis. df = 32, from the overall analysis p <.01

Controlling  FW Compute –And use it for Tukey or related procedure Or apply a Bonferroni or Sidak procedure For example, Week 2 versus Week 3 t = (22 ‑ 9.33)/SQRT(7.2(1/9 + 1/9)) = 10.02, q = * SQRT(2) = For Tukey, with r = 5 levels, and 32 df, critical q.01 = 5.05

Heterogenity of Variance If suspected, use individual error terms for a posteriori comparisons –Error based only on the two levels being compared. –For Week 2 versus Week 3, t(8) = 10.75, q(8) = 15.2 –Notice the drop in df

SAS WS-ANOVA.sas Proc Anova; Class subject week; Model duration = subject week; SAS will use SS error = SS total – SS subjects – SS weeks

SourceDFAnova SS Mean Square F ValuePr > F subject <.0001 week <.0001 SourceDFSum of Squares Mean Square F ValuePr > F Model <.0001 Error Corrected Total

Data in Multivariate Setup data ache; input subject week1-week5; d23 = week2-week3; cards; And data for three more subjects

Week 2 versus Week 3 proc anova; model week2 week3 = / nouni; repeated week 2 / nom; The value of F here is just the square of the value of t, 10.75, reported on Slide 23, with an individual error term. SourceDFAnova SS Mean Square F ValuePr > F week <.0001 Error(week)

proc means mean t prt; var d23 week1-week5;

Proc Anova; Model week1-week5 = / nouni; Repeated week 5 profile / summary printe; Sphericity Tests VariablesDFMauchly's Criterion Chi-SquarePr > ChiSq Orthogonal Components We retain the null that there is sphericity.

Univariate Tests of Hypotheses for Within Subject Effects SourceDFAnova SS Mean Square F Value Pr > FAdj Pr > F G - GH - F week <.0001 Error(week) Greenhouse-Geisser Epsilon Huynh-Feldt Epsilon1.0756

Epsilon Used to correct for lack of sphericity Multiply both numerator and denominator df by epsilon. For example: Degrees of freedom were adjusted according to Greenhouse and Geisser to correct for violation of the assumption of sphericity. Duration of headches changed significantly across the weeks, F(2.7, 21.9) = 85.04, MSE = 7.2, p <.001.

Which Epsilon to Use? The G-G correction is more conservative (less power) than the H-F correction. If both the G-G and the H-F are near or above.75, it is probably best to use the H-F.

Profile Analysis Compares each level with the next level, using individual error. Look at the output. –Week 1 versus Week 2, p =.85 –Week 2 versus Week 3, p <.001 –Week 3 versus Week 4, p =.002 –Week 4 versus Week 5, p =.29

Multivariate Analysis MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no week Effect StatisticValueF Value Num DF Den DF Pr > F Wilks' Lambda <.0001 Pillai's Trace <.0001 Hotelling- Lawley Trace <.0001 Roy's Greatest Root <.0001

Strength of Effect  2 = SS weeks / SS total = / =.774 Alternatively, if we remove from the denominator variance due to subject,

Higher-Order Mixed or Repeated Univariate Models If the effect contains only between- subjects factors, the error term is Subjects(nested within one or more factors). For any effect that includes one or more within-subjects factors the error term is the interaction between Subjects and those one or more within- subjects factors.

AxBxS Two-Way Repeated Measures CLASS A B S; MODEL Y=A|B|S; TEST H=A E=A  S; TEST H=B E=B  S; TEST H=A  B E=A  B  S; MEANS A|B;

Ax(BxS) Mixed (B Repeated) CLASS A B S; MODEL Y=A|B|S(A); TEST H=A E=S(A); TEST H=B A  B E=B  S(A); MEANS A|B;

AxBx(CxS) Three-Way Mixed (C Repeated) CLASS A B C S; MODEL Y=A|B|C|S(A B); TEST H=A B A  B E=S(A B); TEST H=C A  C B  C A  B  C E=C  S(A B); MEANS A|B|C;

Ax(BxCxS) Mixed (B and C Repeated) CLASS A B C S; MODEL Y=A|B|C|S(A); TEST H=A E=S(A); TEST H=B A  B E=B  S(A); TEST H=C A  C E=C  S(A); TEST H=B  C A  B  C E=B  C  S(A); MEANS A|B|C;

AxBxCxS All Within CLASS A B C S; MODEL Y=A|B|C|S; TEST H=A E=A  S; TEST H=B E=B  S; TEST H=C E=C  S; TEST H=A  B E=A  B  S; TEST H=A  C E=A  C  S; TEST H=B  C E=B  C  S; TEST H=A  B  C E=A  B  C  S; MEANS A|B|C;