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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 - PowerPoint PPT Presentation
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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 DataSubject Wk1 Wk2 Wk3 Wk4 Wk5
1 21 22 8 6 62 20 19 10 4 43 17 15 5 4 54 25 30 13 12 175 30 27 13 8 66 19 27 8 7 47 26 16 5 2 58 17 18 8 1 59 26 24 14 8 9
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 SquareA 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.
Error
Treatment
BlocksErrorTreatmentBlocks
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 = = 7893.69.• The total SS is then 11,060 ‑ 7893.69 =
3166.31.
NYYSSTOT
22 )(
SSweeks• From the marginal totals for week we
compute the SS for the main effect of Week as: (2012+ 1982+ 842+ 522+ 612) / 9 ‑ 7893.69 = 2449.20.
• Wj is the sum of scores for the jth week.
CMnW
SS jweeks
2
SSSubjects• From the subject totals, the SS for
subjects is: (632+ 572+ ...... + 812) / 5 ‑ 7893.69 = 486.71.
• S is the sum of score for one subject
CMnSSS i
subjects
2
SSerror• 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.• SSSubjects x Weeks = SStotal – SSsubjects – SS weeks
• = 3166.31 ‑ 486.71 ‑ 2449.2 = 230.4.
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 (Yi Yj) 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 (201 + 198)/18 = 22.17 for baseline, (84 + 52 + 61)/27 = 7.30 for treatment.
t
• The 7.20 is the MSE from the overall analysis.
• df = 32, from the overall analysis• p < .01
21.18
271
18120.7
30.717.22
11
jierror
ji
nnMS
MMt
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 = 10.02 * SQRT(2) = 14.16.• For Tukey, with r = 5 levels, and 32 df,
critical q.01 = 5.05
2tq
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 SSerror = SStotal – SSsubjects – SSweeks
Source DF Anova SS
Mean Square
F Value Pr > F
subject 8 486.7111 60.83888 8.45 <.0001week 4 2449.200 612.3000 85.04 <.0001
Source DF Sum of Squares
Mean Square
F Value Pr > F
Model 12 2935.91111 244.65925 33.98 <.0001Error 32 230.40000 7.200000 Corrected Total
44 3166.31111
Data in Multivariate Setupdata ache; input subject week1-week5; d23 = week2-week3; cards;1 21 22 8 6 62 20 19 10 4 43 17 15 5 4 54 25 30 13 12 175 30 27 13 8 66 19 27 8 7 4And 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.
Source DF Anova SS
Mean Square
F Value Pr > F
week 1 722.0000 722.0000 115.52 <.0001Error(week) 8 50.00000 6.250000
proc means mean t prt;var d23 week1-week5;
Proc Anova;
Model week1-week5 = / nouni;
Repeated week 5 profile / summary printe;Sphericity TestsVariables DF Mauchly's
CriterionChi-Square Pr > ChiSq
Orthogonal Components
9 0.2823546 8.1144619 0.5227
We retain the null that there is sphericity.
Univariate Tests of Hypotheses for Within Subject Effects
Source DF Anova SS
Mean Square
F Value
Pr > F Adj Pr > FG - G H - F
week 4 2449.2 612.30 85.04 <.0001 <.0001 <.0001
Error(week) 32 230.40 7.2000
Greenhouse-Geisser Epsilon 0.6845Huynh-Feldt Epsilon 1.0756
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 theH-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 AnalysisMANOVA Test Criteria and Exact F Statistics for the Hypothesis of no week Effect
Statistic Value F Value
Num DF
Den DF
Pr > F
Wilks' Lambda 0.01426 86.39 4 5 <.0001
Pillai's Trace 0.98573 86.39 4 5 <.0001Hotelling-Lawley Trace
69.1126 86.39 4 5 <.0001
Roy's Greatest Root
69.1126 86.39 4 5 <.0001
Strength of Effect• 2 = SSweeks / SStotal = 2449.2/3166.3 = .774• Alternatively, if we remove from the
denominator variance due to subject,
914.4.2302.2449
2.24492
ErrorConditions
Conditionspartial SSSS
SS
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=AS;TEST H=B E=BS;TEST H=AB E=ABS;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 AB E=BS(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 AB E=S(A B);TEST H=C AC BC ABC E=CS(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 AB E=BS(A);TEST H=C AC E=CS(A);TEST H=BC ABC E=BCS(A);MEANS A|B|C;
AxBxCxS All WithinCLASS A B C S; MODEL Y=A|B|C|S;TEST H=A E=AS;TEST H=B E=BS;TEST H=C E=CS;TEST H=AB E=ABS;TEST H=AC E=ACS;TEST H=BC E=BCS;TEST H=ABC E=ABCS;MEANS A|B|C;