correlated-samples anova
DESCRIPTION
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 PresentationTRANSCRIPT
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Correlated-Samples ANOVA
The Univariate Approach
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An ANOVA Factor Can Be• Independent Samples
– Between Subjects• Correlated Samples
– Within Subjects, Repeated Measures– Randomized Blocks, Split Plot
• Matched Pairs if k = 2
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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.
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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
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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.
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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
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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.
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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.
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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
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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.
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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
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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 )(
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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
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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
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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.
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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.
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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.
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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.
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Mixed Designs• You may have one or more correlated
ANOVA factors and one or more independent ANOVA factors
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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.
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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
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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
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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
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SAS• WS-ANOVA.sas • Proc Anova;• Class subject week;• Model duration = subject week;• SAS will use SSerror = SStotal – SSsubjects – SSweeks
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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
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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
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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
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proc means mean t prt;var d23 week1-week5;
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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.
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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
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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.
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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.
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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
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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
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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
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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.
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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;
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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;
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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;
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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;
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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;