Repeated Measures

 

I.  Why do repeated measures experiments?

 

A.  Power: subjects serve as their own controls; effects of treatments are measured against the mean effect of all the treatments combined on the subject.

            B.  Efficiency

 

            C.  Example:  SR(AF) X BF

                        N=30, a=3, b=3, n=10

            (3 groups of 10 subjects receiving each of 3 treatments)

    

     Source             E(MS)                    df   ___ Error Line

     1. A          nbs2A + bs2S(A)       a-1 = 2            2

     2. S(A)       bs2S(A)              a(n-1)= 27     

     3. B          nas2B + s2S(A)B         b-1 = 2             5

     4. A X B      ns2AB + s2S(A)B        a-1)(b-1)= 4       5

     5. S(A) X B   s2S(AB)               (b-1)a(n-1)=54

 

If different subjects were run in each treatment, one would need 3 X 30= 90 subjects.

 

SR(AF X BF) Factorial              N=90

 

Source              E(MS)               df        _    Error Line

1. A          nbs2A + s2S(AB)       a-1=2               4

     2. B          nas2B + s2S(AB)         b-1=2               4

     3. A X B      ns2AB + s2S(AB)        (a-1)b-1)=4         4       

4. S(A X B)   s2S(A)B               ab(n-1)=81   

 

For the test of B to reach significance in the repeated measures design one would need an F.05(2,54)=3.19.  For this test to reach significance in the factorial design, one would need an F.05(2,81)=3.12.  So, little power is lost at a great savings in subjects.  Also, note that if a smaller number of subjects are run in the factorial design, there is a substantially larger loss of power.  For example, if n=4 and so N=36, the test of B would need an F(2,27) value of 3.34 to reach significance at the .05 level.

 

II.  Model for repeated measures with 1 repeated factor.

 

 

 

Treatment

 

Subject

1

...

j

...

k

 

 

1

y11

 

y1j

 

y1k

 

.

 

 

 

 

 

 

 

i

yi1

 

yij

 

yik

 

.

 

 

 

 

 

 

 

N

yni

 

ynj

 

ynk

.

 

 

.1

 

.j

 

.k

..

 

 

SStotal= SS(yij-..)2

 


SSb.people= kS(i.- ..)2          

 


SSw.people= SS(yij-i.)2

 

SStotal=SSb.people+SSw.people

 

     SStreat= nS(.j-..)2               

         

     SSresidual = SS[(yij-..)-(i.- ..)-( .j-..)]2

 

     SSw.people=SStreat + SSresidual

 

     Model: SR X TF

 

     Source         E(MS)          df            Error Line

     1. S      ts2S                n-1          

     2. T      ns2T + s2ST          k-1             3

     3. S X T  s2ST                 (n-1)(k-1)

 

III. Assumptions of Univariate Repeated Measures

 

            A. ANOVA assumptions of:

                        1. Homogeneity of variances

2. Errors are independently and identically distributed as N(0, s2).

3. Subject effects are independent and distributed as N(0, s2).

            B. Additional assumption of sphericity.

1. Observations may covary, but the degree of covariance must remain the same across treatments( e.g. no carry-over effects).

2. If the covariances are heterogeneous, the error term will generally be an underestimate and F tests will generally be positively biased (cf. Box 1954).

 

IV. Comparisons between treatments.

 

A. Often the hypothesis of interest in a repeated measures design, like hypotheses in factorial designs, does not correspond to the tests for main effects.  In this case use a contrast.

                            MScontrast =     

where n=number of observations in the mean

 

1.  MScontrast can be tested against the standard error term for the corresponding effect  (e.g. MSST).

 

2.  MScontrast can also be tested against deviations from the expected trend.  This method treats all of the variance not attributed to the contrast(s) as error.  For example, to test a linear contrast by the second method, calculate:

SSdev.linear = SSST + (SST-SSlinear),

df(dev.linear)= dfST+dfT-1 = [(n-1)(k-1)]+(k-1)-1 = n(k-1)-1

MSdev.linear= SSdev.linear/df(dev.linear)

 

When the degrees of freedom are large (say over 30), then the two approaches do not differ.  Using this procedure tends to negatively bias F-tests because the denominators are often too large.

 

3.  One could also use a multivariate approach to testing the contrast.  This approach is equivalent to creating a new variable that is a linear combination of the dependent variables using the contrast weights ( y* = ci *yijk ) and then performing an F-test to test whether the mean of  this new variable is different from zero (this is what many computer programs report).

 

V.  Example (Winer p268)

                                                                                                                                               

Person              Drug 1              Drug 2              Drug 3              Drug 4              Mean  

 

1                      30                    28                    16                    34                    27

2                      14                    18                    10                    22                    16

3                      24                    20                    18                    30                    23

4                      38                    34                    20                    44                    34

5                      26                    28                    14                    30                    24.5

           

Mean               26.4                 25.6                 15.6                 32                    24.9

 

SSbetween S= 4(170.2)=680.8

SSwithin S = 811

SSdrugs = 5 (139.64)=698.2

SSSXD = 112.8

SStotal = 1491.8

 

Source               SS                    df                     MS                  F       

Between S        680.8                 4                    170.2  

Within S           811                  15                    54.07

  Drugs               698.2               3                    232.73             24.76

  S X D              112.8             12                     9.4

Total                1491.8             19

 

If we assume that drugs 1-4 are really 4 dosage levels of the same drug, it makes sense to test for trends.

 

                             Drug dosage            

 

                                                26.4                 25.6                 15.6                 32

Contrast                                                           Weights                                                                       

Linear                                       -3                     -1                     1                     3

Quadratic                                 1                     -1                     -1                     1

Cubic                                       -1                     3                     -3                     1

 

Tests against SXD

                                                                                        F=MScontrast/MSSXD

MSlin= 5(6.8)2= 11.56                                                              1.23

                20                  

MSquad= 5(17.2)2 = 369.8                                                        39.34**

                        4

MScubic = 5(35.6)2 = 316.84                                                     33.7 **

                           20    

Total   = 698.2

 

Tests against deviations from trend:

 

SS dev.lin   =    (SSresidual + SSdrugs)- SSlin

            799.4   =112.8   +        698.2  - 11.56

SS dev.quad  = (SSresidual + SSdrugs)- SSlin-SSquad

            429.64  = 112.8       +  698.2    - 11.56-369.8

SS dev.cubic = (SSresidual + SSdrugs)- SSlin-SSquad-SScubic

                      = SSres=112.8

 

Flin = MSlin      =                 11.56         =  .20          df=(1,n(k-1)-1)=(1,14)               

        Msdev.lin                 799.44/14                  

 

Fquad = MSquad       =       369.8            = 11.19**   df=(1,n(k-1)-1-1)=(1,13)

             MSdev.quad           429.64/13

 

Fcubic= MScubic =            316.84        = 33.7**    df=(1,n(k-1)-1-1-1)=(1,12)

             MSdev.cubic         112.8/12

 

Multivariate approach

                                      SS             DF       MS             F                 P

 

      linear          11.560         1     11.560        3.074          0.154

        error             15.040     4        3.760

      quadratic    369.800       1    369.800     26.797          0.007

        error              55.200    4       13.800

      cubic           316.840      1    316.840     29.778          0.005

        error            42.560     4       10.640

 

VI.  Violations of sphericity

 

A. Assumptions

 

For the typical F ratio used in univariate repeated measures analyses to be exact, the data must demonstrate homogeneity of treatment differences; that is, given any two treatments i and j,

s2 (vi-vj) is constant.  If the covariance matrix of the data is symmetrical (i.e., has compound symmetry) then this assumption will be met.  For example, if the covariance matrix for a design with three treatments were to have compound symmetry, it would have the form:

 

                T1       T2       T3      

T1            s2        ps2       ps2

T2            ps2       s2       ps2

T3            ps2       ps2       s2

 

where p=constant.  Although compound symmetry is not necessary for the assumptions of a repeated measures analysis F-test to be met, it is sufficient.  Compound symmetry is violated when one treatment covaries differently with some treatments than with other treatments.  This may occur when treatments are not presented in a counterbalanced order (e.g. a latin square is not used).  If the covariances are heterogeneous, the error term will generally be an underestimate and the F tests will generally be positively biased (cf. Box 1954).

 

B. Testing for violations

 

Compound symmetry can be tested by inspection and by several summary measures, such as Box's (1950) extension of Bartlett's test for homogeneity of variance.

 

C. Box (1950) test of compound symmetry

 

1. Take sample variance/covariance matrix S (pooled over levels of any grouping factors).

 


2. Construct S'=    var...cov...cov

                              .     .     .    .

                             cov...var...cov

                              .     .     .    .

                             cov...cov...var

 

where var=mean of s2i (main diagonal) of S

           cov=mean of s2ij (off diagonal elements) of S

 

3. Calculate:

            M= -(N-p)ln(|S|/|S'|)     | | = determinant of matrix

 

                        C=          q(q+1)2(2q-3)                     

                                     6(N-p)(q-1)(q2+q-4)

 

                        df= q2+q-4

                                      2

where N= number of subjects

             p= number of groups

             q= number of repeated measures treatments

 

4. Evaluate C2(df)=(1-C)M

If this is significant, then the matrix violates the compound symmetry assumption.

 

D.  Correcting for violations

 

1.  Use multivariate techniques: MANOVA, Hotelling T2.  These are extensions of ANOVA and t-test for multivariate data.

            2.  Set bounds on the true significance of the F test (Box 1954).

a. In a repeated measures design, the F ratio for a within-subjects effect is distributed as F[dfnumerator e,dfdenominator e] where 1/dfnumerator < e < 1. e measures the degree to which the covariance matrix deviates from compound symmetry.

            b. If e=1, then F ratio has normal df -- this yields a lower bound on the alpha level.

            c. When e=1/dfnumerator , the numerator of the F ratio has 1 degree of freedom -- this yields an upper bound on the alpha level (maximum heterogeneity of covariances).

d. The true significance of the effect will lie between that provided by these tests.

3.  Example for S X T design

a. F ratio s2T/s2ST distributed as  F[(k-1) e,(k-1)(n-1) e] where 1/(k-1) < e < 1; k is the number of treatments.

            b. If e=1, then F ratio has normal df -- this yields a lower bound on the alpha level.

c. If e=1/(k-1), then the F ratio has 1,n-1 degrees of freedom -- this yields an upper bound on the alpha level.

d. The true significance of the effect will lie between that provided by these tests.  The F(k-1,(k-1)(n-1)) level is too liberal, while the F(1,n-1) level is too conservative.

4.  Estimate the true significance of the F-test (estimate e).        

            a. Box's (1954) e^ aka Greenhouse-Geisser F

e^ =                  

              

where sij= is the i,j element in the sample variance/covariance matrix.

 sii= mean of variances in sample (main diagonal in matrix)

             s..= mean of all variances and covariances in sample

 si.= mean of all variances and covariances with variable i (row of matrix)

 k = number of treatments

 n = number of subjects

 

This procedure tends to underestimate the true e and hence to underestimate the significance of a result, but not by as much as the lower bound approximation of 1/k-1 above.

 

b. Huynh-Feldt e~

 

e~ =                             n(k-1) e^ -2    

                         (k-1)[n-1-(k-1) e^]

                       

In general, e~ tends to overestimate e.  In fact, e~ can be larger than 1.  When this occurs, e~ is set to 1. e~ will usually be larger than e^ so it will increase the power and the possibility of Type I errors in the analysis.