repeated measures design (16.6 olrt)homepage.divms.uiowa.edu/~rdecook/stat5201/notes/4... · a...

Repeated measures design (16.6 OLRT)

STAT:5201

Week 15 - Lecture 1

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Comparison and contrast to a Type I split-plot

As mentioned earlier, some repeated measures designs may beanalyzed essentially the same as a Type I split-plot.

Example: (a modification of OLRT p. 438)

To study the effects of infant formula on growth, 30 infants arerecruited and 10 are randomly assigned to each of 3 distinct formulas.Weight is measured twice: at 1 week of age and again at 4 weeks.

An infant is nested in a formula.An infant is observed over both levels of time (1 week and 4 weeks).

Researchers are interested in a formula effect, a time effect, and/or aninteraction between formula and time.

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Consider the model that allows for correlation due to repeatedmeasures on the same subject:

Yijk = µ+ αi + Sj(i) + βk + (αβ)ik + εijk

Sj(i)iid∼ N(0, σ2S(a)), εijk

iid∼ N(0, σ2),with random effects independent from each other,

i = 1, 2, 3; j = 1, ..., 10; k = 1, 2.

The above model looks the same as the model fitted to Lenth’ssplit-plot laundry example (temp/fabric) we saw earlier.

However, this is not a ‘true’ split-plot because age was notrandomized; indeed, age cannot be randomized. A split-plot designhas two sizes of units and two randomizations. This experiment hastwo sizes of units, but only one randomization.

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Inclusion of random effects induces a correlation structure in the data.

This model coincides with a block diagonal correlation structure asbelow:

Because there were only 2 repeat measurements per subject, theanalysis (or correlation structure) is the same as in the split-plotexample.

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So, when is the analysis of a repeated measures different than asplit-plot? (HINT: more time points)

Let’s now consider the experiment on infant formula but takemeasurements at birth, 1 week, 2 weeks, 3 weeks, and 4 weeks (i.e.at 5 time points).

If we assume all 5 observations on a subject have the same correlationstrength, then the covariance of the 5 observations would like...

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The above covariance structure can be called compound symmetry(SAS) or sphericity (SPSS), depending on the software being used.

Compound symmetry is a block-diagonal variance-covariancestructure (independence between subjects).

A compound symmetry (CS) variance structure is assumed to be truefor a split-plot design (with 2 randomizations).

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Modeling autocorrelation

Does this CS structure seem most appropriate for the 5 time pointsdata collection?

Even if we have ‘time’ as a main effect in our model (linear,quadratic, etc.), we may expect that residuals that are closer in timeare more strongly correlated than those that are farther apart in time.

Can I incorporate this into my correlation structure?

NOTE: If there are three or more levels of a repeated measure factor, then the

assumption of sphericity (or compound symmetry) can be evaluated.7 / 51


Yes. We can include this autocorrelation in modeling the residuals.

Mixed Model Notation: Y = Xβ + Zu + e

where u is the vector of random effects (e.g. subject effects)and e is the vector of residuals

Var(u) = GVar(e) = R

}Var(Y ) = Var(Zu + e) = ZGZ ′ + R

For a true split-plot we assume Var(e) = σ2IN .Or that the residuals are independent.

Autocorrelation can be modeled with R.

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Regardless of how we specify the correlation within a subject, we stillhave correlation within the subject and independence betweensubjects, or a block-diagonal variance-covariance structure...

We are now allowing for a wider variety of structures for W besidescompound symmetry.

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PROC MIXED: RANDOM and REPEATED

Before we use SAS to code our autocorrelation, we will consider twodifferent ways to specify the split-plot compound symmetry structurein PROC MIXED...

The RANDOM statement specifies ‘G’, the covariance structure of therandom (subject) effects.

The REPEATED statement specifies ‘R’, the covariance structure ofthe residuals.

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These two fitted models are the same:

1 PROC MIXED data=dt;

class formula subject;

model weight=formula|time;

random subject(formula);

repeated / type=vc sub=subject(formula); /* default */

run;/* ‘vc’ says variance components or σ2I is used for the residuals,

‘sub’ specifies the ‘whole plot’ experimental unit (EU). */

2 PROC MIXED data=dt;


model weight=formula|time;

repeated / type=cs sub=subject(formula);

run;

/* ‘cs’ says compound symmetry for the residuals on each EU*/

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Recall the earlier example with secretary nested in number of periods,but observed over both time periods.

First way to state the model using RANDOM and G :

proc mixed data=data;

class Periods Secretary TimeDay;

model NumSorted=Periods TimeDay TimeDay*Periods/

ddfm=satterth residual;

random Secretary(Periods);

run;

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OUTPUT FROM FIRST CODING:

Cov Parm Estimate

Secretary(Periods) 334.47

Residual 81.1333

Type 3 Tests of Fixed Effects

Num Den

Effect DF DF F Value Pr > F

Periods 2 12 9.04 0.0040

TimeDay 1 12 11.19 0.0058

Periods*TimeDay 2 12 6.28 0.0136

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Second way to state the model using REPEATED and R:


class Periods Secretary TimeDay;

model NumSorted=Periods TimeDay TimeDay*Periods/


repeated /type=cs sub=Secretary(Periods);

run;

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OUTPUT FROM SECOND CODING:

Cov Parm Subject Estimate

CS Secretary(Periods) 334.47

Residual 81.1333


Num Den


Periods 2 12 9.04 0.0040

TimeDay 1 12 11.19 0.0058

Periods*TimeDay 2 12 6.28 0.0136

⇒ The same model is fitted by both.15 / 51


For data collected over time at uniform points, the first orderautoregressive model AR(1) is a common choice.

This structure allows observations taken closer in time to be morestrongly correlated than those taken farther apart in time.

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Including autocorrelation using the REPEATED statement:



model weight = formula|time/


random subject(formula);

repeated /type=AR(1) sub=subject(formula);

run;

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Example: Repeated measures for sales

Example (Sales)

Advertising Sales ExampleA company is interested in comparing the success of two differentadvertising campaigns. It has 10 test markets, and 5 test markets will berandomly assigned to each campaign. The sales for each test market willbe recorded after 1 months, 2 months, and 3 months.

Response: SalesFactors: Campaign (1 or 2)

Time (1,2,3)Test Market (10 unique markets)

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Example (Sales)

THE DATA:

The SAS System 3

Test

Obs Market Campaign Time Sales

1 1 1 1 958

2 1 1 2 1047

3 1 1 3 933

4 2 1 1 1005

5 2 1 2 1122

6 2 1 3 986

. . . . .

. . . . .

. . . . .

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Example (Sales)

Plot of the observations: Sales vs. time

NOTE: This seems quadratic in time so we will set time as a continuousvariable. 20 / 51


Example (Sales)

1) SAS statements for specifying Compound Symmetry (CS)correlation structure (no time-based autocorrelation here):

We set Var(G ) and Var(e) to have variance-components structure(i.e. diagonal matrices). The overall variance-covariance structure of Y isa compound symmetry form (block diagonal). We will request the G andR matrices from SAS. This model assumes quadratic in time (a differentcurve for each campaign).

proc mixed data=sales plot(only)=residualpanel(conditional);

class TestMarket Campaign;

model Sales=Campaign|Time Campaign|Time*Time/

ddfm=satterth outp=outdata residual;

random Testmarket(Campaign)/ G;

repeated / type=vc sub=TestMarket(Campaign) R;

run;21 / 51


The above gives the same correlation structure as a split-plot analysisusing ‘sphericity’ or ‘compound symmetry’ correlation.

Example (Sales)

The Mixed Procedure

Model Information

Data Set WORK.SALES

Dependent Variable Sales

Covariance Structure Variance Components

Subject Effect TestMarket(Campaign)

Estimation Method REML

Residual Variance Method Parameter

Fixed Effects SE Method Model-Based

Degrees of Freedom Method Satterthwaite

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Example (Sales)

Class Level Information

Class Levels Values

TestMarket 10 1 2 3 4 5 6 7 8 9 10

Campaign 2 1 2

Dimensions

Covariance Parameters 2

Columns in X 9

Columns in Z 10

Subjects 1

Max Obs Per Subject 30

Covariance Parameter Estimates


TestMarket(Campaign) 76284

Residual TestMarket(Campaign) 357.97

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Example (Sales)

Requested G matrix shown below.

G = Var(u)

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Example (Sales)

Requested R matrix (per subject) shown below.

R = Var(ei ) {per subject i}

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Example (Sales)

Fit Statistics

-2 Res Log Likelihood 273.4

AIC (Smaller is Better) 277.4

AICC (Smaller is Better) 277.9

BIC (Smaller is Better) 278.0


Num Den


Campaign 1 9.42 0.48 0.5035

Time 1 16 155.76 <.0001

Time*Campaign 1 16 0.07 0.8007

Time*Time 1 16 173.07 <.0001

Time*Time*Campaign 1 16 0.01 0.9144

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Example (Sales)

1a) SAS statements for specifying Compound Symmetry (CS)correlation structure (no time-based autocorrelation here):

Same model, but only using REPEATED and R matrix to specify.




ddfm=satterth outp=outdata residual;

repeated / type=cs sub=TestMarket(Campaign) R;

run;

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Example (Sales)




Residual TestMarket(Campaign) 357.97

R = Var(ei ) {per subject i}, shows compound symmetry.

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Example (Sales)

Fit Statistics





⇒ This is the same model as the first fit.

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Example (Sales)

2) SAS statements for first order Autoregressive correlationstructure:Same fixed effects, include the random subject effect, but also include anAR(1) correlation structure for the residuals within a TestMarket, or forVar(ei ).





random TestMarket(Campaign)/ G;

repeated / type=ar(1) sub=TestMarket(Campaign) R;

run;

NOTE: The repeated statement is different here.30 / 51


Example (Sales)

The Mixed Procedure

Model Information

Data Set WORK.SALES

Dependent Variable Sales

Covariance Structures Variance Components,

Autoregressive

Subject Effect TestMarket(Campaign)

Estimation Method REML

Residual Variance Method Profile

Fixed Effects SE Method Model-Based

Degrees of Freedom Method Containment

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Example (Sales)

Class Level Information

Class Levels Values

TestMarket 10 1 2 3 4 5 6 7 8 9 10

Campaign 2 1 2

Dimensions

Covariance Parameters 3

Columns in X 9

Columns in Z 10

Subjects 1

Max Obs Per Subject 30

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Example (Sales)




AR(1) TestMarket(Campaign) -0.7119

Residual 273.26

Requested G matrix and R matrix (per subject) shown on next page.

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Example (Sales)

G = Var(u)

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Example (Sales)


AR(1) TestMarket(Campaign) -0.7119

Residual 273.26

R = Var(ei ) {per subject i}

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Example (Sales)

The W in the ’block diagonal’ structure for individual i :

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Example (Sales)

Fit Statistics






Num Den


Campaign 1 10.1 0.45 0.5154

Time 1 8.08 97.86 <.0001

Time*Campaign 1 8.08 0.04 0.8438

Time*Time 1 8 107.03 <.0001

Time*Time*Campaign 1 8 0.01 0.9337

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Comparing models (same fixed effects, differentcovariances) using fit statistics

When comparing models with different covariance structures but the samefixed effects...

If you have ‘nested’ covariance structures, you can use a likelihoodratio test...LRT = −2logλ = −2log L(θ̂0)

L(θ̂)= −2log(Likelihoodsmaller

Likelihoodlarger)

= −2(LLsmaller ) + 2(LLlarger )= −2(Res log likelihoodθ̂0

) − −2(Res log likelihoodθ̂) *on SAS output

Otherwise, you can comparing using fit statistics like AIC and BIC.

Example (Sales)

By looking at the AIC fit statistics (or the BIC) where smaller isbetter, we can see that we gained by including the AR(1) structurebecause the CS model had an AIC of 277.4 and the model withAR(1) had an AIC of 275.5.

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Comparing models (same fixed effects, differentcovariances) using fit statistics

Example (Sales)

Likelihood ratio test for H0 : ρ = 0 vs. H1 : ρ 6= 0 (nested models)

Model 1: -2 Res Log Likelihood = 273.3645Model 2: -2 Res Log Likelihood = 269.5389

−2logλ = −2log L(θ̂0)

L(θ̂)= −2log(L(θ̂0))−−2log(L(θ̂))

= 273.4− 269.5 = 3.8256−2logλ

.∼ χ1 under H0

p-value: P(χ1 > 3.8256) = 0.05047

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Using PROC GLIMMIX

This LRT can actually be requested directly if using PROC GLIMMIX withthe ‘covtest’ statement.

Example (Sales)

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Using PROC GLIMMIX

Example (Sales)

It appears that there is no significant Campaign effect, and Sales is quadratic in

time, and there is some evidence of first-order autocorrelation in the residuals.

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Other covariance structures

If there are only two time points, compound symmetry covers allpossible variance-covariance structures. If there are many repeatedmeasures, other structures should be considered.

Four repeated measurements per subject, could try an ‘unstructured’form as Type=UN in SAS, or a Toeplitz form.

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Other covariance structures

If the time pints in your repeated measures design are not uniformlyspaced among subjects, you can use the ‘spatial power’ covariancestructure where the subject-specific times between observations isutilized (dij is the euclidean distance between the i th and j th

observations on a subject).

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Repeated measures in a blocked design

Example (Pigeons and stimuli)

Consider a situation that looks a lot like a RCBD but is not becausethe treatments are done in sequential order, rather than randomized.

Let’s compare an incorrect RCBD analysis with a more appropriaterepeated measures analysis.

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Pigeon example (repeated measures)

We have data on 6 pigeons. Each pigeon was given a sequence of 8 stimuli, and a response was observedafter each. The stimuli were not randomized, making this a repeated-m,easures study (AKA longitudinaldata). The data are in a separate file with variable names in the first row.

data pigeon;infile "pigeon.dat" firstobs=2;input PIGEON stim Resp;

Here is an initial analysis, treating this like an ordinary block design.

proc mixed;class PIGEON stim;model resp = stim / ddfm = satterth outp = diags;random PIGEON;lsmeans stim / diff;

The Mixed ProcedureConvergence criteria met.

Covariance ParameterEstimates

Cov Parm EstimatePIGEON 153.00Residual 55.8097

Fit Statistics-2 Res Log Likelihood 304.4AIC (smaller is better) 308.4AICC (smaller is better) 308.7BIC (smaller is better) 308.0

Type 3 Tests of Fixed EffectsNum Den

Effect DF DF F Value Pr > Fstim 7 35 4.91 0.0006

Least Squares MeansStandard

Effect stim Estimate Error DF t Value Pr > |t|stim 1 37.9167 5.8992 8.41 6.43 0.0002stim 2 41.8667 5.8992 8.41 7.10 <.0001stim 3 32.9833 5.8992 8.41 5.59 0.0004stim 4 33.7500 5.8992 8.41 5.72 0.0004stim 5 35.9667 5.8992 8.41 6.10 0.0002stim 6 44.2500 5.8992 8.41 7.50 <.0001stim 7 45.5167 5.8992 8.41 7.72 <.0001stim 8 52.7167 5.8992 8.41 8.94 <.0001

Differences of Least Squares MeansStandard

Effect stim _stim Estimate Error DF t Value Pr > |t|stim 1 2 -3.9500 4.3131 35 -0.92 0.3660stim 1 3 4.9333 4.3131 35 1.14 0.2605stim 1 4 4.1667 4.3131 35 0.97 0.3407stim 1 5 1.9500 4.3131 35 0.45 0.6540stim 1 6 -6.3333 4.3131 35 -1.47 0.1509stim 1 7 -7.6000 4.3131 35 -1.76 0.0868stim 1 8 -14.8000 4.3131 35 -3.43 0.0016stim 2 3 8.8833 4.3131 35 2.06 0.0469stim 2 4 8.1167 4.3131 35 1.88 0.0682stim 2 5 5.9000 4.3131 35 1.37 0.1801stim 2 6 -2.3833 4.3131 35 -0.55 0.5841stim 2 7 -3.6500 4.3131 35 -0.85 0.4032stim 2 8 -10.8500 4.3131 35 -2.52 0.0166stim 3 4 -0.7667 4.3131 35 -0.18 0.8599

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< similar for rest >

Here, we point out the all pairwise comparisons between the 8 stimuli havethe same standard error of 4.3131.

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stim 3 5 -2.9833 4.3131 35 -0.69 0.4937stim 3 6 -11.2667 4.3131 35 -2.61 0.0132stim 3 7 -12.5333 4.3131 35 -2.91 0.0063stim 3 8 -19.7333 4.3131 35 -4.58 <.0001stim 4 5 -2.2167 4.3131 35 -0.51 0.6105stim 4 6 -10.5000 4.3131 35 -2.43 0.0202stim 4 7 -11.7667 4.3131 35 -2.73 0.0099stim 4 8 -18.9667 4.3131 35 -4.40 <.0001stim 5 6 -8.2833 4.3131 35 -1.92 0.0630stim 5 7 -9.5500 4.3131 35 -2.21 0.0334stim 5 8 -16.7500 4.3131 35 -3.88 0.0004stim 6 7 -1.2667 4.3131 35 -0.29 0.7707stim 6 8 -8.4667 4.3131 35 -1.96 0.0576stim 7 8 -7.2000 4.3131 35 -1.67 0.1040

With repeated-measures data, we didn’t have the opportunity to randomize like in a regular block de-sign, so this is cause to seriously question the independence of the errors. Thus, before looking too hardat these results, let’s check the autocorrelation of the data, using the Durbin-Watson test. This can be doneusing PROC REG. Only the D-W results are shown.

proc reg data = diags;model resid = / DW;

The REG ProcedureModel: MODEL1Dependent Variable: Resid ResidualDurbin-Watson D 1.319Number of Observations 481st Order Autocorrelation 0.334

We have an autocorrelation of .334, and, roughly, the standard error is 1/p

n = 1/p

48 ⇡ .14, so theautocorrelation is evidently nonzero. This autocorrelation includes transitions between the 8th observationon one bird and the 1st observation on the next bird, which one would guess would tends to decrease it. Itseems like we should take this autocorrelation seriously.

The following code is thwe same as the previous PROC MIXED run, except there’s an additional RE-PEATED statement that specidies a first-order autocorrelation structure (TYPE = AR(1)) for each pigeon(SUBJECT = PIGEON). The RCORR option asks to print the estimated residual correlation matrix.

proc mixed;class PIGEON stim;model resp = stim / ddfm = satterth outp = diags;random PIGEON;repeated / type = AR(1) subject = PIGEON rcorr;lsmeans stim / diff;


Estimated R Correlation Matrix for PIGEON 1Row Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8

1 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.09092 0.060972 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.090923 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.13564 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.20225 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.30156 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.44977 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.67068 0.06097 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000

Covariance Parameter EstimatesCov Parm Subject EstimatePIGEON 145.77AR(1) PIGEON 0.6706Residual 85.6493

2






n = 1/p






1 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.09092 0.060972 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.090923 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.13564 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.20225 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.30156 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.44977 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.67068 0.06097 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000


2

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n = 1/p






1 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.09092 0.060972 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.090923 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.13564 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.20225 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.30156 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.44977 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.67068 0.06097 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000


2






n = 1/p






1 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.09092 0.060972 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.090923 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.13564 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.20225 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.30156 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.44977 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.67068 0.06097 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000


2

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n = 1/p






1 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.09092 0.060972 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.1356 0.090923 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.2022 0.13564 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.3015 0.20225 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.4497 0.30156 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.6706 0.44977 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000 0.67068 0.06097 0.09092 0.1356 0.2022 0.3015 0.4497 0.6706 1.0000


2Fit Statistics-2 Res Log Likelihood 293.1AIC (smaller is better) 299.1AICC (smaller is better) 299.7BIC (smaller is better) 298.4




Effect stim Estimate Error DF t Value Pr > |t|stim 1 37.9167 6.2104 7.65 6.11 0.0003stim 2 41.8667 6.2104 7.65 6.74 0.0002stim 3 32.9833 6.2104 7.65 5.31 0.0008stim 4 33.7500 6.2104 7.65 5.43 0.0007stim 5 35.9667 6.2104 7.65 5.79 0.0005stim 6 44.2500 6.2104 7.65 7.13 0.0001stim 7 45.5167 6.2104 7.65 7.33 0.0001stim 8 52.7167 6.2104 7.65 8.49 <.0001


Effect stim _stim Estimate Error DF t Value Pr > |t|stim 1 2 -3.9500 3.0668 32.3 -1.29 0.2069stim 1 3 4.9333 3.9638 33.7 1.24 0.2219stim 1 4 4.1667 4.4655 25.4 0.93 0.3596stim 1 5 1.9500 4.7725 18.3 0.41 0.6876stim 1 6 -6.3333 4.9678 13.8 -1.27 0.2233stim 1 7 -7.6000 5.0945 11.2 -1.49 0.1634stim 1 8 -14.8000 5.1777 9.5 -2.86 0.0179stim 2 3 8.8833 3.0668 32.3 2.90 0.0067stim 2 4 8.1167 3.9638 33.7 2.05 0.0485stim 2 5 5.9000 4.4655 25.4 1.32 0.1982stim 2 6 -2.3833 4.7725 18.3 -0.50 0.6235stim 2 7 -3.6500 4.9678 13.8 -0.73 0.4748stim 2 8 -10.8500 5.0945 11.2 -2.13 0.0562stim 3 4 -0.7667 3.0668 32.3 -0.25 0.8042stim 3 5 -2.9833 3.9638 33.7 -0.75 0.4569stim 3 6 -11.2667 4.4655 25.4 -2.52 0.0182stim 3 7 -12.5333 4.7725 18.3 -2.63 0.0170stim 3 8 -19.7333 4.9678 13.8 -3.97 0.0014stim 4 5 -2.2167 3.0668 32.3 -0.72 0.4750stim 4 6 -10.5000 3.9638 33.7 -2.65 0.0122stim 4 7 -11.7667 4.4655 25.4 -2.63 0.0141stim 4 8 -18.9667 4.7725 18.3 -3.97 0.0009stim 5 6 -8.2833 3.0668 32.3 -2.70 0.0109stim 5 7 -9.5500 3.9638 33.7 -2.41 0.0216stim 5 8 -16.7500 4.4655 25.4 -3.75 0.0009stim 6 7 -1.2667 3.0668 32.3 -0.41 0.6823stim 6 8 -8.4667 3.9638 33.7 -2.14 0.0400stim 7 8 -7.2000 3.0668 32.3 -2.35 0.0252

Note that the fit statistics (�2 log likelihood, AIC, etc) are all smaller than for the first model, indicatingthat this is a better model. We estimate the first-order autocorrelation to be .67, a fairly large value. TheP value for the overall test of stim is .0072, compared with .0006 from the first model; so it appears thatthe first model exaggerates the significance of that factor. The standard errors of the least-squares meansare somewhat larger in this second model; and the SEs of the differences between least-squares means rangefrom 3.06 to 5.18, whereas in the first model they are all 4.31. Thus, this second analysis is very differentfrom the first. Owing to the positive autocorrelation, we have more power (lower SEs) for comparingadjacent stimuli, and less for comparing distant ones.

3

With this model, treatment (stimulus) is still significant but not quite assmall of a p-value as before.

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3









3

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