prepare a table for fitrm and manova

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I am try to perform a one way repeated measure MANOVA on a set of data. In this data, I have 6 subjects. Each subject is measureed 4 times (t1-t4) for 11 different measures (m1-m11). Each subject went through all 44 measures.
I think the table should look like the table in the attached .mat file, and the model specification should be 'm1-m11~1'.
In the table, each subject has 4 rows, and each row contain the subject ID and 11 measures at a specific time point.
My goal is to see if there is any sifnificance across time witnin any measure.
I am wondering if it is the right table layout and modelspec for my goal.
  4 个评论
Scott MacKenzie
Scott MacKenzie 2021-10-26
I'm not sure what to suggest other than doing the ANOVA on each measure, using time (t1-t4) as a within-subjects factor.
Fan Yang
Fan Yang 2021-10-27
I still believe a one-way repeated Multivariate ANOVA is the best fit for my dataset. It predicts one indepentent measure using multiple correlated outcome measures.
The link above explained the model very well even though the precedure is domonstrated in SPSS.

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Scott MacKenzie
Scott MacKenzie 2021-10-27
编辑:Scott MacKenzie 2021-10-27
Using the example in the documentation for the manova function, here's a MANOVA for your data (attached):
load atb;
Meas = table([1 2 3 4 5 6 7 8 9 10 11]','VariableNames',{'Measurements'});
rm = fitrm(atb,'m1-m11~ID','WithinDesign',Meas);
manova(rm)
ans = 8×9 table
Within Between Statistic Value F RSquare df1 df2 pValue ________ ___________ _________ _______ ______ _______ ___ ___ _______ Constant (Intercept) Pillai 0.45813 1.0991 0.45813 10 13 0.42785 Constant (Intercept) Wilks 0.54187 1.0991 0.45813 10 13 0.42785 Constant (Intercept) Hotelling 0.84545 1.0991 0.45813 10 13 0.42785 Constant (Intercept) Roy 0.84545 1.0991 0.45813 10 13 0.42785 Constant ID Pillai 0.50648 1.3341 0.50648 10 13 0.30768 Constant ID Wilks 0.49352 1.3341 0.50648 10 13 0.30768 Constant ID Hotelling 1.0262 1.3341 0.50648 10 13 0.30768 Constant ID Roy 1.0262 1.3341 0.50648 10 13 0.30768
Above, Wilks' lambda = 0.49352. The corresponding F-statistic is not significant (F[10,13] = 1.3341, p = .30768) implying no significant difference between the eleven measures over the four time points with six IDs (subjects). Your data just look like random numbers, so the result is not surprising. Hopefully, you'll have more interesting results with real data.

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