Does fitrm & ranova support within subject models without between subject factors

I am trying to use fitrm and ranova (Matlab2014a) to run a repeated measures anova on a 2x2 within subjects design. I currently have 4 variables representing the responses for all subjects in each cell of the 2x2 design.

I do not wish to include a between subject factor but rmfit seems to require at least one.

The code I am using is as follows.

alpha_power = randn(24,4); % This is my data but lets generate it for now
t = table(alpha_power(:,1),alpha_power(:,2), alpha_power(:,3), alpha_power(:,4),'VariableNames',{'Y1','Y2','Y3','Y4'}); % Create a table storing the respones

within = table({'A';'A';'V';'V'},{'T';'S';'T';'S'},'VariableNames',{'Attention','TMS'}); % Create a table reflecting the within subject factors 'Attention' and 'TMS' and their levels.

rm = fitrm(t,'Y1-Y4','WithinDesign',within); % Here the code breaks.

fitrm seems to require terms in the model specification which reflect the predictor variables (e.g. 'Y1-Y4~Gender'), but these are automatically specified as between subject factors, I do not have between subject factors.

Is it possible to specify terms as within subject factors?

[Theresa]

This was very helpful. I was wondering if you have figured out how to do a repeated measures anova on that model. I am having trouble finding examples of how to do: ranovatbl = ranova(rm,'WithinModel',WM) with a 2-way repeated measures that's all within. Thank you!

Answer 1

[Tom Lane]

Instead of

'Y1-Y4'

try

'Y1-Y4~1'

You do have to specify a complete formula, but you can include just the constant term on the RHS.

[Abbie]

Let's assume Attention and TMS each have four levels rather than two as above. When I build a model using 'Y1-Y4~1' and run ranova as ranova(rm,'WithinModel','Attention*TMS'), the output is rather confusing to me. As an example, it might look like this...

                              SumSq       DF      MeanSq        F         pValue       pValueGG      pValueHF      pValueLB
                               __________    __    __________    ______    __________    __________    __________    __________

    (Intercept)                4.0584e+05    1     4.0584e+05    422.66    7.1196e-09    7.1196e-09    7.1196e-09    7.1196e-09
    Error                          8641.8    9          960.2
    (Intercept):Attention          3.3996e+05    1     3.3996e+05     390.3    1.0117e-08    1.0117e-08    1.0117e-08    1.0117e-08
    Error(Attention)                   7839.1    9         871.01
    (Intercept):TMS          2.0646e+05    1     2.0646e+05    234.74    9.3829e-08    9.3829e-08    9.3829e-08    9.3829e-08
    Error(TMS)                   7915.5    9          879.5
    (Intercept):Attention:TMS    1.6844e+05    1     1.6844e+05    215.35    1.3643e-07    1.3643e-07    1.3643e-07    1.3643e-07
    Error(Attention:TMS)             7039.5    9         782.17

To start, why are the degrees of freedom only one? Since there are four levels of Attention and four levels of TMS, I expect the degrees of freedom for each of these to be 3. Can anyone provide any insight?

Thanks in advance for the help!

[Jonathan]

I am having similar problems as Abbie.

I tried to do a two way repeated measures Anova with the factors Condition and Trial, with 2 and 30 levels, respectively. I created a data table Y in the wide format with one row for each subject (17 rows) and one column for each level (60 columns). Additionally, I created a table to identify the factor levels for each datapoint:

w1 = cat(1,repmat({'Ground'},30,1), repmat({'Height'},30,1));

w2 = [[1:30]';[1:30]'];

Within = table(w1,w2,'VariableNames',{'Condition','Trial'});

When I test the following model:

rm = fitrm(Y,'Y1-Y60 ~ 1','WithinDesign',Within);

using: ranovatbl = ranova(rm,'WithinModel','Condition+Trial') I got:

ranovatbl =

                             SumSq     DF    MeanSq       F        pValue     pValueGG    pValueHF
                             ______    __    ______    ________    _______    ________    ________

    (Intercept)                1720    1       1720      1.6448    0.32823    0.32823     0.32823 
    Error                    2091.4    2     1045.7                                               
    (Intercept):Condition    60.488    1     60.488    0.082038     0.8015     0.8015      0.8015 
    Error(Condition)         1474.6    2     737.32                                               
    (Intercept):Trial        2693.1    1     2693.1      6.0929    0.13232    0.13232     0.13232 
    Error(Trial)             884.03    2     442.01                                               

                             pValueLB
                             ________

    (Intercept)              0.32823 
    Error                            
    (Intercept):Condition     0.8015 
    Error(Condition)                 
    (Intercept):Trial        0.13232 
    Error(Trial)

In this table you can see that the number of degrees of freedom is 1 for all tested terms, while there are 30 levels for the "Trial" factor. Additionally the factor names Trial and Condition are preceded by "(Intercept):".

Therefore I wonder: Is it possible to do a simple 2 factor within subjects repeated measures Anova using the 2014 statistics toolbox without between subject factors?

Kind regards,

Jonathan

[PIT]

I have the exact same question: I would also like to remove the Intercept from the WithinSubjects terms and to adjust the DFs. Is that possible?

[Logan]

I have the same problem as well and I've seen a similar question posted. Has anyone managed to figure out why the DFs are so low?

Also I've noticed that, like Jonathan and Abbie above, all my pValueGG and HF values are the same as the LB (lower bound). I'm not that familiar with RM ANOVAs but doesn't that mean that the data is completely non-spherical, i.e. as far from a value of 1 as possible???

Answer 2

[Chaoran Zhang]

Let me sum up the critical points for problems like this one. Some of them are mentioned in the above answers.

1. If you want to do a two way repeated measures ANOVA, which means purely within-subject, you need a constant term on the RHS. (by Tom Lane)
2. It is recommanded to use the data type of categorical to mark your within-subject variables. (by Francis Z. Fan)
3. If you don't want you model to contain a constant, add '-1' to your Wilkinson notation, because 'Statistics and Machine Learning Toolbox™ notation always includes a constant term unless you explicitly remove the term using -1'.

If you follow the instructions above, you will get a result exactly the same as SPSS!

Here's my test code:

t = table( ...
    [3 6 4 3]', ...
    [4 6 4 2]', ...
    [5 7 5 2]', ...
    [4 5 3 3]', ...
    [8 9 8 7]', ...
    [12 13 12 11]', ...
    'VariableNames',{'a1b1','a1b2','a1b3','a2b1','a2b2','a2b3'});
Meas = table(categorical([1 1 1 2 2 2])',categorical([1 2 3 1 2 3])','VariableNames',{'a','b'});
rm = fitrm(t,'a1b1,a1b2,a1b3,a2b1,a2b2,a2b3~1','WithinDesign',Meas);
ranova(rm,'WithinModel','a*b-1')

Here's the output of MATLAB:

                       SumSq     DF    MeanSq       F         pValue       pValueGG      pValueHF     pValueLB 
                       ______    __    _______    ______    __________    __________    __________    _________

    (Intercept):a      80.667    1      80.667      72.6     0.0033958     0.0033958     0.0033958    0.0033958
    Error(a)           3.3333    3      1.1111         1           0.5           0.5           0.5          0.5
    (Intercept):b      81.083    2      40.542    153.63    7.0263e-06    0.00094758    0.00072093    0.0011315
    Error(b)           1.5833    6     0.26389         1           0.5           0.5           0.5          0.5
    (Intercept):a:b    56.583    2      28.292    119.82    1.4572e-05    0.00082622    0.00026348    0.0016322
    Error(a:b)         1.4167    6     0.23611         1           0.5           0.5           0.5          0.5

Here's the output of SPSS:

主体内效应检验						
测量:   score 
源		III 类平方和	自由度	均方	F	显著性
a	假设球形度	80.667	1	80.667	72.600	.003
	格林豪斯-盖斯勒	80.667	1.000	80.667	72.600	.003
	辛-费德特	80.667	1.000	80.667	72.600	.003
	下限	    80.667	1.000	80.667	72.600	.003
误差 (a)	假设球形度	3.333	3	1.111		
	格林豪斯-盖斯勒	3.333	3.000	1.111		
	辛-费德特	3.333	3.000	1.111		
	下限	    3.333	3.000	1.111		
b	假设球形度	81.083	2	40.542	153.632	.000
	格林豪斯-盖斯勒	81.083	1.034	78.388	153.632	.001
	辛-费德特	81.083	1.087	74.562	153.632	.001
	下限	    81.083	1.000	81.083	153.632	.001
误差 (b)	假设球形度	1.583	6	.264		
	格林豪斯-盖斯勒	1.583	3.103	.510		
	辛-费德特	1.583	3.262	.485		
	下限	    1.583	3.000	.528		
a * b	假设球形度	56.583	2	28.292	119.824	.000
	格林豪斯-盖斯勒	56.583	1.142	49.535	119.824	.001
	辛-费德特	56.583	1.383	40.914	119.824	.000
	下限	    56.583	1.000	56.583	119.824	.002
误差 (a*b)	假设球形度	1.417	6	.236		
	格林豪斯-盖斯勒	1.417	3.427	.413		
	辛-费德特	1.417	4.149	.341		
	下限	    1.417	3.000	.472

Answer 3

[Jim]

Thank you Tom! That worked perfectly.

Answer 4

[Francis Z. Fan]

before doing ranova, one should convert factors to categorical, like

https://cn.mathworks.com/matlabcentral/answers/140799-3-way-repeated-measures-anova-pairwise-comparisons-using-multcompare

within2 = within; within2.Attention = categorical(within2.Attention); within2.TestCond = categorical(within2.TestCond); within2.TMS = categorical(within2.TMS);