Deconvolution Analysis of FMRI Time Series Data - Waisman ...

1.4.4 General Linear TestsWe continue consideration of Example 1.4.3.5. Recall that the full model parameter vectorwas represented by:b t = b 0 b 1 h R [0] h R [1] h R [2] h M [0] h M [1] h M [2] h E [0] h E [1] h E [2]Example 1.4.4.1 Test of a single regression parameterSuppose that we wish to test whether there is a signicant response to input stimulus m(t)at time lag 1 (TR). As a test of hypotheses, this can be expressed as:H o : h M [1] = 0vs. H a : h M [1] 6= 0In matrix terms, the null hypothesis is specied by H o : C = 0, whereC = 0 0 0 0 0 0 1 0 0 0 0 :Since the null hypothesis involves only s =1linear constraint, matrix C is actually a rowvector. Parameter h M [1] is the 7th parameter in the list of parameters for the full model,so matrix C consists of all 0's, except for a single 1 in the 7th position.Store the above matrix (as arow) into le glt1.mat, and execute the following script:Program 3dDeconvolve Command Line for Example 1.4.4.13dDeconvolve n-input1D LingNoise.1D n-num stimts 3 n-stim file 1 Random.1D -stim label 1 ''Random'' -stim maxlag 1 2 n-stim file 2 Markov.1D -stim label 2 ''Markov'' -stim maxlag 2 2 n-stim file 3 English.1D -stim label 3 ''English'' -stim maxlag 3 2 n-glt 1 glt1.mat -glt label 1 ''h[1] Markov''The output is the same as for Example 1.4.3.5, but with the additional GLT output:General Linear Test: h[1] MarkovLC[0] = 5.0166R^2 = 0.8065 F[1,7] = 29.1811 p-value = 1.0064e-03It is important to note that LC[0] = 5:0166 is equal to h M [1], the estimated lag-1coecient for the Markov stimulus. Also, note that F [1 7] = 29:1811 is the square of thecalculated t;stat for h M [1]. (Can you explain why?)43