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I am struggling with correctly reporting the results of a linear (mixed) basis function model I ran. The model is specified in R as

rlmer <- lmer(Y ~ group * DoG(A) * B + (DoG(A):B | subject), data = rdat)

where group describes the patient group a subject belongs to, and A and B are within-subject variables (A is continuous and uniformely distributed on an interval, B is a factor). DoG(A) is a 1st-derivative-of-Gaussian basis function with fixed hyperparameters mu and sigma. Y is normally distributed.

The results of this model are

anova_lmerModLmerTest(rlmer)
# Type III Analysis of Variance Table with Satterthwaite's method
                          # Sum Sq  Mean Sq NumDF DenDF F value    Pr(>F)    
# group                    0,04679 0,023393     2    51  1,7898 0,1773450    
# DoG(A)                   0,01697 0,016972     1    46  1,2984 0,2603442    
# B                        0,46332 0,231658     2 48978 17,7234 2,021e-08 ***
# group:DoG(A)             0,23074 0,115368     2    46  8,8264 0,0005645 ***
# group:B                  0,16047 0,040117     4 48972  3,0692 0,0154156 *  
# DoG(A):B                 0,29976 0,149881     2    43 11,4669 0,0001004 ***
# group:DoG(A):B           0,42021 0,105052     4    43  8,0372 6,156e-05 ***

summary_lmerModLmerTest(rlmer)
#                               Estimate  Std..Error            df   t.value      Pr...t..
# (Intercept)                   0.005100    0.003328     97.991394  1.532530  1.286140e-01
# group2                        0.004615    0.004863     98.929525  0.949061  3.449023e-01
# group3                        0.000900    0.005000    103.113659  0.179923  8.575664e-01
# DoG(A)                        0.008584    0.005354     46.028930  1.603394  1.156885e-01
# B3                            0.012278    0.002886  49047.324713  4.253883  2.104840e-05
# B2                            0.003812    0.002286  48624.568134  1.668078  9.530671e-02
# group2:DoG(A)                -0.000316    0.007879     47.498923 -0.040063  9.682109e-01
# group3:DoG(A)                 0.008015    0.008106     49.422445  0.988763  3.275952e-01
# group2:B3                    -0.000294    0.004244  48925.767261 -0.069198  9.448322e-01
# group3:B3                    -0.006401    0.004433  49073.238759 -1.443776  1.488084e-01
# group2:B2                     0.007092    0.003356  48656.324449  2.113365  3.457462e-02
# group3:B2                     0.003561    0.003504  48719.213897  1.016284  3.094993e-01
# DoG(A):B3                    -0.048448    0.006818     42.001968 -7.105823  1.026047e-08
# DoG(A):B2                    -0.012040    0.004888     38.356533 -2.463073  1.837358e-02
# group2:DoG(A):B3              0.029324    0.010059     43.466560  2.915261  5.599089e-03
# group3:DoG(A):B3              0.058013    0.010388     45.614588  5.584450  1.238292e-06
# group2:DoG(A):B2              0.006611    0.007215     39.771591  0.916309  3.650276e-01
# group3:DoG(A):B2              0.022203    0.007468     42.114707  2.973015  4.861136e-03

(dof were estimated using Satterthwaite's method in R's lmerTest package).

I am mainly interested in interpreting terms containing DoG(A), indicating weaker or stronger contributions of DoG(A) to responses Y in different conditions of group and B, rather than the effects group, B, and group*B.

I am now wondering how to conduct/report paired comparisons that explain interactions group:DoG(A), B:DoG(A), and group:DoG(A):B. My best solution so far is to sum coefficients containing DoG(A) for each particular condition. This sum of coefficients would describe the amplitude of DoG(A) for each condition. As an example, group3 in condition B3 would show a DoG amplitude of

0.008584 + 0.008015 - 0.048448 + 0.058013, 

corresponding to a sum of coefficients

DoG(A) + group3:DoG(A) + DoG(A):B3 + group3:DoG(A):B3. 

For a group comparison to dummy group1 in condition B3, this value would be compared to

0.008584 - 0.048448, 

corresponding to

DoG(A) + DoG(A):B3

Although I believe this is conceptually correct, I feel like it is quite a non-stand way to report the model statistics. Also, in the above example, I am not sure which would be the correct t- and p-values to report (would it be the ones of group3:DoG(A):B3, t(45.61)=5.58, p=1.24e-06?). I'm stuck as this model is very specific and I don't know how to elegantly compare pairwise amplitudes of the basis function.

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