# Table 6 Three possible methods of averaging ratios between successive score values

equal baseline scores:
S 1,i ratio (R i ) S 2,i log(R i ) C i = (R i -1)/(R i + 1)
50 2 100 0.301 0.333
50 1 50 0.000 0.000
50 0.5 25 0.301 -0.333
Method (1): mean S 1,i  = 50.0 Method (2): mean of R i values: 1.17
mean S 2,i  = 58.3
(mean S 2,i )/(mean S 1,i ) = 1.17
Method (3): mean of log(R i ) values: 0.000 ; mean R: 1.00
(re-transformed mean log-value:10(mean log(Ri))
mean of C i values: 0.000 ; mean R: 1.00
(re-transformed mean C-value: (1 + C)/(1-C))
non-equal baseline scores, largest baseline with smallest R i value:
S 1,i ratio (R i ) S 2,i log(R i ) C i = (R i -1)/(R i + 1)
20 2 40 0.301 0.333
40 1 40 0.000 0.000
60 0.5 30 -0.301 -0.333
Method (1): mean S 1,i  = 40.0 Method (2): mean of R i values: 1.17
mean S 2,i  = 36.7
(mean S 2,i )/(mean S 1,i ) = 0.917
Method (3): mean of log(R i ) values: 0.000 ; mean R: 1.00
(re-transformed mean log-value:10(mean log(Ri))
mean of C i values: 0.000 ; mean R: 1.00
(re-transformed mean C-value: (1 + C)/(1-C))
non-equal baseline scores, largest baseline with largest R i value:
S 1,i ratio (R i ) S 2,i log(R i ) C i = (R i -1)/(R i + 1)
20 0.5 10 0.301 0.333
40 1 40 0.000 0.000
60 2 120 -0.301 -0.333
Method (1): mean S 1,i  = 40.0 Method (2): mean of R i values: 1.17
mean S 2,i  = 56.7
(mean S 2,i )/(mean S 1,i ) = 1.42
Method (3): mean of log(R i ) values: 0.000 ; mean R: 1.00
(re-transformed mean log-value:10(mean log(Ri))
mean of C i values: 0.000 ; mean R: 1.00
(re-transformed mean C-value: (1 + C)/(1-C))
1. S 1,i and S 2,i , subsequent score values of three items ( i = 1..3) at times ‘1’ and ‘2’ respectively. R i , ratio between S 2,i and S 1,i (multiplication factor of S 1,i ). log(R i ), logarithmically transformed R i -values. C i , R i -values transformed as Contrast-values. Three methods of averaging ratio values are illlustrated; Methods (3) includes two variants of data transformation, logarithmically and as Contrast. For further explanation, see text. 