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)) | |||||