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| sgn(T-B) 100[abs(T-B)] / abs(B), T and B have different signs | {{{ sgn(T-B) 100[abs(T-B)] / abs(B) if T and B have different signs or |
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| 100[abs(T) - abs(B)]/abs(B) = 100(SF - 1), T and B have the same sign | 100[abs(T) - abs(B)]/abs(B) = 100(SF - 1) if T and B have the same sign }}} |
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| If we ignore the absolute values in the above formulae we can end up with misleading values due to failing to ignore the signs of B and T. For example we have scores of -3 for B and +2 for T then the total number of errors B+A = -1 where in fact there are 5 errors made in total. | If we ignore the absolute values in the above formulae we can end up with misleading values due to failing to ignore the signs of B and T. For example if we have scores of -3 for B and +2 for T then the total number of errors equal to B+A equals -1 when in fact there are 5 errors made in total. |
Difference in errors adjusted for overall number when the errors have different signs
Suppose we have error scores which can take positive or negative values, of equal importance, on two conditions, baseline (B) and treatment (T) then assuming at least one of B and T are non-zero the difference in the errors adjusted for total number can be expressed using absolute values, abs(), as
sgn(T-B)[abs(T-B)] / [(abs(B) + abs(T)] if T and B have different signs or [abs(T) - abs(B)] / [(abs(B) + abs(T)] if T and B have the same sign
Alternatively, provided B is nonzero, we can express the difference in errors as a percentage change (PC) by
sgn(T-B) 100[abs(T-B)] / abs(B) if T and B have different signs or 100[abs(T) - abs(B)]/abs(B) = 100(SF - 1) if T and B have the same sign
where SF is the scaling factor relating to T and B.
For example if T and B both have the same sign and the number of errors doubles (SF=2) it follows abs(T) - abs(B) = 2 abs(B) so abs(T) = 2 abs(B) which gives a 100 (2abs(B)- abs(B))/abs(B) = 100% increase in errors from baseline.
If we ignore the absolute values in the above formulae we can end up with misleading values due to failing to ignore the signs of B and T. For example if we have scores of -3 for B and +2 for T then the total number of errors equal to B+A equals -1 when in fact there are 5 errors made in total.