Page 48 - Quality control of pharmaceuticals (07-PA 704)
P. 48
Relative error: Er = ( − ) × 100
Or Ea = ( ̅ − ) × 100
Where: Xi is the found value, Xt is the true value, ̅ is the mean value.
Example (1):
The results of six replicate determinations of standard solutions containing 20.00
ppm of iron (ш) analyzed in exactly the same way are: 19.4, 19.5, 19.6, 19.8, 20.1
and 20.3 ppm. Calculate the absolute and relative errors.
The standard solution contain 20.00 ppm of iron (ш), this is the true value.
Mean, ̅̅ ̅ = 19.4+19.5+19.6+19.8+20.1+20.3 = 19.78 ≃ 19.8 ppm
6
Absolute error Ea = ̅ i – Xt = 19.8-20 = 0.2 ppm
Relative error: Er % =( ̅ − ) × 100 = 19.8−20 × 100 = -1.5%
20
The sign (+) or (-) must be retained.
Mathematical Calculation and Statistical Treatment of Analytical Data
1. Standard deviation(S):
