Question

A researcher is interested in determining the detection and quantitation limits of a method for the detection of warfarin. Ten method blanks gave the dimensionless instrument readings: 72.3 , 34.1 , 29.3 , 41.3 , 86.3 , 27.1 , 79.1 , 64.7 , 31.7 , and 72.5 . Ten samples containing a low concentration of warfarin near the detection limit had a mean reading of 189.1 . The slope of the calibration curve is 2.50×109 M−1 . Estimate the signal and concentration detection limits and the lower limit of quantitation for warfarin.

Answer 1

by using all number calculate mean and standard deviation by using formulas

mean = 53.84 = X

standard deviation = 23.22 = S

signal detection limit = Yblank + 3 S

                                 = 189.1 + 3 x 23.22

                                = 258.8

signal detection limit = 258.8

concentration detection limit = 3S / M

                                            = 3 x 23.22 / 2.50 x 10^9

                                             = 2.79 x 10^-8 M

concentration detection limit = 2.79 x 10^-8 M

lower limit of qunatitation = 10 S / M

                                        = 10 x 23.22 / 2.50 x 10^9

                                        = 9.29 x 10^-8 M

lower limit of qunatitation = 9.29 x 10^-8 M