Question

Determine the precision and accuracy of these data for dioxin. Sample 1 Sample 2 Sample 3 25.8 ng/mL 81.8 ng/mL 265 ng/mL Known concentratio 37.1 22.2 27.5 24.3 24.2 28.4 76.2 62.4 75.0 94.3 73.9 77.0 273 257 235 281 227 253 Experimentally determined values (ng/mL) Sample 1 Precision (Rel Std Deviation) Number Sample 2 Precision (Rel Std Deviation) Sample 3 Precision (Rel Std Deviation) Number Number Sample 1 Accuracy Relative Error Sample 2 Accuracy (Relative Error) Sample 3 Accuracy (Relative Error) Number Number Number

Answer 1

Write the expression for Precision and accuracy as follows: Precision = relative standard deviation = - (137) where, standard deviation, s= 1 average, 7 = ? accuracy = relative error accuracy = expert experimental value-actual value - actual value ........2 - - - - - - - - - - - - - - - - - - - - - - - - - - - for Sample (1), 37.1+22.2+27.5+24.3+24.2+28.4 average, = 163.7 = - = 27 28 standard deviation, s = n-1 Iſ (37.1-27.28) +(22.2-27.28)* +(27.5-27.28) +(24.3-27.28)* +(24.2-27.28)* +(28.4-27.28)° ) 6-1 = f141.9084 V5 = 5.327 Calculate the precision% and accuracy% as follows: Precisi on% = x100 % = 3322 <100 % = 19.53% experimental value-actual value 10006 accuracy = actual value 27.28-25.8 *100% 25.8 = 5.74% Therefore, the required solution is Precision% = 19.53% and accuracy%= 5.74% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

for Sample (2), 76.2+62.4+75+94.3+73.9+77 458.8 average, = = 76.47 2x standard deviation, s = 11 1-1 (76.2-76.47)* +(62.4–76.47)* +(75–76.47) (+(94.3-76.47) +(73.9-76.47) +(77 - 76.47) 6-1 _ 524.9934 V5 = 10.25 Calculate the precision% and accuracy% as follows: Precision%= 2x100 % = 0.4x100 % = 13.4% 76.47 experimental value-actual value. accuracy= x100 % actual value 76.47-81.8 x 100% 81.8 = -6.52% Therefore, the required solution is Precision% = 13.4% and accuracy%=-6.52% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - for Sample (3), 273+257+235+281+227 +253 average, 1 = 1526 . = 254.3 3(x-7) standard deviation, s = n-1 (273–254.3)* + (257–254.3)* +(235-254.3)) +(281–254.3)* +(227–254.3)* +(253–254.3)?) 6-1 2189.34 V5 = 20.93 Calculate the precision% and accuracy as follows: Precisi on% = x100 % = 20,93x100 % = 8.23% accuracy%= experimental value-actual values value x100 % actual value 254.3-265 x 100% 265 = -4.04% Therefore, the required solution is Precision%= 8.23% and accuracy% = -4.04%