LOD & LOQ Estimation

[Jumpshooter]

What would you say were the LOQ LOD given the cited data on an ethanol quantification run wherein the STD was serially diluted to obtain the "levels"? A cursory glance would imply that 5 mg/mL = LOQ and 0.005 mg/mL as the LOD. But unsure....
Pls indicate if using the European Pharma ICH method or other source. Thanky.

STD Dilution Level Peak Area Peak Area
(mg/mL ) Ethanol n-Propanol
0.005 67 53
0.005 30 17
0.05 109 22
0.05 101 6
0.5 957 207
0.5 954 167
5 9360 1453
5 9565 1392
50 94498 14043
50 93867 13980

[Don_Hilton]

Do you have signal to noise values for these peaks? Two replicates does not give you much information for estimating a confidence interval - you have only one degree of freedom at each level - and the t vaue is quite large.

My guess is that you are integratign noise at the lowest level of propanol. Note that the areas are larger than the smaller of the smaller of the two values for n-propanol at 0.05 mg/mL - and these are about the same area that you have for the lowest level of ethanol.