MDL = t(n-1, 1-a = .99) (S)
where:
MDL=the method detection
t (n-1, 1-a=.99)=the students' t value appropriate for a 99% confidence level and a standard deviation estimate with n-1 degrees of freedom as given in subd. 2.
S=standard deviation of the replicate analyses.
MDLLCL=0.69 MDL
MDLUCL=1.92 MDL where MDLLCL and MDLUCL are the lower and upper 95% confidence limits respectively based on 7 aliquots.
MDL=2.681 (Spooled)
where 2.681 is equal to t (12, 1-a=.99)
MDLLCL=0.72 MDL
MDLUCL=1.65 MDL
where LCL and UCL are the lower and upper 95% confidence limits respectively based on 14 aliquots.
Number of replicates | Degrees of freedom (n-1) | t (n-1, 1-a= .99) |
7........... | 6 | 3.143 |
8........... | 7 | 2.998 |
9........... | 8 | 2.896 |
10.......... | 9 | 2.821 |
11 .......... | 10 | 2.764 |
16.......... | 15 | 2.602 |
21 .......... | 20 | 2.528 |
26.......... | 25 | 2.485 |
31 .......... | 30 | 2.457 |
61 .......... | 60 | 2.390 |
2.326 |
Wis. Admin. Code Department of Natural Resources NR 252.0365