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L1592_frame_C14 Page 122 Tuesday, December 18, 2001 1:49 PM

st n-1, 0.01

Mean MDL

FIGURE 14.1 EPA definition of the MDL.

Example 14.1

Seven replicate specimens were measured to obtain the following results: 2.5, 2.7, 2.2, 2.2, 3.1,
2
2.6, 2.8. The estimated variance is s = 0.10, s = 0.32, t = 3.143, and the MDL = 3.143(0.32) =
1.0. Another set of seven replicate specimens was analyzed to get a second estimate of the
2
variance; the data were 1.6, 1.9, 1.3, 1.7, 2.1, 0.9, 1.8. These data give s = 0.16, s = 0.40, and
the MDL = 3.143(0.40) = 1.3. A statistical F test (0.16/0.10 ≤ F 6,6 = 4.3) shows that the two
estimates of the variance are not significantly different. Therefore, the two samples are pooled
to give:

0.10 + 0.16
s pooled = --------------------------- = 0.13
2
2
and

s pooled = 0.36

s pooled has ν = 12 degrees of freedom. The improved (pooled) estimate of the MDL is:
MDL = 2.681s pooled = 2.681 0.36) = 0.96
(

This should be rounded to 1.0 µg /L to have a precision that is consistent with the measured values.

Example 14.2

Calculate, using the EPA method, the MDL for the lead data given in Table 14.1. The data come
from a laboratory that did extra work to evaluate the limit of detection for lead measurements.
Known amounts of lead were added to filtered effluent from a municipal activated sludge plant.
Therefore, the aliquots with zero lead added are not believed to be entirely free of lead. By
definition, a minimum of seven replicates containing the analyte is needed to compute the MDL.
Therefore, we can use the 1.25 µg/L and 2.50 µg/L spiked specimens. The summary statistics are:

1.25 µg/L 2.50 µg/L
y 3.07 4.16
s 2 0.55 0.41
s 0.74 0.64
n 20 14

Pooling the sample variances yields:

(
19 0.55) + 13 0.41)
(
s pooled = ------------------------------------------------- = 0.49
2
19 + 13
© 2002 By CRC Press LLC