That is,
and d is_the percent deviation of x from m, the true mean.
d = 100(x- m)/x.
Table 3 gives values of n computed using Equation (1) for a = 0.01 and
0.05 (i.e., 99 and 95% confidence*), d = 10, 25, and 50% and the c's
actually obtained for the five aliquot sizes.
These results assume the
estimated mean x to be normally distributed.
This should be approximately
true (Snedecor and Cochran, 1967, page 51), particularly for the larger
aliquot sizes, even though the individual aliquot concentrations are
clearly not normal.
Table 3 indicates the prohibitively large number of aliquots of 1l-g size
required to estimate the soil sample mean even with only 50% accuracy.
Clearly, the average of 1 or 2 one-g aliquots per sample would give a
very crude estimate of the true mean concentration for each field sample.
On the other hand, if study objectives require an accuracy of 10% with
95 or 99% confidence for each sample, then only 4 to 6 aliquots of 100-¢
size would appear to be sufficient.
Table 3.
Number of Aliquots Required to Be l-a Percent Confident
-That the Estimated Arithmetic Mean of a Field Sample is
Within d% of the True Mean for Aliquot Sizes of 1, 10,
25, 50, and 100 g
Aliquot
Size
Coefficient
of Variation
Percent Accuracy (d)
50%
a = 54]
14154
25%
14
54%
10%
1%
lg
0.79
10
17
{39
66
240
413
10 ¢g
0.27
2
2]
5
8
28
49
25 g
0.30
2
2;
6
10
35
60
50 g
0.12
1
1
1
2
6
10
100 ¢g
0.09
‘1
1);
1
1
4
6
Please note that we are considering here only the number of aliquots per
field sample.
Optimum numbers of field samples are discussed at the end
of the next section.
CHOOSING THE NUMBER AND SIZE OF ALIQUOTS
A number of references discuss and/or derive theoretical mathematical
expressions that, under certain simplifying assumptions, relate variables
such as aliquot size and the volume and density of contaminate particles
*Zy 12 for a = 0.01 and 0.05 is 2.57 and 1.96, respectively.
413