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L1592_Frame_C10 Page 89 Tuesday, December 18, 2001 1:46 PM
the variance of the computed concentration is:
(
(
2
2
2
2
2
Var C() = σ C = ( 20) σ y 2 + σ y 1 ) = 400 σ y + σ y 2 ) = 800σ y 2
2
Suppose that the standard deviation of a burette reading is σ y = 0.02 mL, giving σ y = 0.0004.
For y 1 = 38.2 and y 2 = 25.7, the concentration is:
(
C = 20 38.2 25.7) = 250 mg/L
–
and the variance and standard deviation of concentration are:
2
Var C() = σ C = ( 20) 0.0004 + 0.0004) = 0.32
(
2
σ C = 0.6 mg/L
Notice that the variance and standard deviation are not functions of the actual burette readings.
Therefore, this value of the standard deviation holds for any difference (y 2 − y 1 ). The approximate
95% confidence interval would be:
(
250 ± 2 0.6) mg/L = 250 ± 1.2 mg/L
Example 10.3
Suppose that a water specimen is diluted by a factor D before titration. D = 2 means that the
specimen was diluted to double its original volume, or half its original concentration. This might
be done, for example, so that no more than 15 mL of titrant is needed to reach the end point (so
that y 2 − y 1 ≤ 15). The estimated concentration is C = 20D(y 2 − y 1 ) with variance:
σ C = ( 20D) σ y + σ y 2 ) = 800D σ y 2
(
2
2
2
2
D = 1 (no dilution) gives the results just shown in Example 10.2. For D > 1, any variation in
2
error in reading the burette is magnified by D . Var(C) will be uniform over a narrow range of
concentration where D is constant, but it will become roughly proportional to concentration over
a wider range if D varies with concentration.
It is not unusual for environmental data to have a variance that is proportional to concentration. Dilution
or concentration during the laboratory processing will produce this characteristic.
Multiplicative Expressions
The propagation of error is different when variables are multiplied or divided. Variability may be magnified
or suppressed. Suppose that y = ab. The variance of y is:
σ y = σ a a + σ b b 2
2
2
2
2
and
2 2 2
------ = ------ + σ b
σ y
σ a
------
y 2 a 2 b 2
© 2002 By CRC Press LLC