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Absolute Error
The difference between the measured value
and the true value is referred to as the
absolute error. Assume that analysis of an
iron ore by some method gave 11.1%
while the true value was 12.1%, the
absolute error is:
11.1% - 12.1% = -1.0%
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Relative Error
The relative error is the percentage of the
absolute error to the true value. For the
argument above we can calculate the
relative error as:
Relative error = (absolute error/true
value)x100%
= (-1.0/12.1)x100% = -8.3%
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Relative Accuracy
The percentage of the quotient of observed
result to the true value is called relative
accuracy.
Relative accuracy = (observed value/true
value)x100%
For the abovementioned example:
Relative accuracy = (11.1/12.1)x100% =
91.7%
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Standard Deviation
The standard deviation for a set of data
provides information on the spread of
the values.
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What we mean by spread is clear from the
graphs blue and red. Values used to draw
red graph are not as close to each others
as values in the blue graph. Therefore,
values in the red graph have higher
spread from the mean and have higher
standard deviation. On the contrary,
values in blue graph are close together
and have a lower spread from their mean,
and thus have a lower standard deviation.
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For an infinite or large number of data points
(more than 20) or when the true mean is known,
the population standard deviation is defined
as:
s = ( S (xi - m)2 / N )1/2
Where s is the population standard deviation, m is
the population mean, xi is the individual data
point, and N is the number of data points
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However, in real chemical laboratories where a
sample is analyzed, an experiment is repeated
three to five times and thus a very limited data
points (3-5) is collected. The sample standard
deviation (s) is defined as:
s = ( S (xi - x)2 / (N-1) )1/2
x is the average (mean) of the data points. The
sample standard deviation is also called
estimated standard deviation since it is only
an estimate of s .
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Example
The following replicate weights were obtained for a
sample: 29.8, 30.2, 28.6, and 29.7 mg. Calculate
s, s(mean), RSD, and RSD(mean)
Solution
First, we find x
X = (29.8+30.2+28.6+29.7)/4 = 29.6
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xi
xi – x
(xi – x)2
29.8
30.2
28.6
29.7
0.2
0.6
1.0
0.1
0.04
0.36
1.00
0.01
S = 1.41
s = ( S (xi - x)2 / (N-1) )1/2
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s = (1.41/3)1/2
s = 0.69 mg
S(mean) = s / N1/2
S(mean) = 0.69/(4)1/2
S(mean) = 0.34 mg
RSD or CV = (0.69/29.6)x100% = 2.3%
RSD(mean) = (0.345/29.6)x100% = 1.1%
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It should be recognized that as the number of
experiments is increased, the precision of the
measurement is increased as well. This means
smaller s due to smaller spread. This is because
s a 1/N1/2 which means that decrease in s as N
increases is not linear which implicates that,
after some number of experiments, further
increase in the number of experiments will result
in very little decrease in s (try it on your
calculator), which does not justify extra time and
effort.
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Pooled Standard Deviation (sp)
When the same experiments are done using two different
methods, the standard deviation can be pooled in order
to determine the reliability of the analysis method
(proposed or new).
Sp = {S (xi1 – x1)2 + S(xi2 – x2)2)/(N1 + N2 –2)}1/2
Sp is the pooled standard deviation, x1, x2 are average
values for data set 1 and 2, respectively, N1, N2 are the
number of data points of data set 1 and 2, respectively
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Example
Mercury in a sample was determined using a standard
method and a new suggested method. five replicate
experiments were conducted using the two procedures
giving the following results in ppm
New Method
10.5
9.9
10.4
11.2
10.5
Standard method
10.1
10.3
10.2
10.3
10.4
Find the pooled standard deviation
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First find x1 and x2
X1 = (10.5+9.9+10.4+11.2+10.5)/5 = 10.5
X2 = (10.1+10.3+10.2+10.3+10.4)/5 = 10.3
S(Xi1-X1)2 = { (10.5-10.5)2 + (9.9-10.5)2 +(10.4-10.5)2 +
(11.2-10.5)2 + (10.5-10.5)2}
S(Xi1-X1)2 = 0.86
S(Xi2-X2)2 = { (10.1-10.3)2 + (10.3-10.3)2 + (10.2-10.3)2 +
(10.3-10.3)2 +(10.4-10.3)2}
S(Xi2-X2)2 = 0.06
Sp = {(S(Xi1-X1)2 + S(Xi2-X2)2)/ (N1 + N2 –2)}1/2
SP = {(0.86+0.06)/(5+5-2)}1/2 = 0.34
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