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Data Handling
Error and Standard Deviation

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Logarithms
The digits to the left of the decimal point in a
logarithmic value are not counted since they merely
reflect the log 10x and they are not considered
significant. The zeros to the right of the decimal
point are all significant.
Examples
Log 2.0x103 = 3.30 (two significant figures in both
terms). The blue digit in the answer is not significant
as it comes from the 103 portion which has nothing
to do with expressing the number of significant
figures

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Log 1.18 = 0.072 (three significant figures in
both terms, the blue zero in the answer is
significant)
Antilog of 0.083 = 1.21
Log 12.1 = 1.083 (three significant figures in
both terms, the blue digit in the answer is
not significant)

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Errors
Errors can be classified according to nature of
the error into two types, determinate and
indeterminate errors.
A determinate error (sometimes called a
systematic error) is an error which has a
direction either positive or negative. An
example of such an error is performing a
weight measurement on an uncalibrated
balance (for instance it always add a fixed
amount to the weight).

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An indeterminate error is a random error
and has no direction where sometimes
higher or lower estimates than should
be observed are obtained.
In many cases, indeterminate errors are
encountered by lack of analyst
experience and attention. Indeterminate
errors are always present but can be
minimized to very low levels by good
analysts and procedures.

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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:
AE = 11.1% - 12.1% = -1.0%
The negative sign indicates a negative error

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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%
RE = (-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 above mentioned 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 the 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 the blue graph are
closer 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 number of 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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Data Handling
Standard Deviation
Propagation of Error

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Standard deviation of the mean (s(mean))
S(mean) = s / N1/2
Another important expression of deviation is the
relative standard deviation (RSD) or
sometimes called coefficient of variation (CV)
where
RSD = ( s / x ) X100%
RSD (mean) = ( s(mean) / x ) X100%

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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 0.2 0.04
30.2 0.6 0.36
28.6 1.0 1.00
29.7 0.1 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 is because s a 1/N1/2 which means that
decrease in s as N increases is not linear which
implies that, after some number of experiments,
further increase in the number of experiments
will result in very little decrease in s, which
does not justify extra time and effort.

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Pooled Standard Deviation (sp)
When replicate samples are done using two different
methods, the standard deviation can be pooled in
order to determine the reliability of the analytical
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 sets 1 and 2, respectively, N1 and 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 Standard method
10.5 10.1
9.9 10.3
10.4 10.2
11.2 10.3
10.5 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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Standard Deviation of the
Difference
When multiple samples are analyzed by a proposed
and standard methods, Sd is the calculated standard
deviation for the difference.
Sd = (S ( Di – D )2 / (N-1))1/2
Sd is the standard deviation of the difference, Di is the
difference between a result obtained by the standard
method from that obtained by the proposed method
for the same sample. D is the average of all
differences.

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Example
Mercury in multiple samples was determined using a
standard method and a new suggested method. Six
different samples were analyzed using the two
procedures giving the following results in ppm
Sample New Method Standard method
1 10.2 10.5
2 12.7 11.9
3 8.6 8.7
4 17.5 16.9
5 11.2 10.9
6 11.5 11.1
Find the standard deviation of the difference.

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It is wise to construct a table as below
New Method Standard method Di
10.3 10.5 -0.2
12.7 11.9 +0.8
8.6 8.7 -0.1
17.5 16.9 +0.6
11.2 10.9 +0.3
11.5 11.1 +0.4
_____________________________ ________
S Di = 1.8
D = 1.8/6 = 0.30

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S(Di – D)2 = { (-0.2-0.3)2 + (+0.8-0.3)2 + (-
0.1-0.3)2 + (+0.6-0.3)2 + (+0.3-0.3)2 +
(+0.4-0.3)2 } =
{0.25+0.25+0.16+0.09+0+0.01}
S(Di – D)2 = 0.76
Sd = ( S ( Di – D )2 / (N-1) )1/2
Sd = (0.76/5)1/2 = 0.39

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Propagation of Errors
As seen earlier, each measurement has
some uncertainty associated with it.
During a process of calculations the
uncertainty in the answer can be
calculated from uncertainties in
individual measurements.

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Calculation of error in the answer
depends on whether the mathematical
operation is a summation /subtraction
or multiplication/division.
It should be clear that in a process of
calculating a final answer, as the
number of mathematical operations
increase, error will propagate.

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Addition and subtraction
The absolute uncertainty in the answer Sa can
be evaluated from absolute uncertainties in
individual numbers (b, c, d, .. ) as below:
Sa
2 = Sb
2 + Sc
2 + Sd
2 + …
Where, Sa, Sb, Sc, and Sd are absolute
uncertainties in answer, b, c, and d
(estimated standard deviation in answer, b, c,
and d).

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Example
Three samples were analyzed for iron
content. The average percentage of
iron in the first sample was 65.06, the
second sample contained 56.13, and
the third contained 62.68%. The
estimated standard deviation of each of
the three samples were + 0.07, + 0.01,
and + 0.02%, respectively. What is the
average iron content of the samples
depending on these results?

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% Iron = {(65.06 + 0.07%) + (56.13 + 0.01%) + (62.68 +
0.02%)}/3 = (183.87/3) + Sa %
% iron = 61.29 + Sa %
Sa
2 = (+ 0.07)2 + ( +0.01)2 + (+0.02)2 = + 5.4x10-3
Sa = 7.3x10-2
% Iron = 61.29 + 0.073%
It is clear that we should retain two digits in the
uncertainty as the answer is known to the nearest
one hundredth (to get accurate number of significant
figures).
Therefore, the answer should be reported as:
% Iron = 61.29 + 0.07%

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Multiplication and Division
The absolute uncertainty in calculations involving
multiplication and division can not be estimated
directly. The first step in such operations is to find
the relative uncertainty in the answer from relative
uncertainties in individual measurements as follows
(Sa
2)rel = (Sb
2)rel + (Sc
2)rel+ (Sd
2)rel + …
Where, (Sb)rel = Estimated standard deviation in b (i.e.
uncertainty in b)/ absolute value of b
Sa = Answer x (Sa)rel

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Find the result of the following
calculation, using the correct number
of significant figures:
(2.23 + 0.01)*(3.508 + 0.007) = 7.82 + sa
Since 2.23 is the key number
(Sa)rel = {(+0.01/2.23)2 + (+0.007/3.507)2 }1/2
(sa)rel = 4.91*10-3
Sa = 7.823*4.91*10-3 = 0.0384
Answer = 7.82 + 0.04

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Chloride in a 25 mL sample was
determined by titration with a 0.1167 +
0.0002 M AgNO3 solution. If the titration
required an average AgNO3 volume of
36.78 mL and the standard deviation in
the volume was 0.04 mL, find the
uncertainty in the number of mmol of
chloride contained in 250 mL chloride
sample.

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You should remember that the standard
deviation is the absolute error in
volume of AgNO3
mmol chloride = mmol AgNO3 Since Ag+
reacts with Cl- in a 1:1 mole ratio
mmol AgNO3 = molarity AgNO3 * Volume (mL) AgNO3
mmol AgNO3 = (0.1167 + 0.0002) ( 36.78 +
0.04) = 4.292 + ?

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Since this is a multiplication process we use
the equation for relative uncertainty
(Sa
2)rel = ( +0.0002/0.1167)2 + ( + 0.04/36.78)2
(Sa)rel = +2.03x10-3
Sa = 4.292 x (+2.03x10-3 ) = 8.71x10-3 ( This is
the uncertainty in 25 mL chloride)
Sa in 250 mL chloride = 10 x 8.71x10-3 =
+0.0871 mmol

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If we are to report the number of mmol
chloride in 250 mL sample, the answer
would be
Answer = 42.92 + 0.0871 mmol
The final answer should be 42.92 + 0.09
mmol since only two digits after the
decimal points are allowed here to
express actual uncertainty depending
on the number of significant figures.

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The Confidence limit
The standard deviation of a set of measurements
provides an indication of the precision inherent in
these measurements. However, no indication of how
close the obtained result is from the accurate result
can be deduced from the standard deviation.
The confidence interval presents the range within
which the accurate value might occur. This range is
called the confidence interval. The probability that
the true value occurs within this range is called the
confidence level.

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Confidence limit = x + ts/N1/2
Where; t is a statistical factor which
depends on the confidence level and
the number of degrees of freedom
(number of experiments – 1).
Confidence interval (range) = (x + ts/N1/2)
to (x - ts/N1/2)

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The standard deviation for the analysis of a carbonate
sample was 0.075% for the results 93.50, 93.58, and
93.43% carbonate. Find the confidence limit and
range at the 95% confidence level. t95% = 4.303 (from
table)
Confidence limit = x + ts/N1/2
X = (93.50+93.58+93.43)/3 = 93.50
Confidence limit = 93.50 + 4.303x0.075/31/2
Confidence limit = 93.50 + 0.19%
Range = (93.50 – 0.19%) – (93.50 +0.19%)
Range = 93.31-93.69%

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Tests of Significance
In this section we deal with two tests used for
comparing two analytical methods, one is a
new or proposed method and the other is a
standard method. The two methods are
compared in terms of whether they provide
comparable precision ( the F test ), based on
their standard deviations or variances. The
other test ( t test ) tells whether there is a
statistical difference between results
obtained by the two methods.

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The F Test
The precision of two methods could be
compared based on their standard deviations
using the F test which can be defined as the
ratio between the variances ( the variance is
the standard deviation squared ) of the two
methods. The ratio should always be larger
than unity. That is, the larger variance of
either method is placed in the nominator.
F = S1
2/S2
2 > 1

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Where, S1
2 > S2
2
Values of F ( a statistical factor ) at
different confidence levels which can
be obtained from statistical F tables.
When Fcalculated < Ftabulated this is an
indication of no statistical difference
between precision or variances of the
two methods.

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Example
In the analysis of glucose using a new
developed procedure and a standard
procedure, the variances of the two
procedures were 4.8 and 8.3. If the
tabulated F value at 95% confidence
level at the number of degrees of
freedom used was 4.95. Determine
whether the variance of the new
procedure differs significantly from that
of the standard method

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F = S1
2/S2
2
F = 8.3/4.8 = 1.7
Since Fcalculated < Ftabulated , there is no
significant statistical difference
between the variances of the two
methods (i.e. there is no significant
statistical difference between the
precision of the two methods).

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The Student t Test
To check whether there is a significant
statistical difference between the results of a
new or proposed procedure and a standard
one, the t test is used. As we did above, we
calculate t and compare it to the tabulated
value at the required confidence level and at
the used degrees of freedom. There is no
significant statistical difference between the
results of the two methods when tcalculated <
ttabulated .
There are three situations where the t test is
applied:

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a. When an Accepted Value is Known
The tcalc is calculated from the relation
below and compared to ttab
m = x + ts/N1/2 or more conveniently,
+ t = (x - m) N1/2/s

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Example
A new procedure for determining copper was
used for the determination of copper in a
sample. The procedure was repeated 5 times
giving an average of 10.8 ppm and a
standard deviation of +0.7 ppm. If the true
value for this analysis was 11.7 ppm, does
the new procedure give a statistically correct
value at the 95% confidence level? ttab =
2.776
•

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Substitution into equation below, we get:
+ t = (x - m) N1/2/s
+ t = (10.8-11.7) 51/2/0.7
+ t = 2.9
the tcalc is larger than the ttab. Therefore,
there is a significant statistical
difference between the two results
which also means that it is NOT
acceptable to use the new procedure
for copper determination.

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b. Comparison between two means
When an accepted value is not known and the
sample is analyzed using the new procedure
and a standard procedure. Here, we have two
sets of data, a standard deviation for each
set of data and a number of data points or
results in each set. Under these conditions,
we use the pooled standard deviation for the
two sets. The same equation in a is used but
with some modifications. The t value is
calculated from the relation
(N1 Ns)1/2
(N1+ Ns)1/2
( x1 – xs)
Sp
+ t = *

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Where, x1 and xs are means of
measurements using the new and
standard methods. N1 and Ns are
number of replicates done using the
new and standard methods,
respectively. Sp is the pooled standard
deviation.
In such calculations it is wise to apply
the F test first, and if it passes the t test
is then applied.

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Example
Nickel in a sample was determined using
a new procedure where six replicate
samples resulted in a mean of 19.65%
and a variance of 0.4524. Five replicate
analyses where conducted using a
standard procedure resulting in a mean
of 19.24% and a variance of 0.105. If the
pooled standard deviation was +0.546,
is there a significant difference
between the two methods?

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First, let us find whether there is a significant
difference in precision between the two
procedures, by applying the F test
F = 0.4524/0.105 = 4.31
The tabulated F value is 6.26. Since Fcalculated <
Ftabulated , then there is no significant
statistical difference between the precision
of the two procedures. Therefore, we
continue with calculation of t test.

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+ t = 1.23
The tabulated t value is 2.262. Since tcalculated <
ttabulated for nine degrees of freedom at 95%
confidence level, we conclude that there is
no significant statistical difference between
the results of the two methods.
(N1 Ns)1/2
(N1+ Ns)1/2
( x1 – xs)
Sp
+ t =
*
(6*5)1/2
(6+ 5)1/2
( 19.65 – 19.24)
0.546
+ t =
*

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c. The t Test with Multiple Samples
Till now we have considered replicate
measurements of the same sample. When
multiple samples are present, an average
difference is calculated and individual
deviation from a mean difference is
calculated and used to calculate a difference
standard deviation, Sd which is used in a
successive step to calculate t.
+ t = DN1/2/Sd
Sd = [S ( Di – D )2 / (N-1)]1/2

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Sd is the standard deviation of the difference,
Di is the difference between a result obtained
by the standard method subtracted from that
obtained by the proposed method for the
same sample. D is the average of all
differences.
Example
Mercury in multiple samples was determined
using a standard method and a new
suggested method. Six different samples
were analyzed using the two procedures
giving the following results in ppm:

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Sample No. New Method Standard method
1. 10.3 10.5
2. 12.7 11.9
3. 8.6 8.7
4. 17.5 16.9
5. 11.2 10.9
6. 11.5 11.1
Find the standard deviation of the difference. If the two
methods have comparable precisions, find whether
there is any significant difference between the
results of the two methods at the 95% confidence
level. The tabulated t value for five degrees of
freedom at 95% confidence level is 2.571.

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Sample No. New Method Standard method Di
1. 10.3 10.5 -0.2
2. 12.7 11.9 +0.8
3. 8.6 8.7 -0.1
4. 17.5 16.9 +0.6
5. 11.2 10.9 +0.3
6. 11.5 11.1 +0.4
_______________________________________________
SDi = 1.8
D = 1.8/6 = 0.30

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S ( Di – D )2 = { (-0.2-0.3)2 + (+0.8-0.3)2 + (-0.1-0.3)2 +
(+0.6-0.3)2 + (+0.3-0.3)2 + (+0.4-0.3)2 } =
{0.25+0.25+0.16+0.09+0+0.01}
S ( Di – D )2 = 0.76
Sd = ( S( Di – D )2 / (N-1) )1/2
Sd = (0.76/5)1/2 = 0.39
+ t = 0.30x61/2/0.39 =1.88
The calculated t value is less than the tabulated t value
which means that there is no significant difference
between the results of the two methods.
+ t = DN1/2/sd

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The Q Test
In several occasions, when replicate experiments are
done one of the data point may look odd or faulty.
The analyst is confused whether to keep it or reject
it. The Q test provides a means to judge if it should
be retained or rejected. This can be done by applying
the Q test equation:
Q = a/w
Where a is the difference between the suspected result
and the result nearest to it in value, w is the
difference between highest and lowest results.

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Once again, if the calculated Q value is
less than the tabulated value, then the
suspected data point should be
retained.
In contrast to F and t tests the statistical
value of Q depends on the number of
data points rather than the number of
degrees of freedom.

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Example
In the replicate determination of gold you got
the following results: 96, 99, 97, 94, 100, 95,
and 72%. Check whether any point should be
excluded at the 95% confidence level.
Tabulated Q95% = 0.568 for 7 observations
Arrange results: 72, 94, 95, 96, 97, 99, 100
Q = a/w
Qcalc = (94-72)/(100 - 72) = 0.79
Qcalc > Qtab
The point 72% should be rejected.

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Example
In the replicate determination of gold you got the
following results: 96, 99, 97, 94, 100, 95, and 88%.
Check whether any point should be excluded at the
95% confidence level. Tabulated Q95% = 0.568 for 7
observations.
Arrange results: 88, 94, 95, 96, 97, 99, 100
Solution
Q = a/w
Qcalc = (94-88)/(100-88) = 0.50
Qcalc < Qtab
The point 88% should be retained.

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Linear Least Squares
Frequently, an analyst constructs a calibration curve
using several standards and draws a straight line
among the data points in the graph.
In many cases, the line does not cross all points and the
analyst starts judging where the straight line should
pass.
Human judgment is not perfect and, unfortunately, may
be biased.
The method of linear least squares is a mathematical
method that help us choose the best path of the
straight line.

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It is well known that the equation of a straight line is
mathematically represented by
y = mx + b
Where m is the line and b is the line intercept, x and y
are variables.
The slope, m, can be calculated from the relationship
m = {Sxiyi – [(SxiSyi)/n]}/{ Sxi
2 – [(Sxi)2/n]}
b = y – mx
x, y are average values of xi and yi.

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The standard deviation of any of the yi
points (Sy) is given by the relation
Sy = {([Syi
2 – (( Syi)2/n)] – m2 [Sxi
2 – ((
Sxi)2/n)])/(n-2)}1/2
The uncertainty in slope can then be
calculated from Sy as follows
Sm = {Sy
2/ [Sxi
2 – (( Sxi)2/n)]}1/2

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Example
Using the following data and without
plotting, if the fluorescence of a
riboflavin sample was 15.4 find its
concentration.

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(Sxi)2 = 2.250
x = (Sxi)/n = 1.500/5 = 0.300
y = (Syi)/n = 83.6/5 = 16.72
m = {Sxiyi – [(SxiSyi)/n]}/{ Sxi
2 – [(Sxi)2/n]}
Substitution in the equation above gives
m = {46.6 – [(1.500*83.6)/5]}/ {0.850 –[(2.250/5]}
m = 53.75

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This Excel plot gives the same results for slope and intercept
as calculated in the example.

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To calculate b we use the equation
b = y – mx
b = 16.72 – 53.75*0.300 = 0.60
Now we are ready to calculate the sample
concentration
y = mx + b
15.4 = 53.75*x + 0.60
x = 0.275 ng/L

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Correlation Coefficient (r)
When the points that are supposed to be on a
straight line are scattered around that line
then one should estimate the correlation
between the two variables. The correlation
coefficient serves as a measure for the
correlation of these two variables. This can
be very important if correlation between
results obtained by a new method and a
standard method is required.
r = {nS xiyi – (SxiSyi)}/ {[nSxi
2 – (Sxi)2][nSyi
2 –
(Syi)2]}1/2

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Two graphs with the same straight line
equation??

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Example
Calculate the
correlation
coefficient of the
data :
Solution
First we find Syi
2 and
(Syi)2 from the table
in previous example
Syi
2 = 2554.66
(Syi)2 = 6988.96

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Substituting in the correlation coefficient
equation above:
r = {5*46.6-(1.500*83.6)} / {[5*0.850 –
2.250][5*2554.66-6988.96]}1/2
r = 1.00
The correlation coefficient occurs between + 1.
As the correlation coefficient approaches
unity, correlation increases and exact
correlation occurs when r = 1.
An r value less than 0.90 is considered bad
while that exceeding 0.99 is considered
excellent.

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Currently, many scientists prefer to use
the square of the correlation
coefficient, r2 rather than r, to express
correlation.
Evidently, the use of r2 is a more strict
criterion as a smaller value is always
obtained when fractions are squared.

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Detection Limits
All instrumental methods have a degree
of noise associated with the
measurement that limits the amount
of analyte that can be detected.
1. Detection Limit is the lowest
concentration level that can be
determined to be statistically different
from an analyte blank.

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When a graphical display of results is
obtained, the detection limit of the
instrument can be defined as the
concentration of analyte resulting in a
signal that is twice as the peak to peak
noise (the distance between the two
dashed lines in the schematic below).

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Peak-to-peak noise level as a basis for detection limit.
A “detectable” analyte signal would be 12
divisions above a line drawn through the
average of the baseline fluctuations

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2. Detection Limit is the concentration
that gives a signal three times the
standard deviation of the background
signal.
To calculate the detection limit:
a. Find the average of the blank signal
b. Find the standard deviation of the
blank
c. Find the net analyte signal
analyte conc. * 3*s
analyte signal
DL =

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Example
A blank solution in a colorimetric analysis
resulted in absorbance readings of 0.000,
0.008, 0.006, and 0.003. A 1 ppm standard
solution of the analyte resulted in a reading
of 0.051. Calculate the detection limit.
The standard deviation of the four data
points of the blank can be calculated to
be + 0.0032 and the mean of the blank
is 0.004

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The net reading of the standard = 0.051 – 0.004
= 0.047
The detection limit is the concentration which
results in three times the standard deviation
(3 x 0.0032 = 0.0096).
Detection limits = 1 ppm x 0.0096/0.047 = 0.2
ppm
The absorbance reading of the least detectable
concentration = 0.0096 + 0.004 = 0.014