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## Tests of Significance

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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 ... – PowerPoint PPT presentation

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Title: Tests of Significance

1
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.

2
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 S12/S22

3
• Where, S12 gt S22
•
• Values of F ( a statistical factor ) at
different confidence levels which can be obtained
from statistical F tables. When Fcalculated lt
Ftabulated this is an indication of no
statistical difference between precision or
variances of the two methods.

4
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5
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

6
• F S12/S22
•
• F 8.8/4.8 1.73 (the subscript is because the
answer is less than the key number)
• Since Fcalculated lt 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).

7
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 lt ttabulated .
• There are three situations where the t test is
applied

8
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9
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

10
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
•

11
• 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.

12
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
•

13
• 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.

14
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?

15
• 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
lt Ftabulated , then there is no significant
statistical difference between the precision of
the two procedures. Therefore, we continue with
calculation of t test.

16
•
• t 1.23
•
• The tabulated t value is 2.262. Since
tcalculated lt 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.