English version
11IMPROVING PROBABILITY
In those
cases where the probability to obtain a result within 10% of the analyte
homeostatic condition value is very low when testing a single sample, it is suggested
to test a second sample withdrawn approximately 15 days later, when the patient
clinical conditions allow to do so, being the average of both determinations
the result of the study.
The average of both results, not just:
·
improves, significantly, the probability of a better
biochemical information,
but also:
·
allows error detection in the
preanalytical phase
.
o
most of all in those cases where both two
results cannot be averaged because they are too different (with a discrepancy
greater than the biochemical reference change value
(B
RCV) without a clinical origin explanation. Those cases should be reviewed by
using
a
new sample, in order to determine the true biochemical condition of the
patient.
There are
many factors that can generate doubts by using a single result, in a randomly
taken sample and measured with methods that, even being automated, they show
that can present variations of the systematic error which are independent of
the system precision variations that can affect a specific group of samples
providing erroneous results.
It cannot
passively be accepted that the clinical laboratory results are affected by an
error.
It must be
assured to the patient that any error or doubt in the lab result is
appropriately controlled in such a way that the biochemical information obtained
from the lab report is not modified.
In order to
reduce the result disparity due to the biological variation, for the proposed
analytes in the previous pages, the benefit of the information obtained by
using a second sample will be studied.
Overall, it
can be remembered that:
When a
measurement result is obtained by the average
of various observations, the dispersion of the possible values for that average
is no longer the coefficient variation observed
in each individual observation but corresponds to the quotient between the
coefficient variation and the square root of the number of observations
performed.
The
statistical parameter, as defined, is known as measurement standard error.
When a
measurement result is informed, based on a single observation, the measurement
standard error is equal to the variation coefficient.
It is
frequent referring to the standard deviation, or to the percent variation
coefficient for measurements performed
with a single observation, when the correct way would be to refer to the
measurement standard error or the measurement standard error percent even
though it was performed with a single
observation.
In the
particular case of the Clinical Laboratory measurements, the results dispersion
is affected principally by two factors:
a) Analytical variability of the
measurement process
b) Intraindividual biological
variability
Both define
the biochemical variation, as previously mentioned in this blog.
So that, if:
CV_{A }% is the coefficient variation percent of the measurement
process
CV_{ I }% is the Intraindividual biological variation
coefficient percent
CV_{B }% is the Biochemical variation coefficient
percent
CV_{B }% =
(CV_{ I }% ^{2}
+ CV_{A }% ^{2})
^{0.5 }
When the
lab result is obtained by calculating the average of two observations, the
dispersion of that average is equal to the standard error percent (ES%):
ES % = CV_{B
}% / (n ^{0.5}), where n corresponds to the number of samples
processed.
For two samples:
ES %= CV_{B
}% / (2 ^{0.5}) = (CV_{ I }% ^{2} + CV_{A }% ^{2}) ^{0.5 }/ 1.41 =
=0.707 · (CV_{ I }% ^{2} + CV_{A }% ^{2 }) ^{0.5 }
=0.707 · (CV_{ I }% ^{2} + CV_{A }% ^{2 }) ^{0.5 }
ES % = 0.707 · CV_{B }%
This value is shown as: % CV_{B }/ (2^{0.5})
in the tables below.
Let´s take as an example the CEA determination. It´s been seen that the
probability of the result to be within 10% of the homeostatic condition with a
single determination performed without error is: 57.0%
So,
assuming that:
 The analytical method has a CV_{A }% = 5.0 %, and
 The analyte concentration presents a biological variability =12.7%
*
The
following question arises:
What is
the probability that, with TWO samples taken in different days, the average of
the individual determinations does not differ in more than 10% of the CEA value
corresponding to the homeostatic condition being studied?
Using the same calculations as used in the previous page, but now
considering the analytical variation and biological variation for two
independent observations and also the Standard error value % for two
observations must be known: % CV_{B }/ (2^{0.5})
Calculations for CEA
CV_{A }% =
5.0 %
CV_{ I }% =
12.7 %
CV_{B }% is the Biochemical variation coefficient
percent
CV_{B}^{5%}_{
}% = (12.7_{ }% ^{2} + 5.0_{
}% ^{2 }) ^{0.5 }=
13.6 %
If the
laboratory result is calculated as the average of two determinations, the
dispersion of that average is equal to the standard error percent (ES%):
ES %= CV_{B }^{5%}_{ }% / (2 ^{0.5}) = CV_{B }^{5%}_{ }%
/ 1.41 = 0.707 · 13.6 % =
9.6%^{ }
This value,
expressed as a standard error, is equal to the standard deviation of the
distribution of the averages calculated with two observations.
Knowing:
·
The
standard deviation % for the
distribution of the average between two observations:
(% CV_{B }/ (2^{0.5}) and
·
The
condition that must be accomplished, meaning thath the dispersion of the
average of two results does not differ in more than 10% than the concentration
which defines the homeostatic condition of the analyte number of limits
standard deviations for that condition
can be calculated
Those
limits are statistically known, as mentioned before, as Z score.
As for the
CEA example, the value for % CV_{B }/2^{0.5} is 9.6% so,
the upper and lower limits in terms of number of standard deviations (Z) is
calculated as follows:
Z= 10% / (% CV_{B }/ (2^{0.5}))
Z= 10%/ 9.6 % = 1.04
It is
interpreted, then, that the desirable limits are at ± 1.04 standard deviations
of the distribution of the averages calculated from two results obtained with
two samples from the patient and performed with a method with CV_{A}=
5.0 %
How to
determine the probability that the average of results from two blood withdrawn
samples is within that range?
To
determine probability, the tables for normal distributions by using the Z
previously calculated, must be consulted.
A table
example can be found in the following web site:
http://www.pavementinteractive.org/wpcontent/uploads/2007/08/Normal_table.gif
http://www.pavementinteractive.org/wpcontent/uploads/2007/08/Normal_table.gif
For a ± Z value of = ± 1.04 corresponds a probability of 0.1492
for each one of the tails, one obtained from values greater than + Z the other obtained from values lower than –Z.
Therefore,
for both tails corresponds: 2 x 0.1492 = 0.2984
Since
probability for a distribution is 1, the probability from values from –Z until + Z is equal to =
=1  0.2984 = 0.7016
Expressed
as percent, is: 100 x 0.7016 = 70.2 %
Interpretation:
·
Withdrawing two samples from a
patient and
·
Performing all the necessary
procedures to determine concentration for each one of the samples with a method
with CV_{A}% = 5.0%
The probability that the AVERAGE of both
results, for the CEA concentration to be informed does not differ in more than
10% of the concentration which defines the homeostatic condition of the analyte,
is 70.2%
Let´s compare the value of probability by using
this procedure: 70.2% versus the probability obtained, by using a single sample
procedure with NO operational error, of 57%!!
This calculation has been performed with the
rest of analytes analyzed in this blog, the results are posted in the table below.
Where,
% CV_{I
} :_{ } Intraindividual biological variation
coefficient percent
% CV_{B}^{5% }: Biochemical variation coefficient
percent, calculated with CV_{A} % = 5%^{}
% CV_{B}^{5%}_{ }/2^{0.5 }: Standard Error % of the average, variation
coefficient % of the distribution
of the average calculated with results from two samples.
of the average calculated with results from two samples.
Z _{10% } : limit values of
standard deviation for a dispersion of value of 10%
TABLE : probability
values for each one of the tail greater than Z
%P_{10%}^{5% X 2}: probability that the informed value does not
differ in more than 10%
of the concentration which defines the homeostatic condition
of the analyte, using the average of two determinations
of the concentration which defines the homeostatic condition
of the analyte, using the average of two determinations
analyte

% CV_{I}

% CV_{B}^{5%}

% CV_{B}^{5%}_{
}/2^{0.5}

Z _{10%}

TABLA

%P_{10%}^{5% X 2}

Urea

12.3

13.3

9.4

1.07

0.1423

71.54

Glucose

5.7

7.6

5.4

1.87

0.0307

93.9

Cholesterol

5.4

7.4

5.2

1.92

0.0274

94.5

HDL CHOL

7.1

8.7

6.1

1.63

0.0516

89.7

LDL CHOL

8.3

9.7

6.9

1.46

0.0721

85.6

Urate

9.0

10.3

7.3

1.37

0.0853

82.9

Creatinine

5.3

7.3

5.2

1.94

0.0262

94.8

Protein. total

2.7

5.7

4.0

2.49

0.0064

98.7

Albumin

3.1

5.9

4.2

2.40

0.0082

98.4

Triglyceride

20.9

21.5

15.2

0.66

0.2546

49.1

Iron

26.5

27.0

19.1

0.52

0.3015

39.7

Bilirubin Total

23.8

24.3

17.2

0.58

0.2810

43.8

Sodium

0.7

5.0

3.6

2.80

0.0026

99.5

Potassium

4.8

6.9

4.9

2.04

0.0207

95.9

Chloride

1.2

5.1

3.6

2.75

0.0030

99.4

Calcium

1.9

5.3

3.8

2.64

0.041

91.8

Magnesium

3.6

6.2

4.4

2.30

0.0107

97.9

Phosphate

8.5

9.9

7.0

1.43

0.0764

84.7

GPTALT

24.3

24.8

17.5

0.57

0.2843

43.1

GOTAST

11.9

12.9

9.1

1.10

0.1357

72.9

GGT

13.8

14.7

10.4

0.96

0.1685

66.3

Alkaline phosphatase,

10

11.2

7.9

1.27

0.1020

79.6

LDH

8.6

9.9

7.0

1.42

0.0778

84.4

CPK

22.8

23.3

16.5

0.61

0.2709

45.8

Amylase (pancreatic)

11.7

12.7

9.0

1.11

0.1335

73.3

analyte

% CV_{I}

% CV_{B}^{5%}

% CV_{B}^{5%}_{
}/2^{0.5}

Z _{10%}

TABLE

%P_{10%}^{5% X 2}

17Hydroxyprogesterone

19.6

20.2

14.3

0.70

0.2420

51.6

Androstendione

11.1

12.2

8.6

1.16

0.123

75.4

Cortisol

20.9

21.5

15.2

0.66

0.2546

49.1

Estradiol

18.1

18.8

13.3

0.75

0.2266

54.7

SHBG

12.1

13.1

9.3

1.08

0.1401

72.0

DHEAS

4.2

6.5

4.6

2.17

0.0150

97.0

Testosterone total

9.3

10.6

7.5

1.34

0.0901

82.0

Free testosterone

9.3

10.6

7.5

1.34

0.0901

82.0

analyte

% CV_{I}

% CV_{B}^{5%}

% CV_{B}^{5%}_{
}/2^{0.5}

Z _{10%}

TABLE

%P_{10%}^{5% X 2}

CEA

12.7

13.6

9.6

1.04

0.1492

70.2

AFP

12

13.0

9.2

1.09

0.1379

72.4

ferritin

14.2

15.1

10.6

0.94

0.1736

65.3

CA 153

6.1

7.9

5.6

1.79

0.0367

92.7

CA 199

16

16.8

11.9

0.84

0.2005

59.9

CA 125

24.7

25.2

17.8

0.56

0.2877

42.5

PSA TOTAL

18.1

18.8

13.3

0.75

0.2266

54.7

analyte

% CV_{I}

% CV_{B}^{5%}

% CV_{B}^{5%}_{
}/2^{0.5}

Z _{10%}

TABLE

%P_{10%}^{5% X 2}

LH(males)

14.5

15.3

10.8

0.92

0.1788

64.2

FSH(males)

8.7

10.0

7.1

1.41

0.0793

84.1

Prolactin(men)

6.9

8.5

6.0

1.66

0.0485

90.3

Insulin

21.1

21.7

15.3

0.65

0.2578

48.4

analyte

% CV_{I}

% CV_{B}^{5%}

% CV_{B}^{5%}_{
}/2^{0.5}

Z _{10%}

TABLE

%P_{10%}^{5% X 2}

T 3


8.7

10.0

7.1

1.41

0.0793

84.1


T4

.


4.9

7.0

4.9

2.02

0.0217

95.7


TSH


19.3

19.9

14.1

0.71

0.2389

52.2


T4 free


5.7

7.6

5.4

1.87

0.0307

93.9


T3 free


7.9

9.3

6.6

1.51

0.0655

86.9

Conclusion:
The
probability that a result, expressed as the average of two determinations, does
not differ in more than 10% of the homeostatic condition of the patient
performed with a method with an analytical variability of 5%, increases
significantly comparing to the condition of performing a single determination from
a single sample with an ideal process with 100% accuracy.