IMPROVING PROBABILITY

English version


11-IMPROVING 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analyte homeostatic condition value is very low when testing a single sample, it is suggested to test a second sample withdrawn approximately 15 days later, when the patient clinical conditions allow to do so, being the average of both determinations the result of the study.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 pre-analytical 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)     Intra-individual biological variability

Both define the biochemical variation, as previously mentioned in this blog.

So that, if:

CVA %  is the coefficient variation percent of the measurement process
CV I %  is the Intra-individual biological variation coefficient percent
CVB %  is the Biochemical variation coefficient percent

CVB % =  (CV I % 2  +   CVA % 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 % = CVB % / (n 0.5), where n corresponds to the number of samples processed.

For two samples:


ES %= CVB % / (2 0.5) = (CV I % 2 +   CVA % 2) 0.5   / 1.41 = 
        =0.707 ·   (CV I % 2  +   CVA % 2 ) 0.5  

ES % = 0.707 · CVB %

This value is shown as: % CVB / (20.5) 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 CVA % = 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: % CVB / (20.5)


Calculations for CEA

CVA %   = 5.0 %
CV I %   = 12.7 %
CVB %   is the Biochemical variation coefficient percent

CVB5% % =  (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 %= CVB 5%  % / (2 0.5)  = CVB 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:
      (% CVB / (20.5)  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 % CVB /20.5 is 9.6% so, the upper and lower limits in terms of number of standard deviations (Z) is calculated as follows:



Z=  10% / (% CVB / (20.5))
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 CVA= 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/wp-content/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 CVA% = 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,


% CVI    :   Intra-individual biological variation coefficient percent


% CVB5% :  Biochemical variation coefficient percent, calculated with CVA % = 5%

% CVB5% /20.5    : Standard Error % of the average, variation coefficient % of the distribution 
                                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


  
%P10%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




   analyte
    % CVI
    % CVB5%
% CVB5% /20.5
    Z 10%
TABLA
  %P10%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







GPT-ALT
24.3
24.8
17.5
0.57
0.2843
43.1







GOT-AST
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
    % CVI
    % CVB5%
% CVB5% /20.5
    Z 10%
TABLE
  %P10%5% X 2
17-Hydroxyprogesterone
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







DHEA-S
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
    % CVI
    % CVB5%
% CVB5% /20.5
    Z 10%
TABLE
    %P10%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 15-3
6.1
7.9
5.6
1.79
0.0367
92.7







CA 19-9
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
    % CVI
    % CVB5%
% CVB5% /20.5
    Z 10%
TABLE
    %P10%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
    % CVI
    % CVB5%
% CVB5% /20.5
    Z 10%
TABLE
    %P10%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.