Is it always, a single sample, enough to be analyzed?


9-English version 

Assuming that:

·         the analytical method has a 100% accuracy
·         the concentration of the analyte under study exhibits a 
        specific  biological variation

The following question arises:

With only one determination (or test), what is the probability that a lab result does not differ in more than 10% of the value that corresponds to the homeostatic condition of the analyte under study?

To be able to answer this question, when only one test is performed on a single sample, the calculation to find out which is the probability to obtain one result that does not differ in more than 10% of the value that defines  the homeostatic condition of the analyte under study, must be performed.

The calculation is obtained taking into account only the biological variation of the analyte in the patient, assuming that all the laboratory procedures, even in the pre-analytical, analytical and post analytical phases are error free.

This means that, it is supposed that all the procedures followed by the laboratory yield accurate results regarding the analyte concentration in the sample withdrawn from the patient.

To perform the probability calculation, the biological variation coefficient must be known. Those values can be found in Dr Westagard web site, http://www.westgard.com/biodatabase1.htm

Important note: This database is currently updated.

Knowing:
·       the intraindividual percentage biological variation coefficient  CVI %, (CVw %) and

·   the condition that must be accomplished, which is that the dispersion of results does not differ in more than 10% of the concentration that defines  the homeostatic condition of the analyte,the number of standard deviations for that condition can be calculated.

In statistics,those numbers of standard deviations are known as Z score

For the Glucose example, CVI % has a value of 5.7% as was seen before, therefore the lower and upper limits, in terms of number of standard deviation (Z), can be calculated as follows: (database 2010)

Z=  10% / CVI%
Z=   10%/ 5.7%  =  1.75

It must be interpreted, then, that the expected limits are at ± 1.75 standard deviations of distribution of all glucose concentration that the patient might have.

How can the probability that, a single blood sample withdrawal contains the concentration within that range, be calculated?

To calculate the probability it is a need to refer to  the probability charts for normal distributions, by inserting the Z value previously estimated.

A chart 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 ± 1.75 corresponds a value of 0.0401 for each one of the tails greater to the Z value.

Therefore, for both tails corresponds: 2 x 0.0401 = 0.0802


Since the whole distribution probability value is 1, the probability between –Z  up to 
+ Z is equal to = 1 - 0.0802 =  0.920

Expressed as a percentage is 100 x 0.92 = 92.0 %


Interpretation:

With a single blood sample withdrawn from a patient and performing all the appropriate procedures to determine the concentration without operational error, there is a 92% probability that the glucose concentration result to be informed does not differ in more than 10% of the concentration which defines the analyte homeostatic condition.

The same calculation was performed for a different number of analytes and are posted in the tables below.

   analyte
    % CVI
    Z 10%
TABLE
% Probability 10%





Urea
12.3
0.81
0.2090
58.2





Glucose
5.7
1.75
0.0401
92.0





Cholesterol
5.4
1.85
0.0322
93.6





HDL CHOL
7.1
1.41
0.0793
84.1





LDL CHOL
8.3
1.20
0.1151
77.0





Urate
9.0
1.11
0.1335
73.3





Creatinine
5.3
1.89
0.0294
94.1





Protein. total
2.7
3.70
0
100.0





Albumin
3.1
3.23
0
100.0





Triglyceride
20.9
0.48
0.3156
36.9





Iron
26.5
0.38
0.352
29.6





Bilirubin Total
23.8
0.42
0.3372
32.6





Sodium
0.7
14.29
0
100.0





Potassium
4.8
2.08
0.0188
96.2





Chloride
1.2
8.33
0
100.0





Calcium
1.9
5.26
0
100.0





Magnesium
3.6
2.78
0.0027
99.5





Phosphate
8.5
1.18
0.119
76.2





GPT-ALT
24.3
0.41
0.3409
31.8





GOT-AST
11.9
0.84
0.2005
59.9





GGT
13.8
0.72
0.2358
52.8





Alkaline phosphatase,
10
1.00
0.1587
68.3










LDH
8.6
1.16
0.1230
75.4





CPK
22.8
0.44
0.3300
88.6





Amylase (pancreatic)
11.7
0.85
0.1977
88.6




   analyte
    % CVI
    Z 10%
TABLE
% Probability 10%
17-Hydroxyprogesterone
19.6
0.51
0.3050
39.0





      Androstendione
11.1
0.90
0.1891
62.2





Cortisol
20.9
0.48
0.3156
36.9





Estradiol
18.1
0.55
0.2912
41.8





SHBG
12.1
0.83
0.2033
59.3





DHEA-S
4.2
2.38
0.0087
98.3





Testosterone total
9.3
1.08
0.1401
72.0





Free testosterone
9.3
1.08
0.1401
72.0


   analyte
    % CVI
    Z 10%
TABLE
% Probability 10%
CEA
12.7
0.79
0.2148
57.0





AFP
12
0.83
0.2033
59.3





ferritin
14.2
0.70
0.2420
51.6





CA 15-3
6.1
1.64
0.0505
89.9





CA 19-9
16
0.63
0.2643
47.1





CA 125
24.7
0.40
0.3446
31.1





PSA TOTAL
18.1
0.55
0.2912
41.8



   analyte
    % CVI
    Z 10%
TABLE
% Probability 10%
LH(males)
14.5
0.69
0.2451
51.0





FSH(males)
8.7
1.15
0.1251
75.0





Prolactin(men)
6.9
1.45
0.0735
85.3





Insulin
21.1
0.47
0.3192
36.2


   analyte
    % CVI
    Z 10%
TABLA
% Probability 10%
T 3
8.7
1.15
0.1251
75.0





T4
4.9
2.04
0.0207
95.9





TSH
19.3
0.52
0.3015
39.7





T4 free
5.7
1.75
0.0401
92.0





T3 free
7.9
1.27
0.1020
79.6


The probability calculations, speak for themselves and avoid all type of comments. On top of that the probability values will be even lower taken in consideration the pre analytical, analytical and post analytical phase error, as occurs in the daily practice.