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
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all the procedures followed by the laboratory yield accurate results regarding the analyte concentration in the sample withdrawn from the patient.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
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
+ 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.
