9English 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 preanalytical, 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all the procedures followed by the laboratory yield accurate results regarding the analyte concentration in the sample withdrawn from the patient.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, CV_{I }% 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% / CV_{I}%
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/wpcontent/uploads/2007/08/Normal_table.gif
http://www.pavementinteractive.org/wpcontent/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

% CV_{I}

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

GPTALT

24.3

0.41

0.3409

31.8

GOTAST

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

% CV_{I}

Z _{10%}

TABLE

% Probability _{10%}

17Hydroxyprogesterone

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

DHEAS

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

% CV_{I}

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 153

6.1

1.64

0.0505

89.9

CA 199

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

% CV_{I}

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

% CV_{I}

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.