Editing Reference range (section)

===Interpretation of standard ranges in medical tests===
In case of [[medical test]]s whose results are of continuous values, reference ranges can be used in the interpretation of an individual test result. This is primarily used for [[diagnostic test]]s and [[Screening (medicine)|screening]] tests, while [[monitoring (medicine)|monitoring test]]s may optimally be interpreted from previous tests of the same individual instead.

====Probability of random variability====
Reference ranges aid in the evaluation of whether a test result's deviation from the mean is a result of random variability or a result of an underlying disease or condition. If the reference group used to establish the reference range can be assumed to be representative of the individual person in a healthy state, then a test result from that individual that turns out to be lower or higher than the reference range can be interpreted as that there is less than 2.5% probability that this would have occurred by random variability in the absence of disease or other condition, which, in turn, is strongly indicative for considering an underlying disease or condition as a cause.

Such further consideration can be performed, for example, by an [[Differential diagnosis#Specific methods|epidemiology-based differential diagnostic procedure]], where potential candidate conditions are listed that may explain the finding, followed by calculations of how probable they are to have occurred in the first place, in turn followed by a comparison with the probability that the result would have occurred by random variability.

If the establishment of the reference range could have been made assuming a normal distribution, then the probability that the result would be an effect of random variability can be further specified as follows:

The [[standard deviation]], if not given already, can be inversely calculated by the fact that the [[absolute value]] of the difference between the mean and either the upper or lower limit of the reference range is approximately 2 standard deviations (more accurately 1.96), and thus:

:{{math|Standard deviation (s.d.) ≈ {{sfrac|{{mabs | (Mean) - (Upper limit) }} |2}}}}.

The [[standard score]] for the individual's test can subsequently be calculated as:

:{{math|1= Standard score (''z'') = {{sfrac|{{mabs | (Mean) - (individual measurement) }} |s.d.}}}}.

The probability that a value is of a certain distance from the mean can subsequently be calculated from the [[Standard score#prediction intervals|relation between standard score and prediction intervals]]. For example, a standard score of 2.58 corresponds to a prediction interval of 99%,<ref name=Kirkup2002>[https://books.google.com/books?id=rDsec-JnCAwC&pg=PA111 Page 111] in: {{cite book |author=Kirkup, Les |title=Data analysis with Excel: an introduction for physical scientists |publisher=Cambridge University Press |location=Cambridge, UK |year=2002 |isbn=978-0-521-79737-5 }}</ref> corresponding to a probability of 0.5% that a result is at least such far from the mean in the absence of disease.

====Example====
{{Hatnote|The method is described in further detail at [[differential diagnosis]].}}
Let's say, for example, that an individual takes a test that measures the [[ionized calcium]] in the blood, resulting in a value of 1.30&nbsp;mmol/L, and a reference group that appropriately represents the individual has established a reference range of 1.05 to 1.25&nbsp;mmol/L. The individual's value is higher than the upper limit of the reference range, and therefore has less than 2.5% probability of being a result of random variability, constituting a strong indication to make a [[differential diagnosis]] of possible causative conditions.

In this case, an [[Differential diagnosis#Specific methods|epidemiology-based differential diagnostic procedure]] is used, and its first step is to find candidate conditions that can explain the finding.

[[Hypercalcemia]] (usually defined as a calcium level above the reference range) is mostly caused by either [[primary hyperparathyroidism]] or malignancy,<ref name=Kumar>Table 20-4 in: {{cite book |author1=Mitchell, Richard Sheppard |author2=Kumar, Vinay |author3=Abbas, Abul K. |author4=Fausto, Nelson |title=Robbins Basic Pathology|publisher=Saunders |location=Philadelphia |year= 2007|isbn=978-1-4160-2973-1 }} 8th edition.</ref> and therefore, it is reasonable to include these in the differential diagnosis.

Using for example epidemiology and the individual's risk factors, let's say that the probability that the hypercalcemia would have been caused by primary hyperparathyroidism in the first place is estimated to be 0.00125 (or 0.125%), the equivalent probability for cancer is 0.0002, and 0.0005 for other conditions. With a probability given as less than 0.025 of no disease, this corresponds to a probability that the hypercalcemia would have occurred in the first place of up to 0.02695. However, the hypercalcemia ''has occurred'' with a probability of 100%, resulting adjusted probabilities of at least 4.6% that primary hyperparathyroidism has caused the hypercalcemia, at least 0.7% for cancer, at least 1.9% for other conditions and up to 92.8% for that there is no disease and the hypercalcemia is caused by random variability.

In this case, further processing benefits from specification of the probability of random variability:

The value is assumed to conform acceptably to a normal distribution, so the mean can be assumed to be 1.15 in the reference group. The [[standard deviation]], if not given already, can be inversely calculated by knowing that the [[absolute value]] of the difference between the mean and, for example, the upper limit of the reference range, is approximately 2 standard deviations (more accurately 1.96), and thus:
:{{math|1=Standard deviation (s.d.) ≈ {{sfrac|{{mabs | (Mean) - (Upper limit) }} |2}} = {{sfrac|{{mabs | 1.15 - 1.25 }}|2}} = {{sfrac| 0.1 |2}} = 0.05}}.

The [[standard score]] for the individual's test is subsequently calculated as:
:{{math|1= Standard score (''z'') = {{sfrac|{{mabs | (Mean) - (individual measurement) }} |s.d.}} = {{sfrac|{{mabs | 1.15 - 1.30 }}|0.05}} = {{sfrac|0.15|0.05}} = 3}}.

The probability that a value is of so much larger value than the mean as having a standard score of 3 corresponds to a probability of approximately 0.14% (given by {{math|(100% &minus; 99.7%)/2}}, with 99.7% here being given from the [[68–95–99.7 rule]]).

Using the same probabilities that the hypercalcemia would have occurred in the first place by the other candidate conditions, the probability that hypercalcemia would have occurred in the first place is 0.00335, and given the fact that hypercalcemia ''has occurred'' gives adjusted probabilities of 37.3%, 6.0%, 14.9% and 41.8%, respectively, for primary hyperparathyroidism, cancer, other conditions and no disease.