CLS 2523 BASIC CONCEPTS OF HEMATOLOGY

Quality Assurance, Quality Control, and Statistics 

This teaching syllabus forms the basis of the basic hematology curriculum.   All objectives listed are in the cognitive domain unless otherwise noted.  The student at the end of the instructional period, is responsible for meeting these objectives by achieving a cumulative score of 70% or better on all problem sets, case studies, major exams, quizzes, and library assignments.   Objectives are listed in numerical order.  The student, upon completion of the classroom component of this syllabus will be responsible to successfully: 

01       EXPLAIN WHAT HEMATOLOGY CONSISTS OF.

 Hematology may be simply defined as the study of blood but actually involves the following:
      A.   analysis of the concentration, structure, and functions of
            blood, its elements, and products.
      B.   quantification and differentiation of cells based upon their
            morphology.
      C.   stain specific elements of the cell and detect cellular antigens.

Hematology, in the decade of the 1940's, was characterized by having only a few routine determinations (examples: bleeding time, platelet count, hemoglobin, hematocrit, WBC count, RBC count, and differential), now has become sophisticated to include cytochemical markers, cell surface markers, measurement of cell volumes, identifying reactive alterations in cell populations, and identification of neoplastic changes in cell populations. Coagulation enzymes
studies, once mysterious, are now routine.

 02       WRITE OR DESCRIBE A BRIEF OVERVIEW OF THE
QUANTIFICATION CONCEPT IN HEMATOLOGY.

 QUANTIFICATION infers the precise measurement of a substance that results in a numerical value.  Hematology was at one time characterized by the manual counting and measurement of blood cells, hemoglobin, packed cell volumes (hematocrit), and blood coagulation procedures.  Many of the manual derived values are now obtained by automated cell counters.  Certain manual counts, once relying on Thoma and similar pipets, now employs the Unopette systems.  The hemocytometer was at one time the basis of all counting procedures is now used for spinal fluids, semen counts, platelet counts, and synovial fluid counts.   Some testing methods remain essentially unchanged, only being adapted to an automated system An example is the prothrombin time, once performed by adding patients plasma to a test tube followed by addition of thromboplastin was tilted back and forth until the fibrin clot was observed visually.  This procedure is now performed by an automated instrument that uses an optical density detector.

 03       WRITE OR DESCRIBE THE DIFFERENTIATION CONCEPT IN
HEMATOLOGY.

 Differentiation in hematology generally employs manual discrimination and evaluation of blood cells under a microscope.  There are three classes of cells that are distinguished: leukocytes, thrombocytes, and erythrocytes.  This required preparation of a blood smear that is stained with Wright’s stain or some other special stain.  When observing leukocytes (1) note the size, shape, and color, (2) rank the cells according to their predominance, and (3) what malignant or non-malignant changes are present.  When observing erythrocytes (1) note the size, shape, and color, (2) how much deviation occurs for the typical, biconcave shape, and (3) if immature forms are present.  Thrombocytes are to be observed with the same care.  The smear is to be examined for the presence parasites (examples: malaria, trypanosomiasis, or babesiosis).

 04       STATE THE APPROXIMATE VOLUME OF BLOOD IN AN ADULT.

Approximately 8.0% of the body weight is blood.  A 154 pound male will have an approximate volume of 5.2 quarts or 5.0 liters of blood.   A larger person will have more blood volume than a smaller person.  There are approximately 32 mLs of blood for each pound of body weight.

 05       EXPLAIN QUALITY ASSURANCE (QA) IN THE CLINICAL LABORATORY.

 Quality assurance (QA) is not a test, a record of a result, or a statistical value. It is a continuous process that is maintained by the laboratory and hospital to assure confidence in laboratory and hospital services to guarantee the integrity of its services and products.  QA activities encompass all of the non-analytic activities, those activites that are not part of the clinical testing process.  The laboratory organizes it activities to provide the best possible health care to the patient.  The following are examples of QA activities:
A.    management and monitoring personnel,
B.    using quality materials (reagents, instruments, supplies, etc.),
C.    using established procedures and established statistics (a
       procedure manual),
D.    specimen collection, identification, transport, accession, and handling
        prior to testing,
E.    reporting results,
F.    fee charges for tests performed,
G.   using corrective actions to obtain desired results,
H.   monitoring patient satisfaction.

06    THE IMPORTANCE OF MONITORING FOR ERRORS.

The Clinical laboratory is concerned about quality and accuracy of the tests that are reported to primary care givers.  The laboratory monitors where these errors can appear that will affect the accuracy of test results.   These errors can occur prior to the test analysis and if they manifest, they are called preanalytical errors or variables.  If the error occurs during the testing process, then it become an analytical error.  If the error appears after the test is performed and reported, then it is known as a post-analytical error.

The preanalytical error occurs before the test is performed.   This error source can occur at the beginning of test ordering and flling out the requisition.  Examples of this type of error includes:
01    duplicate or missing requisitions
02    tests omitted from the requisition
03    incorrect ordering of tests
04    patient identification error
05    incorrect blood collection
06    specimen transport error
07    specimen handling/processing in the lab

Analytical errors occur during the testing process.  Examples of these errors are:
01    deteriorated or wrong reagents
02    any instrument malfunction
03    laboratorian error
04    incorrect recording of test results

When the lab determines that the testing process was conducted in a flawless manner and there were no mistakes, the report is ready to be released.  At this point in time, any errors that take place are postanalytical.   Examples of these are:
01    failure to notify the physician of critical values.
02    failure to report test results in a timely manner.
03    placement of report in the chart of the wrong patient 
04    miscommunications that are detrimental to the patient regarding
         the tests performed. 

07       EXPLAIN QUALITY CONTROL (QC) IN THE CLINICAL LABORATORY.

Quality control (QC) encompasses quality assurance as it focuses on analytical activities that are associated with the testing process.  QC consists of:
A.    running control samples with patient samples,
B.    using established statistical methods to determine reliability of
       test procedures and test results,
C.    monitoring instrument and laboratorian performance.

 08       DESCRIBE A QUALITY CONTROL SPECIMEN.

First, it can be either a commercially prepared or in-lab prepared specimen.  Second, it has been assayed several times to determine a range of values.  Third, it resembles the patient specimens, but is called a “control”. 

 09       DESCRIBE THE THREE LEVELS OF CONTROL SPECIMENS.

 Abnormal High Controls.  These are specimens that are comparable to
pathological situations.  Use hemoglobin as an example.  An abnormal
hemoglobin standard should not be 50 g/dL.  A practical value would be 20-24 g/dL.
 Normal Controls.  The test values should fall in the range for that of the average population. Normal average hemoglobin values for the female is 12.0 to 16.0 g/dL and the male at 14.0 to 18.0 g/dL.  An appropriate range for a single normal hemoglobin control would be 12 to 18 g/dL.
 Abnormal Low Controls.  This specimen should mimic a pathologically low condition.  For the hemoglobin control, a low value of 2.6 g/dL is impractical, but a low value that ranges from 6.0 to 10.0 g/dL is more practical. 

10       DEFINE AND/OR EXPLAIN PRECISION.

 Precision (or precise) is a measure of performance.  It is the ability to obtain the same results time after time.  If you are throwing five darts at a target and all five darts fall in a tight pattern somewhere on the target, that is precision.   It may lack in accuracy, but the results are consistent and are reproducible in the sense that the darts fall close together.  Statistically the test results must be reproducible.   This precision must be replicated at any time of the day or night, regardless of the laboratorian that might be doing the testing or the kind of instrument or procedure being used.

 11       DEFINE AND/OR EXPLAIN ACCURACY.

Accuracy is a measure of performance.  It is the ability to obtain the true value of a test.  It is like throwing darts at a target and hitting the exact center of the target each time.

 12       DEFINE AND/OR EXPLAIN SAMPLE.

It is a representation of what is being tested.  A drop of blood from a
capillary puncture on a slide for a differential is a representative of the entire volume of circulating blood.  A sample can be a Hematocrit tube of blood or a tube of blood for coagulation studies.   It represents the subset of a population.        

 13       DEFINE AND/OR EXPLAIN POPULATION.

A population is the total number or mass or volume or whatever is being measured.   The entire blood volume of the body represents a population.  The total number of red blood cells or platelets represent a population.

 14       DEFINE AND/OR EXPLAIN RANDOM ERROR.

Random error is an error, mistake, or inaccuracy that has no set pattern.   It cannot be predicted.  It is an attempt to explain the variation of results when many tests are performed on the same sample.  The use of mathematical manipulations designated as “standard deviation”, coefficient of variation”, and “variance” are designed to measure and detect random error or any other type of error.

 15       DEFINE AND/OR EXPLAIN STANDARDIZATION?

 Standardization is a process in which the laboratory uses specimens and controls of known values to establish the performance of values of a procedure.  For example, if the lab wanted to establish the normal hemoglobin range of male students on the ASU campus, the lab would collect blood from a minimum of 40 healthy males of different heights, weights, and races.  The values obtained would constitute the normal range to compare other test results against.  Statistics would be performed to determine the 2± Standard Deviation units.  If the lab can obtain blood from 100 health male students, the standardization would be more accurate.

 16       DEFINE AND/OR EXPLAIN CALIBRATION. 

The laboratory will select a control specimen (with known values) and use it to operate the instrument.  If the test data that is obtained is consistent with the known values, then it is assumed that the machine is performing properly, which is to say, it is calibrated.  This assures that accurate and precise patient results will be obtained and reported.

17       DEFINE AND/OR EXPLAIN THE TERM DELTA CHECK.

A delta check is quality control check that utilizes a patient’s sample that was tested earlier and performing the test again on the same analyte.  If the comparison of the two analytes fall within the established range, it is deemed that the accuracy and precision of the test is acceptable.  This was formerly known as the 'previous patient’s check'. If there is an unusual difference in the two test results, the difference is dalled a delta. 

One method for calculating the units for a delta check begin by subtracting the first patient's test results from the same patient's test results.  The difference is called delta units.   In the next step, convert the delta units to "% delta" with this formula.
% delta  =  100 ×  (result of #2 ― result of #1) and divide this by 2.

 How does the laboratory establish the criteria for delta limits? 
One method may be as simple as using an empirical system in which the limits are  established based upon the experience of laboratory personnel.   In this case the laboratory and pathologist agree what constitutes acceptable limits.   Another method is where the lab sets up the statistical t-test to establish 95% or 99% confidence limits.
The delta limits may be expressed as [1]  percent of allowable change or [2]  number of delta limits of allowable change.

The following are examples of Delta Check Limits for selected analytes.  Remember that each laboratory will probably have their established limits.
Analyte                Normal Range                         allowed
                                                                                change
Albumin             (4.0  -  5.5 g/dL)                         20 %
RBC count         (4.1  -   5.3 × 10⁶/μL)                 10 %
PT                        (10.5 - 13.0 seconds)           <5 seconds
APTT                  (30 - 40 seconds)                <15 seconds

What happens if a large delta value is obtained during the lab testing process?   The lab must check and determine if there is an error due to a specimen mix-up (mis-idenified the patient specimen) or an analytical / testing error.  The lab must repeat the test, even if they have to go back to the patient and collect a new specimen. 

Delta check data tends to be characterized by a high false positive rate.  When the lab rechecks the cause of the delta check, the usual rule is that there is no errors.  It has been found that were there is a false positive, it is due to some type of aggressive therapy I (as IV therapy or renal dialysis).  Patients who are undergoing aggressive therapy tend to have widely variable test results.  Many tests tend to lend well to delta checks and other do not.  RBC counts, indices, platelet counts, and prothrombin times usually provide helpful delta check information.  WBC counts tend to demonstrate wide variations over time and should not be used for delta checks.

Consider a sample Delta Check problem using Prothrombin.
First test result  =   10.9 seconds
Second test result  =  12.4 seconds
From the above formula    12.9 ― 10.9  =  23.8 and divide this by two and the answer is 11.9.  The change is less than 5 seconds and this is an acceptable delta limit. 

18       DEFINE AND/OR EXPLAIN A REFERENCE RANGE.

The reference range is an acceptable range of values for an analyte in a  healthy person.  For example; the reference range for a WBC count may be accepted to be 5,000 to 10,000/μL.  Reference ranges are established using selected populations of healthy individuals.  It is possible to use as few as 25 individuals, but the minimum number should be 40.  100 individuals is preferred because it provides a better distribution of test results.  There are those who advocate for using 500 people, but that number may be logistically impractical.

 19       DEFINE AND/OR EXPLAIN RANGE.

Range is the difference between the lower and upper limits of a variable.  It may be a set of values in a random sample.  If the values obtained from a population of healthy individuals, then it is a normal or reference range.  Any value outside the limits of this range can be designated as abnormal.

 20       DEFINE AND/OR EXPLAIN RELIABILITY.

 Reliability is the continued accuracy and precision in the instruments,
reagents, control specimens, and procedures.  If everything functions as it should, it is dependable.  It describes the ability of a method to maintain its accuracy, precision, sensitivity, specificity and ruggedness.  If personnel come and go, new instruments replace older models, and the lab changes the brand of reagents; then if the methods continues to still produce the right results, it is reliable.

 21       DEFINE AND/OR EXPLAIN SENSITIVITY.

Sensitivity is the ability of the procedure (and reagents) to detect a small amount of the analyte and do so with reasonable accuracy.   In the decade of the 1960's, the Tallqvist’s hemoglobin test was available.  It required a drop of blood to be placed on white filter paper and visually compared to a color chart.  The values only approximated the actual hemoglobin content of blood, hence it was not very sensitive.  This method was replaced by the Newcomer’s method, followed by the Oxyhemoglobin method, and now the current cyanmethemoglobin method.   Each new method represented an increase in sensitivity.  Also improved was precision, accuracy, and reproducibility.

 22       DEFINE AND/OR EXPLAIN SPECIFICITY.

Specificity it the ability of the procedure (and reagents) to detect a single analyte being tested.  An example from body fluid testing will illustrate this concept.  The glucose oxidase test of the reagent strip will detect only glucose, hence it is specific.  The clinitest tablet detects glucose because of the presence of a reducing group, but also reacts with any other substance that contains a reducing group.  Another example is found in the diagnosis of Gaucher disease.  This disorder is characterized by the absence of the enzyme beta-galactosidase which allows glucocerebrosides to accumulate in all the tissues.  The testing system is specific for beta-galactosidase and if this enzyme is missing, there is no reaction.  There are no other enzymes that will react with this substrate and cause a false positive test.

23       DEFINE AND/OR EXPLAIN RUGGEDNESS.

Ruggedness describes the ability of a test method of maintaining it accuracy and precision despite the appearance of a variety of variables that might have the potential to slightly alter the test results.  Variables would include such things as temperature fluctuation, the reconstitution of a reagent, or the instrument itself. 

 24       DESCRIBE A PRIMARY STANDARD.

 A primary standard is that quality control sample in which its analytes are in their pure form and the concentration of each is known.  It is recognized as unique and is accepted as being the most reliable for analyzing with the patient’s sample.

 25       DESCRIBE A SECONDARY STANDARD.

The secondary standard is a quality control specimen whose analytes are known because they have been compared or referenced against a primary standard.

 26       EXPLAIN THE GAUSSIAN CURVE AND HOW IT “FITS” INTO THE LABORATORY QUALITY CONTROL (QC) TESTING STRATEGIES.

 The Gaussian curve (or normal distribution curve) is a plot of many data points obtained by laboratory analysis of a group of normal individuals.  The data points will tend to cluster around an average value or mean.  These data points may be plotted on a graph so that the frequency values of the number of tests or subjects is plotted against the test values and a distribution curve can be plotted.  The following is an example of a Gaussian curve.
  

 27       BRIEFLY DISCUSS CONFIDENCE INTERVALS. 

Confidence intervals are those intervals that contain a specific portion of data.  In the Gaussian curve,
[1]        68.2% of the test results/data lies within ±1.0 S.D. (Standard
Deviation, a specific confidence interval).  Statistically, approximately 68.2% of any given set of data will fall within this interval.
[2]        95.6% of the test results/data lies within ±2.0 S.D. Statistically,
approximately 95.6% of any given set of data will fall within the 2.0 S.D. interval.
[3]        99.7% of the test results/data lies within ±3.0 S.D.  Statistically, approximately 99.7% of any given set of data will fall within the ±3.0 S.D. interval.
         A.   Most all test data will fall within the ±3.0 S.D. interval.
         B.   Approximately 0.28% to 0.30% of test results/data will fall
               outside the ±3.0 S.D. due to random error or chance.
         C.   Any data that falls outside the ±3.0 S.D. interval is said to be
               suspect and not fitting the norm.

28       DEFINE AND/OR EXPLAIN MEAN (0 ).

The mean is taking all the test values and determining the sum of those values.  Next, the sum of the test values (∑x) is divided by the number of the test values (n) performed.  The mathematical value obtained is called the mean.  It may be written as 0 (also called the X- bar) and referred to as the arithmetic average.  The mean is calculated to measure the central tendency.   The formula is written as follows:
                    ∑x             
         0    =  -----         
                    n            

 29       DEFINE AND/OR EXPLAIN MODE. 

Mode describes the most frequently reoccurring number in a set of data points or values.  For example, if the data points are obtained in the following order: 1, 4, 3, 2, 4, 1, 2, 3, 4, 4, 0, 5, 2, 1,  and 0; then the mode is four.  Note that “0" occurs twice, “1" occurs thrice, “2" occurs thrice, “3" occurs twice, “4" occurs four times, and “5" occurs once.  Mode simply identifies the most frequently occurring variable in the set of data points. 

 30       DEFINE AND/OR EXPLAIN MEDIAN. 

The word median implies a measure of central tendency.  It is obtained by ranking all the data points in numerical order of increasing values (from low to high).  The median is that data point that falls exactly between the high and low points.  For example; in objective 28, there are 15 data points and if numerically arranged, they would appear as: 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, and 5.  Two is the middle value (seventh number),  hence the number 2 is designated as the  median. 

 31       DEFINE, SET UP, AND/OR EXPLAIN STANDARD DEVIATION. 

The standard deviation (written as S or SD) describes how the data falls about the mean.  This distribution is expressed as a statistical measurement.   The formula is written as follows:  
                             
(1)     First step . . . .  find the “mean”
(2)     Step two . . . . determine the difference between 0  and the
         separate values of x.   This will be the (0 - x) in the equation.
(3)     Step three . . . .   square the differences of (0- x)²
(4)     Step four . . . .     total and obtain the sum of the differences of
          the squares,
(5)     Step five . . . .     divide the sum of the differences of the
          squares by n - 1          
(6)     Step six . . . .       extract the square root of the quotient.

 32       WHEN GIVEN DATA, THE STUDENT WILL BE ABLE TO CALCULATE THE STANDARD DEVIATION.  (NOTE: The instructor will work through a sample problem using the following data.) 

The data points are:  5.1, 5.2, 5.3, 5.4, 5.2, and 4.8.  Note that there is only a total
of six data points.
A.     The sum (∑x) of the data points is 31.0
B.     To find the mean, divide sum by the number of replications.
C.     Next step is to find (0 - x)²   
         5.2 - 5.1      =      (0.1)²      =        0.01
         5.2 - 5.2      =      (0.0)²      =        0.0
         5.2 - 5.3      =      (-0.2)²     =        0.01
         5.2 - 5.4      =      (-0.2)²     =        0.04
         5.2 - 5.2      =      (0.0)²      =        0.0
         5.2 - 4.8      =      (0.4)²      =        0.16
         a.   In the column (0 - x), the intent is to determine the
               difference between the mean value and each of the data
               points.
         b.   The differences are then squared to form the set of values
               to yield each independent (0 - x)².
D.      Step four.   Find the sum (∑) of (0 - x)²
          a.   The addition of the column "(0- x)²"  provided ∑(0- x)²  = 
                 0.22
E.      Step five.  Use the formula in Objective 30 to calculate the
          standard deviation.
                 

33       WHEN GIVEN STATISTICAL DATA, DESCRIBE HOW THE ONE, TWO, THREE STANDARD DEVIATIONS DISTRIBUTE IN THE GAUSSIAN CURVE. 

It has been consistently demonstrated in a number of statistical studies that data presents a typical frequency of distribution.  This is called a Gaussian or normal distribution curve.  This curve begin with the frequency of values continuing to increase until the mean is reached and then the frequency of values will begin to fall away producing a mirror image of the increasing slope. 

Consider the following problem where n = 41 and data points are distributed from a low of 4.0 to a high of 32.0.  Note the instructor will present this problem on the board to demonstrate how to interpret a typical statistical Gaussian curve.  Two problem sets that correlate with objectives 31, 32, and 39 are located in Blackboard in the section designated as 'Course Information'. These are to be completed and turned in on the 5^th classroom day following completion of this lecture unit.

Data Points   Frequency of
                         occurrence
     4                   1          In the way that these data points are listed
      8                   3          the value of 4 occurs just one, the value of 8
     14                  7          occurs three times, the value of 14 occurs
     18                  9          seven times, etc. for a total of 41 data     
     20                  9          points.   
     24                  7
     28                  4
     32                  1

A.   The sum (∑) of the data points (n) =   780
B.   The minimum data point is '4' and the maximum data point is '32'.
C.   The range of these data points is from 4 to 32 with a 28 point
       spread.
D.   The mode is 19 because it falls between 18 and 20. 
E.    Median is 20
F.    Mean (0)  =   19.02   (for this course, adjust the decimal point to
      the nearest tenth and since .02 is not sufficient to scale up to .1,
      then drop it).  The adjusted value should now be 19.
G.  The standard deviation (S or SD) is 6.0 which is calculated as
      follows:

Data     Frequency    (0 - x)²                           (0- x)²       Frequency
Points  Occurrence   Step 1            Step 2        Step 3      times (× - x)²
   4             1                  (19 - 4)²           (15)²          225               225
   8             3                  (19 - 8)²           (11)²          121                363
  14            7                   (19 - 14)²         (5)²            25                 175
  18            9                   (19 - 18)²         (1)²              1                    9
  20            9                  (19 - 20)²         (-1)²             1                    9
  24            7                   (19 - 24)2         (-5)²           25                 175
  28            4                   (19 - 28)²         (-9)²           81                 324
  32            1                   (19 - 32)²         (-13)²        169                 169
The ∑ of (× - x)²  is equal to 1449.
The calculation of the standard deviation (S or SD) is as follows:

          
To calculate the SD for the Gaussian curve, if values are to be plotted
to the left of the curve (decreasing values) subtract from the mean as
follows:
A.   For -1.0 SD, subtract 6 from 19 and the value becomes 13.
B.   For -2.0 SD, subtract 12 (two SD equivalents) from 19 and the value
       becomes 7.
C.   For -3.0 SD, subtract 19 (three SD equivalents) from 19 and the
       value becomes 1.

To calculate the SD for right side of the Gaussian curve (increasing
values), add to the mean as follows:
A.    For +1.0 SD, add 6 to 19 and the value becomes 25.
B.    For +2.0 SD, add 12 (two SD equivalents) to 19 and the value
       becomes 31.
C.    For +3.0 SD, add 18 (three SD equivalents) to 19 and the value
       becomes 37.  Examine the following Gaussian curve and notice how it
       conforms to the calculations.

34       EXPLAIN THE PURPOSE OF THE COEFFICIENT OF VARIATION (CV) IN STATISTICS AND HOW IT IS CALCULATED.

The coefficient of variation (CV) is a mathematical way to compare variability between analytes.  It attempts to measure the degree of precision in testing by depicting the Standard Deviation as a percentage of the mean..  The smaller the percentage the more precise the procedure/testing.  It is expressed mathematically as:
                                                SD
                                    CV  =  ----  (x) 100
                                                0

To calculate the CV, assume that the SD =  0.2 and 0  =  5.2 :
                 CV = 0.2/5.2  (x)  100
                        CV = 0.038  (x)  100
                                CV = 3.85%
This is not a large percentage value and it therefore indicates a reasonably precise procedure for testing.

35       EXPLAIN THE CONCEPT OF THE STATISTICAL TERM, “VARIANCE” AND ITS USE IN LABORATORY STATISTICS.

Variance (s²) is a mathematical attempt to statistically measure variability (the degree of precision in testing).  It is represented by the equation ∑(0 - x)²/n - l.  This concept provide little useful information about testing precision and is seldom used.

 36       EXPLAIN HOW “SKEWNESS” IN DISTRIBUTION CURVES CAN PROVIDE QUALITY CONTROL INFORMATION. 

If test values do not distribute appropriately about the mean, the resulting curve is said to be skewed.  This means that frequency distributions will present curves that are not a typical Gaussian curve and/or the curve extends in the wrong direction.  If the curve is located to the right of the mean, it is called positively skewed.  This means that the curve is biased toward maximum values. If the curve
is to the left, it is negatively skewed and is biased toward minimum values..  This means that the mode, median, and mean have shifted in relationship to each other.  Note.... Skewedness is a lack of symmetry.
       

 37       EXPLAIN WHAT IS MEANT BY SYSTEMATIC ERROR. 

Systematic error is that which occurs within the test system.  It may be caused by such things as  [ 1 ] calibration error, [ 2 ] malfunction of the instrument (a reagent line stops up or a component fails), or [ 3 ] defective reagents.  This is not to be interpreted as  random error or an interfering substance. 

38       EXPLAIN WHAT IS MEANT BY CONSTANT SYSTEMATIC ERROR. 

There is something about the testing system that is causing a constant bias to be added to the test results.  Two examples that would describe this problem are [ 1 ] a weak photocell that needs replacement, and [ 2 ] a control reagent that has a component is too dilute or  too concentrated. 

39        EXPLAIN HOW AN INTERNAL QUALITY CONTROL SYSTEM DIFFERS FROM AN EXTERNAL ONE. 

An internal quality control consists of:
[1]        Running assayed QC controls with patient specimens.
[2]        Plotting data from the QC controls on graphs to detect shifts,
               trends or outliers.
[3]        When opening new controls, that the “new” values be compared
               to the “old” values in order to validate their reliability.   

 [4]    If the internal QC detects outliers (these are values that fall
         outside the acceptable limits of the test), [ 1 ] make a record that
         explains the problem and [ 2 ] describe the action that was taken
         to detect the error.

 An external quality control consists of:
[1]    Lab A has an agreement with another lab ( B ) that uses the same
        controls and instruments that are employed in Lab A.   QC results
        of both labs are compared to each other.
[2]    If there is a significant difference between the two, neither lab
        will substitute the other’s results for theirs.  The difference will
         be resolved by locating the problem.

 40       DESCRIBE A LEVEY-JENNINGS (L-J) CHART AND HOW IS USED IN THE LABORATORY.

 The Levey-Jennings (L-J) chart employs the mean and standard deviations of the Gaussian curve and plots these two parameters (vertical axis) against time (horizonal axis).  This chart will appear as follows:
           

Notice how the left end of this chart looks like the Gaussian curve placed on its side.

 41       ILLUSTRATE A NORMAL CONTROL PLOT USING A LEVEY-JENNINGS (L-J) CHART.

 Note that the chart is set up so that all the data points fall within the ±2.0 SD ranges.  Any data point that falls in this range is generally considered to be acceptable performance.
           

 42       DISCUSS THE WESTGARD’S MULTI-RULE SYSTEM.

 This is a assessment system to identify out-of-control quality control (QC) results.  James O Westgard proposed six rules to determine acceptance or rejection of QC results in a Levey-Jennings chart.  It is based upon the following symbol: A_NS.  A = rule designation that infers the number of control results.  N = number of SD’s, and S = standard deviation.

Rule 1_3S            One control value is in question and it value falls outside the 3SD limit.  This violation suggests that it is a possible random error.  The laboratorian should determine the cause for this result and determine if the test run is
acceptable.  Generally the control run is not accepted.
Rule 1_2S            This is one control value and it falls between the 2 and 3 SD limits.  This may be a warning of a possible instrument malfunction.  It also is suggestive of a systematic error (see Objective 37.)
Rule 4_1S           There are four consecutive control value data points plotted between 1 and 2 SD on the same side of the mean on the Levey-Jennings graph.  This is a warning that a possible trend or shift may be occurring or possibly systematic error.   The control run should be rejected and investigated to determine the possible cause.
Rule R_4S           This is a situation where there are two controls in the same run and the two consecutive values are more than four standard deviations apart.  One control value is outside the + 2 SD limit and the other control value is outside the ­2 SD limit.  This may be a possible random error.  Investigate
Rule 2_2S            One control is being used and its data points have fallen between the same 2 and 3 standard deviation limits on two consecutive days or at different times of the same day.  Investigate as possible random error.
10_×   When using a single control and ten consecutive control values fall on the same side of the mean.  This is an indicator of systematic error.  If the lab is using two controls and each falls on five consecutive sides, then investigate.

43       ILLUSTRATE A LEVEY-JENNINGS CHART THAT DEMONSTRATES A SHIFT.  LIST THE TWO MOST PROBABLE CAUSES FOR THESE RESULTS.

 Shifts tend to appear suddenly.  There is a rapid change away from the mean, producing a bias that is consistent in one direction.  In this first example, notice how the plots are shifting upward.  This example represents the shift and it is occurring on one side of the mean.  Shifts differ from trends by suddenly appearing on one side of the mean or other.  Also included are three examples of Westgard's rules violations.
               
This example would not be seen in a laboratory since corrective action would have taken place on day eleven or twelve.             

      This second example of a shift that occurs within the two standard deviations limit.
                 
In this second example, the shift stay within the ±2 SD's .  The laboratory would closely monitor this movement of the data points and if the control test results did not fall downward on the mean or below it, they would take corrective action.

Instrument malfunction (weakening of photolamps and/or photo-cells), methodology, and defective controls (through contamination or deterioration) are the more common causes of shifts.   Other factors that are known to affect shifts are power surges, improper calibration, and technical error.

A shift is generally considered (by many labs) to occur when there is the placement of six control value plots on one side of the mean.   Remember that a shift usually occurs abruptly.

If the shift occurs on the positive side of the Levey-Jennings Chart it is said to be an “up” shift.   “Up shifts” can be caused by:
      [1]      partial electric failure in the instrument.
      [2]      over diluted new standard.
      [3]      reagent that undergoes some degree of deterioration then
                     stablilizes.
      [4]      contaminated reagent(s).
      [5]      inaccurate timer (yielding slower times).
      [6]      over heating of reaction mixtures.
 If the shift occurs on the negative side of the Levey-Jennings Chart, it is said to be a “down” shift.   “Down shifts’ can be caused by:
      [1]      evaporation of reagents causing a concentrated effect.
      [2]      contamination of reagent(s).
      [3]      inaccurate timer (yielding faster times).
      [4]      under heating of the reaction mixture.
      [5]      standard is too concentrated
      [6]      contaminated glassware.

NOTE: Instrument malfunction, methodology, defective controls, power surges, improper calibrations, and technical error can contribute to shifts and/or trends.

44       ILLUSTRATE A LEVEY-JENNINGS CHART THAT DEMONSTRATES A TREND.  LIST THREE  PROBABLE CAUSES FOR THESE RESULTS.

Trends (may be referred to as drifts) may be gradual changes (and may be subtle changes) that drift in one direction over time, usually presenting when six consecutive plots occur in the same general direction occurring on the chart.  Trends may be due to [1] slowly deteriorating reagents, [2] problems with the instrument tubing, or [3] a weakening in the light detector component of the photometer unit.  Trends can be caused by the same factors that causes shifts.  The following is a textbook example of a trend that shows the following;
      A.     Two Westgard rules violations assuming that corrective
               action were not taken which allowed the trend continued.
      B.     A data point that falls on the +2 SD which would alert the
              laboratory to take corrective action and look for the factors
              that may be causing this consistently upward movement. 
      C.     This example would not expected to be seen in a laboratory
               as they would have noted that there was an upward trend
               starting on day 7 and would have initiated corrective action
               on day 12 when the control test results failed to fall below
               the mean.                    
        D.    When a data point falls outside the acceptable limits, the
                laboratory will initiate corrective action and repeat the
                test.  In this example, preventive maintenance corrected
                the problem and the repeated control test resulted in an
                acceptable data point.

Second example of a trend and notice the downward movement of the data points across the mean:
                   

45       ILLUSTRATE A LEVEY-JENNINGS CHART THAT DEMONSTRATES DISPERSION AND LIST TWO POSSIBLE CAUSES FOR THIS RESULT.

Dispersion occurs when there is an increase in random errors or an increase in lack of precision.  Two examples of causes might be a [1] fluctuating electrical voltage (stability problem) or [2] poor mixing of control specimens (inconsistency in technique).   Look for widely scatter data points.
                   

46       DESCRIBE AND/OR ILLUSTRATE A LEVEY-JENNINGS CONTROL CHART R_4s RULE VIOLATION.

 This chart is plotted using two levels of controls.  When two consecutive data plots are more than 4 standard deviations units apart (one plot beyond the - 2 SD limit and the other beyond the + 2 SD limit) then the chart is in violation of the 4_RS rule.

47       DESCRIBE HOW TO USE A TWIN PLOT CHART FOR QUALITY CONTROL MONITORING. 

Also known as the Youden plot, this is a graph that is plotted using two controls.  This plot chart is two Levey-Jennings charts placed at right angles to each other.  The controls can be any combination of high, normal, or low concentrations.  Plot control results on a daily basis. 
01    If the data points tend to cluster at corners B or D then a shift or      trend is developing. 
02    If the data points fall outside the ±2.0 SD limits, yet close to the limits, systematic error is likely. 
03    If the plots fall away from the diagonal line may be an indicator of random error.  
04    If a data point falls within any of the shaded areas (A, B, C, D), this represents a reading beyond the ±2.0 SD limits and required immediate corrective action. 
05    If the data points fall within the central square (dots), which represents 1.0 SD, then there is zero differences in the two controls and the data is highly accurate.

       
        Note that this Q.C. plot technique is better performed with a computer program.     

48            DISCUSS OR DESCRIBE THE "P" VALUE OR "PROBABILITY LEVEL.

The "P" value, also called the "probability level" is the probability of obtaining a value of a test result that is as large as or larger than the table value calculated for that statistical system.  The "P" value is a decimal which represents the probability expressed as a percentage.  The "P" value as utilized ina testing system for the "t" test for "F" test referes to the correctness of the assumption that the null hypohtesis is true.  NOTE:  The "null hypothesis" states that there is NO significant difference between two sets of test data.  If something is evaluated at the 0.01 probability level an it is found that the test statistic obtained is larer than that allowed by the table value, then this indicates that in one out of one hundred times (or 1.0 % of the time) thre will be no difference between the two sets of data.   It also means that 99% of the time there will be a difference between the two sets of data.   The recommendation then becomes that the null hypothesis NOT be accepted.  If the test value occurs at the 0.05 probability level, where the test values are larger than the table value, then there will be an occurrence of five times out of one hundred  (5% of the time), where there will be no difference in the data, but 95 times, the null hypothesis is wrong and there is a difference.  Any time the "t"  or "F" test values exceed the table values for the probability level that is selected, then the differences between the sets of data is significantly different.

49        BRIEFLY DESCRIBE THE “t” TEST AND ITS PURPOSE IN STATISTICS.

The “t” test is also called the “paired ‘t’ test”, "t" distribution, and the student test.  It is a statistical tool to compare the accuracy of two testing methods by comparing the mean values of two procedures.  This test is based upon the premise that there are no differences in the two tests.  This statistical concept is called the null hypothesis.  The “t” value that is obtained in the analysis must be compared to an established table of “t” values.  If the calculated “t” value is equal to or larger than the “t” value in the table, the null hypothesis is declared invalid and there is a significant difference in the two procedures.  If the calculated “t” value is less than the table’s “t” value, there is no difference between the two procedure, therefore the null hypothesis is valid. The common practice is to accept or reject the null hypothesis at either the 0.01 (1%) or 0.05 (5%) probability level, also designated as the "P" or "critical level".  The 0.05 probability level is used most often.

The "t" test is a mathematical tool to compare a Gaussian-type curve developed from a small sample of data with a normal Gaussian curve.  The normal Gaussian curve is developed from a large sample of data which tends to minimize error.  The "t" test represents a family of curves  and the shape of each curve is a function of the sample size.  The "t" tables provide information about the information that is inherent in each curve that is determined by the size of the sample. 

50     BRIEFLY DESCRIBE THE “F” TEST AND ITS PURPOSE STATISTICS.

 The “F” test is a statistical method to compare the precision of two methods.  It uses the SD’s of two procedures in the following formula:

                                    F = (SD)²/(SD)²

The SD (the numerator) is from the method with the larger variance.  The SD (the denominator) is from the method with the smaller variance.  The value of F, obtained by this division process is compared to an established table of critical values of “F”.  If the calculated “F” value is equal to or higher than the established critical “F” value, the procedure with the smaller SD value will be more precise.  If the calculated “F” value is less than the established critical “F” value, the procedures are not different.  The "F" test only determines if the variances are different or similar.

The following example shows how the "F" test can be used to compare two procedures whether on two different testing instruments or two test kits.   Assume that the following results were obtained on two test kits and the values shown are in mg/dL.
        Date        Current          New
                      Test Kit       Test Kit
         1              22.7            23.2            Note:  Four tests were not
         1              -----            22.9             performed with the current 
         2              23.1            24.3             kit on days 1, 3, and 5.
         2              24.8            23.0
         3              -----            24.1
         4              23.0            23.1
         5              -----            25.3
         5              -----            24.3
         5              22.5            22.7
         6              23.1            25.3
         7              24.5            24.3
         8              22.6            25.0
         9              22.8            24.0
         9              25.1            22.5       
         10             22.4            22.8
         11             23.8            22.9

        ∑               280.5           405.1       The "F" Test is a test of a null
        n                12                16          hypothesis which states that
        mean          23.32            23.83     there is no difference 
        SD (S)           0.96             0.99      between the two kits.
        SD²                  0.917            0.986
        "F" Test  =  0.986 divided by 0.917
        "F" Test  =  1.08

To evaluate the "F" test value, it is required that a "F" Test Degree of Freedom Table be available.  The current test kit tests presents 12 sets of data points.  To find the 'degrees of freedom' (d.f.), use the formula "n - 1" and 'n' equals 12.  For the new kit, 16 sets of data points were presented and 'n' becomes 15.  Using the 'degree of freedom' table, locate the degrees of freedom (denominator) row and locate the number '12'.  Match this with the degrees of freedom (numerator) column and locate the critical value number for either 0.01 or 0.05 probability.  Assume that you elect to compare at the 0.05 probability level, then the critical value will be 2.70 for the "F" test.  The calcuated "F" test value is 1.08 which is smaller than the critical value.  This means that the 'null hypothesis' is correct and there are no differences between the two testing methods.   NOTE:  If the calculated "F" test value had been larger that the critical value of the 0.01 or  0.05 probability levels, the null hypothesis would have to be rejected because the large difference between the two variables supports the hypothesis that the two tests are significantly different.

51       EXPLAIN HOW TO DETERMINE A NORMAL RANGE AND ESTABLISH A NORMAL REFERENCE INTERVAL.

The purpose of establishing a normal range is to determine what are acceptable lab values to be used for identifying abnormal patients.  When setting up the range of values, the following factors must be considered: [1] age, [2] geographical location, [3] gender, [4] test method, [5] state of health, [6] body mass, and [7] diet.  Once the population has been identified, ideally select a group of 100 people.  Perform and record the test results for each person.  Perform the following calculations to determine the statistical data: mode, range, minimum and maximum values, mean, and standard deviation.   Evaluate your results and ask yourself the following questions:
            [1]   Does approximately 68.2% of the test subjects data fall
                    within ±1 SD?
            [2]   Does approximately 95.6% of the test subjects data fall
                   within ±2 SD?
            [3]   Does approximately 99.5% of the test subjects data fall
                   within ±3 SD?

If you find that the data that you have collected is statistically satisfactory, then it may be interpreted as follows:
[1]        If the patient’s test results fall within ±2.0 SD, then the
               patient’s values are NOT abnormal
[2]        If the patient’s test results fall between ±2.0 and ±3.0 SD, then
              the patient’s values may be abnormal.  The determination will
              be a professional judgment and may be in error.
[3]        If the patient’s test results are outside the ±3.0 SD limits, the
               patient’s results may be considered abnormal.

NOTE:   If the test results that are derived from the normal/healthy population produce a skewed Gaussian distribution curve, then a more sophisticated statistical procedure is required.  See a statistician to help you set up a normal curve.

This web site is maintained by Whitney Williams, wwilliam@astate.edu

This page last updated 07/28/08