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Quality Control

Quality Control Procedures put into place to monitor the performance of a laboratory test with regard to accuracy and precision. Question What is the difference between accuracy and precision?. Accuracy – measure of how close experimental value is to true value.

alohilani-matthew
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Quality Control

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  1. Quality Control Procedures put into place to monitor the performance of a laboratory test with regard to accuracy and precision

  2. QuestionWhat is the difference between accuracy and precision? Accuracy – measure of how close experimental value is to true value Precision –measure of reproducibility

  3. Ways to estimate true value • Mean (average) (X) X = Σxi/n xi - single measured value n- number of measured values 2.Median – Order xi values, take middle value (if even number of xi values - take average values of two middle values) 3. Mode – most frequent xi value If above is to estimate the true value what does this assume?

  4. Proposedway to measure precision Average Deviation = Σ(X – xi)/n Does this estimate precision? No – because the summation equals zero, since xi values are less than and greater than the mean

  5. Ways to Measure Precision • Range (highest and lowest values) • Standard Deviation

  6. Σ(xi – X)2 s = (n-1) Standard Deviation s – standard deviation xi - single measured value X – mean of xi values n - number of measured values

  7. Variations of Std Dev • Variance - std. dev. squared (s2) Variances add, NOT std deviations. To determine total error for a measurement that has individual component standard deviations for the measurement s1, s2, s3, etc [i.e., random error in diluting calibrator (s1), temperature change (s2), noise in spectrophotometer (s3), etc.] (stotal)2 = (s1)2 + (s2)2 + (s3)2 + …

  8. Variations of Std Dev (cont.) 2. Percent Coefficient of Variation - (%CV) %CV = 100 * S/X

  9. + + - - Monitoring Performance with Controls 1.Values of controls are measured multiple times for a particular analyte to determine: a) “True value” – usually X b) Acceptable limits - usually 2s 2. Controls are run with samples and if the value for the control is within the range X 2s then run is deemed acceptable

  10. Determining Sample Mean and Sample Std Dev of Control (Assumes Accurate Technique) Methodology for Analyte End Data Analysis Result s X Sample Std Dev Sample Mean Control with Analyte Repeat “n” times

  11. Determining True Mean and True Std Dev of Control (Assumes Accurate Technique) Methodology for Analyte End Data Analysis Result σ μ True Std Dev True Mean Control with Analyte ∞ Repeat times

  12. There is another way!!! Statistics

  13. Sample (of population) Population Take finite sample X s μ σ Gives range around X and s that μ and σ will be with a given probability Statistics

  14. Rather than measuring every single member of the population, statistics utilizes a sampling of the population and employs a probability distribution description of the population to “estimate within a range of values” µ and σ

  15. Number or frequency of the value Value Probability Distribution Continuous function of frequency (or number) of a particular value versus the value

  16. Number or frequency of the value Value Properties of any Probability Distribution • Total area = 1 • The probability of value x being between a and b is the area under the curve from a to b a b

  17. The most utilized probability distribution in statistics is? Gaussian distribution Also known as Normal distribution Parametric Statistics – assumes population follows Gaussian distribution

  18. µ - µ - µ + µ - µ + µ + 3σ 2σ 3σ 2σ 1σ 1σ Gaussian Distribution • Symmetric bell-shaped curve centered on μ • Area = 1 3.68.3% area μ + 1σ (area = 0.683) 95.5% area μ + 2σ (area = 0.955) Number or frequency of the value 99.7% area μ + 3σ (area = 0.997) μ x (value)

  19. 0.683 0.955 µ - µ - µ - µ + µ + µ + 0.997 3σ 2σ 2σ 3σ 1σ 1σ What Gaussian Statistics First Tells Us Area under the curve gives us the probability that individual value from the population will be in a certain range These are the chances that a random point (individual value) will be drawn from the population in a given range for Gaussian population Number or frequency of the value 1) 68.3% chance between μ + 1σ and μ - 1σ 2) 95.5% chance between μ + 2σ and μ - 2σ 3) 99.7% chance between μ + 3σ and μ - 3σ μ x (value)

  20. 1 f(x) = 2 πσ2 µ - µ - µ - µ + µ + µ + µ 1σ 1σ 2σ 3σ 3σ 2σ Gaussian Distribution Equation e Number or frequency of the value [f(x)] -(x - µ)2 2σ2 x (value)

  21. 1 e f(x) = 2 πσ2 Gaussian curves are a family of distribution curves that have different µ and σ values A. Changing µ -(x - µ)2 2σ2 B. Changing σ

  22. µ - µ - µ - µ + µ + µ + µ 1σ 1σ 2σ 3σ 3σ 2σ To determine area between any two x values (x1 and x2) in a Gaussian Distribution x2 e 1 dx Area between = x1 and x2 f(x) = 2 πσ2 Number or frequency of the value [f(x)] -(x - µ)2 2σ2 x1 x1 x2 x (value)

  23. 1 dx e Area = 2 πσ2 Any Gaussian distribution can be transposed from x values to z values x value equation x2 z2 1 -(z)2 -(x - µ)2 2 π 2 2σ2 x1 z = (x - µ)/σ z1 z value equation e Area = dz

  24. To determine the area under the Gaussian distribution curve between any two z points (z1 and z2) z2 1 e Area between z1 andz2 = 2 π dz -(z)2 2 z1

  25. Transposition of x to z z = (x - µ)/σ The z value is the x value written (transposed) as the number of standard deviations from the mean. It is the value in relative terms with respect to µ and σ. z values are for Gaussian distributions only.

  26. 1 dx e Area = 2 π(2.5)2 At this point, we can use Gaussian statistics to determine the probability of selecting a range of individuals from a population (or that an analysis will give a certain range of values). What is the probability that a healthy individual will have a serum Na concentration between 141 and 143 mEq/L (σ = 2.5 mEq/L)? 143 -(x - 140)2 Normal range of [Na] in serum 2(2.5)2 141 140 135 145 You could theoretically do it this way, however the way it is done is to transpose and use table [Na] mEq/L (x)

  27. Normal range of [Na] in serum -2 -1 0 1 2 z 135 140 145 1.2 0.4 x To do this need to transpose x to z va and use the table What is the probability that a healthy individual will have a serum Na concentration between 141 and 143 mEq/L (σ = 2.5 mEq/L)? Transpose x values to z values by: z = (x – μ)/σ Which for this problem is: z = (x – 140)/2.5 Thus for the two x values: z = (141 – 140)/2.5 = 0.4 z = (143 – 140)/2.5 = 1.2

  28. Normal range of [Na] in serum -2 -1 0 1 2 z 135 140 145 1.2 0.4 x To do this, need to transpose x to z values and use the table What is the probability that a healthy individual will have a serum Na concentration between 141 and 143 mEq/L? So to solve for area: 1. Determine area between z=0 to z = 1.2 Area = 0.3849 (from table) 2. Determine area between z=0 to z=0.4 Area = 0.1554 (from table) 3. Area from z=0.4 to z=1.2 0.3849 – 0.1554 = 0.2295 Answer: 0.2295 probability

  29. Our goal: To determine μ Cannot determine μ What can we determine about μ ?

  30. The Problem Establishing a value of μ of the population The Statistics Solution 1. Take a sample of X from the population. 2. Then from statistics, one can make a statement about the confidence that one can say that μ is within a certain range around X

  31. Sample (of population) Population Take finite sample X s μ σ Gives range around X and s that μ and σ will be with a given probability Statistics

  32. Distribution of Sample Means How Statistics Gets Us Closer to μ

  33. μ X1 X2 X3 XN Distribution of Sample Means – Example of [Glucose]serumin Diabetics Sample means are determined Population of Diabetics For this example: n=25 N=50 n - sample size (# of individuals in sample) N – number of trials determining mean

  34. By theory, the distribution of sample means will follow the Central Limit Theorem

  35. Sample means (X) of taken from a population are Gaussian distributed with: • mean = μ(μ true mean of the population) • std dev = (σ is true std dev for the population, n is sample size used to determine X) • [called standard error of the mean (SEM)] σ/ n Central Limit Theorem • Conditions: • Applies for any population that is Gaussian [independent of sample size (n)] • Applies for any distributed population if the sample size (n) > 30 • Assumes replacement or infinite population

  36. μ + Number or frequency of X 1σ/ n μ - μ + μ + μ - X 1 SEM 2 SEM 1 SEM 2 SEM μ + 2σ/ n μ - μ - μ 1σ/ n (Sample Means) 2σ/ n Central Limit Theorem μ is true mean of the population σ is true std dev for the population n is sample size used to determine X)

  37. SEM =σ/ n μ - μ + μ - μ + X 1 SEM 2 SEM 1 SEM 2 SEM μ (Sample Means) The absolute width of the distribution of sample means is dependent on “n”, the more points used to determine X the __________ the width. smaller ?

  38. Larger sample size “n” SEM =σ/ n Smaller sample size “n” X (Sample Means)

  39. SEM =σ/ n -(X - µ)2 1 2SEM2 e f(X) = Transposing: z = (X - µ)/SEM 2 π SEM2 μ - μ - μ + μ + μ -(z)2 1SEM 2SEM 1SEM 2SEM X (Sample Means) f(z) = 2 e -3 -2 -1 0 1 2 3 z value 1 2 π

  40. For the sample mean distribution of X values, z= z = (X – μ)/SEM What does a z value mean? The number of standard deviations from the mean. z values are for Gaussian distributions only. For the population distribution of x values, z= Std dev = σ and mean = μSo z = z = (x - µ)/σ Std dev = SEM and mean = μSo z =

  41. How the distribution of sample means is used to establish the range in which the true mean μ can be found (with a given probability or confidence) 1) An experiment is done in which ONE sample mean is determined for the population 2) Because the distribution of sample means follows a Gaussian distribution then a range with a certain confidence can be written

  42. μ - μ + μ - μ + Area = 0.955 1 SEM 2 SEM 1 SEM 2 SEM μ – 2SEM < X < μ + 2SEM μ X (Sample Means) There is a 95.5% chance (confidence) that the one determination of X will be in the range indicated. This range can be written mathematically as: However this does not answer our real question, we want the range that μ is in!

  43. μ – 2SEM < X < μ + 2SEM – 2SEM < X - μ < + 2SEM -X– 2SEM < - μ < - X + 2SEM Subtract X from each part of the expression +X +X X -2SEM < μ < +2SEM X We have are the 95.5% confidence limits for X What we want are the 95.5% confidence limits for μ We get this by simply rearranging the expression Subtract μ from each part of the expression Multiply each part of the expression by -1 +2SEM > +μ > - 2SEM Writing so range is given as normal (going from lower to upper limit)

  44. -2SEM < μ < +2SEM This 95% confidence range for μcan be written as the following + expresion: +2SEM X X X

  45. +?SEM +? SEM +? SEM X X X A range for μ can be written for any desired confidence 99.7 % confidence? 68.3 % confidence? 75.0% confidence? What z value do you put in?

  46. For 75% confidence need area between +/- z value of 0.750 0 z value

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