Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 Precision –measure of reproducibility Ways to estimate true value 1. 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? Proposed way 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 Ways to Measure Precision 1.Range (highest and lowest values) 2. Standard Deviation Standard Deviation s= Σ(xi – X)2 (n-1) s – standard deviation xi - single measured value X – mean of xi values n - number of measured values Variations of Std Dev 1. 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 + … Variations of Std Dev (cont.) 2. Percent Coefficient of Variation (%CV) %CV = 100 * S/X 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 Determining Sample Mean and Sample Std Dev of Control (Assumes Accurate Technique) Methodology for Analyte Result End Data Analysis X Control with Analyte Repeat “n” times s Sample Sample Std Dev Mean Determining True Mean and True Std Dev of Control (Assumes Accurate Technique) Methodology for Analyte Result End Data Analysis μ True Mean Control with Analyte Repeat ∞ times σ True Std Dev There is another way!!! Statistics Sample Population μ σ (of population) Take finite sample Statistics X s Gives range around X and s that μ and σ will be with a given probability 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 Probability Distribution Number or frequency of the value Continuous function of frequency (or number) of a particular value versus the value Value Properties of any Probability Distribution 1. Total area = 1 Number or frequency of the value 2. The probability of value x being between a and b is the area under the curve from a to b a b Value The most utilized probability distribution in statistics is? Gaussian distribution Also known as Normal distribution Parametric Statistics – assumes population follows Gaussian distribution Gaussian Distribution 1. Symmetric bell-shaped curve centered on μ Number or frequency of the value 2. Area = 1 3. 68.3% area μ + 1σ (area = 0.683) 95.5% area μ + 2σ (area = 0.955) 99.7% area μ + 3σ (area = 0.997) µ µ µ - - - 3σ 2σ 1σ μ µ µ µ + + + 1σ 2σ 3σ x (value) 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 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 Number or frequency of the value These are the chances that a random point (individual value) will be drawn from the population in a given range for Gaussian population 0.683 0.955 0.997 µ µ µ - - - 3σ 2σ 1σ μ µ µ µ + + + 1σ 2σ 3σ x (value) Gaussian Distribution Equation f(x) = 1 2 πσ2 e Number or frequency of the value [f(x)] -(x - µ)2 2σ2 µ µ µ - - - 3σ 2σ 1σ µ µ µ µ + + + 1σ 2σ 3σ x (value) Gaussian curves are a family of distribution curves that have different µ and σ values -(x - µ)2 f(x) = 1 2 πσ2 e A. Changing µ 2σ2 B. Changing σ To determine area between any two x values (x1 and x2) in a Gaussian Distribution x2 1 Area f(x) = between = x1 and x2 2 πσ2 e Number or frequency of the value [f(x)] -(x - µ)2 dx 2σ2 x1 µ µ - - x1 3σ 2σ µ 1σ x2 µ µ µ µ + + + 1σ 2σ 3σ x (value) Any Gaussian distribution can be transposed from x values to z values x value equation x2 -(x - µ)2 1 Area = 2 πσ2 e 2σ2 dx z = (x - µ)/σ x1 z value equation z2 -(z)2 1 Area = z1 2π e 2 dz To determine the area under the Gaussian distribution curve between any two z points (z1 and z2) z2 -(z)2 Area between = z1 and z2 1 2π z1 e 2 dz 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. 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)? Normal range of [Na] in serum 143 -(x - 140)2 Area = 1 2 π(2.5)2 e 2(2.5)2 dx 141 You could theoretically do it this way, however the way it is done is to transpose and use table 135 140 [Na] mEq/L (x) 145 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: Normal range of [Na] in serum 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 -2 z = (143 – 140)/2.5 = 1.2 -1 0 z 1 2 0.4 135 140 x 1.2 145 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: Normal range of [Na] in serum 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 -2 -1 0 z 1 2 0.4 Answer: 0.2295 probability 135 140 x 1.2 145 Our goal: To determine μ Cannot determine μ What can we determine about μ ? 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 Sample Population μ σ (of population) Take finite sample Statistics X s Gives range around X and s that μ and σ will be with a given probability Distribution of Sample Means How Statistics Gets Us Closer to μ Distribution of Sample Means – Example of [Glucose]serum in Diabetics Population of Diabetics Sample means are determined X1 μ X2 X3 XN For this example: n=25 N=50 n - sample size (# of individuals in sample) N – number of trials determining mean By theory, the distribution of sample means will follow the Central Limit Theorem Central Limit Theorem Sample means (X) of taken from a population are Gaussian distributed with: 1)mean = μ 2)std dev =σ/ n (μ true mean of the population) (σ is true std dev for the population, n is sample size used to determine X) [called standard error of the mean (SEM)] Conditions: 1) Applies for any population that is Gaussian [independent of sample size (n)] 2) Applies for any distributed population if the sample size (n) > 30 3) Assumes replacement or infinite population Central Limit Theorem Number or frequency of X μ is true mean of the population σ is true std dev for the population n is sample size used to determine X) μ μ - - 2σ/ n 1σ/ n μ μ - - 2 SEM 1 SEM μ μ μ + + 1σ/ n 2σ/ n μ μ + + 1 SEM 2 SEM X (Sample Means) The absolute width of the distribution of sample means is dependent on “n”, the more points used to determine X the smaller ? __________ the width. SEM =σ/ n μ μ - - 2 SEM 1 SEM μ μ μ + + 1 SEM 2 SEM X (Sample Means) Larger sample size “n” SEM =σ/ n Smaller sample size “n” X (Sample Means) -(X - µ)2 1 f(X) = e 2SEM2 2 π SEM2 SEM =σ/ n Transposing: z = (X - µ)/SEM μ μ - - μ 2SEM 1SEM μ μ + + 1SEM 2SEM -(z)2 X (Sample Means) -3 -2 -1 0 z value 1 2 3 f(z) = 1 2π e 2 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 = μ So z = z = (x - µ)/σ For the sample mean distribution of X values, z= Std dev = SEM and mean = μ So z = z = (X – μ)/SEM 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 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: Area = 0.955 μ μ - - 2 SEM 1 SEM μ μ μ + + μ – 2SEM < X < μ + 2SEM 1 SEM 2 SEM X (Sample Means) However this does not answer our real question, we want the range that μ is in! 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 μ – 2SEM < X < μ + 2SEM Subtract μ from each part of the expression – 2SEM < X - μ < + 2SEM Subtract X from each part of the expression -X– 2SEM < - μ < - X + 2SEM Multiply each part of the expression by -1 +X +2SEM > +μ > +X - 2SEM Writing so range is given as normal (going from lower to upper limit) X - 2SEM < μ < X + 2SEM X - 2SEM < μ < X + 2SEM This 95% confidence range for μ can be written as the following + expresion: X + 2SEM A range for μ can be written for any desired confidence 99.7 % confidence? X + ? SEM 68.3 % confidence? X + ? SEM 75.0% confidence? X + ? SEM What z value do you put in? For 75% confidence need area between +/- z value of 0.750 0 z value General Expression for Range μ is Within with Specified Confidence Estimator of μ + (Confidence Coefficient) x (SD of Estimator Distribution) z value X [chose z value whose area between the +Z and –z value equals the probability (confidence) desired ] SEM (σ/ n) σ – population true std dev n – size of sample used to determine X Problem What range would µ be within from a measured X of 159 mg/dL (sample size =25) if σ = 10 mg/dL with a 76% confidence? With a 95% confidence?