Download presentation

Presentation is loading. Please wait.

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

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

4
Ways to estimate true value 1.Mean (average) (X) X = Σx i /n x i - single measured value n- number of measured values 2. Median – Order x i values, take middle value (if even number of x i values - take average values of two middle values) 3. Mode – most frequent x i value If above is to estimate the true value what does this assume?

5
Proposed way to measure precision Average Deviation = Σ(X – x i )/n Does this estimate precision? No – because the summation equals zero, since x i values are less than and greater than the mean

6
Ways to Measure Precision 1.Range (highest and lowest values) 2. Standard Deviation

7
Standard Deviation Σ(x i – X) 2 (n-1) s = s – standard deviation x i - single measured value X – mean of x i values n - number of measured values

8
Variations of Std Dev 1.Variance - std. dev. squared (s 2 ) Variances add, NOT std deviations. To determine total error for a measurement that has individual component standard deviations for the measurement s 1, s 2, s 3, etc [i.e., random error in diluting calibrator (s 1 ), temperature change (s 2 ), noise in spectrophotometer (s 3 ), etc.] (s total ) 2 = (s 1 ) 2 + (s 2 ) 2 + (s 3 ) 2 + …

9
Variations of Std Dev (cont.) 2.Percent Coefficient of Variation - (%CV) %CV = 100 * S/X

10
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 - + - +

11
Determining Sample Mean and Sample Std Dev of Control (Assumes Accurate Technique) Control with Analyte Methodology for Analyte Result End Data Analysis Repeat “n” times X s Sample Mean Sample Std Dev

12
Determining True Mean and True Std Dev of Control (Assumes Accurate Technique) Control with Analyte Methodology for Analyte Result End Data Analysis Repeat times μ σ True Mean True Std Dev ∞

13
There is another way!!! Statistics

14
Population μσ Sample (of population) Xs Take finite sample Statistics Gives range around X and s that μ and σ will be with a given probability

15
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

16
Continuous function of frequency (or number) of a particular value versus the value Probability Distribution Number or frequency of the value Value

17
1.Total area = 1 2.The probability of value x being between a and b is the area under the curve from a to b Properties of any Probability Distribution Number or frequency of the value Value a b

18
The most utilized probability distribution in statistics is? Gaussian distribution Also known as Normal distribution Parametric Statistics – assumes population follows Gaussian distribution

19
1.Symmetric bell-shaped curve centered on μ 2.Area = 1 Gaussian Distribution 3. 68.3% area μ + 1σ (area = 0.683) μ 95.5% area μ + 2σ (area = 0.955) 99.7% area μ + 3σ (area = 0.997) x (value) Number or frequency of the value 1σ 1σ µ-µ- 1σ 1σ µ+µ+ 3σ 3σ µ-µ- 3σ 3σ µ+µ+ 2σ 2σ µ-µ- 2σ 2σ µ+µ+

20
Area under the curve gives us the probability that individual value from the population will be in a certain range What Gaussian Statistics First Tells Us μ x (value) 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 1σ 1σ µ-µ- 1σ 1σ µ+µ+ 0.997 3σ 3σ µ-µ- 3σ 3σ µ+µ+ 0.955 2σ 2σ µ-µ- 2σ2σ µ+µ+ 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

21
Number or frequency of the value [f(x)] Gaussian Distribution Equation f(x) = 1 2 πσ 2 e -(x - µ) 2 2σ22σ2 µ 1σ µ-µ- 2σ µ-µ- 3σ µ-µ- 1σ µ+µ+ 3σ µ+µ+ 2σ µ+µ+ x (value)

22
Gaussian curves are a family of distribution curves that have different µ and σ values f(x) = 1 2 πσ 2 e -(x - µ) 2 2σ22σ2 A. Changing µ B. Changing σ

23
Area between = x 1 and x 2 Number or frequency of the value [f(x)] To determine area between any two x values (x 1 and x 2 ) in a Gaussian Distribution f(x) = 1 2 πσ 2 e -(x - µ) 2 2σ22σ2 µ 1σ µ-µ- 2σ µ-µ- 3σ µ-µ- 1σ µ+µ+ 3σ µ+µ+ 2σ µ+µ+ x (value) x1x1 x2x2 x1x1 x2x2 dx

24
Any Gaussian distribution can be transposed from x values to z values x value equation z value equation z = (x - µ)/σ e Area = 1 2 π -(z) 2 2 z2z2 z1z1 Area = 2 πσ 2 1 x2x2 x1x1 e -(x - µ) 2 2σ22σ2 dx dz

25
To determine the area under the Gaussian distribution curve between any two z points (z 1 and z 2 ) z1z1 z2z2 1 2 π e -(z) 2 2 dz Area between z 1 and z 2 =

26
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.

28
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). 135 140 145 [Na] mEq/L (x) 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 Area = 2 π (2.5) 2 1 143 141 e -(x - 140 ) 2 2 (2.5) 2 dx You could theoretically do it this way, however the way it is done is to transpose and use table

29
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)? Normal range of [Na] in serum 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 z 2-2 01 0.4 1.2 135140145 x

31
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? Normal range of [Na] in serum 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 z 2-2 01 0.4 1.2 135140145 x

32
Our goal: To determine μ Cannot determine μ What can we determine about μ ?

33
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

34
Population μσ Sample (of population) Xs Take finite sample Statistics Gives range around X and s that μ and σ will be with a given probability

35
Distribution of Sample Means How Statistics Gets Us Closer to μ

36
Distribution of Sample Means – Example of [Glucose] serum in Diabetics μ Population of Diabetics X1X1 For this example: n=25 N=50 n - sample size (# of individuals in sample) N – number of trials determining mean Sample means are determined X2X2 X3X3 XNXN

38
By theory, the distribution of sample means will follow the Central Limit Theorem

39
Sample means (X) of taken from a population are Gaussian distributed with: 1)mean = μ (μ true mean of the population) 2)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)] Central Limit Theorem σ / n 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

40
X(Sample Means) μ 1σ/ n μ-μ- μ+μ+ 2σ / n μ-μ- μ+μ+ Central Limit Theorem 2 SEM μ-μ- 1 SEM μ+μ+ μ-μ- 2 SEM μ+μ+ 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)

41
μ 2 SEM μ-μ- 1 SEM μ+μ+ μ-μ- 2 SEM μ+μ+ The absolute width of the distribution of sample means is dependent on “n”, the more points used to determine X the __________ the width. SEM =σ/ n smaller? X(Sample Means)

42
X Larger sample size “n” Smaller sample size “n” SEM =σ/ n

43
μ 2SEM μ-μ- 1SEM μ+μ+ μ-μ- 2SEM μ+μ+ SEM =σ/ n X (Sample Means) f(X) = 1 2 π SEM 2 e -(X - µ) 2 2SEM 2 e f(z) = 1 2 π -(z) 2 2 0-2213 -3 z value Transposing: z = (X - µ)/SEM

44
What does a z value mean? The number of standard deviations from the mean. For the population distribution of x values, z= Std dev = σ and mean = μ So z = z = (x - µ)/σ Std dev = SEM and mean = μ So z = z = (X – μ)/SEM z values are for Gaussian distributions only. For the sample mean distribution of X values, z=

45
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

46
μ 2 SEM μ-μ- 1 SEM μ+μ+ μ-μ- 2 SEM μ+μ+ X (Sample Means) There is a 95.5% chance (confidence) that the one determination of X will be in the range indicated. Area = 0.955 This range can be written mathematically as: μ – 2SEM < X < μ + 2SEM However this does not answer our real question, we want the range that μ is in!

47
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+X+2SEM>+μ+μ+X+X>- 2SEM X < μ X<+ 2SEM Writing so range is given as normal (going from lower to upper limit)

48
X- 2SEM < μ X<+ 2SEM This 95% confidence range for μ can be written as the following + expresion: X+ 2SEM

49
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?

50
For 75% confidence need area between +/- z value of 0.750 z value 0

52
General Expression for Range μ is Within with Specified Confidence X z value [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 Estimator of μ + (Confidence Coefficient) x (SD of Estimator Distribution)

53
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?

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google