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

Σ(xi – X)2

(n-1)s =

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)

Control with Analyte

Methodology for Analyte

Result End Data Analysis

Repeat “n” times

X sSampleMean

SampleStd Dev

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 σTrueMean

TrueStd Dev

∞

There is another way!!!

Statistics

Population

μ σ

Sample (of population)

X sTake finite sample

StatisticsGives 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

Continuous function of frequency (or number) of a particular value versus the value

Probability Distribution

Nu

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or

fre

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en

cy

of

the

val

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Value

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

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Valuea b

The most utilized probability distribution in statistics is?

Gaussian distribution

Also known as Normal distribution

Parametric Statistics – assumes population follows Gaussian distribution

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)

Nu

mb

er o

r freq

ue

ncy

o

f the

valu

e

1σ

µ

- 1σ

µ+

3σ

µ

- 3σ

µ+

2σ

µ

- 2σ

µ+

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)

Nu

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ncy

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f the

valu

e

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

µ

- 1σ

µ+

0.997

3σ

µ

- 3σ

µ+

0.955

2σ

µ

-2σ

µ+

1) 68.3% chance between μ + 1σ and μ - 1σ

2) 95.5% chance between μ + 2σ and μ - 2σ3) 99.7% chance between μ + 3σ and μ - 3σ

Nu

mb

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ncy

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f the

valu

e [f(x

)]Gaussian Distribution Equation

f(x) = 1

2 πσ2e

-(x - µ)2

2σ2

µ 1σ

µ

- 2σ

µ

- 3σ

µ

- 1σ

µ+

3σ

µ+

2σ

µ+

x (value)

Gaussian curves are a family of distribution curves that have different µ and σ values

f(x) = 1

2 πσ2

e

-(x - µ)2

2σ2

A. Changing µ

B. Changing σ

Area between =x1 and x2

Nu

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f the

valu

e [f(x

)]To determine area between any two x values

(x1 and x2) in a Gaussian Distribution

f(x) = 1

2 πσ2e

-(x - µ)2

2σ2

µ 1σ

µ

- 2σ

µ

- 3σ

µ

- 1σ

µ+

3σ

µ+

2σ

µ+

x (value)

x1 x2

x1

x2

dx

Any Gaussian distribution can be transposed from x values to z values

x value equation

z value equation

z = (x - µ)/σ

eArea =

1

2 π

-(z)2

2

z2

z1

Area =2 πσ2

1

x2

x1

e

-(x - µ)2

2σ2dx

dz

To determine the area under the Gaussian distribution curve between any two z points (z1 and z2)

z1

z2

1

2 πe

-(z)2

2 dzArea between z1 and z2

=

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).

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

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 serumTranspose 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 0-1 1

0.4 1.2

135 140 145x

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 serumSo 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 0-1 1

0.4 1.2

135 140 145x

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

Population

μ σ

Sample (of population)

X sTake finite sample

StatisticsGives 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

μ

Population of Diabetics

X1For this example:

n=25

N=50

n - sample size (# of individuals in sample)

N – number of trials determining mean

Sample means are determined

X2

X3

XN

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

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) > 303) Assumes replacement or infinite population

X (Sample Means)

μ1σ/ n

μ

-1σ/ n

μ+

2σ/ n

μ

-2σ/ n

μ+

Central Limit Theorem

2 SEM

μ

-1 SEM

μ+

1 SEM

μ

-2 SEM

μ+

Nu

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f X

μ is true mean of the population

σ is true std dev for the population

n is sample size used to determine X)

μ2 SEM

μ

-1 SEM

μ+

1 SEM

μ

-2 SEM

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)

X (Sample Means)

Larger sample size “n”

Smaller sample size “n”SEM =σ/ n

μ2SEM

μ

- 1SEM

μ+

1SEM

μ

- 2SEM

μ+

SEM =σ/ n

X (Sample Means)

f(X) =

1

2 π SEM2

e

-(X - µ)2

2SEM2

ef(z) = 1

2 π

-(z)2

20-2 -1 21 3-3

z value

Transposing: z = (X - µ)/SEM

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=

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

μ2 SEM

μ

-1 SEM

μ+

1 SEM

μ

-2 SEM

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!

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

X - 2SEM < μ X< + 2SEM

Writing so range is given as normal (going from lower to upper limit)

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% confidenceneed area between +/- z value of 0.750

z value

0

General Expression for Range μ is Within with Specified Confidence

Xz 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)

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?