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Example of uncertainty calculation

This example is translated from ISO 98-3:2008, modified and supplemented with additional theoretical information for a better understanding. The ISO 98-3 norm describes the calculation of uncertainty, the standard is translated into Russian in the document GOST 54500.3

Example of uncertainty

Problem statement

The length of the calibrated part with a nominal length of 50mm is determined by comparison with another sample, in advance of known length. Comparing the two samples, we get the length difference, which is expressed by the formula:

(1) d = l(1 + αθ) - ls(1 + αsθs)
  • l - the length of the calibrated part at a temperature of 20 °C
  • ls - the length of the sample at a temperature of 20 °C, indicated in the relevant documentation
  • α, αs - coefficients of thermal expansion of the part and sample
  • θ, θs - the standard deviation of the measurement at a temperature of 20 °C

Additional information to the task condition

Based on 25 independent measurements using the instrument used, an experimental standard deviation of 13 nm was determined. The value of the effective number of degrees of freedom is 18. Calibration certificate of a measuring instrument for comparing two lengths indicates that based on six measurements, the uncertainty due to random errors is 0.01 µm s the confidence level is 95%, the uncertainty in case of systematic errors is 0.02 microns at the level of three RMS deviations (25% confidence). The coefficient of thermal expansion of the sample: αs = 11.5 × 10-6 °C-1, the uncertainty is set square distribution with boundaries ± 2 × 10-6 °C-1. The temperature in the measuring chamber (19.9 ± 0.5) °C. Temperature difference of part and sample is in the range ±0.05 °C.

Measurement model

(2) l = [ls(1 + αsθs) + d] / (1 + αθ) = ls + d + ls(αsθs - αθ) + ...

If we use the sample temperature and the part temperature in the measurement model, then it will be necessary take into account the correlation between these values. To avoid complications, we introduce the difference temperatures of the part and sample: δθ = θ - θs and we introduce the assumption that θ and δθ do not correlate with each other. The same is true for the difference in expansion coefficients δα = α - αs

Assume that δθ and δα are equal to zero, but the corresponding uncertainties are not equal to zero.

(3) l = f(ls,d,αs,θ,δα,δθ) = ls + d - ls[δα·θ + αs·δθ]

It follows from equation (3) that a statistical estimate of the length of the part, l, can be obtained from a simple equation ls + dμ, where ls is the length of the sample indicated in the documentation when 20°C, dμ - statistical estimate of the mean obtained by calculating the arithmetic mean of five (n=5) independent measurements.

Total uncertainty

According to the formula of total certainty, the uncertainty of formulas of the measurement model (3):

(4) u2c(l) = c2s·u2c(ls) + c2d·u2c(ld) + c2αs·u2c(αs) + c2θ·u2c(θ) + c2δα·u2c(δα) + c2δθ·u2c(δθ)
where
cs = ∂f/∂ls = 1 - (δα·θ + αs·δθ) = 1
cd = ∂f/∂d = 1
cαs = ∂f/∂αs = -lsδθ = 0
cθ = ∂f/∂θ = -lsδα = 0
cδα = ∂f/∂δα = -lsθ
cδθ = ∂f/∂δθ = -lsα
From where follows:
(5) u2c(l) = u2(ls) + u2(d) + l2sθ2u2(δα) + l2sα2su2(δθ)

Sample uncertainty u(ls)

The documentation for the sample indicates an extended uncertainty U = 0.075 microns with an overlap coefficient k=3. Hence:

u(ls) = (0.075 microns) / 3 = 25 nm

Uncertainty of the length difference u(d)

The experimental standard deviation of comparing the difference of lengths l and ls is based on the result is 25 independent measurements and it is equal to 13 nm. In this example, we have made 5 measurements, from where the standard uncertainty of the average value of the measurement data is:

u(dμ) = s(dμ) = (13 nm)/√5 = 5.8 nm

The uncertainty of the random error of the measuring instrument, according to the Student's distribution with the stpenu freedom ν = 6 - 1 = 5, with an overlap coefficient k = t95(5) = 2.57:

u(d1) = (0.01microns) / 2.57 = 3.9 nm

Uncertainty of the systematic error of the measuring instrument:

u(d2) = (0.02 microns) / 3 = 6.7 nm

Total uncertainty:

u2(d) = u2(dμ) + u2(d1) + u2(d2) = 93 nm2
u(d) = 9.7 nm

Uncertainty of the coefficient of thermal expansion u(αs)

u(αs) = (2 × 10-6 °C-1) / √3 = 1,2 × 10 -6 °C-1

Uncertainty of thermal expansion of the sampleu(θ)

The maximum temperature deviation during the measurement, Δ = 0.5°C. Suppose that the temperature changes within the given limits according to the cyclic law, according to the sine wave, then:

σt = √[∫(t-tμ)2sin(t)dt] = √2
u(Δ) = (0,5 °C) / √2 = 0,35 °C

The uncertainty of the average temperature in the measuring chamber follows from the standard deviation of the average value:

θμ = 19,9°C - 20°C = -0,1 °C
u(θμ) = ±(-0,1 °C) = 0,2 °C.

The standard deviation θ can be taken as the standard deviation of the mean θμ, hence the uncertainty:

u2(θ) = u2(θμ) + u2(Δ) = 0,165 °C2
u(θ) = 0,41 °C

Uncertainty of the difference in thermal expansion coefficientsu(δα)

The statistical expectation δα is 1 × 10-6 °C-1 with equal probability, that the value of δα will be within the specified limits, hence the standard uncertainty:

u(δα) = (1 × 10-6 °C-1) / √3 = 0,58 × 10 -6 °C-1

Uncertainty of temperature differenceu(δθ)

u(δθ) = (0,05 °C) / √3 = 0,029 °C

Standard total uncertaintyuc(l)

The total uncertainty is calculated by the formula (5):

uc2(l) = (25 nm)2 + (9.7 nm)2 + (0.05 m)2(-0,1 °C)2(0,58 × 10-6 °C-1)2 + (0.05 m)2(11,5 × 10-6°C-1)2(0.029°C)2 = 1002 nm2
uc(l) = 32 nm
The calculations show that the main contribution to the uncertainty is the uncertainty of the sample

If the measurement model Y = f(X1, X2, ..., Xn) has a nonlinearity on in the measured area, it is necessary to include higher-order uncertainties, then:

uc2(l) = (25 nm)2 + (9.7 nm)2 + (0.05 m)2(-0,1 °C)2(0,58 × 10-6 °C-1)2 + (0.05 m)2(11,5 × 10-6°C-1)2(0.029 °C)2 + (0.05 m)2(0,58 × 10-6°C-1)2(0.41 °C)2 + (0.05 m)2(1,2 × 10-6°C-1)2(0,029 °C)2 + 1140 nm2
uc(l) = 34 nm

Measurement result

From the sample certificate we have ls = 50,000623 mm at 20°C. The average value of the length difference as a result of five independent measurements is 215 nm. The length of the part l =ls + dμ at 20 °C is 50,000838 mm.

Calculation of relative uncertainty

Relative uncertainty is calculated as the ratio of uncertainty to the measurement result. Measurement result l = 50,000838 mm, with total uncertainty uc = 32 nm. Then the relative total uncertainty is uc /l = 6,4 × 10-7

Calculation of effective degrees of freedom

The calculation of effective degrees of freedom is performed according to the Welch–Satterthwaite equation:

νeff = uc4(y) / Σni=1ui4(y)/νi
If u(xi) is an uncertainty of type B, then, as a rule, the number of degrees of freedom tends to to infinity, otherwise the number of degrees of freedom for n dimensions is calculated by the following algorithm: if the statistical estimate the average is calculated using the arithmetic mean formula, then ν = n - 1 if the statistical estimate is determined by by the least squares method using m independent factors, then ν = n - m.

The number of degrees of freedom of the sample is set by the calibration certificate, νeff(ls) = 18.

The value dμ was obtained based on five measurements, but the uncertainty value was obtained based on 25 measurements, the number of degrees of freedom dμ: ν(dμ) = 25 - 1 = 24. The number of degrees of freedom d1: (d1) = 6 - 1 = 5. The number of degrees freedom for ν(d2) = 8. Where is the number of degrees of freedom νeff(d) = 25.6. To calculate the uncertainty of the difference in the expansion coefficients, we will set the significance level to 10%, in this case u(δα) = 50. To calculate the uncertainty of the temperature difference, we will set the significance level to 50%, then u(δθ) = 2.

veff(l) = (32 nm)4/[(25 nm)4/18 + (9.7 nm)4/25.6 + (2.9 nm)4/50 + (16.6 nm)4/2] = 16,7
ui(x) Source of uncertainty Value of uncertainty ci∂f/∂xi ui(l) ≡ |ci|u (xi) (nm) Number of degrees of freedom
u(ls)Sample calibration25 nm12518
u(d)Measuring the difference between the length of the part and the sample9.7 nm19.725.6
u(dμ)Independent measurements5,8 nm24
u(d1)Random errors3.9 nm5
u(d2)Systematic errors6,7 nm8
u(αs)Thermal expansion coefficient of the sample1,2× 10-6 °C-100
u(θ)Measuring chamber temperature0.41°C-100
u(θμ)The average temperature of the measuring chamber0.2 °C
u(Δ)Cyclic temperature change in the measuring chamber0.35 °C
u(δα)Difference of expansion coefficients0.58 × 10-6 °C-1lsθ2,950
u(δθ)Difference of expansion coefficients0.029°C-lsαs16,62
uc2(l) = Σui2(l) = 1002 nm2
uc(l) = 32 nm
νeff(l) = 16
Table 1. Total uncertainty and its components

Extended uncertainty

Suppose that it is necessary to obtain an extended uncertainty U99 =k99 uc(l) with a confidence interval of approximately 99%. The effective value of the degrees of freedom of the standard total uncertainty, equal to 16.7 we round it up exclusively in the smaller direction. The value of the Student's distribution, according to the table, is t99(16) = 2.92, from where U99 = t99(16)uc(l) = 2.92 × (32 nm) = 93 nm.

l = (50,000 838 ± 0.000 093) mm, where the error is calculated as the extended uncertainty U =kuc. uc is an extended uncertainty with a coverage coefficient k = 2.92, which was determined from Student distributions for 16 degrees of freedom and a confidence interval of 99%. Corresponding extended relative uncertainty U/l = 1,9 × 10-6

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Effective degrees of freedom