Estimation of influence of random errors resulting from limited number of direct observations.        

   All discussed above, including the confidence interval calculation, is fair only, if there is a sufficient volume of information, obtained from a large amount of observations, that is, when n→∞.

However in practice the amount of observations in each series is limited, that considerably reduces authenticity of the information presented. So is it possible in this case to use the computed confidence interval?

Known English mathematician Gosset (pseudonym, Student) has given the positive answer. But then the confidence interval should be broadened, being multiplied by factor t (n, P), known as the Student’s factor.

     This factor depends on the amount of observations, that is, on the volume of information. At a very large amount of observations, when n→∞ , t (n, P) → 1. When decreasing the amount of observations, the Student’s factor increases (Tab.5).

               Table 5.

VALUE OF THE STUDENT’S FACTOR

The Student’s factor also depends on a selected confidence coefficient, that is, on the claims on veracity (see Tab.5).

At resonable claims on veracity (P = 0,95) the factor t (n, P) is small. At the same time under the excessive claims on veracity (for example, at P = 0,999, when it is authorized no more than one value of error from a series of 1000 observations to leave the computed confidence interval), the Student’s factor is increased sharply.

Values of the Student’s factor for the confidence coefficients P = 0,95; 0,99; 0,999 at different amounts of observation n are represented in the Tab. 5[6].

Thus, taking into account the Student’s factor, a half-width of the confidence interval is:

The most widespread operational procedure at processing the correct measurement results (which do not contain systematic errors) of direct equally accurate observations is following.

     1. After performance of n observations we have the results:

               x₁, x₂, x₃, ..., x_n

                2. We calculate an arithmetic mean value :

    3. We calculate “n” remainder errors n_i:

n_I = X_i -,

    4. We calculate σ by the Bessel formula ESD:

    5. We specify the confidence coefficient P.

At technical measurements it is quite enough P = 0,95. At very responsible measurements, concerned with air safety, distribution of material valuables, etc., we specify the larger values of the confidence coefficient, depending on the measurement importance.

Knowing the amount of observations “n” and the confidence coefficient P, it is possible to find the Student’s factor, using Tab.5.

      6. We calculate a half-width of the confidence interval :

7. We record the measurement result in the form, which contains three numbers:

-the mean value of the measurand;

-the confidence interval;

     -the confidence coefficient,

     that is:

For example, we have gained the result of voltage measurement:

                       U = 100 V ± 3 V; P= 0,95.

It means, that after multifold observations we have detected, that 100V is the most authentic value of the voltage. And, at least in 95 observations from 100 outcomes the result is not less than 97 V (that is 100 V - 3 V) and no larger than 103 V (that is 100 V + 3 V).

If the measurement errors are distributed under the normal law, the confidence coefficient P=0,95, and the confidence interval is symmetrical, the abbreviated record of the measurements result, without indicating the confidence coefficient can be used:

For example:

U = (100 ± 3) V.

Questions for self-verification:

1. Is it possible to avoid the random errors in measurement results by an exploratory way? Why?

 2. What criterion is used for characterizing and comparing accuracy of different series of observations?

3. How does it look also what axioms the normal distribution law respond to?

      4.  How to compute the ESD by Bessel’s formula?

5. How to compute the ESD of arithmetical means of many series of observations values ? What is it necessary to do for a reduction of an influence of the random errors of the results of multifold observations, for example, in 10 times?

6. What approximations of random errors distributions laws do you known? What examples of practical implementation of these approximations can you give?

7.  What is a histogram? How can it be constructed?

8. What is a Student’s factor? What does it depend on? Where can we find it?

9. What are the confidence interval and the confidence coefficient? How are they connected with the Student’s factor?

10. What algorithm of the limited amount of observations data processing do you know? In what form is it necessary to present the measurement result?

DETERMINATION OF INDIRECT MEASUREMENT RESULT ERROR

Result of indirect or inferential  measurement is determined after execution of mathematical operations with the data obtained by direct measurements:

where

yis a result of indirect measurement;

 fis a known function;

are the results of direct measurements.

If the absolute errors of all the direct measurements are known, it is possible to determine an absolute error of indirect measurement resultaccording to the formula:

Where:                are the partial derivatives of the indirect measurement function on each from the arguments.

For example, we determine an electric current power Р, having measured its value I with error I and knowing object resistance R with error.

Power:

               thus: 

 Accurasy parameters of the every quantity direct measurment result may be represented not by their absolute errors, that is in a so-called interval form, but through ESD, σ_xi. It is more informative form for respensentation of the measurement accuracy parameters.

The relation between the interval and by means of ESD ways of the measurement accuracy parameters representation is stipulated by error distribution law. For the normal distribution law the confidence coefficient P = 0,95 corresponds to the confidence interval , and the confidence coefficient P = 0,9973 corresponds to the confidence interval .

So let us assumed that in the same way as before an indirect measurement has been executed:

Direct measurement results of its components:  and their ESD: are known.

Then ESD of the indirect measurement result is:

ERROR SUMMATION

Earlier, when we analysed errors of measurements, we decomposed them into the segregate components. Some of them were caused by the imperfection of the measurement method, other were brought due to the imperfection of the MIs used, their influence on the object, because of enviroment influencing factors variations, etc.

Now we pose a problem of summation all these components, to determine a total measurement error.

Let us be known "n" of the error components, which arise from the different causes and are not interconnected (their correlation is equal to zero). Let us consider also, that each from these components has the normal distribution law that frequently occurs in practice.

If confidence intervals of the error components are chosen by the rule "of three sigma", that is , it practically determines their largest possible values. Thus, the confidence coefficient P = 0,9973 means, that after 10 000 measurements not more than 27 errors can leave the  limits.

There are two ways of the total error determination: an arithmetic and a geometrical summation of their segregate components.

Arithmetic summation. In this case the total error

where  is the largest possible size of the i-th error component absolute value .

As can be seen, from the formula, as a result of calculations we shall obtain the largest of possible in practice errors. It is possible only when all the error components have the greatest of possible values and acts all together with the same sign.

The probability of such coincidence of circumstances is very small. If in the known fable a swan, a crab and a pike cannot synchronously pull a cart with the most power, and else in the same direction, moreover, the error components all together cannot have the same sign and the largest size simultaneously.

    Therefore, as a rule, the experimentally determined error is much less than calculated by the arithmetic summation formula.

 So in such a way it is necessary to calculate an error only in extremely critical cases, concered with the danger for life, possibility of catastrophes and so on.

In everyday engineer practice, and also in your course and diploma projects calculations on this formula is not recommended, therefore it gives very poor, but mainly else the low-probable accuracy parameters.

Geometrical summation. This method is based on the central theorem of the probability theory, and for the total error calculation it proposes the formula:

This formula usage gives considerably better results. The total error is considerably smaller, than in the previous case,that, as a rule, is closer to the desirable values of the accuracy parameters of the designed products.

But insofar motivated and honoured for a demanding to itself engineer is an usage of this formula instead of the previous one? The calculations with usage of the body of mathematics of a probability theory demonstrate, that a confidence coefficient of a confidence interval, which is computed with this formula, P = 0,9973. It is a very high probability, which demonstrates, that no more than in 0,27 % cases the experimentally determined error can exceed the computed by the geometrical summation formula one.