7 random measurement errors rev 4 090324

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7 Measurement Errors: Random 0908341 Measurements & Instrumentation

Copyright held by the author 2009: Dr. Lutfi R. Al-Sharif Page 1 of 24

Chapter 7Measurement Errors: Random

(Revision 4.0, 24/3/2009)

Anyone who attempts to

generate random numbers bydeterministic means is, ofcourse, living in a state of sin.

John von Neumann,1903-1957,

Hungarian-American Mathematician

Imagine!!

You are an engineer working in a factory. You are asked by the managingdirector to report the correct value of the voltage in an electronic circuit at acertain point. You take your digital voltmeter and take a number of readings.

You find that they vary slightly, so you take exactly 10 readings and recordthem. You find that there are small variations in the readings, despite the factthat you know that the actual voltage has not changed during your readings.

You decide to take the average of the 10 readings and report theaverage. Is this correct? Have you reported the whole truth?

What do you say to the managing director?What result do you give him?

1. IntroductionAs discussed in earlier chapters, we can think of the error in the measurementbetween the measured value and the true value as the sum of three errors:Systematic, Random and Noise. This is summarised in the equation shownbelow (where the errors can either be positive or negative):

)()()()()( tetetetptq noisesystrand +++=

The use of the random errors as a term is a representation of our ignorance.They are only random because we do not understand them; we do not knowtheir sources. Had we had full knowledge of their reasons, we would have

been able to classify them as systematic errors. There would be no need inour model for a term that accounts for random errors

1. This is similar to the

argument that probability is merely a method for representing our ignorance.With systematic errors, we understand the reason for them, and so we

can eliminate them, reduce them or at least account for their presence. Inother words we can deal with them in a deterministic fashion. We canquantify the magnitude of systematic errors.

Due to the fact that we do not understand the causes andmechanisms for random errors, we cannot eliminate them. Thus we have to

1

The only entity that knows these errors and their source is the omniscient, God Almighty.So for Him, there are no random errors; all errors are systematic.


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deal with them in a different way to systematic errors. We deal with randomerrors in a probabilistic fashion. We cannot quantify the exact value of therandom error, but we can state that it lies within a range of values with acertain level of confidence.

2. Approach to dealing with random errorsLet us assume that systematic and noise errors have been eliminated, theequation above reduces to the following:

)()()( tetptq rand+=

We need to develop an approach to eliminate the random error term above.An alternative definition of systematic and random errors is as follows

[1]:

Systematic errors are always to one side of the true value, whilerandom errors are at both sides of the true value.

This is not necessarily the best definition of these two types of errors, but canbe usefully used to illustrate the technique to deal with random errors.

If we accept that random errors are both sides of the true value, andwe further assume that the arithmetic sum of the deviation above and belowthe true value are zero, then we can state this as follows:

( )( ) 0)(lim =

temean randnn

In other words, if we take a large number of samples, n, then the randomerrors will cancel out and we end up with:

( ) )()(lim tptqnn

=

In effect this gives us the approach of dealing with the random errors bytaking a number of sample n of the measured value, q(t) and averaging overn.

It is worth noting the following two important points:

1. It is assumed that the true value of the measurand (i.e., measuredvariable) will not change during the process of measurement.Otherwise, the process above will be invalid.

2. In practice, it is impossible to take an infinite number of samples, n.We have to limit n to a practical value. The number of sample n iscalled the sample size. The total number of possible samples is calledthe population. We are effectively selecting a sample out of the totalpopulation. The selection of the size of the sample is important and

has an effect on the confidence level we have in the final result. This isdiscussed in more detail in the rest of this Chapter.


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3. Mean and Standard DeviationAs discussed in the last section, the most effective tool used to eliminaterandom errors is to find the mean of a number of samples, n.

However, in order to quantify the confidence level in the result we need

to understand how dispersed the readings (sample values) are from themean. For this we have to find a suitable dispersion measure.

3.1 The MeanLet us assume that we have taken n measurement of a certain variable, sothe sample size is n. For the purpose of this analysis, we shall ignore anysystematic and noise errors and assume that all errors are due to randomeffects.

The mean (also referred to as the expected value or the average) of the nmeasurement is:

n

xx

n

ii

== 1

Note that as the number of readings increases, the mean of all the n readingsstarts approaching the true value of the variable that is being measured(provided that the only errors in the measurement process are random, andthat all systematic and noise errors have been identified and eliminated):

truenn

xx =

lim

In relation to the measured value, q(t), and the true value, p(t), this equationcan be expressed as follows:

)()(lim tptqnn

=

In other words, as the number of samples approaches infinity, the mean of themeasured value approaches the true value

2(assuming that systematic errors

and noise errors have been eliminated or accounted for).This is a very important statement. As discussed in Chapter 1, the aimof any measurement system is to minimise the error between the measuredvalue and the true value.

The analysis above assumes an infinite number of samples. Inpractice however, it is not possible to take an infinite number ofmeasurements (or even a very large number). It is more realistic to take alimited number of measurements (say 10, 30 or 100).

Example 1

2The same is not true of systematic errors. No matter how many samples are taken, the

magnitude of the systematic error will not reduce.


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Assume that we use two different voltmeters to measure the voltage in acircuit, whereby the true value does not change during the measurements.Both measurement processes are subject to random errors caused by noise.The results of the ten samples from each system in V were as follows:

System A (V) 10.1 9.4 9.7 10.6 10.9 10.8 9.1 9.3 9.9 10.2System B (V) 10.4 9.3 9.2 10.6 11.4 11.3 9.1 8.6 9.3 10.8

Find the mean value of the two systems and comment on the results andcomment on the result if you are told that the true value of the voltage in thecircuit is actually 10 V.

SolutionThe average for each system is found by summing the reading and dividingby their number (10 in this case). This gives an average reading of 10 V forboth systems. This implies that both systems are accurate. However, thisresult is misleading as it ignores the imprecision that is present in System Bcompared to system. In other words this result does not comment on theprecision of the two systems.

The precision of the two systems can be better compared graphicallyby examining the figure below. The readings from both systems have beenplotted on a scale. It is clear that the readings from system B are moredispersed that system A, and thus system A is more precise. However, themean for both systems is the same. This is an unfair representation of theresults.

Figure 1: Distribution of samples for example 1.

This example shows the limitation of using the mean of the readings as a soleindication of the results of a measurement system. An additional measure isneeded that provides an indication of the dispersion of the results (or theimprecision of the system).

3.2 Dispersion measures (sum of absolute deviations


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Using the mean as an only measure of the true value is misleading. Let usassume we have two different systems that give the same mean value of ameasurand, but the second system is much more precise than the firstsystem. It will be unfair to treat both answers with the same level ofconfidence. We need to find a method to account for the higherprecision of

the second system compared to the first. The higherprecision indicates lowerrandom errors.

The fact that the readings of the first system have large random errorsshows that they are dispersed. We thus need a measure ofdispersion.

The most obvious measure is to take the sum absolute value of thedeviations of each reading from the mean of the samples (sample size nreadings).

( )=

=n

i

ixxdeviationstotal1

N.B.: We must take the absolute value of the deviations; otherwise the resultwill always be zero (by definition, the total deviation of all samples from themean is zero).

As we are taking n samples, it will be preferable to normalise3 thisresults by dividing by the number of samples, n (assuming that we have alarge number of sample n).

( )

n

xx

deviationsnormalised

n

i

i=

= 1

Example 2Two measurement systems A and B are used to take 10 readings of thecurrent in a circuit in A. The results are shown, and it is assumed that the truevalue did not change during these measurements. Any changes are due torandom errors in the process and noise.

Compare the mean value from both systems. Plot the readings fromboth systems on a horizontal scale and compare the precision of bothsystems. Find the sum of absolute deviations from the mean for both systemsand comment on the results.

System A (A) System B (A)13 10.4

6.5 9.3

9.9 9.2

10.4 10.4

10.5 11.4

10.2 11.3

9.5 9.1

9.9 8.6

9.9 9.5

3Normalisation is a process by which we divide by a parameter within a population or system

to allow us to compare results between different populations or systems.


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10.2 10.8

SolutionThe values from both systems are shown in Table 1 below. The table alsoshows the sum of absolute deviations and the sum of the square of the

deviations. The samples are also graphically plotted in Figure 2. The figureclearly shows that the samples from A are less dispersed that B, except fortwo extreme readings. This a great disadvantage for system A compared tosystem B as these two extreme deviations are presenting a large error.

Figure 2: Distribution of the samples for example 2.

Examining the sum of the absolute deviations (8.6 A for both systems) we seethat both systems have the same value. So the absolute sum of deviationsdoes not provide a good representation of the disadvantage present in systemA (the two large deviations). If we look at the sum of the square of thedeviations (22.02 A

2for A compared to 8.76 A

2for B), we can see that this

clearly penalises system A compared to system B. The larger value is causedby the squaring of the two large deviations that A has.

Table 1: Sum of deviatons and square of the deviations for systems A and B.

System A (all values in ampere) System B (all values in ampere)

Reading (A)Abs. dev. From

mean (A)Square of dev.

From mean (A2)

Reading (A)Abs. dev. From

mean (A)Square of dev.

From mean (A2)

13 3.0 9.0 10.4 0.4 0.16

6.5 3.5 12.25 9.3 0.7 0.49

9.9 0.1 0.01 9.2 0.8 0.64

10.4 0.4 0.16 10.4 0.4 0.16

10.5 0.5 0.25 11.4 1.4 1.96

10.2 0.2 0.04 11.3 1.3 1.69

9.5 0.5 0.25 9.1 0.9 0.81

9.9 0.1 0.01 8.6 1.4 1.96

9.9 0.1 0.01 9.5 0.5 0.2510.2 0.2 0.04 10.8 0.8 0.64


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Total 8.6 22.02 Total 8.6 8.76

This example shows that a better dispersion measure to the use of the sum ofthe square of the deviations from the mean compared to the sum of theabsolute values of the mean.

3.3 Dispersion measures (sum of the square of the deviations)As discussed earlier, the problem with the sum of the absolute value of thedeviations from the mean is that it treats large deviations with the sameweighting (importance) as small deviations. It is fairer to penalise largerdeviations from the mean compared to smaller deviations. Squaring is a toolthat allows us to do that. Squaring also obviates the need to take the absolutevalue.

So squaring the deviations produces the so called variance:

( )n

xxv

n

i

i= = 12

The variance is a measure of the spread of the readings around the meanvalue. The unit of the variance is the square of the unit of the measurand. Inorder to express the spread with a tool that has the same units as themeasurand, we can take the square root of the variance. This is called thestandard deviation and is calculated as follows:

( )n

xxv

n

i

i

= == 12

The standard deviation is a good measure of the spread of themeasurements. It has the same units as the measurand and its mean.Taking the square of the deviation penalises large deviations from the mean,and the effectively uses the absolute value of the deviations.

3.4 n or n-1?The discussion in the last subsection has assumed the following:

1. The sample size is large (i.e., n is large).2. As the sample size is large, we have assumed that the mean value of

the n samples is near enough to the mean of the whole population (i.e.,

)(tpxxxn == ).

However, this is only true ifn is large. When n is small, we have the followingtwo discrepancies:

1. The mean of the sample size n is different from the mean of thepopulation (and is thus different from the true value,p(t)).


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2. When calculating the variance and the standard deviation, thedeviations are calculated from the mean of the sample rather than themean of the population. We cannot use the mean of the population aswe do not know it, and we have to use the mean of the n samples thatwe measured. This leads to an error.

Statistically is has been shown that the variance and the standard deviationwill be nearer their true value if we divide by n-1 rather than n. In order todistinguish between the sample and the population, we shall use thesubscripts n and n-1 when describing the variance and standard deviation.Thus,

nv will refer to the variance of the population

1nv will refer to the variance of the sample

n will refer to the standard deviation of the population

1n will refer to the standard deviation of the sample

And the sample variance and standard deviation will be calculated as follows:

( )

1

1

2

1

=

=

n

xx

v

n

i

in

n

( )

11

2

11

==

=

n

xx

v

n

i

in

nn

Notice that the term nx has been used to emphasise the fact that we are using

the mean of the sample rather than the mean of the population, due to thesmall value ofn.

In general, it is accepted that ifn is less than 30 (or 20) it is consideredsmall and the n-1 methodology is used.

For completeness, we shall show the population variance and standarddeviation equation with the correct notation:

( )

n

xx

v

n

i

i

n

=

= 1

2

( )

n

xx

v

n

i

i

nn

=

== 1

2


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UnitsThe unit of the mean of a distribution and the unit of the standard deviation ofany distribution are the same as the unit of the original variable in thedistribution. Otherwise we would not be able to add and subtract the standarddeviation to/from the mean (we can only add and subtract quantities that have

the same units). The unit of the variance is the square of the unit of themeasured variable in the original distribution.

4. The Gaussian distributionIn to be able to express our answers later on this chapter with levels ofconfidence we need to use the Gaussian distribution. We shall assume thatthe random errors in our measurement are caused by random variables thathave a Gaussian distribution of its amplitude.

Many processes follow a standard distribution. White noise alsofollows a Gaussian distribution. Note that the probability distribution of anyvariable describes the amplitude of the signal and tells us nothing about itsfrequency content.

The Gaussian distribution is shown in Figure 3 below. The probabilityof a certain variable being smaller than z is equal to the area F(z) (i.e., thearea under the curve from to the value ofz).

Figure 3: Gaussian probability distribution curve.

The function F(z) cannot be evaluated from a function and is tabulated. Thetable is shown at the end of this Chapter. The Gaussian distribution is widelyused in predicting future measured values as well as quality control, as thefollowing two examples show.

Example 3You are given a pile of 10,000 resistors of nominal value 470 ohms. Youdecide to pick a sample of 100 resistors and measure them to get their meanand standard deviation. You find that the mean of your sample is 470 ohms

and the standard deviation is 23.5 ohms. You will reject all resistors that areoutside the +/- 5% tolerance of the nominal value.


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a) How many resistors will you reject?b) If you later discover that your multi-meter has a systematic error of +5

ohms (i.e., it shows a value that is 5 ohms higher than the correctvalue), what is the correct value of the mean and standard deviation of

the sample you took?Hint for part b): Examine the formula used to calculate the mean andstandard deviation.

Solutiona) The absolute deviation you will accept is 5%470 = 23.5

So your deviation will be one standard deviation above the mean andone standard deviation below the mean. So the value of z you will usein the table is 1. The value from the table for z=1 is 0.8413. Thismeans that 15.87% of all resistors will be more than 1.05 of thenominal value, and 15.87% will be below 0.95 of the nominal value. So

the percentage of resistors rejected will be: 0.1587+0.1587=31.74%

So out of the total 10,000 resistors, 3,174 resistors will be rejected.

b) Based on a systematic error of +5 , the true mean will be 465 . Thestandard deviation however will still be the same at 23.5 , as all thereadings as well as the mean will be shifted by the same value, so thedifferences between them will remain the same.

Example 4A factory manufactures chrome plated steel rods. A quality control check isbeing carried out on the rods by measuring the diameter of the rod. The rodis designed to a diameter of 12.000.75 mm (i.e., the diameter should bebetween 11.25 to 12.75 mm). The diameter of the manufactured rods followsa Gaussian distribution and the standard deviation of the diameter of themanufactured rods is found to be 0.5. Calculate the percentage of rods thatwill be rejected by the quality control process.

SolutionThe standard deviation of the process is 0.5. As the deviation in eachdirection is 0.75 mm, then we can say that the deviation in each direction is

1.5 of the standard deviation or 1.5

. Looking up the value of 1.5 in theF(z) table gives an area (for A and B) as 0.9332.So the area of B (that is equal to the area of C) is 1-0.9332=0.0668.

The total rejected items will be equal to the areas of B and C, or 2 x 0.0668 =0.1336=13.36%.

5. Standard Error of the MeanUp to now we have only used the mean and standard deviation to predictfuture measurement value, or in quality control. We still cannot express theanswer of a measurement process with any level of confidence. This can bedone by the use of the standard error of the mean, which is introduced in this

section.


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5.1 Parameters of a sample sized nIn effect we have now arrived at a method to describe the outcome of

measuring n samples of a population. By using the mean, nx , the standard

deviation, 1n , and the number of readings, n, we have now three parameters

that can be used to express a probabilisticanswer to the original question:What is the true value ofp(t) (xTRUE)?As discussed earlier, due to the nature of random errors we can only

express this answer in a probabilistic manner (and not in a deterministicmanner). This is discussed in detail in the next sub-section on the standarderror of the mean.

5.2 Standard Error of the MeanLet us assume that all the students of the University of Jordan were asked tocarry out a measurement of the height of statue at the centre of the universitycampus.

Each student is asked to take 10 readings and provide the 10 readingsto an operator who enters the 10 readings on a PC. If we construct adistribution of all the readings from all the students, it is very likely that thedistribution is a Gaussian distribution (taking into consideration all the randomerrors that the students introduce when taking the measurements).

As the number of readings (10 times the number of students in theuniversity) is large, we expect the average of all the readings to approach thetrue value. The standard deviation of this distribution would be arepresentation of the random errors. Let us assume that the resulting meanfrom all these readings is 2104.1 mm and that the standard deviation is 3.5mm. The mean of these readings is a very good approximation of the true

height of the statue.A rough plot of the resulting distribution from all the students

measurements is shown Figure 4 below.

Figure 4: Probability distribution function of all readings (x).

If we now take a range that provide a deviation of zx from the calculatedmean (above and below), this provides us with a so-called confidence intervalwithin which we can say that a certain percentage of all readings lie. Forexample if we take zas 2 (i.e., we deviate 2 standard deviations above andbelow the mean) we can say that 95.4% of all readings lie within this range

Measured height of statue,x(mm)

Mean of all readings, (mm)

Probability/Frequencyof reading

Standard deviation, x (mm)

2x (mm)2x (mm)

Area equal 95.4% of allreadings


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(2104.17mm). The area of 95.4% can be indirectly found by plugging thenumber 2 into the F(z) table (Gaussian distribution table, a copy of which hasbeen provided at the end of this Chapter).

5.3 Standard Error of the Mean

As discussed earlier, each student was asked to take 10 readings (not justone reading). All of the 10 readings went into the PC and contributed to thefinal distribution.

Now let us assume that instead of taking all 10 readings from eachstudent, we ask him/her to calculate the mean of his/her 10 readings andprovide the result. We now plot the distribution of these means, a rough plotof which is also shown in Figure 5. We note that the distribution is alsoGaussian but with a smaller standard deviation. The ratio of the standard

deviation of the x (mean) to the standard deviation ofx is the ratio 1:n ,

where n is the number of sample in each set of readings that each studenttook. The standard deviation of the mean is called the standard error of themean and is equal to the standard deviation of the original distribution dividedby the square root of the sample size (n). This is also referred to as . Thiscan be summarised as:

10

11 === nnxn

Figure 5: Sampling distribution of the mean of each 10 readings.

Measured height of statue,x(mm)

Mean of all readings, (mm)


Standard deviation, x (mm)

2x (mm)2x (mm)

Area equal 95.4% of allreadings

Mean of the 10 readings of the height of statue, x (mm)

Mean of all means, (mm)

Standard deviation of themean, (mm)

2 (mm)2 (mm)

Area covers 95.4% of allmeans


Sample size (n)=10

10

xxx

n

===


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As the sample size increases (i.e., n, the number of samples that eachstudent takes increases) then the standard error of the mean decreases andthe curve become narrower. This is demonstrated graphically by usingstatistical software that is shown in Figure 6. It shows how the increase in

sample size from n=2 to n=25 has gradually reduced the standard error of themean and thus narrowed the curve.


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Figure 6: Demonstration that shows the effect of sample size on the distribution of themean (source: http://onlinestatbook.com/simulations/CLT/clt.html).


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5.4 Confidence IntervalsWe now ask one of the students, Samira, to express her opinion about thetrue height of the statue by just knowing the standard deviation of the originaldistribution. Samira obviously knows the 10 readings that she took and shethen calculates the mean of these 10 readings. The average of her 10

readings is 2105.2mm. We tell her that the standard deviation of all thereadings carried out by the students is 3.5 mm.

Samira then calculates the standard error of the mean by dividing 3.5by the square root of 10, to get the standard error of the mean of 1.107 mm.In order to express her answer with 95.4% confidence level, she will use aconfidence interval with a deviation of 2 (i.e., 2x1.107=2.214mm). So shesays:

I am 95.4% confident that the height of the statue lies between2105.22.214mm (i.e., 2103.0-2107.4).

Her answer is shown graphically in Figure 7.

Figure 7: Graphical representation of Samira's confidence interval.

So to summarise:

The standard deviation of the distribution of a variable x( 1n ) applies to the

original readings ofx. The standard deviation of the mean ( ) of thesereadings when taking samples of n each time is called the standard error ofthe mean. It applies to the distribution of the mean of each sample and is alsoreferred to as . The two standard deviations are related by the size of the

sample n, as follows:n

n 1=

. To calculate a confidence interval forxwe

use 1n ; to calculate a confidence interval forx we use .

Mean of the 10 readings of the height of statue, x (mm)

2104.1 mm(true mean)

2 (mm)2 (mm)

ProbabilityDensityfunction

Sample size (n)=10

mmn

xxx 107.1

10

5.3

10=====

2105.2 mm (average ofSamiras 10 readings)(Samiras mean)

Confidence interval

2x


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Example 5A voltmeter was used to measure the voltage in a circuit and gave thefollowing 10 readings (assuming that the real value of the voltage did notchange during the measurements and that systematic errors had beeneliminated):

36.1, 36.3, 35.9, 35.2, 35.7, 35.6, 36.1, 36.2, 36.4, 35.8

a) Express the voltage reading as a range with a 95.4% confidenceassuming the data above follows a Gaussian distribution.

b) If the number of readings was increased to 30 and the mean andstandard deviation did not change, what would the new answer be?

SolutionUsing MS Excel, we can develop a table to calculate the mean value, theabsolute value of the deviations from the mean and the square of the

deviations. By adding the square of the deviations this and dividing this by n-1 which is 9 in this case, this gives us the variance. Taking the square root ofthe variance gives us the standard deviation.

# |x-x(mean)| (x-x(mean))2

x(mean)

1 36.1 0.17 0.0289 35.93

2 36.3 0.37 0.1369

3 35.9 0.03 0.0009 variance= sum of the square of the deviations/9

4 35.2 0.73 0.5329 variance 0.133444444

5 35.7 0.23 0.0529 Standard deviation 0.365300485

6 35.6 0.33 0.1089

7 36.1 0.17 0.0289

8 36.2 0.27 0.0729

9 36.4 0.47 0.2209

10 35.8 0.13 0.0169

Total 1.201

As can be seen from the spreadsheet shown above, the mean is 35.93 V (totwo decimal places) and the standard deviation is 0.365 (to three decimalplaces). Thus the standard error of the mean can be calculated as:

115.010

365.01 ===

n

n

a) In order to get a confidence interval with a confidence level of 95.4%, weneed to check the normal/Gaussian distribution table for the number ofstandard deviations (or standard error of the mean) that need to be includedin the confidence interval in order to achieve the required level of confidence.

We need to find the numberzthat corresponds to a confidence intervalof 95.4% or 0.954. This corresponds to the area A shown in the figure below(the yellow area). But the value in the table F(z) corresponds to the areasA+B (i.e., the orange + yellow areas). So the orange area will be equal to:

Orange area = Blue area = (1-0.954)/2=0.023


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z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7703 0.7734 0.7764 0.7793 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8906 0.8925 0.8943 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

1.8 0.9641 0.9648 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890

2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916

2.4 0.9918 0.9920 0.9922 0.9924 0.9926 0.9928 0.9930 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952

2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964

2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974

2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981

2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9986 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993

3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995

3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996

3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998

3.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998

3.6 0.9998. 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999

Standard Gaussian Distribution Table (F(z))

So the area covered by A + B is equal to:

F(z)=1-0.023=0.977

This is the area that we look up in the table of the Gaussian distribution tableat the end of this document. Against an area 0.977 (F(z)) we have z= 2.0

Note that zis the number of multiple of the standard deviation from themean (i.e., it is the deviation from the mean, normalised by dividing it by thestandard deviation). For this reason, zis always dimensionless.

So the confidence interval will include 2 (or 2 for the standard errorof the mean).


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So the confidence interval for the value ofxwill be (we will refer to the truevalue ofxasxTRUE):

VxxTRUE 23.093.35115.0293.352 ===

In other words, we are 95.4% confident that the true value ofx lies between35.70 V and 36.16 V

b) If the number of readings changes from 10 to 30 and the values of themean and standard deviation do not change, then the value of the standarderror of the mean changes to:

067.030

365.01 ===

n

n

So the new confidence interval becomes:

VxxTRUE 134.093.35067.0293.352 ===

Example 6The following measurements were taken with a frequency meter of thefrequency of a square wave signal (the circuit was in steady state andtherefore, although the measurements varied due to random errors, thefrequency of the signal was actually constant). Assume that the amplitude ofreadings taken follows a Gaussian distribution. Ten readings were taken (unitin Hz):

100.1, 100.3, 99.9, 96.5, 101.1, 102.3, 98.7, 97.2, 103.1, 100.8

Answer the following questions:

a) Calculate the mean, variance, standard deviation and the standarderror of the mean of the readings.

B

A

C

p.d.f.

Ma nitudeValue from table=A+B 2


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b) Assume we take a large number of readings and that the standarddeviation and mean are unchanged, what is the probability that thefrequency measured is outside the range of 96.9 Hz to 103.1 Hz.

c) Express the final answer for the frequency measured as a confidenceinterval with a confidence level of 99.72%.

Solutiona) Mean= 100.0 Hz

Standard deviation= 2.07 HzStandard error of the mean = 2.07/sqrt(10)= 0.65 Hz

b) It is very important to identify whether we are dealing with x or x . In

this case we are looking at the future readings of x (future values of the

reading). Hence we shall use 1n rather than .

The deviation above the mean of 100 Hz is:103.1-100 =3.1 HzThe deviation below the mean is:100-96.9 = 3.1 HzSo the deviation above and below the mean is equal.

The deviation as a multiple of the standard deviation is:z= 3.1/2.07=1.5

F(z) from the table is equal to 0.9332

So the probability that the frequency is more than 103.1 Hz is1- 0.9332= 0.0668But this is only the area representing the deviation above 103.1 Hz.We also need to include the probability that the frequency is less than96.9 Hz, which is also equal to 0.0668.

So the total probability the frequency will be outside the range of 96.9Hz to 103.1 Hz is 2 x 0.0668 = 0.1336 or 13.36%

c) It is very important to identify whether we are dealing with x or x . In

this case we are asked to express our confidence in the true value (i.e.,x). Hence we shall use rather than1n .

As we are using a confidence period around the mean, we must usethe standard error of the mean rather than the standard deviation (i.e.we must use the standard deviation of the mean rather than thestandard deviation of the variable itself).

A confidence level of 99.72% is equivalent to a deviation around themean of

F(z)=1-((1-0.9972)/2)=0.9986

So zfrom the table is 3.


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In other words the confidence interval will be represented by adeviation around the mean of 3 times the standard error of the mean.

Hzxx 95.110065.031003 ===

Example 7Two students carry out measurements on the value of temperature in achemical reaction in a factory. Student A takes 100 samples and student Btakes 11 samples. Bothstudents get the following results:

The mean value of their readings is 45 C.

The standard deviation of their readings is 3.2 C.

You can assume that the readings follow a Gaussian distribution. The

managing director of the factory asks each student the following question:

How confident are you in percentage terms that the true valueof the temperature of the chemical reaction lies in the range of44 C to 46 C?

What would the answer from each student be?

SolutionAs we are talking about the true value, we use the standard error of the mean(i.e., the standard deviation of the mean).

For student A, his standard error of the mean is:

C32.0100/2.3x

==

As for student B, his standard error of the mean is:

C96.011/2.3x

==

For each student, the deviation from the mean is 1 C.

Thus z for student A will be: 1/0.32= 3.125And z for student B will be: 1/0.96= 1.04

Using the Gaussian table gives us a confidence level for student A of:

Confidence for student A = 1- 2(1- F(3.125))= 1- 2(1-0.9991)= 99.82%

Confidence for student B = 1 2 (1- F(1.04))= 1- 2(1-0.8508)=70.16%

References


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[1] Measurement & Instrumentation Principles, Alan S. Morris, Elsevier,2001.

[2] Guidelines for Evaluating and Expressing the Uncertainty of NISTMeasurement Results, NIST Technical Note 1297, 1994 Edition, BarryN. Taylor and Chris E. Kuyatt.

[3] http://onlinestatbook.com/simulations/CLT/clt.html

Further ReadingMechanical Measurements, Beckwith, Marangoni, Lienhard, 6

thEdition,

Perason International Edition (Chapter 3 contains an excellent treatmentto random error and data presentation).

Instrumentation Measurement and Analysis, B.C. Nakra & K.K. Chaudhry,Tata McGraw Hill Publishing Company Limited, 1985. (Comprehensiveapproach in Chapters 17 and 18)

Measurement & Instrumentation Principles, Alan S. Morris, Elsevier, 2001.(Section 3.5)

Theory and Design for Mechanical Measurements, Figliola & Beasley, 2ndEdition, John Wiley & Sons, Inc., 1995 (Chapters 4 and 5).

Experimental Methods for Engineers, J. P. Holman, McGraw HillInternational Edition, 7

thEdition, 2001 (Chapter 3).

Mechanical Measurements and Control, D.S. Kumar, Metropolitan (NewDelhi), Third Revised and Enlarged Edition (Section 3.3).

Measurement Systems: Application and Design, Ernest O. Doebelin, FifthEdition, McGraw Hill, 2003. (Pages 45-54).

Principles of Measurement Systems, John P. Bentley, Pearson PrenticeHall, Fourth Edition 2005. (Section 2.3 and section 3.2)

Modern Electronic Instrumentation and Measurement Techniques, Albert D.Helfrick & William D. Cooper, Prentice Hall International Edition, 1990(Section 1.5 and 1.6).

A course in mechanical measurements and instrumentation, A.K. Sawhney,Dhanpat Rai & Sons. (Chapter 7)

Problems1. A 1000 resistors of nominal value of 820 ohms are being measured. Theresistors must be within a 5% tolerance of the nominal value of 820 ohms. Ifthe value of the resistors follows a Gaussian distribution and the standarddeviation is 20.5, calculate the numberof resistors that would be rejected by

the quality control process (any resistors deviating by more than 5% above orbelow the nominal value will be rejected).

2. An electronic scale is used to weigh a bag of flour. Ten measurements arecarried out giving the following results (in gm):

1201, 1203, 1110, 1118, 1200, 1201, 1205, 1117, 1202, 1204

a) Calculate the mean, standard deviation and standard error of themean of the weight of the bag of flour.

b) Hence express the mean value of the bag with a confidence levelof 68%, 95.4% and 99.7%.


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3. A large group of students is asked to measure the current flowing in anelectric circuit (where the current stays constant during these measurements).A sample of 10 readings is taken from the students readings in order to drawconclusions about the true value of the current in the circuit. The 10 samples

were in mA:

6.31, 6.34, 6.27, 6.29, 6.32, 6.31, 6.36, 6.23, 6.25, 6.32

Answer the following questions. Assume that the current amplitude follows aGaussian distribution.

a) Calculate the mean, standard deviation and standard error of themean.

b) What is the probability that a future reading is more than 6.38 mA.c) Express the true value of the current to a confidence level of 86.6%.

d) If the number of samples taken is increased to 100 samples, and themean and standard deviation remain unchanged, what would be thenew answer to part (c) above.

4. The following measurements were taken with an analogue meter of thecurrent flowing in a circuit (the circuit was in steady state and therefore,although the measurements varied due to random errors, the current flowingwas actually constant):

Readings in mA:21.5, 22.1, 21.3, 21.7, 22.0, 22.2, 21.8, 21.4, 21.9, 22.1

a) Calculate the mean, variance, standard deviation and the standarderror of the mean of the readings.

b) Hence express the measured value as a confidence interval thatincludes the true value with a confidence level of 89.04%, assumingthat the readings follow a Gaussian distribution.

5. (Variation on Example 7): A student carries out measurements on thevalue of temperature in a chemical reaction in a factory. She takes 11samples and gets the following results:

The mean value of his readings is 45 C.

The standard deviation of is 3.2 C.

You can assume that the readings follow a Gaussian distribution. Themanaging director of the factory asks the student the following question:

How confident are you in percentage terms that the true valueof the temperature of the chemical reaction lies in the range of44 C to 46 C?

a) What would the answer from the student be?


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b) The factory manager is not happy with the students answer and says thathe needs to be 95.4% confident of the answer. What does the student needto do to be that confident?


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C i ht h ld b th th 2009 D L tfi R Al Sh if P 24 f 24

Standard Gaussian Distribution Table (F(z))

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7703 0.7734 0.7764 0.7793 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8906 0.8925 0.8943 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

1.8 0.9641 0.9648 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890

2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916

2.4 0.9918 0.9920 0.9922 0.9924 0.9926 0.9928 0.9930 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952

2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964

2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974

2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981

2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9986 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993

3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995

3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996

3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.99983.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998

3.6 0.9998. 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999

7 random measurement errors rev 4 090324

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