Errors and Treatment. Topping. 1972

ERRORS OF OBSERVATIONAND THEIR TREATMENT

byJ. TOPPING, PH.D., F.INST.P.

Formerly Vice-Chancellor, Brunei University

,. FOURTH EDITION

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PREFACEThis little book is written in the first place for students in technicalcolleges taking the National Certificate Courses in Applied Physics;it is hoped it will appeal also to students of physics, and perhapschemistry, in the sixth forms of grammar schools and in theuniversities. For wherever experimental work in physics, or inscience generally, is undertaken the degree of accuracy of the, measurements, and of the results of the experiments, must be ofthe first importance. Every teacher of experimental physics knowshow "results" given to three or four decimal places are often inerror in the first place; students suffer from "delusions of accuracy."At a higher level too, more experienced workers sometimes claim adegree of accuracy which cannot be justified. Perhaps a considera-tion of the topics discussed in this monograph will stimulate instudents an attitude to experimental results at once more modestand more profound.T~e mathematical treatment throughout has been kept as simple

as possible. It has seemed advisable, however, to explain thestatistical concepts at the basis of the main considerations, and itis hoped that Chapter 2 contains as elementary an account of theleading statistical ideas involved as is possible in such small compass.it is a necessary link between the simple introduction to the natureand estimation of errors given in Chapter I, and the theory of errorsdiscussed in Chapter 3. Proofs have usually been omitted butreferences to other works are given in the text. There is also a listof books for furtber reading.I am much indebted to other writers, which will be obvious, and

to many groups of students particularly those at The Polytechnic,Regent Street, London, who bore patiently with my attempts to getthem to write every experimental result as x ± y. I am also muchIndebted to friends and old students who have helped me with thenrovision of suitable data;' arid. r am specially grateful t~ Mr.Norman Clarke, F.lnst.P., Deputy Secretary of The Institute ofPhysics, who has kindly read the manuscript and made many helpfuluggestions.The author of a book of this kind must always hope that not too

many errors, accidental or personal, remain.",cIon J. TOPPING

July, 1955.

5

PREFACE TO THIRD EDITIONOpportunity has been taken to make one or two corrections anda few slight additions.I am grateful to all those who have written and made suggestions.

It is pleasing that the book has found acceptance in universitiesand other institutions, both in this country and overseas.

J. TOPPINGBrunei College of Technology,

London, W.3.October, 1961.

PREFACE TO FOURTH EDITIONWith the adoption by Britain of the system of S.1. units appropriatchanges have been made throughout the book.

Some other small revisions have also been made.October, 1971. J. TOPPIN

6

CONTENTSChapter Page

l. ERRORS OF OBSERVATION 91 Accidental and systematic errors; 2 Errors and fractionalerrors; 3 Estimate of error; 4 Estimate of the error incompound quantities; 5 Error .in a product; 6 Error in aquotient; 7 Use of the calculus; 8 Error in a sum ordifference.

2. SOME STATISTICAL IDEAS

9 Frequency distributions; 10 The mean; 11 Relativefrequency; 12 The median; 13 Frequency curves; 14 Mea-sures of dispersion; 15 The range; 16 The mean deviation;17 The standard deviation; 18 Evaluation of standarddeviation, a; 19 Sheppard's correction; 20 Charlier's checks;

• 21 The mean and standard deviation of a sum; 22 Certainspecial frequency distributions; 23 The binomial distribution;24 The Poisson distribution; 25 The normal distribution;26 Relation between a normal and a binomial distribution;27 The mean deviation of a normal distribution; 28 Areaunder the normal error curve; 29 Sampling, standard errorof the mean; 30 Bessel's formulre: 31 Peters' formulee ;32 Fitting of a normal curve; 33 Other frequency distri-butions.

3. THEORY OF ERRORS

34 The normal or Gaussian law of error; 35 Applicability ofthe normal law of error; 36 Normal error distributions;37 Standard error of a sum or difference; 38 Standard errorof a product; 39 Standard error of a compound quantity;40 Method of least squares; 41 Weighted mean; 42 Stan-dard error of weighted mean; 43 Internal and externalconsistency; 44 Other applications of the method of leastsquares, solution of linear equations; 45 Solution of linearequations involving observed quantities; 46 Curve fitting;47 Line of regression; 48 Accuracy of coefficients; 49 Othercurves.

REFERENCES

BIBLIOGRAPHY

INDEX

7

29

72

115

116

117

r

CHAPTER 1

ERRORS OF OBSERVATION"And so we see that the poetry fades out of the problem, andby the time the serious application of exact science begins weare left with only pointer readings."

EDDINGTON

1. Accidental and systematic errorsAlthough physics is an exact science, the pointer readings of thephysicist's instruments do not give the exact values of the quantitiesmeasured. All measurements in physics and in science generallyare inaccurate in some degree, so that what is sometimes called the"accurate" value or the "actual" value of a physical quantity, suchas a length, a time interval or a temperature, cannot be found.Ho~ever, it seems reasonable to assume that the "accurate" valueexists, and we shall be concerned to estimate limits between whichthis value lies. The closer these limits the more accurate themeasurement. In short, as the "accurate" value is denied us, weshall endeavour to show how the "most accurate" value indicatedby a set of measurements can be found, and how its accuracy canbe estimated.Of course the aim of every experimentalist is not necessarily to

make the error in his measurements as small as possible; a cruderresult may serve his purposes well enough, but he must be assuredthat the errors in his measurements are so small as not to affect theconclusions he infers from his results.The difference between the observed value of any physical quantity

and the "accurate" value is called the error of observation. Sucherrors follow no simple law and in general arise from many causes.Even a single observer using the same piece of apparatus severaltimes to measure a certain quantity will not always record exactlythe same value. This may result from some lack of precision oruniformity of the instrument or instruments used, or from thevariability of the observer, or from some small changes in otherphysical factors which control the measurement. Errors of observa-tion are usually grouped as accidental and systematic. although it issometimes difficult to distinguish between them and many errorsare a combination of the two types.

9

CHAP. 1 ERRORS OF OBSERVATION

Accidental errors are usually due to the observer and are oftenrevealed by repeated observations; they are disordered in theirincidence and variable in magnitude, positive and negative valuesoccurring in repeated measurements in no ascertainable sequence.On the other hand, systematic errors may arise from the observeor the instrument; they are usually the more troublesome, forrepeated observations do not necessarily reveal them and even whentheir existence or nature has been established they are sometimesdifficult to eliminate or determine; they may be constant or mayvary in some regular way. For instance, if a dial gauge is used inwhich the pivot is not exactly centred, readings which are accuratelyobserved will be subject to a periodic systematic error. Again,measurements of the rise of a liquid in a tube using a scale fixed tothe tube will be consistently too high if the tube is not accuratelvertical; in this case the systematic errors are positive and pro-portional to the height of liquid. Further, measuring devices maybe faulty in various ways; even the best possible instruments rarelimited in precision and it is important that the observer shoulappreciate their imperfections.Errors peculiar to a particular observer are often termed persona

errors; we sometimes speak of the "personal equation." Errors 0this kind are well authenticated in astronomical work. Bessel, foinstance, examined tile estimates of time passages in astronomicaobservations and fouad that systematic differences existed amongsthe leading astronomers of his time. That similar differences exisamongst students and observers today will be familiar to teacherand scientific workers alike.More fundamentally there is the "error" introduced by the ver

process of observation itself which influences in some measure thphenomenon observed. In atomic physics this is specially important and is enshrined in the uncertainty principle due tHeisenberg, but in macroscopic phenomena with which we shall bmainly concerned it can be neglected. Further, there are manphenomena in physics, often included under the general term"noise," where fluctuations arise due to atomic or sub-atomicparticles which set a natural limit to the accuracy of measurements.These fluctuations are, however, usually very much smaller thathe errors which arise from other causes. For instance, moleculabombardments of the suspension of a suitably sensitive galvanmeter produce irregular deflexions which can be recorded.

10

ERRORS OF OBSERVATION CHAP. 1

example of such deflexions obtained using a torsion balance isshown in Fig. 1.

.----------------_ ..__ ... -

30sec

Fig. 1. Record of the deflexions of a supersensitive torsion balanceshowing irregular fluctuations in time due to the Brownian motion of the

instrument. (From an investigation by E. Kappler.)

•2. Errors and fractional errorsIf a quantity Xo units is measured and recorded as x units we

shall call x - Xo the error in xo, and denote it usually bye. Itmight be positive or negative but we shall assume throughout thatits numerical value is small compared with that of xo; this is usuallywritten iel«IXol·We can write

x=xo+e= xo(1 + I)

where 1= e/xo and is known as the fractional error in Xo. AlsolOOe/xo is called the percentage error in Xo. Of course e/xo can bewritten as e]x approximately when lel<lxol, that is, when 1/1<1.

We note that ~= 1 +1Xo

and Xo 1 1 I if III Ix=l+I':::. - «

e eXo x

~=~(l+f)':::.:'Xo x x

11

Also,

CHAP. ERRORS OF OBSERVATION

3. Estimate of errorIf only a single measurement is made any estimate of the error

may be widely wrong. There may be a large personal or accidentalerror, as well as that due to the lack of precision of the particularinstrument used.To take a simple example, a student timing the oscillations of a

simple pendulum may count 49 swings as 50, and the stop-watchhe uses may be accurate to perhaps O· 1 second. Also his "reactiontime," in using the stop-watch, may be different from that of anotherstudent, but perhaps in a timing such as this the effect of the "reactiontime" may be neglected as it is reasonable to assume it is the same atthe beginning as at the end of the time interval. In this case, assum-ing he counts the swings correctly the error in the measurement isdictated by the accuracy of the stop-watch. The fractional errorin the timing may be reduced by increasing the number of swingstimed but this tends to increase the error in counting the numberof swings, unless some counting device is used. The followingtimes were obtained by ten different students using the samependulum and the same watch: 37'2, 37'0, 36'9, 36'7, 36'8, 36·2,35'4,37'2,36'7,36,8 seconds for 20 swings and 73·8,74'3,74'0.74'2,74'4,74'0,73'0,74'1,73'6,74'7 seconds for 40 swings.To obviate or reveal accidental errors, repeated measurements of

the same quantity are made by the same observer, whenever possible.(If the phenomenon is unique the measurements cannot be repeated;for example, we cannot repeat the measurements of an eclipse. Ofcourse in a fundamental sense all measurements are unique; theyall refer to a particular instant.) A set of repeated measurementsmight be 10'1, 10'0, 10'0, 10·2, 12'3,10'1,10'1,10'0,10'1,10,2.It seems quite possible that some mistake was made in recording12·3 and it is reasonable to reject it on these grounds. (Resultsshould however not be rejected indiscriminately or without duethought. Abnormal or unusual results, followed up rather thandiscarded, have in the hands of a great scientist often led to im-portant discoveries.) The above measurements indicate that themeasured quantity lies between 10·0 and 10'2, and as the arithmeticmean of the nine measurements is 10+ (1/9)(0'8) ~ 10'1, we cansay that the measured quantity is 10'1 ± 0·1 to indicate the scatterof the measurements about the mean. In fact, in stating that thevalue of the measured quantity is 10'1 the numerical error is likely

12


to be less than 0'1, and indeed it is possible on certain reasonableassumptions to calculate the probability that the error is not greaterthan some assigned magnitude. We shall discuss this more fullylater.We note here that if a quantity Xo units is measured n times and

recorded as x" X2, - - - xn units, we can write x, = Xo + e, wheree, is the error in the measurement x.. The arithmetic mean x of then measurements is

XI + X2 + - - - + xn _e!...1~+_e=-2_+_-_-_-_+---"en=xo +n nand as some of the errors el>e2, - - - en may be positive and somenegative the value of (el +e2+ - - -+ en)/n may be very small. In anycase it must be smaller numerically than the greatest value of theseparate errors.Thus if e is the largest numerical error in any of the measurements

wa have

lei + e2 + -n- - + enl .;;;;e

and consequently [e - xol .;;;;e.Hence in general x will be near to Xo and may be taken as the "best"

value of the measured quantity which the measurements provide.In general the larger the value of n the nearer x approaches xo.It will be noticed that it is not possible to find e" e2' - - - en or e

since Xo is not known. It is usual therefore to examine the scatteror dispersion of the measurements not about Xo but about x. Wehall discuss this more fully later, but if we write x, = X + d, then

d, denotes the deviation of xr from x: it is sometimes called theresidual of x..We have x, = Xo + e, = x + d,

. 0 that e; - dr = x - Xoand el + e2 + - - - + en = n(x - xo)whereas dl + d2 + -- - + dn = O.

We note then that by repeated measurements of the same quantityuccidental errors of the observer may be corrected in some degree,but systematic errors peculiar to him cannot thus be obviated noran any lack of accuracy in the instrument itself. To take an

13


obvious example, with a scale wrongly marked as 9 instead of 10an observer might record measurements of 9 '0, 9 '1, 9 '0, - - -instead of 10· 0, 10'1, 10· 0, - - - and the arithmetic mean of theformer readings, however many there be, will have a "systematic"error of 1'0. This example is not so absurd as at first sight it mayseem, for every instrument records m "instead of M where M - mor (M - m)/M is a measure of the accuracy of the instrument. Itis important that the observer should know what degree of accuracyhe can achieve with the instrument he is using. Indeed thisinformation helps him to decide whether the instrument is theright one to use, whether it is appropriate to the particular endhe has in view.

How accurately can a length of about 50 em be measured usinga metre rule? Is this sufficiently accurate for the purpose? If not,should a cathetometer be used? These are the sort of questionswhich any scientist must ask and answer. The answers to thesequestions dictate the experimental apparatus he uses. It is of listleavail measuring one quantity with an accuracy of 1 in 1000, say, ifanother quantity which affects the result equally can only bemeasured with an accuracy of 1 in 100.

Here the distinction that is drawn between the terms accuracyand precision might be noted. * Accuracy refers to the closenessof the measurements to the "actual" value or the "real" value ofthe physical quantity, whereas the term precision is used to indicatethe closeness with which the measurements agree with one anotherquite independently of any systematic error involved. Using thenotation introduced earlier in this section we say that a set ofmeasurements Xl, X2 - - -, Xn are of high precision if the residuals d,are small whatever the value of x - Xo, whereas the accuracy ofthe measurements is high if the errors e r are small in which case'x - Xo is small too. Accuracy therefore includes precision butthe converse is not necessarily true. Sometimes more is knownof the precision of an instrument than of its accuracy.

It will be clear that the assessment of the possible error in anymeasured quantity is of fundamental importance in science. Itinvolves estimating (a) the accidental error, (b) the systematic orpersonal error of the observer, and (c) the systematic error of the

• I am grateful to Dr. H. S. Peiser who kindly brought to my noticeDr. Churchill Eisenhart's article in Photogrammetric Engineering, Vol.XVIII, No.3, June 1952.

14


instrument. Of these the accidental error is usually assessed byapplying certain statistical concepts and techniques as we shallexplain in Chapter 2. The other two errors, (b) and (c), are some-times merely neglected or are assumed to be smaller than (a) whichis not always true. Indeed systematic errors are often the mostserious source of inaccuracy in experimental or observational workand scientists have to devote much time and care to their eliminationand estimation. Various devices are used depending upon thenature of the measurements and of the apparatus used. There isno common or infallible rule. Certain systematic errors may beeliminated by making the measurements in special ways or inparticular combinations; others may be estimated by comparingthe measurements with those obtained when a quite differentmethod or different instrument is employed. On the other handthe personal error of an observer is usually treated as constantand is determined by direct comparison with an automatic recordingdevice or with the results of other observers. Once determined itis used to correct all the readings made by the observer. Variousother corrections may be applied to the readings to take accountof estimated systematic errors. In all cases the aim is so to correctthe readings as to ensure that all the errors remaining are accidental.Sometimes the experiments are so designed and the measurementso randomized that any remaining systematic errors acquire the

nature of accidental errors. This device is used particularly in manybiological experiments.

Of course, even when all this has been done systematic errorssometimes remain. Birge has pointed out, for instance, that"atomic weight determinations prior to 1920 were afflicted withunknown and unsuspected systematic errors that averaged fullyten times the assumed experimental errors." Also, the history ofthe measurement of fundamental physical constants such as thevelocity of light, the electronic charge and Planck's constant recordshow various systematic errors have been revealed and corrected.For example, in 1929 the accepted value of the electronic charge was4.7700 X 10-10 e.s.u.; later it was revised to 4·8025 X 10-10 e.s.u.,the difference between these two values being much greater than theuccepted accidental error of the earlier value.

Students are recommended to read some of the accounts, par-ticularly the original papers, of experimental work of high accuracy.If one instructive example may be selected, E. C. Bullard'!' has

15


discussed some gravity measurements made in East Africa. Afterexamining very carefully both the accidental and systematic errorsinvolved he concludes that the measurements form a consistent setwith a probable error of about 0·00001 ms-2• However, he cau-tiously adds: "While it is believed that the discussion of the errorsincludes all those large enough to be of any practical importance itmust be remembered that many apparently irreproachable gravitymeasurements have in the past been found to be subject to unexpected and unexplained errors, and until the source of these discrepancies has been found it would be unwise to be dogmatic aboutthe errors in the present work."

4. Estimate of the error in compound quantitiesOnce the error in a measured quantity has been estimated it is a

fairly simple matter to calculate the value of the consequentiaerror in some other quantity on which it depends. "If y is some function of a measured quantity x, the error in

due to some error in x can be found by using some simplemathematical techniques which we shall now explain.

5. Error in a productIf a quantity Q is expressed as the product of ab, where a and

are measured quantities having fractional errorsfJ and/2 respectively,we can write

Q = ab

= ao(l + fJ) x bo(l + I~~ aobo(l + fJ + I~

so that the fractional error in Q is 11 + 12 approximately, that isthe sum of the fractional errors in the two quantities of whichis the product.

EXAMPLE

If the measured lengths of the sides of a rectangle have errors of3% and 4%, the error in the calculated value of the area is 7%approximately, if the errors in the sides have the same sign, or 1%if they have opposite signs.

16


The above result can easily be extended. First we note that byputting a = b it follows that the fractional err~r in a2 is t",:ice t~efractional error in a. Alternatively, the fractional error In a ISone-half the fractional error in a2, that is, the fractional error inNt is one-half the fractional error in N.

Thus if we write

we must have

N= NoO +I)Ni ~ NJO + tf)

This follows of course from the result

O+f)i~l+tf

which can be established independently.Again, if Q = abc - - - the fractional error in Q is approximately

the sum of the fractional errors in a, b, c, - - - .IIIn particular, the fractional error in an equals approximately n

times the fractional error in a. This is true for all values of n,positive or negative.

6. Error in a quotientIf a quantity Q is expressed as the quotient aib where again a

and b have fractional errors 11 and h respectively, we have

Q = alb

aoO + 11)bo(l +I~

= ~(1 +/1)(1 - 12 + - - -)bo

ao~"bcp+/1 -/~Thus the fractional error in alb is approximately the difference ofthe fractional errors in a and b.Also if Q = (abc - - -)/(lmn - - -) the fractional error in Q is theum of the fractional errors in a, b, c, - - - less the sum of thefractional errors in /, m, n, - - - .

17

CHAP. 1 ERRORS OF OBSER VA TlON

EXAMPLE 1If the current i amperes in a circuit satisfying Ohm's law is

calculated from the relation i= E/R where E volts is the e.m.f. inthe circuit and R ohms is the resistance, then the fractional errorin i due to fractional errors IE and IR in E and R respectively isIE - IR approximately.

EXAMPLE 2If g is calculated from the simple pendulum formula g = 41T21/T2,

and we write 1= 10(1 + fi), T = To(1 +I~ where fi.J2 are thefractional errors in I and T respectively, we have

41T2/0 1+ fig = TJ x (1 + 1~2

41T2/0 1 + fi~ TJ x 1 + 2/2

41T2/0 (1 + fi - 2/~~ T:2 x 1o

~ go(1 + fi - 2/~.

Thus the fractional error in g is fi - 2/2, Since fi and 12 may bepositive or negative, it follows that the numerical value of thefractional error in g may be as large as lfil + 21121 or as small aslfil '" 21121·

If, however, as is often the case in practice, fi and 12 are notknown exactly but may have any value between certain limits, thegreatest value of the fractional error can be estimated. For instance,suppose fi lies between - FI and +FI, whilst 12 lies between - F2and +F2; then fi - 2/2 can be as large numerically as FI + 2F2•Indeed we can write

lfi - 2/21 <: FI + 2F2. which gives the greatest fractional error in g.

Instead of this greatest value of the fractional error, a smallerquantity, sometimes called the "most probable" value, is oftenquoted. It is given by the square root of the sum of the squares

18

..

ERRORS OF OBSER V A TION CHAP. 1

f the greatest values of the separate fractional errors, that is,(Fi + 4F~)~, which clearly is greater than FI or 2F2, but less thanFI + 2F2·

1

7. Use of the calculusThe calculus can be-used in the estimation of errors. For suppose

x is a measured quantity and y is a quantity calculated from theformula y = I(x). If ox is the error in x the corresponding errorin y is oy where

Iim oy = dyBx-*O ox dx

e. Oy dy if"" 11 h h . h .Therefore SX ~ dx oX IS sma enoug ; t at IS, t e error ill y

. . I b dy..,<!lIvenapproxunate Y y dx ox.

As an example, suppose y is the area of a circle of radius x, so that

y = 1Tx2

thereforedydx = 21TX

oy. ..,8x ~ 21TX, if oX is small

oy ~ 21TXOX

Hence

nd

Thus the error in the calculated value of the area due to a smallrror ox in the measured value of the radius is 21TXOX, and this

may be positive or negative depending on the sign of ox. It is,of course, the area of the annulus having radii x and x + 8x.

The fractional error in the area is

oy _ 21TXOX = 20Xy - 1TX2 x

that is, approximately twice the fractional error in the radius, inordance with the result proved in Section 5.

Of course, most simple examples including that above hardly19


need the use of the calculus; a little algebra is all that is necessaryas we have shown earlier, but the calculus does facilitate the solutionof more complicated problems.If a quantity Q is a function of several measured quantities

x, y, z, - - - the error in Q due to errors ox, oy, oz, - - - inx, y, z, - - - respectively is given by

oQ oQ oQoQ ~ - ox + - oy + - OZ + - - -

oX oy oz

The first term ~~ ox is the error in Q due to an error ox in x only

(that is, corresponding to oy, OZ, - - - all being zero), and similarly

the second term °o~ oy is the error in Q due to an error oy in y only.

This result is often referred to as the principle of superposition oferrors. •Also, if we suppose that ox, oy, OZ, - - - can have any value

between -el and +e" -e2 and +e2, -e3 and +e3, - - - respec-tively, then the "most probable" value of oQ is given by

oQ )2 oQ 2 sc )2(OQ)2 = (ox X e, + (Oy x e2) + (OZ x e3 + - - -

that is, 0Q is the square root of the sum of the squares of thegreatest errors due to an error in each variable separately.Taking as an example the simple pendulum formula used earlier

we have

therefore

g = 47T21/T2

og ~ og 0/ + og sr()/ oT~ 47T2 0/ _ 87T2[sr- T2 T3

andog 01 sr- ~- -2-g I T

20

ERRORS o·p OBSER VATlON CHAP. 1

that is Ii - 2h, as we found in Section 6. This result is obtainedmore simply by taking logarithms first, so that

and hence

logg = log 47T2+ log 1 - 210g T

~ ~ ~/ _ 20Tg - I T

EXAMPLE 1A quantity y is expressed in terms of a measured quantity x by

the relation y = 4x - (2/x). What is the percentage error in ycorresponding to an error of 1% in x?

We have dy[dx = 4 + (2/x2)

so that oy ~ [4 + (2/x2)]ox

ercentage error in y = (oy/y)l00

~ 1~2/x)[4 + (2/x2)]ox

~ 100(4x2 + 2)ox- x(4x2 - 2)

Thus if ox/x = 1/100, the percentage error in y is (4x2 + 2)/(4x2 - 2).It might be noted that this percentage error varies with x and is

approximately 1% when x is large. Further, it is obviously largewhen 4x2 - 2 is small, that is, when x lies near to ± 2-t; this isnot surprising since y = 0 when x = ±2-t. Actually usingoy ~ [4 + (2/x2)]ox, we find that when x = ± 2-t, oy ~ 80x.

EXAMPLE 2A quantity y is calculated from the formula y = Cx/(1 + x2).

If the error in measuring x is e units, find the corresponding errorin y and show that for all values of x it does not exceed Ce units.

Now dy 1 + x2 - x(2x) 1 - x2dx = C (1 + X2)2 = C ,- . -'"

21

CHAP. I ERRORS OF OBSERVATION

therefore 1 - X2oy ~ C/o . ,,,OX

If the error ox in x is e unit, the error in y is

1 -x2oy ~ G__ . ".e

This can be written oy ~ GeE

where 1 - x2E = (1 + X2)2

~il

I'

How does E vary with x? When x =0, E =1; when x =I,E = 0, and when x > 1, E < 0. Also E -+ ° as x -+ co. Further,E does not change in value when x is replaced by + x and henceits graph is symmetrical about the axis of E and must have theform shown in Fig. 2.

E

A\as-I 1 2-3 -2 3 x../ 01 '--.....----

-o.s

-IFig. 2. Graph of E = (1 - x2)/(1 + x2)2

E has a maximum value when x = ° and minimum values whenx = ± 3t. Thus E lies between 1 (when x = 0) and -t (whenx = ± 3t). Hence the error in y lies between Ce and -GeI8.

22

.\ EXAMPLE 3The viscosity of water, 1), is calculated using Poiseuille's formula

giving the quantity of water Q flowing through a cylindrical tubef length I and radius a in time t under a pressure difference p.Find the error in 1) due to errors in the measured quantities Q, I, aandp.

ERRORS OF OBSERVATION CHAP. ,'\

'T1'pa4t1) = 81QWe have

therefore ,log 1) = log ('T1'18) + logp + 410g a + log t - log 1 - log Q

01) op 40a ot 01 sa-=-+-+-----1) pat I Q

Thus the fractional error in 1) equals a combination of the fractionalrrors in p, a, t, I and Q. The term 40ala is usually the mostimportant since a is very small and hence for an accurate deter-mination of 1) special attention must be paid to the accuracy ofmeasurement of the radius a. Indeed to ensure an accuracy of 1%in 1), the radius must be measured with an accuracy of at least1 in 400.

8. Error in a sum or differenceIf a quantity Q is expressed as the sum of two quantities a and b,

having errors el and e2 we have

Q =a +b= ao + bo + el + e2

So writing the error in Q as e and the fractional error in Q as Iwe have

e = el + e2

und 1= el + e2 = aofi + bo/2ao +bo ao + bo

where II and 12 are the fractional errors in a and b respectively.If el and e2 are known, e and Ican be calculated, but as we have

noted earlier all that is usually known is that el may have anyvalue between -EI and +EI say, whilst e2 may have any value

23

\

CHAP. 1 ERRORS OF OBSERVATION 'I

between - E2 and +E2. What then are the limits between whiche and j' may lie?Clearly e can have any value between -(EI + E2) and

+(EI + E2J. or lei lies between 0 and E[ + E2. It is usual to takethe "most probable" value of lei as (Er + EDt (see Section 6).which is less than E[ + E2 but bigger than either E[ or E2•On the other hand the fractional error I depends on the values

of 00 and bo as well as on fi and 12. and varies between wide limits.

Writing 1= fi + _b+o b-(f2 - fi) =12 + 0+0b (fi - 12)°0 0 00 0

I:it follows that if 00 is numerically much bigger than bo then I equalsI[ approximately. or if 00 is numerically much smaller than bothen I equals h approximately. Again if ao and bo are approxi-mately equal and 01 the same sign I is approximately ·Hfi + 12J.whereas if 00 and bo are approximately equal but 01 opposite sign(so that ao + bo is small) I may be large; indeed in general it willbe large unless it should happen that fi - 12 is very small. I

Again the "most probable" value of III may be taken to be

Veer + E~)100 + bol

EXAMPLE 1

If two lengths of 11 and 12 are measured as 10'0 and 9·0 withpossi ble errors of 0·1 in each case find (i) the greatest error and(ii) the greatest fractional error in the values of 11 + 12 and 11 -/2.We can write

I[ + 12 = (10'0 ± 0·1) + (9'0 ± 0·1)= 19·0 ± 0'2

The greatest error in 11 + 12 is 0·2 and the greatest fractional error is0'2/19'0 = 0·01. Also, 11 -/2 = 1·0 ± 0·2 so that the greatest errorin 11 -12 is 0·2 and the greatest fractional error is as high as 0'2/1·0 = 0·2. We note that the "most probable" fractional error in11 -/2 is v'[(0·1)2 + (0·])2]/1'0 =0'14.

24

J~

CHAP. 1ERRORS OF OBSERVATION

EXAMPLE 2The viscosity TJ of a liquid is measured by a rotation viscometer.

The cylinders are of radii a and b, and a torque G is applied to therotating cylinder so that

G (1 1)TJ = 47T,Q Q2 - lJi.

where ,Q is the angular velocity of rotation. Calculate the fractionalerror in '1 given that a = 0·04 m, b = 0·05 m, that the greatest errorin measuring both a and b is 0·000] m and that the error in G/D.maybe neglected.

Now, G (2 2)OTJ= 47T,Q - (}a + '[)job

so lhat writing oa/a = fi and ob/b = 12 we have

OTJ= (- ~ fi + ~12)/(~ - .!.)Tj a2 b2 a2 b2

= 2(a2h - b2fi)/(b2 - a2)

Hence in this case oTJ/TJ= 2(1612 - 25fi)/9

where the greatest values of I[ and 12 are 0'01/4 and 0'01/5respectively.Hence 10Tj/TJImay be large as

2(16 x 0,01 + 25 x 0'01)9 5 4

that is, 0·021 or about 2%.The "most probable" value of oTJ/TJmay be taken as

2v[(16F2J2 + (25FI)2] = ~(0'07) = 0'0169 9If there were also errors in G and ,Q the fractional error in G/,Qwould have to be added to the value of 101)/TJI calculated above.

25

t

ERRORS OF OBSER VATION CHAP. 1CHAP. 1 ERRORS OF OBSERVATION

EXAMPLE 3In Heyl's method the gravitational constant G is calculated from

the formula

G _ 4172/ (1 1 )- Al - A2 Tl - Tl

where T, and T2 are times of oscillation, I is the moment of inertiaof the system about the axis of suspension and AI> A2 are constants.Estimate the fractional error in G due to errors in TI and T2 of

0·1 s when TI = 1750 sand T2 = 2000 s.

Now since G = 4172/ X (T2 - TI)(T2 + T1)

Al - A2 T?Tl

using the method of Section 7

so O(T2 - T1) O(T2 + TI) 20TI 20T2-= + ----G ~-~ ~+~ ~ ~

The first term on the right-hand side may be as large as ~(2~0)

and is the most important term. Putting OT2= - OTI = 1/10we get the largest value of OG/G is

2(1 1 1) 110 250 + 1750 - 2000 == 1200

The values of TI and T2 quoted above are those given by Heyl(2)but the values of oTJ and OT2 are hypothetical. Heyl did not givean estimate of the error in the value of G, but it is interesting tonote that the value of G he adopted as a result of experiments withgold, platinum and glass spheres was 6·670 X 10-8 em! g.2 s-2"with a precision, as measured by the average departure from themean, of 0·005."

26

EXERCISES 1

1. If y = x2/(l + x2) find oy/y when (a) x = 3, Ox = 0·1 and(b) x = 2, Ox = 0·05.

2. If y = sin (2wt + IX) find the fractional error in y due to anerror of 0·1 % in t when (i) t = 17/2w, (ii) t = 17/W. Find alsothe values of t for which the fractional error in y is least.

3. Find the fractional error in eX corresponding to an error Oxin x. If x = 0·012 is correct to two significant figures showthat eX may be calculated for this value of x correct to foursignificant figures.

4. The mass m grammes of an electron moving with velocity u m s" 'is mo/vlp -- (v2/e2)] where em S-1 is the velocity of light. Showthat the fractional error in m is approximately v2/e2 times the

• fractional error in v, if v/e is small compared with unity.S. If i = k tan 8 find the value of 8 for which (i) the error in i is

least, and (ii) the fractional error in i is least, for a given error ein measuring 8.

6. Given T2 = h + (lOO/h) find the value of h for which the errorin T is least if the error in h is a constant e. Find also the valueof h for which the fractional error in T is least.

7. The diameter of a capillary tube is found by weighing andmeasuring the length of a thread of mercury inserted into thetube. Estimate the error in the calculated diameter of the tubein terms of the errors in the length, mass and density of themercury thread.

8. Using Kater's pendulum g is given by

g = 47T2(hJ + hz)/T2

The length of h, + h, is measured as 1·0423± 0·000 OSm and thetime of oscillation T as 2·048 ± 0·0005 s. Calculate g and thegreatest fractional error.If h, + h« is measured as 1·042± 0·0005 m, how accurately must

T be measured in order that the error in g may be less than 1 in 1000?

27

Table 1..•. IISC Lib 8'lore Class Frequency Class Frequency." •

511.43N72 0-9 2 50-59 32

1111111111111111111111111J

10-19 5 60-69 2520-29 6 70-79 10

79638 30-39 14 80-89 240-49 22 90-99 2

28 29

7963.8


9. The surface tension y of a liquid of density p is found byinserting the liquid into a U-tube of which the two limbs haveradii '1 and r2 respectively. The difference of height h in thetwo limbs is measured and y is calculated from the formula

. .' 1 1 1y.(- - -) = -gph.'1 r2 2

Estimate the fractional error in y if h = 1'06 ern, rl = 0·07 ern,r2 = 0·14 em, and the error in each of these measurements isnot greater than 0 :005 em.

10. The deflexion dot a beam under certain conditions is given byd = 4We3/317Ea4• Find (i) the maximum fractional error,(ii) the "most probable" fractional error in the value of Young'smodulus E calculated from this formula if the error in d is±0·1 %, the error in e is ±0·05 % and the error in a is ±0'1 %.

11. A coil of n turns of radius r carries a current J; the magneticfield at a point on its axis at a distance x from the centre isH = 27Tnr2J(r2 + x2)-3J2. If the error in measuring x is e,find the corresponding error in the value of the field. If e isa constant find for what value of x the error in H is greatest.

12. A large resistance R is measured by discharging a condenser ofcapacitance C charged to potential Vo; the time t taken for thepotential to fall to V is noted. R is given by tl C log (Vo/V).Find the fractional error in R due to errors in t, C, Vo and V.If Vo and V are measured by a ballistic galvanometer, then

••• Vol V = dold where do and d are the corresponding deflexions"~·.of the instrument. Assuming the errors in do and d are equal• in magnitude, show that the greatest value of the corresponding

. :.fractional error in R is fo( 1 + ~o)flOg -? where fo is the

. fractional error in Vo Find the value of Vol V for which this". is least.

CHAPTER 2

SOME STATISTICAL IDEAS"The experimental scientist does not regard statistics as anexcuse for doing bad experiments."

L. HOGBEN

9. Frequency distributionsNumerical data, including scientific measurements as well asindustrial and social statistics, are often represented graphically toaid their appreciation.The first step in dealing with such data, if they are sufficiently

numerous, is to arrange them in some convenient order; this isoften done by grouping them into classes according to theirmagnitude or according to suitable intervals of a variable on whichthey depend. For instance, the percentage marks obtained in anexamination by a number of students could be grouped by countingthe number of students who had marks between 0 and 9, 10 and19, - - - , 90 and 99, thus dividing them into 10 classes. The datacould then be tabulated as shown in Table 1, in which the marksof a sample of 120 students have been used.The number of data in each class is usually called the frequency

for that class. Table 1 shows what is called the frequency distri-bution. The pairs of numbers written in the columns headed"class," for example, 0 and 9, 10 and 19 and so on, are usuallycalled the lower and upper class limits. The width of any class 15the difference between the first number specifying that class .andthe first number specifying the next, that is, 10 for each of .theclasses shown in Table 1. For some groupings, however, .thewidths of the classes may be unequal. ..

CHAP. 2 SOME STATISTICAL IDEAS SOME STATISTICAL IDEAS CHAP. 2

The classification shown in Table 1 obviously helps us tappreciate the distribution of marks amongst the students; we casee at a glance, for instance, how many students have fewer than40 marks and how many have 70 or more. But a graphical repre-sentation can make it possibly even clearer. The data are plottedin Fig. 3, where the marks are represented along the horizontal

40

30>-~ 20(])::::J0-

J 10

24·5 44·5 64·5 84·5Marks

Fig. 3. Frequency polygon of examination marks.

axis and the frequencies along the vertical axis. The points obtainedby plotting the frequency against the mid-value of the correspondingclass, namely, 4·5, 14·5, - - -, 94·5 are joined by straight linesand the resulting figure is known as a frequency polygon.A different method is used in Fig. 4 where a series of rectangles

unit area30~ =

AI ~D

20

10

o 20 40 60 80 100Marks

Fig. 4. Histogram of examination marks.30

Ire constructed of width equal to the class width and of area equalto the frequency of the corresponding class, that is, the rectanglesIn Fig. 4 have areas equal to 2, 5, 6, 14, 22, 32, 25, 10, 2, 2 unitsrespectively. The figure obtained is called a histogram, and thetotal area of the histogram in this case is 120 units equal to thetotal number of students.Since the area of each rectangle in a histogram represents the

frequency in the corresponding class, the heights of the rectanglesare proportional to the frequencies when the classes have equalwidths. In this case the mean height of the rectangles is pro-portional to the mean frequency.

to. The meanThe mean frequency is usually not of particular significance, but

what is often important is the mean of the data. This is defined asfollows: if fi, 12, - - - In are the frequencies in the various classesof which XI, X2, - - -, Xn are the mid-values of the variable, themean value of the variable is given by

(fix! + 12X2 + - - - + /',<)/(/1 + 12 + - - - +IJ (1)

This is the weighted mean of XI, x2, - - - ,xn' the weights beingthe frequencies in the corresponding classes.It can be written as

n n

~ Isxs/~ Is or [Ix]/[/].=! .~I

and is often denoted by s,To evaluate x using the expression (1) directly can sometimes be

laborious, but the arithmetic can be minimized by using the followingsimple device.

Let us write x. = x~+ m where m is some constant. Then

fix! +12X2 + - - - + I nXn= fi(x; + m) +Hx;' + m) + - - - + In<x~ + m)= fix; + hX2 + -- - + InX~ + m(fi + 12 + ... +fJ

31

CHAP. 2

therefore

SOME STATISTICAL IDEAS

fixl + 12x2 + -- - + .f"Xn = fix~ + hx; + - - - + f,,x~ + mfi+/2+---+ln 11+/2+---+ln

or x=x'+m

where X' is the mean of the quantities x;.By choosing m conveniently we can make the evaluation of x

simpler than the evaluation of x. It is clear that X' will be smallif m is chosen near to x; m is often called the working mean or theassumed mean.As a simple example columns one and two of Table 2 give values

of Xs and the corresponding values of f.. In column three angiven the values of l.xs' so that on addition the mean value ofis given by

x=172=3~~3'944 11

Table 2x, Is Isxs x; = Xs - 3·5 Isx;0'5 1 0'5 -3 - 3i 5 5 7·5 -2 -102'5 7 17·5 -1 - 73'5 9 31'5 0 04'5 10 45·0 1 105'5 8 44·0 2 166·5 4 26'0 3 12

sum 44 172'0 18

However, it is simpler to proceed as follows: an examination ofthe data suggests that the mean is somewhere between 3· 5 and4· 5, and so taking m = 3·5 the values of x; are tabulated in columnfour of the table and the values of f,x; calculated as shown in thelast column.

32

SOME STATISTICAL IDEAS CHAP. 2

18It follows that i' = 44 and hence

18x = 3·5 +-44 3~11

(2 IS found directly.It is clear that in this way the amount of arithmetic, and con-

sequently the likelihood of error, is reduced. In many practicalcases the economy is considerable.

II. Relative frequencyIf the classes specified by Xl, X2' - - - , Xn occur with frequencies

),h, - - - ,In' the relative frequency with which Xl occurs is;/(fi + 12 + -- - + 1,'), or generally the relative frequency withwhich XI occurs is fi/"L,f.. If we denote the relative frequency ofXI by rl' the mean of the observations can be written as

1/

X =}.:; r».s=l

We note that rl is represented in a histogram by the area of therectangle corresponding to the class of which XI is the mid-valuedivided by the total area of the histogram. If the scales are sochosen that the total area is unity then each rectangle representsthe corresponding relative frequency.

12. The medianIf a set of observations are arranged in ascending or descendingrder of magnitude the observation in the middle of the set is calledthe median. More precisely, if the number of observations is odd,say 2n + 1, the median is the (n + l)th value; if the number ofobservations is even, say 2n, the middle values of the set are the11thand (n + l)th, the arithmetic mean of which is taken as themedian.For example, the numbers 10, 12, 13, 7, 20, 18, 9, 15, 11 when

arranged in ascending order of magnitude are 7, 9, 10, II, 12, 13,15, 18, 20 of which the median is 12. If, however, the last value 20had not been present the middle terms of the set would have been11 and 12 and the median would have been taken as 11·5.

33

CHAP. 2 SOME STATISTICAL IDEAS

Sometimes, of course, the data are grouped into classes as fcexample in Table 1. In this case the total number of students i120; if the marks were set out in ascending order of magnitudthe middle values would be the 60th and the 6lst. To find tbmedian we therefore find the marks of the fictitious "individual'number 60·5 in the set. To do this we note that 49 students havmarks under 50 and 49 + 32 = 81 students have marks under 601Student 60· 5 is therefore in the class 50-59, and the median .taken to be

50 + 60'5_.::- 49 x 10

that is, 50 + 3'6 = 53'6.

13. Frequency curvesIt is clear that in general the shape of a histogram depends 0

the widths of the classes chosen in grouping the data. For instanwhen the data used in Table 1 are grouped in classes of half tbwidth the results are as shown in Table 3. The correspondin

Table 3Class limits Frequency Class limits Frequency

0-4~}2

50-54 15} 325- 9 55-59 1710-14

i}560-64 14} 25

15-19 65-69 1120-24

~}670-74

~} 1025-29 75-7930-34

~} 1480-84

~}235-39 85-8940-44 12} 22 90-94

~}245-49 10 95-99

histogram is shown in Fig. 5 which should be compared with toriginal histogram shown in Fig. 4.The area of ABCDEF in Fig. 5 is (15 + 17) units, the same

the area of the corresponding rectangle ABCD in Fig. 4. Thistrue for all corresponding portions of the two histograms, so thalthough they have different shapes they have equal areas.

34


unit area1st IZI

10

5

Fig. 5. Histogram of examination marks.

tal area equals in each case the total frequency which, of course,unchanged by any change of class width.In practice the width of the classes is dictated by the nature of

the data, in particular by their number as well as their accuracy;classes either too wide or too narrow do not reveal the general trendof the data. For example, in dealing with the marks shown inTable 1, the smallest possible class width is 1 mark since fractionalmarks are not given, but in fact a class width of less than 5 marks isprobably meaningless as few examiners would claim to be able tomark within 5%. On the other hand the maximum class width is100, in which case the histogram would consist of a single rectanglefrom which no useful information could be derived.Again, if the heights of a group of men were measured to the

nearest 0'5 em, all the men of recorded height 180'5 cm could be putIn the class 180·5 and the class-width would be 0·5 cm. This classwould include all the men with heights between 180·25and 180'75 ern,which are known as the class-boundaries. On the other hand, usingIt class-width of 2 ern the class 162-163'5 might include all the menwith recorded heights of 162, 162'5, 163, 163'5 em; strictly in thatease it would include all those with heights between 161·75 em and163·75em, whieh would be the class-boundaries, and the mid-valueof the class would be 162'75 cm. However, in the class 162-164 theremight be included the men with recorded heights of 162'5, 163,163·5em and half of those with recorded heights of 162 and 164 cm ;

35


regarded as a sample taken from an infinite number distributed in, manner indicated by the frequency curve. The histogram of thetrnple is an approximation to the frequency curve, the degree ofipproximation depending in general on the number of observationsand on the class width. The histogram tends more and moreclosely to the frequency curve as the number of observationsIncreases and the class width decreases, provided the unit of area isNOchosen that the total area under the histogram is always theame as that under the frequency curve, which as stated above isusually taken as unity.If the equation of the frequency curve is known, that is, f is

expressed as some function of the variable x, the mean value ofthe variable can usually be found by expressing it in terms ofIntegrals, namely,

b b

X = J fxdx/ J fdxa a

where a and b define the range of x. For example, consider thedistribution such that the frequency f is given by

Variable. xFig. 6. Frequency curve.

The area under a frequency curve or under any part of it hspecial significance. The scales of the frequency curve are oftechosen so that the total area under the curve is unity. This totalarea being finite does not give the total frequency which is infinitbut the area under any part of the curve gives the relative frequenfor the corresponding range of the variable. The value of the variabfor which the ordinate divides the area under the curve into tequal parts is therefore the median. On the other hand the mocorresponds to the maximum value of the frequency.

We shall take some examples of frequency curves later.merely note here that any finite number of observations might

36

J1 2J1 dx 2[ -1 J1f dx = - -- = - tan x = 171' x2 + 1 71' -1

-1 -1


the mid-value of the 'Classwould then be 163 em. It is important tnthere should be no doubt where the boundaries of the classes anand that there should be no overlap or gaps between successiclasses.It will be noted that the mean of any number of observatio

depends in general on the width of the classes into which tobservations are grouped, since the mean is defined (see Section 1in terms of the mid-values of the classes. However, provided tclass width is not too large the effect of grouping is small andusually neglected. For example, it is found that the mean of tmarks given in Table 1 is 51' 25; if these marks are groupedshown in Table 3 the mean is found to be 51'17. Again, us'only one class covering the whole range, 0--99, the mean is 49·5.With a very large number of data grouped into classes of sm

width it is clear that the outline of the histogram approximatesa continuous curve, as is illustrated in Fig. 6. Such a curve is knoas a frequency curve.

--:;.:.uc

Ju..f = 2/7T(X2 + 1)

Irorn x = - 1 to x = + 1, the factor 2/71' being chosen so thatthe area under the frequency curve is unity; that is,

1

The mean value of the variable = J f xdx = ~[IOg (x2 + I)J ~ 1

-1

which is zero. This is obvious from the symmetry of the curveibout the axis x =O. The median is also x = 0 and so is the mode.

14. Measures of dispersionA very important characteristic of a set of data is the dispersion

or scatter of the data about some value such as the.mean.37


The numbers 40, 50, 60, 70, 80 of which the mean is 60 havedispersion or scatter of 20 an each side of the mean; they arewithin the limits 60 ± 20. Similarly, the numbers 5, 6, 7, 8, 10,have a mean of 8 and are within the limits 8 - 3 and 8 +Different sets of data have, in general, different means and differedispersions. The data corresponding to' the frequency curve A i:Fig. 7 have a smaller dispersion than those represented by tcurve D, whilst the mean of the data A is greater than that of tdata D. Various parameters are used to' measure the dispersioWe shall mention only three, namely, the range, the mean deviatland the standard deviation.

ecV::J

f

Variable

Fig. 7. Frequency distributions with different means and dispersions

15. The rangeThe range of a frequency distribution is defined as the differen

between the least and greatest values of the variable. This isvery simple measure of dispersion and has, of course, limitationsbecause of its simplicity. It gives the complete interval of thvariable over which the data are distributed and So' includes thodata, usually at the ends of the interval, which occur very .frequently. It may be much mare useful in certain cases to' knothe portion of the range af the variable within which a givfraction, say 50%, of the data lie. .

Far example the range of the marks given in Table 1 is 0-9:but 93/120 or 78% of the marks lie between 30 and 69 inclusive.

38


16. The mean deviationIf Xl, X2, - - - Xn are a set of data of which .e is the mean, the

deviations of the data from the mean* are (x, - x), (X2 - x),• - - (xn - x) respectively, which we will write as d., d2, - - - dn•Some of these deviations are positive and same negative; in facttheir sum is zero, that is,

dl + d2 + - - - + d; = Xl + x2 + - - - + xn - nx = O.

The mean deviation (sometimes called the mean absolute deviation)of a set of data is defined as the mean of the numerical values ofthe deviations,

Therefore the mean deviation is

Idd + Id21 + - - - + Idnl = ~f Idsl·n ns~l

Far example, the numbers 7, 5,8,10, 12,6 have a mean of 8 anda mean deviation of (1/6)(1 + 3 + 0+ 2 + 4 + 2) = 2.If the data Xl' X2, - - - Xn have frequencies ii, 12, - - - In respec-

tively, the mean deviation is given by

iildll + 121d21 + - - - + Inldnlii+/2+---+ln

}:f.fxs - xl}:fs (3)

EXAMPLE

Consider the data given in Table 2. The mean was found to' be

3 :~ ~ 3·9. In Table 4 below values of Idsl are given in column

three and values of f.ldsl in column four, from which it follows that

I d .. . 58'0 I 32uie mean eviation IS 44 = . .

• deviations are sometimes taken from the median.

39

.,~il,r.~~Y --..-,I,('~:f~{· tI'O'(Y~\ ~\£>"I'\" )~..Iir


Table '4

X = 3'9, d. = x, - 3'9XJ is Idsl IMsl0·5 1 3'4 3'41· 5 5 2·4 12·02'5 7 1'4 9·83'5 9 0'4 3'64·5 10 0'6 6'05'5 4 1·6 12·86'5 8 2·6 10'4sum 44 58'0

17. The standard deviation

The most imt>ortant measure of dispersion is the standadeviation, usually denoted by a. It is defined in terms of the squanof the deviations from the mean as follows:If d., d., - - -', t::ln are the deviations of the data xl, X2, - - -, X,

from the mean x, then *

G-2 =(dr + d~ + -- - + d~)/n

that is, [1 n J!- [1 n J!a '= - ~ d} = - ~ (x, - x)2ns=1 ns=1

If x" x2, - - -, ~n have frequencies Ji, 12, - - -, In respectively the

a2 = (Jid? + 12d~ + - - - + Ind~)/(Ji + fz + - - - + In>

= ~ !,.(J"s - x)2/ ~ Is.s _1 s=1

a2 is known as the variance of the data. The standard deviationis the root mean square deviation of the data, measured from thmean.If the frequenci~/are known for all values of X over a continuo• The reader is ~ked to note the modification introduced in Section 3,

40


runge (a, b), then the summations can be replaced by integrals andwohave

a2 = r I(x - x)2dx / r fdxa a

EXAMPLE 1Find the standard deviation of the numbers 5, 6, 7, 8, 10, 12.The mean of these numbers is 8, and so the deviations are3, -2, -1, 0, 2, 4 respectively; the sum of the deviations is,f course, zero.

Now a2 = (1/6)(9 + 4 + 1 + 0 + 4 + 16) = 34/6 = 5·67

therefore a = 2·38.

It might be noted that a coefficient of variation is sometimes used.It is defined as 100 a/x and is expressed as a percentage. In theibove example the coefficient of variation is therefore

100(2'38)/8::::: 30%

(4

EXAMPLE 2

If the frequency of the variable x is given by I = Ce=t« for allpositive values of x, find the mean and standard deviation of thedistribution.The mean value of x is given by

x = loocx e-x/a dx/ looc e-x/a dxo 0

which reduces to x = a.Also the standard deviation a is given by

a2 = looC(x - a)2 e-x/a dx / looc e-x/a dxo 0

which reduces to a = a.41


18. Evaluation of standard deviation, aIn cases more complicated than the simple example abo

(Example 1), considerable economy in the arithmetic can be madby using a suitable working mean m, instead of x,As shown earlier

"£fs(xs - m) = "£I.xs - m"£fs

h "£fs(xs - m) "£I.xs -so t at = -- - m = x - m

"£1. "£ I.The summation on the left-hand side, involving deviations fro

the working mean m, can thus be used to find x (see Section 10)Further,

'E.fs(xs - x)2 = "£1. [(xs - m) + (m - x»)2= "£fs(xs - m)2 + 2(m - x)"£fs(x. - m) + (m - x)2,,£

Hence, dividing by "£1. and using equation (5) above -2 "£1S<xs- x)2 "£fs(xs - m)2 + 2( -)( - ) +( -)2a = = m-x x-m m-x"£1. "£1. •

= "£fs(xs - m)2 :..- (m _ x)2"£1.

= fL2 - (m - x)2 (6)We note that "£fs(xs - m)2 is the sum of the squares of th

deviations from the working mean m. It is usually easier, ifinding a2, to choose a suitable working mean m and calculafL2 = ,,£fs(x.-m)2/,,£fs and then use

a2 = fL2 - (m - x)2

The special case when m = 0 is of some importance. Also:one further interpretation of equation (6) might well be noted herethe least value 01 fL2 is a2 and this arises when m = x, that is, thosum of the squares of the deviations of any set of data, measuredfrom any value m, is least when m equals the mean of the data.

EXAMPLE

Calculate the mean and standard deviation of the followinobservations, where 1denotes the frequency of the observation

x 0 1 2 3 4 5 6 7 8 9 101 10 40 72 85 78 55 32 18 7 2 1

42

SOMB STATISTICAL IDEAS CHAP

Using a working mean of 3 the necessary working is tabulatedbelow, where d now denotes the deviation from the working mean,that is, x - m with m = 3.The last two columns are ca.lculated for checking purposes as

explained later in Section 20.

Table 5x f d fd fd2 fed + 1) fed + 1)2

0 10 -3, -30 90 -20 401 40 -2 -80 160 -40 402 72 -1 -72 72 0 03 85 0 0 0 85 854 78 1 78 78 156 3125 55 2 110 220 165 4956 32 3 96 288 128 5127 18 4 72 288 90 4508 7 5 35 175 42 2529 2 6 12 72 14 9810 J 7 7 49 8 64sum 400 228 1492 628 2348

Hence, from equation (5)x - 3 = 228/400

therefore • x = 3·57Also fL2 = 1492/400 and hence from equation (6)

1492a2 = - - (0-57)2 = 3'73 - 0·32 = 3'41

400

therefore a = 1'85

We note that in this case the coefficient of variation equals

(1'85/3-57) x 100% = 52%

19. Sheppard's correctionIf we regard the data given in Table 5 as grouped into classes of

unit width Sheppard's correction should be applied in the evaluationof a2• As we have stated earlier (Section 13) the effect of the classwidth on the mean is usually negligible, but Sheppard has shown

43


that to allow for the class width c the value of ,.,,2 as calculated abovshould under certain conditions be reduced by c2/12.

Since c = 1 in this case, we have,.,,2 = 3'73 - 0'08 = 3·65

a2 = 3'65 - 0·32 = 3'33a = 1·82

and hencetherefore

20. Charlier's checksAs a check on the arithmetic involved in the calculation of x

and a we can use the results}:.f(d + 1) = }:.fd +}:.f

sr« + 1)2 = }:.fd2 + 2}:.fd +}:.f

which are known as Charlier's checks.In the above example }:.f(d + 1) and }:.f(d + 1)2 hav.e been

calculated independently as is shown in the last two columns 0Table 5. The calculated values agree with the values obtained'using Charlier's checks, for these give

}:.f(d +1) = 228 + 400 = 628

and }:.f(d + 1)2 = 1492 + 2(228) + 400 = 2348

These checks are only partial but are usually worthwhile.

21. The mean and standard deviation of a sumIt is sometimes necessary to find the mean and standard deviation

of the sum of two or more sets of quantities, or of two or morefunctions. It is possible to obtain expressions for the mean andstandard deviation of the sum in terms of those of the componentsforming the sum.

Suppose XI' x2, - - - xn and xi, x;, - - - x;" are two sets of valuesof a variable x. Further, suppose the mean values of the two setsare x and x respectively. The mean value of the sum or aggregateof the two sets of values is(XI + x2 + - - - + xn + xl + xi - - - + x;")/(n + m)

= (nx + mx')/(n + m)Similarly if a and a/ are the standard deviations of the two sets

of values, the standard deviation of the aggregate of values can44


be found, by using equation (6) of Section 18, in terms of/I, m, x, X', a and a'.

Another important example of a sum, however, is where avariable z is expressed as the sum of two or more independentvariables x, y, - - -, say z = x + y or more generally z = ax + bywhere a and b are constants.

Suppose x has the values Xl, X2, - - - xn and y has the valuesy" h, ---Ym' Then if z = ax + by it follows that z has thevalues, mn in number, given by ax; + by} with i = 1, 2, 3, - - - nand j = 1, 2, 3, - - - m. If the means of z, x, yare denoted by

, X, Y and the standard deviations by az, ax, ay respectively, it canbe shown that

i = ax + byand ai = a2a~ + b2a;

In particular if z = X + Y we have i = x + y and a~ = a~ + a.~.For example, suppose X has the values 10, 11, 12 and y has the

values 6, 8, 10 so that x = 11 and y = 8, whilst

a~ = (12 + 0 + 12)/3 = 2/3

and a; = (22 + 0 + 22)/3 = 8/3

Then if z = x + y, z has the nine values obtained by adding thethree values of X to each of the three values of y, namely ,

16 18 2017 19 21 .

18 20 22

It follows that i = 19 verifying i = x + y.Also, a~ ~ (32 + 22 + 2 . t2 + 0 + 2. 12 + 22 + 32)/9 = 10/3

verifying that a~= a~+ a;.Again, if z = x - y the values of z are

o22

344

5 6

so that i = 3 verifying i = x - y.Also, a~= (32 + 22 + 2.12 + 0 + 2.12 + 22 + 32)/9,= 10/3 so

that again a~= a~ + a;.45


EXERCISES 2

1. Find the mean and the median of the following;(a) 10,12,14,16,17, 19,20,21(b) the quantities given in (a) when they occur with frequenciesl

2, 3, 4, 5, 6, 8, 10, 12 respectively.2. (a) Find the mean deviation of the following;

15,21,19,20,18,17,22,23, 16,25(b) Find the mean deviation and the standard deviation of the

data given in (a), if their frequencies are respectively;1, 11, 12, 14,9,6,9,5,3,1

3. Thirty observations of an angle B are distributed as followsf being the frequency of the corresponding value. Calculate themean and mean deviation of these readings.

(J J500 36' 20" 1500 36' 21" 2500 36' 23" 4500 36' 24" 6500 36' 25" 8500 36' 26" 5500 36' 27" 3500 36' 28" 1

4. Using (i) 40 and (ii) 50 as the assumed mean find the mean ofthe variable x which occurs with frequency f as follows;

x 20 30 40 50 60 70 80f 53 75 95 100 70 35 22

Find also the median.5. Find the standard deviation of the distribution given in

question 4. Apply Charlier's checks and Sheppard's correctionfor grouping.

6. Find the mean value of the data given (a) in Table 1, (b) inTable 3.

7. Observations on the scattering of electrons in 0.5 plates led tothe following results, which give the relative intensity of electronsin tracks of given radii of curvature. Find the mean radius ofcurvature of the tracks.

46

\ 5·6 5·7 5·8 5·9 6'036'5 40·5 31 9·5 0

SOME STATISTICAL IDEAS CHAP. 2,

Radius ofcurvature (em)

Relative intensity5'4 5'5o 11

ji

B. Draw the frequency curve (the triangular distribution) givenby f = x from x = 0 to x = 1 and f = 2 - x from x = 1 tox = 2. What is the mean value of x?Find the median for the range x = 0 to x = 2, and for therange x = 0 to x = 1.Find also the standard deviation for the range x = 0 to x = 2.

9. Sketch the frequency curve given by f = Cx2(2 - x). Choosethe constant C so that the area under the curve, for which fand x are both positive, is unity; and find the mean value ofx over this range.Find also the standard deviation.

10. Find the mean deviation and the standard deviation of thedistrihution given by f = t7T sin t7TX from x = 0 to x = 4.

11. The mean of 100 observations is 2· 96 and the standard deviationis O' 12, and the mean of a further 50 observations is 2· 93 witha standard deviation of O' 16. Find the mean and standarddeviation of the two sets of observations taken together.

12. The mean of 500 observations is 4 ·12 and the standard deviationis 0 ·IB. One hundred of these observations are found to havea mean of 4·20 and a standard deviation of 0·24. Find themean and standard deviation of the other 400 observations.

13. Find the mean and standard deviations of each of the followingdistributions;

o1

22

1 2 3 4 5 63 8 12 20 18 16345 6785 12 18 26 20 12

7 88 49 104 1

xfxf

Deduce the mean and standard deviation of the sum of the twodistributions.

14. If nl quantities have a mean x and standard deviation CTI andanother n2 quantities have a mean y and standard deviation CTz,

47

\

1CHAP. 2 SOME STATISTICAL IDEAS

show that the aggregate of (n! + n2) quantities has a variancgiven by

nlO'I + n20'~ + nln2(x - y)2nl + n2 (n! + nz}2

15. If the variable x has the values 0, 1, 2, 3,4 and the variablehas the values - 3, -1, 1, 3, find from first principles the meatand standard deviations of the function given by (i) x + y,(ii) 2x - y, (iii) 3x + 4y, and (iv) 3y - 2x. Verify the generaresults quoted in Section 21.

22. Certain special frequency distributionsThe distributions of industrial and social statistics, of scientifi

observations and of various games of chance are manifold. However, many of them approach closely to one or other of three speciallimportant types of frequency distribution which can be deriveand expressed mathematically using the theory of probability.These are known as the binomial, Poisson and normal distributions

There are many others.23. The binomial distributionOn tossing a coin there are two possibilities; we get either heads

or tails. We therefore say that the probability of getting heads onanyone toss is t, as is the probability of getting tails.If we toss a coin a large number of times, say n, we shall find

that the number of times we get tails is tn approximately. In onetrial of 1000 tosses heads were obtained 490 times and tails 510times. Of course, if n is small, the number of heads in anyonetrial may well differ considerably from tn; indeed there may beno heads at all. For instance, if we toss a coin 10 times, we mayfind we get heads 3 times. If we toss the coin another 10 times,we may get heads 8 times. The question arises: if we toss a coinn times, what is the probability of getting heads m times? (ofcourse ° <: m <: n).It can be shown that on tossing a coin n times the probabilities

of getting heads 0, I, 2, - - - n times are given by the successiveterms in the binomial expansion of (t + t)n. For example, ontossing a coin 10 times the probability of getting heads 3 times is

10 x 9 X 8(1)7(1)3 151 x 2 x 3:2 :2 = 128 -, . I

which is 1 in 8 roughly.

48

SOME STATISTICAL IDEAS CHAP. 2,

More generally James Bernoulli (1654-1705) showed that if theprobability of a certain event happening is p and the probabilitythat it will not happen is q (so that p + q = 1), the probabilitiesthat it will happen on 0, 1,2, 3, - - - n out of n occasions are givenby the successive terms of the binomial expansion of (q + p)n,namely,

n(n - 1)qn + nqn-lp + qn-2p2 + -I- pn1.2

Thus the probabilities of getting heads 0, 1, 2, - - - 10 times ontossing a coin 10 times are given by the successive terms of theexpansion of (t + wo, that is,1 1 IOx9 1 1210' 10 X 210' 1><2 X 210' - - -, 210

1= 210(1, 10,45, 120,210,252, 210, 120,45, 10, 1)

These probabilities are represented diagrammatically in Fig. 8.The sum of the probabilities is, of course, unity.

280o 240N200

>< 160£ 120:g 80~ 40l£

012345678910Values of n

Fig. 8. Binomial distribution.

A distribution of data having relative frequencies given by theterms of the expansion of (q + p)n where p + q = 1 is known as abinomial distribution.In practice, relative frequencies conforming approximately to

these theoretical values are only likely to be obtained if a largenumber of trials is involved.It is instructive to represent graphically the frequency distributions

corresponding to the terms of the expansion of (q + p)n for different49

,---,r-- I--

r- I--

ri.,

\


values of p (and q) and of n. We note that in such distributiothe variable n is not continuous; it can only assume positive integravalues. Further it can be proved that the mean of a binomialdistribution is np and the standard deviation is (npq)t.

EXAMPLE 1Verify that the data given below, where f denotes the frequenc

with which an event occurs n times, form a binomial distributionand find the mean and standard deviation.

n 012345f 1 10 40 80 80 32

Since the first and last terms of the expansion of (q + p)rI anqn and pn, if the above are distributed binomially then (Plq)5 musequal 32 or plq = 2, that is, q = t and p = ,.Expanding (t + 1)5 we get

;p + 10 + 40 + 80 + 80 + 32)

which shows that the above distribution is binomial.Hence the mean equals np = 5(t) = 3t and the standard

deviation equals (npq)t = (5 x t x W = toO)t.To find the mean and standard deviation independently we can

use an assumed mean of 3; the working is given below.n f d fd fd2

o 1 -3 - 3 91 10 -2 -20 402 40 -1 -40 403 80 0 0 04 80 1 80 805 32 2 64 128sum 243 81 297

Hence the mean = 3 + 2~~ = 3t and the standard deviation

= [297 _ (!)2J! = (270)t = !(1O)t243 3 243 3 .

EXAMPLE 2A large batch of articles produced by a certain machine a

examined by taking samples of 5 articles. It is found that tb50


numbers of samples containing 0, 1, 2, 3, 4, 5 defective articles are58,32, 7, 2, 1,0 respectively. Show that this distribution is approxi-mately binomial and deduce the percentage of articles in the batchthat are defective.If a large batch of the products contains a fraction p that are

defective, the probability of choosing at random a defective articlemay be taken to be p. Consequently if we choose at random asample of n articles, the. probability that this sample will contains articles which are defective is given by the (s + l)th term of theexpansion of (q + p)",The mean of the above distribution is

(0 + 32 + 14 + 6 + 4 + 0)/100 = 0'56If we equate this to np with n = 5 we get p = 0 ·112. Takingfor simplicity p = i the expansion of (q + p)n in this case is(~+W = 9-5 (85 + 5 X 84 + 10 X 83 + 10 X 82 + 5 x 8 + 1)

32768 + 20480 + 5120+ 640 + 40 + 159049

= 0·5549 + 0·3468 + 0'0867 + 0·0108 + 0·0007 + 0·0000.Hence we should expect the 100samples to be distributed as follows:55,35,9, 1,0,0, which agree well with the numbers actually found.The distribution is therefore approximately binomial and the

percentage of defectives in the batch is likely to be about 11%.

24. The Poisson distributionThe form of the binomial distribution varies considerably

depending upon the values of p and n. One important practicalcase is when p is very small, that is, the probability of the eventhappening is very smail, but n is large, so large that np is notinsignificant.Itcan be shown that if p is small whilst n is large and such that

np = m, the binomial expansion of (q + p)n approximates closelyto the series

(m m2 m3 mn)

e-m 1 +T + fX2 + 1 x 2 x 3 + -- - + n!

where e = 2·71828 correct to 5 decimal places. This is known asthe Poisson series, and any distribution which corresponds to the

51

CHAP. 2 SOME STATISTICAL IDEAS SOME STATISTICAL IDEAS CHAP. 2successive terms of this series is called a Poisson distritution. It i'so named after the French mathematician Poisson (1781-1840who in 1837 developed the theory.

What characterizes a distribution of the Poisson type is that trelative frequency r with which an event happens n times variaccording to the following table:

n 0 1 2 3 --_ semr 1 m m2/1 X 2 m3/1 X 2 x 3 - - _ nrls;For a true probability distribution the sum of the probabilit»

must be unity. For a Poisson distribution the sum of the relatifrequencies is

( m m2 m')e-m 1 +_ + __ + + _1 1 x 2 s!

= e-m x e+m (approximately if s is large enough)= 1 (approximately).

To the same degree of approximation it can be shown that the meaof the distribution is m and the standard deviation is mi,

In Fig. 9 are drawn histograms of Poisson distributions comsponding to (a) m = 1 and (b) m = 4.

(0) (b)Fig. 9. Poisson distributions having (a) mean = 1, (b) mean = 4.

Data conforming approximately to Poisson distributions occurwidely in science, particularly in biology. They also arise in man:industrial problems. The indispensable conditions are that the evem

52

emr emr

m=1

05

0'1, , 'I! '=T'-O ' - - . -

hould happen rarely (p small) and that the number of trials shouldhe large, or strictly so large that the mean number of occurrences

appreciable.

EXAMPLE 1

The following table gives the frequencies f of the number n ofuccesses in a set of 500 trials. Find the mean of the distributionmd verify that the distribution is roughly of the Poisson type.Verify also that the variance equals the mean approximately.

n 0 1 2 3 4 5 6 7 8 9 sumf 24 77 110 112 84 50 24 12 5 2 500nf 0 77 220 336 336 250 144 84 40 18 1505

The mean of the number of successes may be found by calculatingthe values of nf as shown in the table, whence the mean equals1505/500 = 3'01.

The terms of a Poisson series with this value of mare

e-3-0I(1 + 3·01 + 3'012/2 + 3'013/6 + ---)On multiplying by 500 the successive terms become 24· 6, 74' 2,

111'6, 112'0, 84'3, 50'8, 25'4, 11'0, 4'0, 1'4 which agree verywell with the values of f given in the table above.

The variance is found to be 3 ·05. The working, using an assumedmean of 3, is given below, where d = n - 3.

12+ m=4o8642

n d [d [d2

0 -3 -72 2161 -2 -154 3082 -1 -110 1103 0 0 04 1 84 845 2 100 2006 3 72 2167 4 48 1928 5 25 1259 6 12 72

sum 5 1523

53

012345n


5Hence, mean = 3 + 500 = 3 '01

. 1523and vanance = 500 - (0'01)2

= 3'0459~ 3·05

EXAMPLE 2The number of dust nuclei in a small sample of air can be estimat

by using a dust counter. The following values were found in a se:of 400 samples.Number 0/ particles 0 1 2 3 4 5 6 7Frequency 23 56 88 95 73 40 17 5The mean number of particles is found to be 1170/400 ~ 2·9

The terms of the Poisson series with m = 2·9 aree-2-9 (l + 2·9 + 2'92/2 + 2'93/6 + - --)

that is, 0·055 + 0·160 + 0·231 + 0'224 + 0·163 -I- 0'094+ 0'046 + 0'019 -I- 0'007 + - --

On multiplying by 400 the successive terms become 22, 64, 92,90, 65, 38, 18, 8, 3, which agree well with the values of the frequencygiven above. The degree of departure from the Poisson distributionhas been discussed by Scrase. (3)It might be noted concerning these measurements that the number

of dust nuclei in the air is large, and the probability of any onenucleus being within the small sample examined is very small.

EXAMPLE 3A large batch of articles produced by a certain machine are

examined by taking samples of 5 articles. It is found that thenumbers of samples containing 0, 1, 2, 3, 4, 5 defective articles are58, 32, 7, 2, 1, 0 respectively. Show that this is approximately aPoisson distribution (see Example 2, Section 23).The mean number of defective articles is O· 556 ~ 5/9 and the

terms of the Poisson series with m = 5/9 are

-m(1 5 ~ 1~ 6~ )e +9+162+2187+26244+---

54


that is, 0·574 + 0·319 + 0·089 + 0·016 + 0·002 + - - -. Onmultiplying by 100 thy successive terms become 57, 32, 9, 2, 0, 0,which agree well with the actual values.

25. The normal distributionThe normal distribution was first derived by Demoivre in 1733

when dealing with problems associated with the tossing of coins.It was also obtained independently by Laplace and Gauss later.It is therefore sometimes referred to as the Gaussian distribution orthe Gaussian law of errors, because of its early application to thedistribution of accidental errors in astronomical and other scientificdata. Its basic importance in physics and statistics cannot beveremphasized. *The equation of what is known as the normal error curve is of

the form y = A e-iJ2(x-m)2 where A, h, m are constants.The shape of the curve is shown in Fig. 10. There is a maximum

value when x = m and the curve is symmetrical about the line x = m,y

A'

OMO'2A

m-O' m m+O' xFig. 10. Normal error curve.

y = A e-h2(x-m)2 and h2 = 1/2a2.

When x = m, y = A and when x = m ± alh, y = A e-a2 whichis as small as A/IOO when a = 2·15 approximately.The whole area under the curve is given by

coJ A e-iJ2(x-m)2 dx-co

* "The role of the normal distribution in statistics is not unlike that ofthe straight line in geometry. "(4)

55


and this equals (Ay'7T)/h. If A is taken equal to h/y'7T the anunder the curve is unity and the equation of the curve then is

hy =_ e-h'{x-ftVy'7TThis is the frequency curve of what is called a normal or Gaussi

distribution. Strictly, such a distribution is one for which thirelative frequency of the observations having a value betweenand x + ox is yox where y is given by equation (7). Another waof expressing this is to say that for a normal distribution tbprobability that an observation will lie between x and x + ox .

h-- e-k'{x-m)2 oxy'7TThe value of y above, equation (7), is therefore sometimes calledthe probability density or the relative frequency density of thedistribution.It will be noted that any normal distribution is determined by

two parameters hand m. We can show that m is the mean of thedistribution (this is obvious from the symmetry of the curve inFig. 10) and that h, sometimes called the precision constant, isrelated to the standard deviation a of the distribution by therelation 2a2h2 = 1.

Writing m = x and 172 = 1/2a2 we have1y = e-(x-x)2/2a2

ay'(27T)

This then represents a normal distribution of which the mean is xand the standard deviation is a. The shape of the curve for thesame value of x and for different values of a is shown in Fig. 11.

When x = x ± a we have

y = _1_ e-t _ 0·607ay'(27T) - ay'(27T)

whilst when x = x ± 2a we have

1 -2 0·135y=--e ----ay'(27T) - ay'(27T)

56

CHAP. 2SOME STATISTICAL IDEAS

..12(2rr1

y

_1(2rri

o· x ~-I-~ xFig.11. Normal error curvesy = ~ e_(x-.x')2/2a2av'(27T)

26. Relation between a normal and a binomial distributionIt can be verified that the histogram of a binomial distribution

corresponding to (q + p)n approximates very closely, when n islarge, to a normal error curve. Indeed the normal distribution canbe derived mathematically from the binomial. If the terms of theexpansion of (q + p)" are plotted as ordinates against integral valuesof x from 0 to n the points are found to lie approximately, if n islarge enough, on the normal curve

1y = e-(x-m)2/2a2ay'(27T)

with m equal to the mean np of the binomial distribution, and a2

equal to the variance npq.In Fig. 12 are drawn the histogram corresponding to (t + t)10,

and the normal curve with m = 10/3 and a2 = 20/9.Of course there is one very important difference between the two

distributions. The binomial is discontinuous consisting of a setof discrete values corresponding to integral values of n, whereasthe normal distribution is continuous extending over all values ofthe variable. A distribution of discrete values may, of course,approximate closely to a normal distribution.

57


y'20 y'e59·05y

Jr~\\

15

10

,r-j-I

5

~bLo I 2 3 4 5678910 x

Fig. 12. Histogram of binomial distribution given by G+ ~rO,andi1

(dotted curve) the normal error curve y = v(2 ) e-(x-m)2/2a2with m = 10/3 and (12 = 20/9. (1 17

27. The mean deviation of a normal distributionIn Section 16 we defined the mean deviation of any set of data.

If we apply this definition to a normal distribution we find that themean deviation, denoted by T), is given by

T) = a/Y(;;TT) ~ 0·80a

Thus for a normal distribution T) equals 4a/5 approximately.The extent to which any other distribution departs from normalityis sometimes estimated by calculating the ratio T)/a and comparingit with 4/5.For example, if we use the distribution shown in Table S

(Section 18) we find that :E/ldl = 592, whenceT) = 592/400 = 1·48

But a = 1·85 and hence "J/a = 1·48/1·85 = 0·80.28. Area under the normal error curveWe have stated earlier that for a normal distribution th~

probability that an observation will lie between x and x + Ox is58

\

\

rSOME STATISTICAL IDEAS CHAP. 2

1___ e-(x-x)2/2a2 oxay(2TT)

where x and a are the mean a~d standard deviation respectively ofthe distribution. This expression is represented by the area underthe curve

1y = e-(x-x)2/2a2

ay(2TT)

between the ordinates x and x + ox.Further, the probability that an observation will lie between two

values XI and x2 is represented by the area under the curve betweenthe ordinates XI and X2, and equals

1 X2--- f e-(X-»2,2n2dx0"v'(217) Xl

Writing (x - x)/a = 1 this expression becomes

1 J'2__ e-t,2dly(2TT)

'1where tl = (XI - x)/a and 12 = (X2 - x)/a. It can therefore beexpressed as the difference of two integrals of the type

1 IT__ e-t,2dly(2TT) o

This integral has been evaluated for different values of T and isgiven in mathematical tables.It might be noted that the function given by

1 JT 2 ITYTT e-,2 dt = yTT e-,2 dt-T 0

is known as the error function, because of its importance in thetheory of errors, and is denoted by erf(T).

T 1 Tv'2

Hence Y(~TT)l e-t,2dt = YTTI e-,2dt = terf(Ty2)o 0

59

\


This quantity is represented by the shaded area

1Y = -_e-!12V(217)

shown in Fig. 13. The total area under this curve is unity.

y

_1(211)2

-3 -L -I 6 T I 2 3 t

Fig. 13. Graph ofy = v'(~7T) e-!12. Shaded area equals t if T = O. 6745.

it can be shown that the area under this curve from t = _ T tot = Tis t if T = 0'6745.This last result can be interpreted as follows: in a normal

distribution the probability that an observation will lie betweenx ± 0'6745a is t. In other words, the quantity 0'6745a is thedeviation from the mean which is just as likely to be exceeded asnot, and is sometimes called the probable error. It is not a goodterm and is falling into disuse.We give in Table 6 the value of

1 JT~ e-!12dtV(27T)

-T

for a few important values of T. From this table it follows thatin a normal distribution the probability that an observationlies within 3a of the mean is O' 9973, that is, the range of a distri-bution conforming to the normal type is effectively about 6a.Again, the probability that an observation will lie within the range

60


To0·6745123co

Table 6erf (Tv'2)o0-50'68270'95430·99731

g ± 1·96a is 0'95, so this is sometimes called the' 95% zone.Similarly the range x ± 3'09a is the 99'8% zone; the probabilitythat an observation will lie outside this zone is 1 in 500.

29. Sampling, standard error of the meanIt is usual to regard a finite set of data as a sample taken from

a much larger, or infinite, set known as the population. Forexample, 100measurements of a physical quantity, such as a lengthor a time interval, may be thought of as a sample selected fromthe very large number of measurements which could be made andwhich are referred to as the population. The term population isused generally in this way, and not limited to the biological examplesto which it was first applied. The importance in statistics of thisconcept of population cannot be over-emphasized.Unlike the binomial and Poisson distributions, in which the

variable assumes only integral values, the variable in a normaldistribu tion ranges continuously from - co, to + co, and an infinitepopulation or number of observations is implied. In practice,however, the number of observations is limited and it is importantto know whether such a finite sample approximates to the normaltype and, if so, to be able to find the parameters of the normallaw which best fit the observations.A finite number n of observations will conform to a normal

distribution if the frequency of the observations that lie betweenx and x + ox is

nh- e-h2(x-m)1 BxV7T

for all values of x. The practical problem is to test this and tofind the values of m and h appropriate to the given set of observa-

61


tions. It is clear that a perfect fit is unlikely, but it can be shothat the most probable value of m is the mean of the observatioand the most probable value of h2 is 1/2s2 where s is the standardeviation of the observations. We have pointed out earlier that faan infinite population of the normal type m = x and 112 = 1/2where X and a refer to that infinite population.Of course, if we select at random" a sample of n data from

infinite normal population it is obvious that, in general, the mea.of the sample will not be the same as the mean of the who]population; if n is large the two means may not differ very mucbut if n is small they may. It is possible to express mathernaticallthe manner in which the means of different samples of given siare distributed. In fact it can be shown that this distribution iitself normal and such that its mean equals the true mean of tbwhole population, whilst its standard deviation equals a/n! whenn is the number of data in each sample and a is the standacdeviation of the whole population. Fig. 11 illustrates how thdistribution of the means of samples retains its normal form budecreases in dispersion as the size of the samples increases. Idealing with any set of data it is therefore usual to find the mex and to write it in the form x ± ex where ex = oln» and is calledthe standard error of the mean. It should be noted that the quantityO·6745a/v' n (cf. Section 28) which is called the probable error 01the mean is sometimes quoted but the use of the standard error isgenerally recommended and is now widely adopted.Similarly the standard error of the standard deviation is,

a/v'(2n), so that the standard deviation may be written as

a[ 1 ± v'~2nJ (8)

The usefulness and significance of the standard error of thomean may be illustrated by taking a particular example. Supposethe mean of a set of 100 measurements of a certain physical quantityis 2· 341 and that the standard deviation is O' i8. Then the standarderror of the mean is 0'18/100t = 0'018, and we write the resulas 2·341 ± 0·018. Using Table 6 of Section 28 we infer that ifmuch larger number of measurements had been made the mea• Random sampling is defined as such that in selecting an individu

from a population each individual in the population has the same chanof being chosen.

62


would be likely to lie between 2·341 - 0·018 and 2·341 + 0·018with a probability of 0·68. Further, the probability of the meanlying between 2·341 - 0·036 and 2·341 + 0·036 is 0·95. Theseprobabilities vary with the value of n and hence it is important toquote the number of observations, especially if it is small, namely,

x = 2·341 ± 0·018 (100 observations)Alternatively, the uncertainty in the value of a can be indicated

by using expression (8) and writing the standard error of themean as 0'018(1 ± 0·07).

EXAMPLE 1The mean height of 200 men is 1·705 ± 0·002 m and the mean

height of another 300 men is 1·752 ± 0·001 m. Find the mean heightof the 500 men and its standard error.The mean height of the 500 men is

(1·705 x 2 + 1·752 x 3)/50·01 + 0·156

1'70 + = 1·733 m5

Also, the variance of their 500 heights about this new mean maybe found by using relation (6) as follows.Variance of the 200 heights about the new mean

= 200(0'002)2 + (0'028)2= 0·0008 + 0·000 584 = 0·001 384

Variance of the. 300 heights about the new mean= 300(0,001)2 + (0'019)2= 0·0003 + 0·000 361 = 0·000 661

Variance of the 500 heights about the mean

= : (0'001 384) + ~ (0,000661) = 0·000 9505 5

Standard error of the mean = v(O'OOO 95/500)= vO'ooo 001 9 = 0·0014

.'. Mean height of the 500 men = 1·733 ± 0·001 m.EXAMPLE 2

A set of 400 data has a mean of 2' 62. Test whether this can beregarded as a random sample drawn from a normal populationwith mean 2'42 and standard deviation 1·24.

63


The standard error of the mean of a random sample of 400 takfrom the normal population = 1.24/V 400 = 0·062.But the difference between the two means is O·20 which is mo

than three times the standard error of the mean. Hence the sarnplis not likely to be drawn from the given population.

30. Bessel's formula

We have not yet discussed how the standard deviation a ofinfinite population can be derived from the standard deviation sa sample of n data. It can be shown that the best estimate wemake of a2 is given by [n/(n - 1)]s2. Thus, if the sampleXI, x2, - - -, xn the best estimate of a2 is given by

1 n 1 n-- ~ (x, - x)2 = -- ~ d~n - 1 r-I n - 1 r=1

where x is the mean of the sample, and d, = x, - x.This formula differs from that for s2 in that (n - 1) replaces

This is known as Bessel's correction. It is insignificant when nlarge.If the sample consists of the values XI, x2, - - -, xn with f

quenciesjj, 12, - - -, In respectively, the best estimate of a2 is given

i: J;.(xr - x)2/(N - 1) = £, I,. d;/(N - 1)r-I r-I

where N = 'i:,J;. and x is the mean of the sample.Further, it can be shown that the standard error of the stand

deviation is a/v [2(N - 1)]. The standard error of the mean ishowever, a/vN.

EXAMPLE

Returning to the data discussed in Section 18, the mean x wufound to be 3· 57. If we apply Bessel's correction the best estimateof a2 is given by

or

a2 = 1492 - 400(0, 57)2---;3'""9""9 ---'-

1492 - 130 = 1362 = 3,41~~~ 399

(1 = 1,85

64


that, in this case, the correction leaves unchanged the value ofIr, correct to two decimal places.Using Sheppard's correction we get

a2 = 3·41 - 0·08 = 3·33nnd hence a = 1'82.The standard error of the mean is therefore 1'82/400! = 0·091.

The uncertainty in this value is I/V798, that is, about 3· 5%, soIhat the standard error of the mean is 0'091(1 ± 0'035).We can write

x = 3·57 ± 0·0931. Peters' formulaWe have stated earlier in Section 27 that the mean (absolute)

deviation, 7J, of a normal distribution equals a(2/7T)!, that is, 4a/5approximately.When we have to deal with a sample of N data drawn from a

normal distribution, the mean deviation of the distribution is givenby

n

7J= ~ I.ld.l/v[N(N - 1)]s-I

nwhere N = ~ f.. Hence the standard deviation a of the distribution

.=1

is estimated by

a = 7JV(t7T) = ~[2N(; _ 1)J X 'i:,I.ldsl

= 1·25 'i:,I.ldsl/v[N(N - 1)]

Also the standard error of the mean of the distribution is

a 1 I[ 7T ] 'i:,I.ld.1N! = N X 'Y ZeN_ 1) X 'i:,I.ldsl = 1·25iV• IfiV _ "

These are known as Peters' formulas. As they involve Idsl andnot d/, they are more readily computed than the formulee due toBessel used in the previous paragraph.If we take the example discussed in Section 18 we have from

Table 5 that 'i:,l.ld.1= 592.65


Hence (T = 1·25 x 592/y(400 x 399) = 1'85 in agreement wi!the value obtained in Sections 18 and 30.

32. Fitting of a normal curve

It is possible to find the equation of the normal curveNy = e-(x-:I?)'/2a'

(Ty(217)which' best fits a given frequency distribution by finding the mx and the standard deviation (T of the given distribution in tway explained above. The total area under this curve is N whimay be taken equal to the total number of data, that is, the sof the frequencies.

To test how well the curve fits the data it is useful to calculathe values of ySx with Sx equal to the width of the classes inwhich the data are grouped and x equal to the mid-value of eacclass. These values can then be compared with the given fnquencies. We should, as always, not expect reasonable agreemeunless the number of observations is large.

As an example let us consider the data discussed above (Sectio18 and 30), of which the mean x was found to be 3' 57 and tstandard deviation (T was l :82. Since N = 400 we calculate

400___ e-(x-3'57)2j6'66ox1· 82,\/(217)

for x = 0, 1, 2, 3, - - -, 10 with Sx = 1.The values obtained are given as f(calc.) in the following tabl

which also shows. the observed values f(obs.)x 0123456789

f(calc.) 13 33 60 84 85 65 36 15 5 1f(obs.) 10 40 72 85 78 55 32 18 7 2

These results are represented graphically in Fig. 14.The agreement seems reasonably good. It is possible to test "t

goodness of fit" by using a special technique, known as the X2-teswhich enables the significance of the departure of the data frothe assumed type of distribution (in the above example, the normdistribution) to be assessed. We cannot go into this here: referenshould be made to a text-book of statistics.

66 i


100- -,,

I ,,

I,

I,

I\

II

,I

I

f

50

o I 2 3 4 5 6 7 8 g 10 xFig. 14. Histogram and the corresponding normal curve.

\3. Other frequency distributionsAlthough we have restricted our considerations to the binomial,

Poisson and normal distributions it should be emphasized thatthere are many others. Some of these, such as the triangular and themedian distributions, have been included in the examples. Further,the binomial distribution is a special case of a more general multi-nomial distribution which has many forms. Also the normal distribu-lion which represented by

y = A e-(x-m)'j2a'

Is such that 1 d y __ x - mydx- Ci2is a special case of more general distributions for which

1 d y x - ay dx = - bo + bjx +b2x2

These are known as Pearson's typeS.(6)Many of these other distributions are approximately realized in

practice.It might also be noted here that sometimes what is quite an

asymmetrical frequency curve, referred to as skew, becomes

(10)

67


approximately a normal curve when the frequency is plotted agaithe logarithm of the variable rather than against the variable itseSuch a distribution is often called a lognormal distribution.plotting such distributions it is convenient to use logarithmic papehaving a logarithmic scale for the variable. Two types of whatknown as probability paper are also available for plotting noand lognormal distributions. Arithmetic probability paper is suthat if the cmnulative relative frequency of a normal distributionplotted against the variable a straight line is obtained. Moreovethe mean of the variable is given by the point on the straight licorresponding to the cumulative relative frequency of O·5, whi]the standard deviation equals the projection on the variable axof the part of the straight line between the points corresponding tcumulative relative frequencies of O:16 and O·5. Logarlthmlprobability paper has, however, a logarithmic scale for the variabland consequently a lognormal distribution is represented bystraight line.

EXERCISES 3

1. Draw the histograms of the binomial distributions corresponing to (q + p)n when (a) q = s, p = t and n = 6, (b) q =p = t and n = 5.

2. Calculate from first principles the mean x and the standadeviation a of the two distributions given in question 1Verify that x = np and a2 = npq.

3. Find the frequencies of the Poisson distribution of which thmean is 2. Draw the corresponding histogram. Verify that thvariance equals the mean.

4. Show that the following distribution is roughly of the Poissontype, and find its variance.

nOl 2345/ 21 29 22 12 5 1

5. Write down the binomial distribution corresponding(t + t)IO, and fit a normal curve to it. Draw the normal curand the histogram of the distribution.

68


6. Make a table of the values of the probability of getting 6 + xheads, when 12 coins are tossed, for values of x from 0 to 6.

Show graphically that an equation of the form y = a e-bJl2

fits these values very closely if a and b are chosen suitably.Find a and b.

7. Fit a normal curve to the following distribution of intelligencequotients (I.Q.).I.Q. limits 54- 64- 74- 84- 94- 104- 114- 124- 134--144Frequency 1 2 9 22 33 22 8 2 1Draw the normal curve and the histogram of the distribution.Find the ratio "Jla for the data.

8. The number of stoppages on 400 consecutive shifts in an in-dustrial undertaking were recorded as follows:

Stoppages per shift 0 1 2 3 4 5Number 0/ shifts 245 119 30 4 2 0

Show that the distribution is approximately of the Poisson type.

9. Rutherford and Geiger(5) using the scintillation method countedthe number of ex-particlesemitted per unit of time by polonium.Their results are given below; / is the number of times Nex-particles were observed. Show that the mean number ofex-particles emitted is 3· 87 and find the Poisson distributioncorresponding to this mean. Compare this with the observedvalues.N 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14/ 57 203 383 525 532 408 273 139 45 27 10 4 0 1 1

lO. The results of certain measurements on the breakdown voltagesof one hundred insulators are given below; n is the number ofinsulators with a breakdown voltage less than EkV.Breakdownvoltages, E 110 120 130 140 150 160 170 180 190 200 210Number 0/insulators, n 0 3 7 15 26 55 78 92 96 99 100Find the mean and standard deviation, and verify that thedistribution is roughly normal. (Check your results by usingarithmetic probability paper.)

69


11. Assuming that the following distribution is approximanormal, find the mean and its standard error. Use (a) Besseformula, (b) Peters' forrnule.x 25 26 27 28 29 30 31 32 33 341 1 5 17 49 85 52 25 11 4 1

12. If the value of a quantity x and its standard error are given10'23 ± 0'12, what is the probability that the accurate valof x will lie between (i) 10· 11 and 10' 35, (ii) 10'11 and 10,4'(iii) 10'11 and 10'31, and (iv) 10'0 and 10·5.

13. One thousand electric lamps are tested, and it is found ththey have an average life of 950 burning hours with a standadeviation of 150 hours. What number of lamps may be expect,to have a life of (a) less than 650 hours, (b) between 800 an1100 hours, and (c) between 1100 and 1250 hours.

14. A sample of 20 data was found to have a mean of 80'1 wistandard deviation 2· 2. Another sample of 40 had a mean81'4 with standard deviation 3 ,4. Assuming the two samplwere taken from the same normal population, use the 60 dato estimate the mean of the population and its standard err,

15. The diameters of the spores of lycopodium can be found by aainterference method. The results of an experiment are givebelow, where k = 5880 when the diameters are measured .centimetres.k X diameter 14 15 16 17 18 19 20 21 22Number 01spores 1 1 8 24 48 58 35 16 8Find the mean diameter of the spores and the standard erronRepresent the results by a histogram and draw the corresponing normal error curve.

16. Plot the frequency curve for the distribution given by 12/[7T(X2 + 1)] from x = - 1 to x = 1. Find approximately ~standard deviation of this distribution by using the values offrom 0 to 1 at intervals of 0·2.

17. The median law is defined by 1= (l/2a)e-1xlla for all valuof x. Sketch the frequency curve and show that the area undit is unity. Find the standard deviation. I

70


18. The results of examining 213 discharge tubes and noting thevoltage associated with a particular operating characteristicare given below. Verify that the distribution is approximatelynormal. Sketch the normal curve and the histogram of thedata.

oltages 97 98 99 100 101 102 103 104 105 106 107 108 109Number 01

tubes 1 8 10 25 32 42 42 27 15 4 3 3

19. Results of a count of warp breakages during the weaving ofparticular lengths of cloth are given below. Show that thedistribution is approximately Poissonian.Warp breaks per length ofcloth 0 1 2 3 4 5 6

Frequency 15 26 21 19 8 3 0

20. Tests made on electrical contacts to examine how often thecontacts failed to interrupt the circuit due to their weldingtogether yielded the following results.Number of welds per test 0 1 2 3 4 5Frequency 10 13 13 8 4 2·Show that the distribution is roughly Poissonian, and find its

variance.21. According to the kinetic theory the velocities of the molecules

of a gas are such that the component velocities in anyonedirection are distributed normally (Maxwellian distribution).Thus the probability that a molecule has a velocity-component,measured parallel to the axis of x, lying between u and u + ouis A e-hmu2ou where A, hand m are constants. Show thatA = (hm/1T)t, and find the mean value of the velocity-componentu.

22. Using the result quoted in question 21, show that the probabilitythat a molecule has a velocity lying between c and c + ocwhere c has components u, v, w parallel to the co-ordinate axes is

(hm/1T)3/2e-hmc2 ou ov ow

Find the mean value of c and of c2•

71

CHAPTER 3

THEORY OF ERRORS"Everybody believes in the exponential law of errors; theexperimenters because they think it can be proved bymathematics; and the mathematicians because they believeit has been established by observation."(7)

34. The normal or Gaussian law of error

The function y = 1aV(27T) e -(x-x)2/202

which defines a normal frequency distribution is often called tluGaussian law of error. This "law" states that measurements ofgiven quantity which are subject to accidental errors are distribute,normally about the mean of the observations. More precisely, tlaw infers that any set of measurements of a given quantity mayregarded as a sample taken from a very large population-taggregate of all the observations that could be made if the instments and time allowed-and that this population is normal.

The two parameters x and a characterizing this normal distributio:can be estimated by calculating the mean and standard deviatio:of the given set of measurements, as explained in Section 30. Indeedthe mean of these measurements is the best estimate of the value 0:the measured quantity, whilst the standard deviation provides tbbest estimate of the accuracy so obtained. The value of the measure,quantity is usually written as x ± ccwhere cc = a/VI1 is the standard'error of the mean and n is the number of measurements. (As wehave mentioned earlier some writers use not the standard error butthe probable error of the mean which is 0'6745a/VI1, that is,approximately.)

EXAMPLE

Certain values of the velocity of light obtained using differemethods are given below in kilometres per second. Find the meaand its standard error.

72

THEORY OF ERRORS CHAP. 3

Velocity d d2

299782 2 4299798 18 324299786 6 36299774 - 6 36299771 - 9 81299776 - 4 16

sum 7 497

A working mean of 299780 may be used. In the second columnIII the above table are given the values of the residuals d, relativehI the working mean, and in the third column the squares, d2.

Mean value = 299780 + 7/6 = 299781

a2 = 497 - 6(7/6)2 = 985

lurther

therefore a = 9·9

tnndard error of the mean = 9'9/V6 = 4

velocity of light = 299 781 ± 4 km S-1 (6 observations)

As there are only six values the uncertainty in the value of the(andard error is considerable, namely, 1/10+, so that the standardrror might be written as 4(1 ± O' 32).

15. Applicability of the normal law of errorSome groups of observational data in science satisfy closely the

normal error law, but it is by no means universally true. Perhapsthis is not surprising since any theoretical derivation of the law isbused on special assumptions which mayor may not correspond toobservational conditions.This so-called Gaussian law of error was first deduced theoretically

hy Laplace in 1783; he started from the assumptions that the,leviation of anyone of a set of observations from the mean of theIt is the resultant of a large number of very small deviations due to

Independent causes, and that positive and negative deviations ofthe same size are equally probable. (Cases can be quoted, however,where a large number of independent causes, each giving rise to amall deviation, do not result in deviations from the mean conform-

73

olumn the frequency of its occurrence in a total of 300 determina-lions. It will be noted that positive and negative errors are groupedtogether, and so we can only assume they are equally divided.onsequently the mean of the distribution is zero.

Limits of error Frequency f fx2

The third column of the table gives the product of the frequencyand the square of the deviation x. We measure x from the meanto the mid-point of each class, that is, O·OS, 0 ·15,- - -,0·95.

Hence we have0-2 = 15-629/299 = 0-05227

Applying Sheppard's correction we get0-2 = 0-05227 - (0·1)2/12 = 0-05144

or 0- = 0-227Hence the normal error function corresponding to the data is

300y = e-x2/0-10290·227Y(27T)

= 527 e-9·72x2

In column four of the table above are given the calculated valuesof f = 0 -2y for the values of x corresponding to the mid-pointsof each class. The factor 2 is necessary to include negative as wellas positive values of x and the factor O· 1 because of the class width,The normal error curve and the histogram of the data are drawn inig. 15.

CHAP. 3 THEORY OF ERRORS

ing to the normal law, (8» Subsequently, Gauss gave a proof bon the postulate that the most probable value of any numberequally good observations is their arithmetic mean.

Both Laplace and Gauss discussed examples of measuremewhich suggested a wide applicability of the normal law todistribution of accidental errors and its truth was generally accepBut more recently its validity has been challenged on varigrounds. In 1901,Karl Pearson showed that some series of measuments which had been deemed to support the normal law shosignificant departures from it. Later, with collaborators, he msix series of observations resembling those of the determinationright ascension and declination using a transit circle, and he fouagain substantial departures from the normal law. Jeffreys'?'re-examined Pearson's data and confirmed the main result;addition he has shown that the data satisfy a law of Pearsotype VII..It is now generally agreed that the normal law of error is n

universally valid. Nevertheless it is used widely unless thereevidence to show that some quite different probability distributiapplies. It is worthy of note that even when the parent populatiis not normally distributed, the distribution of the means of sampof a given size is usually closer to a normal distribution thanpopulation itself. Jeffreys(6) concludes: "The normal law of ecannot be theoretically proved. Its justification is that in represing many types of observations it is apparently not far wrong,is much more convenient to handle than others that might orrepresent them better."

Of course the applicability of the normal or any other lawbe examined using the X2-test but the physicist, unlike the bioloseldom uses this, perhaps because of the small number of observatiowith which he usually has to deal. However, certain large grouof physical measurements do exist and some have been carefexamined.

36. Normal error distributionsA much discussed example of observational data satisfying

normal error lawwas given originally by Bessel.(lO) He providedfollowing data concerning the errors involved in measuring the riascension of stars. In the first column are given the magnitudethe error of the observation in seconds of time, and in the seco:

74

THEORY OF ERRORS

0-0-0·1 114 0·2850-1- 84 1·8900-2- 53 3·3120·3- 24 2·9400·4- 14 2-835a-5- 6 1·8150·6- 3 1·2680-7- 1 0-5620-8- 1 0·7220·9- 0 0sum 300 15-629

75

CHAP. 3

Calculatedfrequency

1038557321562ooo

300

CHAP. 3 TH EOR Y OF ERRORS

f70

310

110-08 ':(Yo-:04---=0-2 0 02 04 06 0 - x

Fig. 15. Histogram of Bessel'sdata and the corresponding normal curIt is clear that the agreement is very good. Hansmannt'D h

however, examined the data and shown that there are two tyof Pearson curve that give even a better fit than the normal curJeffreys(9) has also discussed the data and in particular examinhow closely they follow the normal error law; he commen"This agreement is therefore surprisingly good, indeed a little tgood, because it suggests that this series has been selected becait gives an unusually good agreement with the law and that otbthat may have disagreed violently with it may have been suppressThis danger of selection makes it undesirable to make use 0:published series to test the law, unless there is some definite reasoto believe that no suppression has taken place."

There are other difficulties in attempting to test the law, nleast for the physicist that observational data in physics are seldosufficiently numerous. But even when the number of observatiois large they are sometimes invalidated in various ways. F,example, Michelson, Pease and Pearson obtained 2885 valuesthe velocity of light from observations made on about 165 differenloccasions extending over about two years. The data are sym-metrically distributed about the mean value of 299774 km S-1 buthe distribution of the residuals is not normal. Birge(12)points outhat a very good fit can be obtained if one takes the sum of tnormal error curves, with standard deviations of 5 and 15 km srespectively, which suggests, as one possibility, two groupsobservations of unequal degrees of accuracy. He adds: "The poiis that if a large assortment of observations are not of equal reliabilitheir residuals cannot be expected to follow a normal error cur

76

THEORY OF ERRORS

!lnd J am more and more convinced that the deviations from suchII curve found so commonly in large groups of physical measure-ments are due usually just to this cause." Several attempts toprovide suitable measurements have therefore been made. Besidesthose of Karl Pearson mentioned earlier, Birge(13)made a series of00 cross-hair settings "on a very wide but symmetrical solarpectrum line, under conditions as favourable as possible to equal

reliability for all observations." His results are shown in Fig. 16

60

so

.40

s'

JO

-6

Fig. 16. The distribution of residuals for 500 measurementsof a spectralline, compared with the Gaussian error curve evaluated by secondmoments. Abscissa represents residual Ii, in 0'001 mm units. Ordinaterepresentsthe number of residuals of magnitude Ii.

(Reproduced from Physical Review.)

where the abscissa unit is 0·00 1 mm and the ordinate v' representsthe number of residuals of magnitude v. The smooth curve drawnIn Fig. 16 corresponds to the normal error curve

hy = _ e-h2x2VTr

with h, as calculated from the observations, equal to 0 ·1965 andl = 500 y. It is clear that over the whole range the observationsfollow the Gaussian curve very closely.

A similar series of observations, 1026 in number, have beenmade by Bond who viewed an illuminated slit using a travelling

77

CHAP. 3

CHAP. 3 THEOR Y OF ERRORS

microscope slightly out of focus. The conditions resembled thin the measurement of a spectrum line. Bond's data were originalgiven as an example of the normal error law, which they seemedsatisfy very well, but Jeffreys'P! subsequently examined them anifound systematic departures from the normal law.

37. Standard error of a sum or difference.We discussed in Chapter 1 (Section 7) the resultant error in a

quantity which is a function of a number of quantities all subjto experimental error.

It is now necessary to consider the estimation of the standaerror of any quantity when the standard errors of the quantities 01which it depends are known. For instance, suppose a quantityis a function of two measured quantities x and y, say, Z = f(x, Y.If the quantities x are distributed about their mean x with a standanerror ex and the quantities yare distributed about their meanwith a standard error f3, how are the quantities z, obtained bevaluating f(x, y) for the measured values of x and y, distributed

To answer this question we use an important property of tnormal error law, sometimes referred to as its reproductive propertIt may be stated as follows:Any deviation, which can be expressed as the sum of a set

deviations each of which satisfies the normal law, itself satisfies t,normal law.

Suppose a deviation D can be written as a linear function ofindependent deviations dl, dz, - - - d" in the form

D = kldl + k2dz + - - - + k"d"where kl' kz, - - - k; are constants.

Suppose further that each of the deviations d, satisfies the normlaw so that the probability that d. lies between x and x + ox ilgiven by

1---,-;:::--:- e-x2/2ai ox°iV(27T)

Then it can be shown(l4) that the probability that D lies betwx and x + ox is given by

1___ e-x2/2a2 oxoy(27T)

where (J2 = kroi + k~o~+ - - - + k~o; (178


In other words, D satisfies a normal law determined by the standarddeviation c,

Certain particular cases are specially important.

(i) If D = dl + d2 we get

02 = or + o~

Also if D = d, - dz, then again 02 = or + o~. We haveverified these results earlier (cf. Section 21) in a few simple cases.We note that 0 is greater than either 01 or 02 but less than 01 + Oz·

More generally, ifD = dl + d2 + -- - + d"

we get 02 = ai + a~+ -- - + o~

If at = a2 = -- - = an then a = otyn. But if dJ = d2 = -- - = d"Ihe deviations are not independent and so the above result does nothold. However, in this case D = nd, and hence a = nat·

(ii) If D is the arithmetic mean of d), d2, - - -, d" so thatD = (d) + d2 + - - - + d,,)/n we get

02 = (ay + a~ + - - - + a~)/n2

When al = 02 = - - - = an this reduces to a2 = ar/n or0' = ot/yn, which accounts for the value of the standard error ofthe mean quoted in Section 29.

(iii) It follows from cases (i) and (ii) that if a set of nl data hasstandard deviation a), and another set of /1z data has standarddeviation Oz, then the standard error of the sum or difference of the

(a2 Oz)t

means of the two sets of data is -1 + -1-n) n2

It follows that if two quantities are given as a ± ex and b ± f3where ex and f3 are the standard errors of a and b respectively, thesum of the two quantities is a + b ± y(cx2 + (32) and the differenceof the two quantities is a - b ± y(ex2+ (32).

EXAMPLE 1A sample of 250 observations has a standard deviation of 3 and

another sample of 400 observations of the same quantity has astandard deviation of 5. Calculate the standard error of the mean

79


of each sample and the standard error of the difference of themeans.

The standard error of the mean of the first sample is3/y'250 = 0·19

The standard error of the mean of the second sample is5/y'400 = 0·25

The standard error of the difference of the two means

= ~[(y'~50)2 + (y'~)2J = ~~ = 0'31

Thus if the two means have values ml and m2 the differencethe means can be written as

ml - m2 ± 0'31

The ratio 1m, - m21/0'31 is obviously of some practical impotance. From Table 6 (Section 2e) we note that if 1m, - m21 >3(0'31iit may be concluded that the two samples are not likely to beloto the same population; the probability that they so belong is 1<than 0,2%.

EXAMPLE 2Determinations of elm for the electron by two different grou

of methods give the following results: 1·75880 ± 0·000 42 a1·75959 ± 0·00036, the unit being 10" C kg ' '. Do the two grouof measurement differ systematically'[The difference between the two mean values is

10-5[79 ± y'(422 + 362)]

that is, 0·00079 ± 0·00055.Thus the difference of the means is about 1·4 times its stand a

error, and so the existence of a systematic error is not ruled out.Birge comments: "In view, however, of the many serious sourc

of systematic error that have been revealed, from time to time,almost every method of determining elm, I think that the smremaining discrepancy should not be taken too seriously."

80


It might be added here that in trying to assess the significanceof the difference of the means of two samples, especially smallsamples, use may be made of what is usually called the t-distribution.We have stated earlier (Section 29) that the mean x of a sample

of n taken from a normal population, of mean 111 and standarddeviation a, is such that the deviation x - m is distributed normallywith mean zero and standard deviation aly'n. This implies thatthe ratio (x - m) : aly'n is distributed normally about zero withunit standard deviation. However, when the standard deviation ais estimated from a given sample of n as explained in Section 30,the distribution of the ratio t = (x - m)y'n/a is not normal, butit has been evaluated and is given in many books(lS) on statistics.From the table of t can be obtained, for any value of n, theprobability that a value of t will be exceeded in random sampling.

38. Standard error of a productSuppose z = x y where x and yare measured quantities of which

the means are }(and y respectively. Suppose the standard error ofx is ex and of y is {3.If we write x = x + x' and y = y + y' the mean of x' is zero

and its standard error is IX, whilst the mean of y' is zero and itsstandard error is {3.We have z = (x + x')(ji + y')

= xy + xy' + yx' + x'y'

The mean of z, denoted by z, is therefore xy. Further, thedeviation of z from xy is xy' + yx' approximately, and henceusing equation (11), Section 37, the standard error of z is y where

y2 = (x{3)2 + (yex)2

We can write z = xy ± y

and we note that (ylxy)2 = (ex/x? + ({3IYP

ex/x is what might be called the fractional standard error of x, sothat the above result indicates that the square of the fractionalstandard error of the product equals the sum of the squares of thefractional standard errors of the factors forming the product.

81


For example, suppose the sides of a rectangle are given as,20·00 ± 0· 04 cm and 10·00 ± 0· 03 ern. The area of the rectanglcan be written as 200'00 ± y ern? where

therefore

that is

(L)2 = (0'04)2 (0'03)2200 20 + 10

y2 = 0·16 + 0·36 = 0'52

Y= 0·72

39. Standard error of a compound quantityCollecting together and extending the results of the previous tw

paragraphs we have:If a number of measured quantities have means ml, m2, - - -, mil

with standard errors aI' a2, - - -, a" respectively then the standarderror of

(i) the sum ml + mz is y(ar + a~)(ii) the difference ml - m2 is y(ar + a~)(iii) the multiple km, is kal(iv) the product mlm2 is y(mra~ + m~aT)(v) the product mlm2m3 is a where

(a/mlm2m3)2 = (adml)2 + (a2/m~2 + (a3/m3)2(vi) the power mf is a where

a/m~ = padmlor a = (pm~-I)al

(vii) any function of ml' m2, - - -. mn>namely,/(ml' m2, - - -, mn),

is a where

Of 2 (0/ 2 0/)2a2 = (om) aT + om) a~ + - - - + (om

na;

Of course (i) to (vi) are special cases of (vii).

EXAMPLE 1The radius r of a cylinder is given as 2·1 ± 0·1 em and tb

length I as 6· 4 ± 0·2 cm. Find the volume of the cylinder and itslstandard error.

82

THEOR Y OF ERRORS CHAP. 3

The volume V cm3 of the cylinder is given by

V= 71(2'1)26'4 ± a

where (;)2 = (~1)2+ (1/and al and a2 are the standard errors of r and I respectively

therefore (~)2 = (~)2 + (~)2V 2·1 6·4

4 1= 441 + 1024

that is

=0,0091 +0,0010 =0·0101

a/V= 0·10

V= 71(2'1)26'4(1 ± 0'10)

= 88·7 ± 8·9

Hence

EXAMPLE 2

The refractive index n of the glass of a prism is given by n =sin teA + D)/sin tA where A is the angle of the prism and D is theangle of minimum deviation. The results of several measurementsof A and D are given as A = 60° 5'2' ± 0·2' and D = 46° 36'6'± 014'. Find the refractive index of the prism for the particularwavelength of light used.Taking teA + D) = 53° 20·9' and tA = 30° 2·6' we have

n = 0·802280/0'500655 = 1·60246.To find the standard error we have

on t sin tA cos teA + D) - t sin teA + D) cos tAoA = sin? tA

sintD2 sin2 tA

sintD1- cos A

83

sin 23° 18·3'1 - cos 60° 5' 2' = - 0' 79


Alsoon _ 1: cos -!(A + D)

oD - sin-!A

= 0·597 = 0.601·00

O:A = ± 0·2' = ± 0·000058 radian

O:B = ± 0·4' = ± 0·000 116 radian

cos 53° 20·9'2 sin 30° 2·6'

But

and

Hence the standard error IX of n is given by

0:2=(0'79 x 0·58 x 10-4)2+(0'60 x 1·16 x 10-4)2

= 0·69 x 10-8

therefore 0: = ± 0·83 X 10-4

Hence the refractive index of the glass of the prism= 1·60246 ± 0·00008

EXERCISES 4I

1. Certain observations of the right ascension of Polaris have beengrouped(16) as follows, where x denotes the deviation in secondsof time from a value near the mean of the observations and ydenotes the number of observations having a deviation x, Fita normal error curve to these observations. Plot the curve anddraw the histogram of the observations on the same figure.

x -3'5 -3'0 -2'5 -2,0 -1'5 -1,0 -0,5y 2 12 25 43 74 126 150

x 0 0'5 1·0 1'5 2·5 3·0 3'5y 168 148 129 78 33 10 2

- 2. Twenty measurements of the acceleration due to gravity werefound to have a mean value of 9·811 m s -2 with a standard devia-tion of 0·014. Another thirty had a mean value of 9·802 m·s-1with a standard deviation of 0·022. Assuming the two sets 0measurements belong to the same normaf population, find tbmean of the fifty measurements and its standard error. Whaare the limits of uncertainty in this standard error?

84 '


3. Fifteen measurements of the surface tension of water have amean value of 0·072 52 N rn" with a standard deviation of0.00064. Find the standard error of the mean and its uncertainty.Given that the accepted value of the surface tension at the tem-perature of the laboratory is 0·073 05 Nm -I show that there islikely to be a systematic error in the measurements.

4. The results of determinations of a physical quantity by twodifferent methods were quoted as 15 + x units, where 100 x hadthe following values:

(a) 37, 39, 38, 37, 39,40,41, 37, 36,39(b) 40, 39, 41, 42, 40, 43, 38, 41, 40, 38.

Is there evidence that the two sets of values differ systematicallyfrom one another?

5. If m = 0·850 ± 0'012 find m2, m3 and 11m.

6. If a = 1·16 ± 0'08 and b = 2·54 ± 0'12 find a + b, ab, alb,(b - a)la and log. a.

7. If v = 0·4107 ± 0·0003 m, u = 0·3513 ± 0·0002 m and1//= l/v+ l/u, calculate/and its standard error.

8. The refractive index n of the glass of a double convex lensis calculated from the formula 1//= (/1 - 1)(1/r, - Ilr,).If r, = 0'312 ± 0·001 m, r z = 1-490 1. 0·005 m and/ = 0·501 ± 0·002 m, find 11.

9. Given thatelm = (1'7592 ± 0'0005) x 10 It C kg- 1

e = (J ·60203 ± 0·000 34) x 10-19 Ccalculate the electronic mass m.

10. Given that

c = (2'99776 ± 0·000 04) x 108 m S-I

Roo = 10973730 ± 5 rn>'m = (9·1091 ± 0·0005) x 10-3' kge = (1·60210 ± 0·00052) x 10-19 C

Eo = (8·854 18 ± 0·000 04) x 10-12 kg " m" 3 s'e'find Planck's constant h, using the relation

h3 = me4/8rdJR~c.

85


40. Method of least squaresThe principle of least squares, which was first formulated b

Legendre, may be expressed as follows:the most probable value of any observed quantity is such thathe sum of the squares of the deviations of the observations frothis value is least.

If Xl' Xl, - - -, x; are observed values of any given quantity, the:according to the principle of least squares the most probable valuof this quantity is X chosen so that

(xl - X)2 + (x2 - X)2 + - - - + (xn - X)2is least.

Now if x is the mean of Xl' Xl, - - -, xn so that L:xs = nx 0L:(xs - x) = 0 we have

L:(x, - X)2 = L:[(xs - x) + (x - X))2= L:(xs - x)2 + n(x - X)2

which is clearly least when X = x.Thus applying the principle of least squares we derive the resul

that the most probable value of a measured quantity is the arithmetimean of the· observations.

More generally, if the observations Xl' Xl, - - -, xn occur witfrequencies ft, h, -- -'!n respectively, then applying the principof least squares the most probable value of the measured quantiis X, such that

ft(x, - X)2 + h(xl - X)l + -- - + fn(xn - X)l

is least.If now we write x = L:fsx,/L:fs, that is, the mean of the observa

tions each "weighted" according to the frequency of its occurrence,we have

L:fs(x, - x) = 0and hence L:fs(xs - X)2 = L:fs[(x, - x) + (x - X»)2

= L:fs(xs - x)2 + (x - X)2L:!

This is clearly least when X = s.Thus the most probable value of the measured quantity is tb

weighted mean of the observations, that is,(fix, + !2x2 + -- - + !nx,)IUi + i2 + - - - + j,)

In such an expressionj', is known as the weight of the observation X

86

THEORY Of ERRORS CHAP. 3

If all the weights are multiplied by the same constant factorthe value of the weighted mean is unchanged. In the abovederivation the weight equals the frequency of occurrence of theobservation, but other significances are given to the weight oftendepending upon the personal assessment of the observer. Some-times the weight attributed to any observation is determined, asexplained later, by the accuracy of the observation.

41. Weighted meanBoth Laplace and Gauss using different techniques established the

principle of least squares mathematically. We have mentionedearlier that Gauss, assuming that the mean of the observations isthe most probable value of a measured quantity, deduced thenormal error law. Conversely, we can use the normal error law todeduce that the most probable value is the mean. The "proof"proceeds as follows:

The probability of the occurrence of an observation x, is given by(h/V7T) e-h'(Xl-x)2

where X is the magnitude of the measured quantity.Hence the probability of the occurrence of the observations

Xl' X2, - - -, xm being the product of the probabilities of theoccurrence of each separately, equals

(h/V7T) n e-h'l:(x.-x)2

if we assume that all the observations belong to the same Gaussianpopulation specified by the precision constant h.

Thus the probability is a maximum if L:(xs - X)2 is a minimum,in keeping with the principle of least squares. As we have shownabove this occurs when x is the mean of Xl>X2, - - -, xn'

It may be, of course, that the measurements Xl' X2, - - -, XIIbelong to different Gaussian populations; the accuracy of measure-ment may be different in each case.

We can then write the probability of the observation x. as

(hs/V7T) e-hHx.-x)'and hence the probability of the occurrence of the observationsXl' x2, - - -, Xn as

h(h2 - - - hn7T-~n e-Eh~(x._x)2

87


This is greatest when L.h~(x. - x? is least, which occurs whenis given by the weighted mean L.h~x$/L.h~. The proof is exactlygiven above.If we write h~ = 1/2a; as quoted in Section 25, the most probablel

value of a measured quantity deducible from the observationsXI' x2, - - -, xn is given by

L.h;x, _ L.xs/a;L.h; - L.1/a~

This reduces, of course, to the arithmetic mean if 0'1 = a2 =- -- = an'Combining this result with that obtained above, it is clear that

if the observations Xl X2, - - -, X" occur with frequencies J;, f2' - - -, Inor if Xs is the mean of Is observations, then the most probable valueof the measured quantity is

L.hU.xs L.f.xs/a; L.Xs/IX;L.h~1s = L.1s/a; = L.1/IX;

where IX.= a/v Is and represents the standard error of the mean(cf. Section 29) corresponding to the observations xs' Thus eacobservation x, can be given a weight proportional to the reciprocal 0the square of its standard error. Alternatively, since the probablerror is a constant multiple of the standard error a weight pro-portional to the reciprocal of the square of the probable error maybe used.

42. Standard error of weighted meanQuite generally, if the observations XI' x2, - - -, xn are given the

weights WI> W2, - - -, w" respectively the weighted mean isx = L.w.x./L.Ws. As proved earlier, the sum L.w.(xs - X)2 isleast when X equals this weighted mean, X.To estimate the accuracy of the weighted mean we find the

quantity a2 given by

n0'2 = 2: w.(xs - x)2/(n - 1)

s-I

This is the variance with which the quantities w; (x, - x) are88


distributed about zero. It can be shown that the standard errorof the weighted mean is given by a/(L.ws)!' that is,

[L.w.(Xs - X?Ji

(n - 1)L.w.

If all the weights are equal this reduces to

[L.(Xs - x?]!

(n - l)n

in keeping with the usual formula. (Section 30.)

EXAMPLE 1Find the most probable value of a quantity of which the following

Ire observed values: 9'4, 9'3, 9'4, 9'5, 9'6, 9'3, 9'7, 9'5,9·2,9'4.Treating the observations as of equal weight, the most probable

value is given by the arithmetic mean, which equals 9 + -fo (4'3) =9'43.To find the standard error of the mean we calculate 0'2; using

o .4 as the mean we get0'2 = 0·28/9 = 0·031

If we used 9 ·43 as the mean we would get

a2 = 0·031 - (0'03)2 = 0·030

The standard error of the mean = y'0'030/y'10 = 0·055.The mean is therefore 9·43 ± O'06, but in view of the accuracy

of the given values one might be inclined to write the mean as9·4 ± 0·1. This would be misleading, however, especially if thenumber of observations were not quoted. It has to be rememberedthat the accuracy 01 the standard error is not very high when thenumber of observations is small. For as we have stated earlier(Section 30) the standard error of the standard deviation is/y'[2(n - 1)], so that the standard error of the mean, a/y'n, issometimes written as

~{1 + 1 }y'n - y'[2(n - 1)]

which in this case equals O'055(1 ± 0·24).89


It is therefore usually recommended that in computations ofkind two doubtful figures should be retained and the standard (Iprobable) error expressed to two significant figures. Accordinthe above result should be written 9· 430 ± O'055 (10 observationor 9'430 ± 0'055(1 ± 0'24). Jeffreys has written "if anybowants to reduce a good set of observations to meaninglessnesscan hardly do better than to round the uncertainty to one fiand suppress the number of observations."

EXAMPLE 2

The observations Xs and their weights Ws are given below. Fithe most probable value of the measured quantity and the standaerror.

Xs Ws ws(xs - 10) d, d2 wsd~s10·25 1 0'25 -0,06 0·0036 0·003610·32 2 0'64 0·01 0·0001 0·000210·43 4 1·72 0·12 0·0144 0·057610·27 3 0'81 -0,04 0·0016 0·004810·16 2 0'32 -0'15 0·0225 0·0450sum 12 3'74 0·1112

The most probable value is the weighted mean given 'by'x = l:wsXs/l:ws> that is, 10+ 3'74/12 = 10·312.Also the standard error of the mean is given by a/(l:w.)t, whe

n

a2 = ~ wsd;J(n - 1)s=1

d, = x. - xand n is the number of observations.Hence from the above table taking x = 10'31 we have

a2 = 0·1112/4 = 0·0278

a = 0·167therefore

and the standard error of the mean = 0'167/VI2 = 0·048.We can write the measured quantity as 10·31 ± 0 ·05 or preferab

10·312 ± 0'048 (5 observations).

90


EXAMPLE 3Values of the ratio elm for the electron as determined by different

methods are given below along with the probable error of eachvalue. Find the most probable value of elm and the standard error.

(e/m) x 10-11 Probable error1·76110 10·0 x 10-41·75900 9·0 X 10-41'75982 4·0 X 10-41·75820 13'0 X 10-41.75870 8'0 X 10-4

We will take the weight of each determination as inversely pro-portional to the square of the probable error; these values Ws areiven in column 1 below. In column 2 are the values of~,= (e/m) X 10-11 - 1'75800 corresponding to a working mean of1 ·75800. The necessary working is given below.

Ws

1·01·26·30·61·610·7

Xs X 1053101001822070

WsXs X 1053101201147121121701

ds X 105151

- 5923

-139- 89

d~ X 108228355

19379

w,dj X 1082284232116126544

Hence omitting the factor 10-11 we have:. 0·01701

weighted mean = 1'75800 + --ro:7 = 1·75959

Therefore, writing ds=(e/m) x 10-"-1'759 59we get as shown above

a2 = 0·00000544/4 = 0·00000136Standard error of the mean = v(0·00000136/10'7) = 0·00036.

Hence elm = (1·75959 ± 0·000 36) X la" C kg " '.

43. Internal and external consistencyIf the weight Ws assigned to the observation Xs is proportional to

i/rx;, where rxs is the standard error of xs' the standard error rx ofthe weighted mean x may be found as explained in Section 42, andis given by

91


~(x - X)2/1X21X2 = 3 S

(n - I)D/IX~

But it is possible to derive another expression for the standerror as follows. We have

x = ~k.xs where k, = ws/~ws

and hence using equation (11) of Section 37, the standard errorx is IX where

1X2 = ~k~lX~ = ~W~IX~/(~ws)2

If Ws is proportional to 1/1X: this reduces to1X2 = (~IM)/(~I/IX~)2 = 1/(~I/IX~) (l

Now expressions (12) and (13) are apparently quite different. 'note that expression (13) depends entirely on the standard errorsthe separate observations, whereas expression (12) depends also upo:the differences between the observations. The latter is thereforefunction of what might be called, following Birge, the "externalconsistency of the observations, whereas expression (13) depenupon the "internal" consistency. We shall denote the standaerrors obtained using expressions (12) and (13) by IXc andrespectively. For samples taken from an infinite normal populatiit may be shown that IX, and IX; are equal, or more strictly, the raIX./IX; is unity with standard error 1Iv' [2(n - 1)]. We note t

Z = ':! = [S~l (xs - X)2/IX~JtIX; (n - 1)

In practice it will be found that Z does not equal unity. Thisnot surprising, for it depends on the values of IXs, some or allwhich may be considerably inaccurate especially if they have beencalculated from few observations. Further, Z depends as well onthe values of (x, - "i)2 and so will be affected by any systemaerrors present.If, however, this ratio is found not to differ from unity significantl

having in mind the uncertainties in the standard errors used, ththe observations as a whole may be regarded as consistent.such a case it is perhaps safer to choose the larger of the values!Xe and lXi as the standard error of the weighted mean.If, on the other hand, IX./IX; differs from unity by an amount mUI

92

THEORY OF 'ERRORS CHAP. 3

reater than is to be expected on the basis of statistical fluctuations,It may be concluded that systematic errors are likely to be present.In such a situation the weights based on the standard errors shouldhe discarded and replaced by others. The assignment of newweights must be somewhat arbitrary, depending inevitably upon thejudgment of the individual concerned and his knowledge of theexperimental conditions.In Example 3 above we have shown that equation (12) leads to

the result IX. = 0·00036. If we use equation (13) we have~1/a~ = r2 X 108(0'010 + 0'012 + 0·063 + 0'006 + 0'016)

= r2 x 108 x O·107where r = O'6745, the ratio of the probable error to the standardsrror.Hence lXi = r-I x 1O-4/v'0'107 = 0·00045Consequently Z = 36/45 = 0·8, from which we may conclude

that the values are consistent, for with n = 10 the uncertainty in ZIs about 0'24. We might therefore safely take a equal to 0'00045,that is, elm = (1'75959 ± 0·00045)107e.m.u.g-l•

EXERCISES 51. Show that the mean of p equally good observations has a weight

p times that of one of them.2. Find the area under the curve y = x2 from x = 0 to x = 8 by

calculating y when x = 0, 2, 4, 6, 8 and giving these ordinates(i) equal weights, (ii) weights I, 2, 2, 2, 1 respectively, and(iii) weights 1,4, 2, 4, 1 respectively.

3. The refractive index of a glass prism was found successively tobe 1'53, 1'57, 1'54, 1'54, 1'50, 1'51, 1'55, 1'54, 1'56 and1.53. Find the mean and its standard error if the observationsare given (a) equal weights, (b) weights 1, 2, 3, 3, 1, 1, 3, 3, 2, 1respectively.

4. Ten observations of a quantity have a mean of 9· 52 andstandard error O: 08; another 20 observations of the samequantity have a mean of 9' 49 and standard error O·05. Findthe mean and standard error of the 30 observations usingweights (i) proportional to the number of observations, and(ii) inversely proportional to the square of the standard errors.

93


:%

5. Two groups of determinations of elm for the electron givefollowing values: 1·75880 ± 0·00042 and 1,75959 ± 0·0003the unit being 1011C kg " '. Find the weighted mean of thevalues and its standard error.

6. Two determinations of the Faraday constant are: 96497·6 ± 6,and 96506·6 ± 7·7 C mol" '. Find the difference of these tWIvalues and its standard error.

Find also the weighted mean of the two values and istandard error. Test for internal and external consistency.

7. Test the following results for external and internal consistenc(a) 10 ± 1, 11 ± 2, 12 ± 1 and 15 ± 2.(b) 10·1 ± 1'2, 11·4 ± 2'4,12·2 ± 1·3 and 14,9 ± 1,7.• '/ rY

\',/ 8. Four determinations of Planck's constant h were rounded aand given with their probable errors as

6·557 ± 0·0066·554 ± 0,0076·546 ± 0·010

and 6,544 + 0·009- ~the unit being 10- 34 Js. Find the most probable value of h andits standard error.

Repeat the calculations using the original values6· 5568 ± 0·00636·5539 ± 0·00726, 5464 ± 0 '00956·5443 ± 0·0091

~~ \

and

9. A precision method of finding the diffusivity p of nickel resultedin the following values with their probable errors:

0·0042184 ± 0·0000021 S-1

42068 ± 7842213 ± 1842093 ± 3542281 ± 5742148 ± 7142135 ± 30

Find the most probable value of p and its standard error.94

10. Values of the velocity of light in kilometres per second ascorrected by Birge(12)are given below:

Author Epoch Corrected result Probable errorRosa-Dorsey 1906'0 299784 10Mercier 1923.0 299782 30Michelson 1926' 5 299 798 15Mittelstadt 1928' 0 299786 10Michelson, Pease

and Pearson 1932· 5Anderson 1936·8Huttel 1937·0Anderson 1940.0

Find the weighted mean and itsinternal and external consistency.

II. Twelve precision values of elm for the electron obtained byseven essentially different methods are given below, the unitbeing 10" C kg " '. Find the weighted mean and standard errorof (i) all twelve values, and (ii) of the first six (the "spectro-scopic" values). Test for internal and external consistency ineach case.

THEORY OF ERRORS

elm1·759131·747971'759141·758151·760481·757001·760061·761101·759001,759821'758201·75870

123456789

101112

CHAP. 3

299774299771299771299776

standard error.

410106

Test for

Probable error X 1()43·75·05·06·05·87·04·0

10,09'04·0

13,08·0

12.. Given that the probability of the occurrence of the set ofmeasurements Xl' X2, - - -, xn is proportional to

P = (hlvTr)n e-h2E(x,-:r)2

where x is the arithmetic mean of the measurements, find thevalue of h for which P is a maximum.

95


44. Other applications of the method of least squares; solutionlinear equations

Legendre applied the method of least squares to the follow'problem.Suppose we have a number of linear equations in the t

variables x and y of the form

a.x + bsY = k,where as. b., k, are constants. Suppose there are n equations when > 2.Values of x and y can be found that satisfy any two of t

equations. These values may not satisfy all the equations. thatthe equations may not be consistent. In this case the questiarises: what are the values of x and y that satisfy all the equatioas closely as possible?The method of least squares can be applied to the solution

this problem. For writing

a.x + bsY - k, = e.we can choose x and y such that the sum of the squares of t"errors" es is least.

Thusn~ (a.x + bsY - ks)2s=1

must be a minimum.Differentiating partially with respect to x and y we get as necess

conditions for a minimum:~a.(a.x + b.y - k,) = 0 ,

•and 'i:.bs(a.x + bsY - ks) = 0

From these two equations the values of x and y can be found.These are the.i'rnost probable" values.The above equations are usually written in the form

[aa]x + [ably - [ak] = 0[ab]x + [bb]y - [bk] = 0

(1(1

nwhere [ab] denotes the summation ~ a.b..

s=1They are known as the normal equations.

96

THEORY OF ERRORS

In determinantal notation we can writex _ y

l[ak][Qb]I-I[aa] [ak]1[bk ] [bb ] [ab ] [bk ] I[aa] [ab]1

[ab] [bb]

CHAP. 3

If different weights can be assigned to the given equations andIVs is the weight of the equation a.x + bsY = ks• then the normalequations become

[waa]x + [wab]y - [wak] = 0[wab]x + [wbb]y - [wbk] = 0

It is clear from these equations that if each of the original equationsismultiplied throughout by the square root of its weight the equationscan be regarded as of equal weight.

EXAMPLE 1Find the most probable values of x and y from the equations2x + y = 5· I. x - y = 1'1. 4x - y = 7' 2 and x + 4y = 5' 9We first construct the following table from the coefficients

as. b•• k, of the given equations.as b, k, a2,2 1 5·1 41 -1 1'1 I4 -1 7·2 161 4 5·9 1

sum 8 3 19·3 22

The most probable values are x = 2·05 and y = 0·97. With morecomplicated examples certain arithmetical checks are introduced.

97

b~1I11619

ask,10·2I ·1

28·85'946'0

a.bs2

-1-441

Thus the normal equations are22x + y = 46'0x + 19y = 20'4

. 874·0 - 20'4 853·6Solving we get x = 418 _ 1 = 417- = 2·047

448·8 - 46·0 402·8and y = "n • = 417 = 0·966

bsks5 '1

- 1'1- 7·223·620'4

CHAP. 3 TH EOR Y OF ERRORS

EXAMPLE 2

Find the most probable values of x and y from the equatio

2x + y = 5 '1, x - y = 1'1 and 4x - y = 7· 2

given that these equations have weights 1, 3 and 2 respectively.We construct the following table:

Ws as b, ks wa3 wa.b, wb2 wa.k,1 2 1 5 ·1 4 2 1 10·23 1 -1 1. 1 3 -3 3 3'32 4 -1 7·2 32 -8 2 57'6sum 39 -9 6 71 '1

Thus the normal equations are

39x - 9y = 71·J

-9x + 6y = - 12'6

Solving we get x = 2'05 and y = 0'97

The most probable values are therefore x = 2·05 and y = 0'97.

EXAMPLE 3The lengths AB, BC, CD, DE along a straight line are measu

as 24'1, 35·8, 30·3 and 33·8 em, but it is known that ADaccurately 90 em and BE is 100 ern. Find the most probable valof AB, BC, CD, DE.Let the lengths of AB, BC, CD, DEbexl, x2,x3,x4 cm respectively.

Then writing

Xl = 24·1 + el, X2 = 35·8 + e2, X3 = 30·3 + e3' x4 = 33·8 +we have eJ + e2 + e) = - O· 2

e2 + e3 + e4 = 0·1

We choose el, e2, e), e4 so that er + e~ + e~ + e~ is a minimwhence

eldel + e2de2 + e)de3 + e4de4 = 098


But from equations (I) and (2) we have

del + de2 + de) = 0

de2 + de3 + de4 ••• 0

(19)

(20)

On multiplying- equations (19) and (20) by the undetermined multi-pliers AI and A2 and adding we get

Aldel + (AI + Avde2 + (AI + A2)de3 + A2de4 = 0

Comparing this with equation (18) we have

so that

el _ e2 _ e3 __ e4AI - Al + A2 - AI + A2 - A~e3 = e2 and e4 = e2 - el

Hence on substituting in equations (16) and (17) we get

el + 2e2 = - 0·2

- el + 3e2 = O· 1Hence el = - 0'16 and e2 = - 0·02

therefore e3 = - 0·02 and e4 = 0'14

Hence the most probable values arc

XI = 23'9, X2 = 35'8, x3 = 30'3, x4 = 33'9

45. Solution of linear equations involving observed quantitiesIn Section 44 we considered the problem of finding the most

probable values of x and y from n(> 2) linear equationsa.x + b,y = ks'Now let us suppose that the constants as, b, are known accurately

but the constants k, are subject to accidental errors of observation.Also let us suppose the equations have been multiplied by the squareroots of their weights, so that the weight of each may be taken asunity.

• In mathematics this method of solution is often referred to as..tRe~·~'- ',,-method of undetermined multipliers. £'''n~UTE"CF ~,

99 .\. \~ ,--,-.~-,<;~ r-';-1""~.e(, \\\R~t\'f

\ ~ •• / ••••• 1

- -- - Cl~·,lk:·,lAt r~~'i;·1 -:

CHAP_ 3 THEORY OF ERRORS

Then, as before, the 1)10stprobable values of x and yare tbsolutions of the normal equations

[aa]x + [ablY = [ak][ab]x + [bb]y = [bk]

Gauss(17) and others have discussed the problem of estimating theweights that may be assigned to the values of x and y so obtained.Denoting these values by Xo and Yo and writing a"xo + b,yo - k., ""d., that is, d1 d2, - - -, d; are the residuals when the most probablvalues Xo and Yo are substituted in the given equations, it has been.shown that the standard error a to be expected in any expressioasxo + bsyo - k, is given by"

a2 = [dd]/(n - 2)

Also if ax and ay denote the standard errors inrespectively, then it can be shown that

a2 a2 a2.:»: =~=_,[bb ] [aa ] t"

where t" is the determinant

1

[aa] [ab]1[ab] [bb]

If, as an example, we consider the equations solved in Example I,Section 44, namely,

2x + y = 5-1, x - y = I-I, 4x - y = 7 -2 and x + 4y = 5 -9

the normal equations are22x + y = 46-0x + 19y = 20-4

leading to x = 2-05 and y = 0-97_The residuals are therefore -0-03, -0-02,

that [Jd] = 0-0031 and a2 = 0-00155_

a~ _ a;_ 0-0015519 - 22 - ----c4C"7C17=---

0-03 and 0-03 sa

Hence

therefore ax = 0'008 and ay = 0-009

•• If there are m unknowns x, y, - - - the denominator is( n - m),

100

THEOR Y OF ERRORS CHAP_ 3

We can write x = 2-05 ± 0-01 and y = 0-97 ± 0-01.As a further example, let us suppose that by direct observation

it has been found independently thatx = 1-00 ± 0-10 and y = 0-90 ± 0-07

Further, suppose that a third independent observation leads tox + 2y = 3-00 ± 0'07

What are the most probable values of x and y?We will give the equations weights inversely proportional to the

squares of their standard errors, namely, 1, 2, 2_ So multiplyingthe equations by I, y2, y2 respectively we get

x = 1-00, yy2 = 0-90y2 and xy2 + 2yy2 = 3 -OOY2each of which can now be regarded as having unit weight. Wetherefore have

46. Curve fittingAnother useful application of the method of least squares is the

fitting of a curve, or a theoretical formula, to a set of experimentaldata.

a.1o

y2

bso

y22y2

ks1-000-9OV23-00y2

b2so2810

ashsoo44

ask,.1-00o6-007-00sum

a2

1o23

The normal equations are3x + 4y = 7-00

4x + 10y = 13-80

leading to x = 1,057 and y = 0-957_Using x = 1-06 and y = 0-96 we get

[dd] = 0-0036 + 0-0072 + 0-0008 = 0-011

Hence a~/l0 = a;/3 = 0-0116/14

leading to ax = 0-09 and ay = 0-05

Hence x = 1-06 ± 0-09 and y = 0-96 ± 0-05

IOJ

bskso1-8012-0013-80


Suppose YI Y2, ••• , Y~ are the values of a measured quantity(or combination of measured quantities) corresponding to tvalues XI' X2, ••• , xn of another quantity x.

Let us assume, for simplicity, that there are experimental erroin the values of Ys but not in the values of xs' Conditions closelapproaching these, where the errors in one of the variables mayneglected, often occur in practice. We assume too there exists al

linear relation between X and y, namely,

y = ax + b

On substituting x = xs' the value of y will not in general equal y,there will be an "error" of amount

ax, + b - Ys

To obtain the linear relation (or the line) which best fits the dawe choose a and b such that the sum of the squares of the "errors'is least, that is,

~(axs + b - Ys)2is least.

The conditions obtained by differentiating partially with resto a and bare

and

~x,,<axs + b - Ys) = 0

~(axs + b - Ys) = 0

a[xx] + b[x] = [xy]

a[x] + bn = [y]

n[xy] - [x][y]a=-7-~~~

n[xx] - [x][x]

b = [y][xx] - [x][xy]n[xx] - [x][ x] (2·

Hence (2

(22)

which give (23)

and

We note that if b = 0 we have y = ax

where [xy] [y]a=-=-[xx] [x]

102

(2


EXAMPLE

Fit a linear law to the values of x and y given in the followingtable, assuming the values of X are accurate.

x y xy xx0 4,6 0 0I 7·1 7 '1 12 9'5 19·0 43 11'5 34'5 94 13'7 54·8 165 15·9 79'5 256 18,6 111·6 367 20·9 146·3 498 23·5 188'0 649 25·4 228,6 81

sum 45 150,7 869,4 285

If the relation is y = ax + b then from equations (23) and (24)we have

8694 - 45 X 150·7a= 2850-452

= 1912'5 = 2,32825

andb = 150'7 x 285 - 45 x 869'4

825= 3826'5 = 4,64

825therefore

y = 2·32x + 4·64

It is instructive to plot the values of x and y, and to draw the straightline that fits the points best.

103


47. Line of regressionIt is of interest to note that the second of the normal equations

equation (22) above, can be written as

)x] + b = [y]n n

This indicates that the point ([x]/n, [y]/n) lies on the liney = ax + b, that is, the line passes through the point (x, ji) whex and ji are the arithmetic means of the values of x and y respectively

In the example given above x = 4· 5 and ji = 15·07, and it cabe verified that these co-ordinates satisfy very closely the equatioy = 2·32x + 4·64.It is therefore useful to write

x = x + X and y = ji + Y

whenceY = aXand hence from equation (25)above a = [XY]/[XX].The relation between x and y can therefore be written

_ [XY] _y - y = [XX] (x-x)

If we write [XX] = na3,., [YY]=na} and 1'2 = [XYF/[xx][yy-ji x-x

we have -- = 1'--ay axThis is known as the line of regression of y on x. We note thathas the same sign as [XY] and hence the same sign as the gradient aOf course, it may happen that both sets of quantities x and

are liable to experimental error, or indeed that the linear relationbetween them is only approximately true. In such a case we canproceed as follows.Assuming that the values of x. are accurate we can derive the

line of regression of y on x in the formy-ji x-x Y X--=r-- or - =1'-

ay ax ay axSecondly, we can assume that the values of Ys are accurate anderive the line of regression of x on y in the form

x-x y-ji X Y--=/'-- or -=r-

ax ay ax ay104

TH EOR Y OF ERRORS CHAP. 3

Fhese equations are identical if r = 1. In general r is not equalto unity and the lines are not coincident; we choose the bisectorof the acute angle between the lines as the line that best fits thedata. It can be shown that r2.;;;; 1. If r = 0 the two lines ofregression are parallel to the axes of x and y, that is, in general~ and yare independent of one another. When r2 = 1 all thepoints (xs' Ys) lie on the coincident lines of regression and there isperfect correlation between x and y. The quantity r is known asthe coefficient of correlation of x and y,. For the example discussed above the value of r may be found

ns follows, using x = 4· 5 and ji = 15' 1.

x X XX Y Y yy XY0 -4'5 20'25 4·6 -10'5 110·25 47'251 -3'5 12·25 7 '1 - 8·0 64·00 28·002 -2'5 6'25 9·5 - 5·6 31'36 14·003 -1'5 2·25 11'5 - 3·6 12·96 5'404 -0'5 0'25 13·7 - 1·4 1·96 0'705 0·5 0·25 15·9 0·8 0·64 0'406 1·5 2'25 18'6 3·5 12·25 5'257 2'5 6'25 20'9 5·8 33'64 14'508 3·5 12·25 23'5 8'4 70'56 29·409 4'5 20·25 25'4 10'3 106·09 46·35

sum 45 0 82'50 150·7 - 0'3 443·71 191' 25Hence, r = 191·25/'\1(82'50 x 443'71) = 0·999. We note also

that [XY]/[XX] = 191'25/82'50 = 2'32 so that the relationbetween x and y is

y - 15'1 = 2' 32(x - 4' 5)or y = 2'32x + 4·66In keeping with the equation found earlier.

48. Accuracy of coefficientsIt is clearly of some importance to be able to estimate the accuracy

of the values of a and b derived by the method outlined above.This can be done by applying the results discussed in Section 45.

For using the values YI' Y2, - - -, Yn and XI' x2, - - -, xn and themost probable values of a and b, we can calculate the residuals d,

105

CHAP. 3 THEOR Y OF ERRORS

given by ax. + b - y.. Then the mean square error <x2 inexpressions ax. + b - Ys is given by

<x2 = [dd]/(n - 2)

Also, if <Xa and <Xb denote the standard errors in the valuesa and b we have, using the normal equations (21) and (22).

a.~/n = <xU[xx] = <x2/tJ.where tJ. is the determinant

I [xx] [x]1 = n[xx] - [x)2[x] n

In the example considered above the normal equations are

285a +45b = 869·445b + lOb= 150'7

leading to a = 2·32 and b = 4'64.

Hence we have <x~/1O= <x~/285= 1X2/825

where 1X2= [dd]/8 and [dd] = 0,3500 as shown below.

x y d dd0 4·6 0'04 0'00161 7·1 -0,14 0·01962 9·5 -0,22 0·04843 11'5 0·10 0'01004 13,7 0,22 0,04845 15·9 0·34 0·11566 18·6 -0'04 0·00167 20·9 -0,02 0·00048 23'5 -0'30 0'09009 25'4 0'12 0·0144

sum 45 150'7 0,3500

HencelXa = 0·023

We can therefore writey = (2'32 ±0'02)x + 4·64 ± 0'12

106

and <Xb = 0,123


An alternative method has been given by Bond.(lS) If we writethe relation y = ax + b in the form Y = aX where Y = y - yandX = x - x we have shown above that a = [XY]/[XX].Hence if the values of Xs are accurate and the standard error of the

values of y, (and therefore of the values of Ys) is IX we can writeformally

a = [X(Y ± <x)]/[XX]which, using standard error (i) of Section 39, leads to

[XYJ ± <xy[XX] [XY] IX

a= [XX] =[XX]±y'[XX]

Therefore the standard error of a is lXa = IX/Y[XX]. This resultis in keeping with that given above since

[XX] [xx] -2--=- -xn n

that is, n[XX] = n[xx] - [xJ2IX~ _ 1X2n - 'n[:-xx·]---r[x"F

Further, since b = Y -- a,r and the standard error of y is <x/v'nwe can write

and hence

b = y ± <x/yn - (a ± IXJX

b = y - ax ± ,J (:2 + <x~2)<x2

<xl = - + 1X~2n

or

so that

= ~ + IXUX)2n n2

=<x2 [xx]n[xx] - [x)2

as given above.It is perhaps more convenient to write the equation in the form

Y ± IX/yn = (a ± lXa)X

107


where

y = y - y, X = x - X, eeo = ee/v[XX] and ee2 = [DD]/(n - 2),D. being the value of Y. - aXs:

Thus in the example considered above we have found thaa = [XY]/ [XX] = 191.25/82'50=2· 32, and hence as shown belo[DD] = 0'3580.

X y D DD-4'5 -10'5 -0'06 0'0036-3'5 - 8·0 0'12 0·0144-2'5 - 5·6 0'20 0'0400-1'5 - 3·6 -0'12 0·0144-0,5 - 1·4 -0'24 0·05760'5 0·8 -0'36 O'lf961'5 3'5 0·02 0·00042'5 5·8 0 03'5 8·4 0·28 0'07844'5 10·3 -0,14 0·0196

sum 0·3580Hence

ee2 = O'3580/8therefore

ee~ = O'3580/(8x 82' 50)= 0'00054

thereforeeeo = 0·023 as before.

Also,....::..= /0' 3580

Hence the relati . V n 'Y 80""" = 0·067won IS

Y ± 0'07 = (2'32 ± 0'02)X

This can be written in terms of x and y as follows

(y - 15'!) ± 0·07 = (2'32 ± 0'02)(x - 4'5)108


therefore Y = (2'32 ± 0·02)x + 4·66 ± v[(0'07)2 + (0'09)2]= (2'32 ± 0'02)x + 4'66 ± 0'12

in keeping with the result found earlier.It might be added that if the observed values x., Ys are given the

weight w. the equations (21) and (22)above becomea[wxx] + b[wx] = [wxy]a [w,x] + blw] = [wy]

a = [w](wxy] - [wx](wy][w][wxx] - [wxF

b = [wy][wxx] - [wxH.wxy][w] [wxx] - lwx F

Also the standard errors of a and b are given by

so that

and

ee2 ee2 ee2......!!...=_b_= ._

[w] [wxx] [w][wxx] - [wxPwhere ee2 = [wdd]/(n - 2).

49. Other curvesIt often happens that the relation between the two variables x

and y is not linear. More generally, the relation may be of the form

y = ao + alx + a2x2 I ••• + Gmx'"involving m + I constants.If n corresponding values of x and yare known, say, (xs' yJ with

s = I, 2, 3, •• " nand n> m + I, the values of the constantsao, at> a2, - - -, am may be chosen such that the sum of the squaresgiven by

n~ (y. - ao - alx. - a2x; - • - - amx,;,)2s ~ I

is least.A particular example will illustrate the method.

If y = ao + alx + a2x2 we choose ao, ai' a2 so that

i:, (Y. - ao - alx. - a2x~)2.~I

is least.109

CHAP. 3 THEORY OF ERRORS THEOR Y OF ERRORS CHAP. 3

Differentiating partially with respect to ao, ai' a2 respectively weobtain the necessary conditions:

We assume that y = ao + a, X + a2X2, where to simplify thearithmetic we take X = 5x - 3 and tabulate the working as follows.

L(ys - ao - a,xs - a2x;) = 0

1X Y X2 X3 X4 Xy X2y y(calc) d d2

LXs(Ys - ao - a,xs - a2xJ) = 0 -3 1·0 9 -27 81 -3,0 9·0 1·05 -0,05 0'()()25-2 2'4 4 - 8 16 -4·8 9·6 2·22 0'18 0'0324

LXJ(ys - ao - a,xs - a2xJ) = 0 -I 6·6 I - I 1 -6,6 6·6 6·69 -0,09 0'()()810 14·2 0 0 0 0 0 14'46 -0'26 0·0676

The normal equations are therefore I 25'7 1 I 1 25·7 25'7 25·54 0·16 0·0256[y] - nao - [x]a, - [xx]a2 = 0 2 4{)·1 4 8 16 80·2 160·4 39·93 0·17 0'0289

3 57·5 9 27 81 172·5 517'5 57·62 -0'12 0·0144[xy] - [x]ao - [xx]a, - [xxx]a2 = 0 i sum o 147'5 28 0 196 264·0 728·8 0'1795

[xxy] - [xx]ao - [xxx]al - [xxxx]a2 = 0

the solutions of which give the most probable values of ao, a, and a2'Again, if we assume that the values of Xs are accurate and the

values of Ys are subject to experimental error, the standard errorsof the values of ao, aJ> a2 can be estimated by the method outlinedin Section 45 above.When the number of constants is large, the solution of the norma

equations may be laborious. Special methods of solution have beedevised.U''- 20)

The normal equations might more conveniently be written

soao + sJal + S2a2 = toSiao + S2a, + S3a2 = tlS2aO+ S3a, + S4a2 = t2

wheren n

Sk = ~ x~ and tk = ~ x~Y,,-I r=1

The general form of these equations is obvious.

EXAMPLE

Fit a parabolic curve to the following data:

0·825·7

1 ·040·1

o1-0

0'22'4

0'4 0·66·6 14·2

110

xy

1·257·5

Hence the normal equations are7ao + 28a2

28al= 147' 5= 264·0

+ 196a2 = 728·8

9'428 and a2 = 1·652.

28ao

leading to ao = 14'463, alHence we have

y = 14'4631 9'428X + l'652X2= 14·463 I 9'428(5x - 3) + I .652(5x - 3)2

. = 1'05 - 2'42x + 41'3x2

The uncertainties in the values of 00, ai' a2 can be estimated asfollows. We first find the residuals d given by y - Go - a, X - a2X2for the various values of X; these are tabulated above. We thencalculate d2 and find

a2 = [dd]/(n - 3) = 0'1795/4

If the standard errors of ao, ai' a2 are denoted by ao, aJ, a2respectively, we have

a2o aI _ a~ _ a2

1

7 ~1-1701-1-=-7 ---'0~28"28 196 0 28 0 28 0

28 0 1961

28 01o 196

11 I


therefore(X~ _ (Xi _ (X~ _ 0'1795/4112 - I2 - 4 - 336

± 0'122, (Xl = ± 0'040, (X2 = ± 0'023therefore (xo =

Hencey = 14·463 ± 0·122 + (9'428 ± 0'040)X + (1'652 ± 0'023)X2

The simplification of this expression is complex, since (Xo, (XI' (X2

are not independent. If, however, we treat them as independent weget

y = 1·05 ± 0·27 - (2·42 ± O'72)x + (41·3 ± 0·6)x2

The standard error 0·27 is overestimated, but the others aresatisfactory.

EXERCISES 61. From the equations

3x+y=2'9, x-2y=0'9 and 2x-3y=i'9find the most probable values of x and y.

2. The equationsx-3y=-5'64x + y = 8'1

and 2x - y = 0·5

have weights 1, 2, 3 respectively. Find the most probablevalues of x and y, and their standard errors.

3. Independent sets of observations led to the resultsu = 1·23 ± 0'06, v = 2·17 ± 0·08

and u + v = 3· 50 ± O' 12.Find the most probable values of u and v and their standarderrors.

4. Find the most probable values of x, y and z that satisfy theequationsx+y+z=4'01, 2x-y+z=l'04,

x + 3y - 2z = 5' 02 and 3x + y = 4· 97assuming the equations have equal weight. Find also thestandard errors of x, y and z.

112


5. Find the most probable position of the point of which themeasured distances from the points (1, 0), (3, 1) and (- I, 2)are respectively 3 '1, 2·2 and 3' 2 units. Estimate the uncer-tainties in the co-ordinates of the point.

6. The three angles of a plane triangle are measured and foundto be

A = 48° 5' 10", B = 60° 25' 24" and C = 70° 42' 7".

Find the most probable values of A, Band C assuming thatthe measurements have (i) equal weights, and (ii) weights 1, 2, 3respectively.

7. Plot the values of x and y given in the example discussed inSection 46 and draw the best straight line through them. Findthe equation of the line drawn.Find the values of a and b using only the values (i) x = 0, 1,

2, 3, 4, 5 and the corresponding values of y, and (ii) x = 5, 6,7, 8, 9 and the corresponding values of y.

8. Two quantities D and d are measured as follows:

d (in.)D (in.)

! i1'19 1·31

I1'42

Ii1'52

111·64

Ii1'76

21·87

Show that they satisfy approximately a relation of the formD = ad + b, find the most probable values of a and b, andestimate the errors in those values.

9. Fit a linear relation to the following data and estimate the errorsin the values of the constants obtained.

1011'0

127·6

136·2

17-0,1

19-3'2

20-5'0

xv

10. Values of the surface tension of water, y N m -1, at differenttemperatures, to C, are given below. If y = a - be, where e isthe temperature on the Kelvin scale, find the most probable valuesof a and b.

1074·22

20 3072-75 71-18

113

4069'56

5067-91

tY X 103

6066·18


11. Values of the surface tension of bromo benzene, y N m - 1, atdifferent temperatures, to C, are given below. Fi t a law of theform y = yo[1 - (8/670)]" where 8 is the temperature in degreesKelvin.

t 15 42 49 78 90 105 125r x 103 37·92 34·92 33·78 30·73 29·30 27·62 25'30

12. Using the values of the velocity of light c given in question 10of Exercises 5, use the method of least squares to show that cis related to the epoch D by the relation

c = 299777·27 ± 2·03 - (0'381 ± 0'234)(D - 1930)May it be concluded that there is experimental evidence oflinear variation of c with time?

13. Fit a parabolic curve to the data given in the example discussedin Section 49, but excluding the value x = 1·2. Use thecalculated expression for' y to find y when x = 1·2.

14. Values of the viscosity of water, 1'/ Ns m ->, at different. temperatures, to C, are given below. Fit a law of the form1-1 = a + bt + ct2•

t 10 20 30 40 50 60 701'/ X 103 1·308 1·005 0·801 0·656 0·549 0·469 0'406

15. Assuming that TJ = Aek/T where T is the temperature on theKelvin scale, use the data given in question J 4 above to findthe most probable values of A and k.

J 14

REFERENCES

(I) BULLARD, E. c. Phil. Trans A, 235, pp. 445-531 (1936).

(2) HEYL, P. R. Bur. Stand. J. Res. (Washington), 5, p, 1243 (1930).

(3) SCRASE,F. J. Quart. J. Roy. Meteorol. Soc .. 61, pp. 368-378(1935).

(4) AITKEN, A. C. Statistical Mathematics, p. 72 (Edinburgh:Oliver and Boyd Ltd., 1952).

(5) RUTHERFORD,E., and GEIGER, H. Phil. Mag., 20, pp. 698-707(1910).

(6) JEFFREYS,H. Theory 0/ Probability (Oxford: Clarendon Press,1948).

(7) WmITAKER, E. T., and ROBINSON,G. Calculus a/Observations,p. 179 (London: Blackie and Son Ltd., 1929).

(8) See ref. (7), p. 177.

(9) JEfFREYS,H. Phil. Trons A, 237, pp. 23/-271 (1938).

(10) BESSEL,F. W. Astron. Nachr .• IS, Nr. 358-359.

(1 I) HANSMANN,G. H. Biometrika. 26, pp. 128-195 (1934).

(12) BeRGE, R. T. Rep. Progr. Phys., 8, p. 95 (1941).

(13) BIRGE, R. T. Phys. Rev., 40. pp. 207--227 (1932).

(14) See ref. (7), p. 175.

(I5) WEATHERBURN,C. E. Mathematical Statistics, p. 188 (London:Cambridge University Press, 1952).

(16) See ref. (7), p. 174.

(I7) See ref. (7), p. 243 and later.

(18) BOND, w. N. Probability and Random Errors, p. 96 (London:Edward Arnold and Co. Ltd., 1935).

(19) HARTREE, D. R. Numerical Analysis (Oxford: ClarendonPress, 1952).

(20) MILNE. W. E Numerical Calculus (Princeton UniversityPress, 1950).

115

BIBLIOGRAPHY

The following is a list of works on statistics and theory of errowhich, it is hoped, might serve as some guide to students through!a very extensive literature. The works range from the most,elementary to the very advanced and are quoted in their approxi-mate order of difficulty.

Statistics and statistical mathematics

MORONEY. Facts from Figures (London: Penguin Books Ltd.,1951).

ArrKEN. Statistical Mathematics (Edinburgh: Oliver andBoyd Ltd., 1952).

WEATHERBURN. Mathematical Statistics (London: CambridgeUniversity Press, 1952).

LEVY and ROTH. Elements of Probability (London: OxfordUniversity Press, 1951).

JEFFREYS. Theory of Probability (Oxford: Clarendon Press,1948).

Errors of observation and related topics

BOND. Probability and Random Errors (London: EdwardArnold and Co. Ltd., 1935).

BRADDICK. The Physics of Experimental Method (London:Chapman and Hall Ltd.),

WHIlT AKERand ROBINSON. Calculus of Observations (London:Blackie and Son Ltd., 1929).

DEMING and BIRGE. Rev. Mod. Phys., 6, p, 119 (1934).LYON. Dealing with data (Oxford: Pergamon Press, 1970).SMART. Combination of Observations (London: CambridgeUniversity Press, 1958).

79638

116

Accuracy 14Accuracy of coefficients 105Aitken, A. C. 55. 115, 116Approximations 16Area under error curve 58Average, see Mean

Bernoulli, J. 49Bessel'scorrection 64forrnula 64

Bessel. F. W. 10, 64, 74, 115Binomial distribution 48. 57Birge, R. T. 15.76,77,92,115BOlld. W. N. 77,107,115Bullard, E. C. 15, 115

Charlier's checks 44Classboundaries 35limits 29widths 29

Coefficient ofcorrelation 105variation 41.43

Consistency. internal and external91

Correction, Sheppard's 43Correlation. coefficient of 105Cumulative frequency 68

Demoivre A. 55Density. probability 56Deviation 38mean 38, 39, 58standard 38, 40, 44

Dispersion 13, 37Distributionbinomial 48Cauchy 37,67lognormal 68median 67multinomial 67

INDEX

Distribution-continuednormal 55, 56, 57Pearson's types 67. 74Poisson 51triangular 67

Dust counter 54

Eddington, A. S. 9Electronic charge 80, 85, 91Equations, normal 96Error,curve 55function 59in a product 16in a quotient 17in a sum or difference 23probable 60standard 62

Errors,accidental 9, 14estimate of 12fractional 11Gaussian law of 55percentage 11personal 10. 14superposition of 20systematic 9, 14theory of 72

Fittinof a normal curve 66of a parabolic curve 109of a straight line 101

Frequency 29curves 34distributions 29,48, 67polygon 30relative 33

Gauss, K. F. 55, 74, 87, 100Gaussian law 55, 72Gravitation, constant of 26Gravity 14Grouping, effect of 36, 43

Hansmann, G. H.1I7

76, 115 ~'_'"'-"'" II:+. \~·ntl!TEDr }; ".g.'+i r .•__.-·_;vtih.

~ r: ,'<:"...•(LIBRAPy:

Hartree, D. R. 110, 115Heisenberg; W. 10Heyl, P. R. 26, 115Histogram 31,58, 76Hogben, L. 29

Infinite population 61

Jeffreys, H. 74, 76, 78, 90, 115,116

Kinetic theory of gases 71

Laplace, P. S. 55, 74, 87Least squares 86, 96, 101Legendre, A. M. 96Linear equations, solution of 96Lognormal distribution 68

Mean 31assumed or working 32deviation 38. 58of a sum 44

standard error of 62, 64, 72weighted 87

Median 33Median distribution 67, 70Method of least squares 86,96,101Milne, W. E. 110,115Multinomial law 67

"Noise" 10Normal distribution 55, 56, 57fitting of 66

Normal equations 96Normal errorcurve 55distribution 74law 72applicability of 73

Pearson, K. 74Pearson types of curve 67, 74Personalequation 10error 10, 14

INDEXPeters' formulae 65POisson, S. D. 51distribution 51series 51

Population 61Precision 14Precision constant 56Probability 48density 56paper 68

Probable error 60, 88of mean 62, 72

Producterror in a 16standard error of 81

Quotient,error in a 17standard error of 82

Random 62Range 38Ratio ?)/a 58Regression, line of 104Relative frequency 33Residual 13Root mean square deviation 40Rutherford, E. 69, 115

Sample 61Sampling, random 62Scatter, 13, 37Scrase, F. J. 54, 1J 5Sheppard's corrections 43Skew 67Solution of linear equations 96Standard deviation 38, 40evaluation of 42of a sum 44

Standard errorof compound quantities 82of mean 62, 64, 72of product 81of standard deviation 64of sum or difference 78of weighted mean 88

118

r-distribution 81Tests of normality 58, 63, 66Theory of errors 72Triangular distribution 67

Uncertainty principle 10Variance 40Variation, coefficient of 41,43-Weatherburn, C. E. 81,1l5

INDEXWeight 86Weighted mean 87. standard error Oy··'Ss.........Weights of observ;uions ~8Whittaker, E. T,'and Robinson, G

72,74,78,84, 100, 1151X2 -test 66, 74one,95% 6199·8% 61

119

Documents

Errors and Treatment. Topping. 1972