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erroranalysis - ERROR ANALYSIS FOR PHYS375 by C C Chang...

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Unformatted text preview: ERROR ANALYSIS FOR PHYS375 by C. C. Chang, University of Maryland 1 Definition Error: In a scientific measurement, an error means the inevitable uncertainty in the measured results. As such, errors are not mistakes. You can not avoid them by being careful. The best you can hope to do is to ensure that errors are as small as reasonably possible, and to have some reliable estimate of how large they are. Discrepancy: If two measurements of the same quantity disagree, then we say that there is a discrepancy. Note: One of the measurements could have been the so-called accepted value, value based on previous measurements as the ”true” value, or a theoretically pre- dicted value. 2 Two Types of Errors Random or Statistical Errors: Experimental uncertainties that can be revealed by repeating the measurements are called random or statistical errors. Systematic Errors: Experimental uncertainties that cannot be revealed by repeating the measurements are called systematic errors. As an example, let’s measure the well-defined width of a table top with a ruler. Uncertainty caused by needing to interpolate between scale markings is a random error. This is because when interpolating, one is probably just as likely to overestimate as to underestimate. On the other hand, uncertainty caused by the distortion of the ruler is a systematic error. This is because if the ruler has stretched, we always underestimate; if the ruler has shrunk, we always overestimate. The treatment of random errors is quite different from that of systematic errors. The statistical methods to be discussed later give a reliable estimate of the random uncertainties, and, as we shall see, provide a well-defined procedure for reducing them. On the other hand, experienced scientists have to learn to anticipate the possible sources of systematic error, and to make sure that all systematic errors are much less than the required precision. Doing so will involve, for example, checking the instruments against accepted standards (or calibrated ones), and correcting them, or buying better instruments if necessary. 3 The Mean and Standard Deviation Error Analysis for PHYS375 2 Suppose we need to measure a quantity 1: and have identified all sources of systematic errors and reduced them to a negligible level. Since all remaining sources of uncertainty are random, we should be able to detect them by repeating the mea- surement several times. Suppose we make N (where N —> 00) measurements of the quantity 1: (all using the same equipment and procedures), and find the N values: m1,m2,...,mN The best est1mate for a: is the average of $1,152, - - - ,mN, i.e., We will also define the following quantities: 1 N 2 : 1 t. I Z l.- — '_ — 2 0'3, popu a ion var1ance Nip;O 12:51:, 1:) 0'3, : population standard deviation or rms (root-mean-square) deviation 1 N _ - Z ._— 2 _ Jig; (VI-21(ml m) For finite N, it is more appropriate to define: 2 .92 : sample variance 2 — 2(151' — 102 q | 2 0'3, 2 s : sample standard deviation || 2 | ._\ m :3 | RI V N) The factor (N — 1) is used in the sample variance and standard deviation instead of N. This is because we have to use data to find the mean 5:. In a certain sense, this left only (N — 1) independent measured values. For large N, it does not make any difference either N or (N — 1) is used. For now, we will use 0'3, to mean the sample standard deviation. As an example, let’s assume we have m,- : 71, 72, 72, 73, 71. 71+72—l—72-l—73-l—71 a—c : f : 71.8 71.8—712 1. — 22 2. a: : w : fl : 0.56 5 5 0'3, 2 0.7 .52 2 0.7 :> write it as 0': s 2 0.8 :> write it as 0'3, 4 Meaning of the Standard Deviation - the Uncertainty in a Single Mea- surement Error Analysis for PHYS375 3 If we were to plot the above result as a histogram, we would have: Instead of just making 5 measurements, if we were to make many measurements of m, we would get a limiting distribution as follows: And the distribution could be represented by the so-called normal or Gaussian dis- tribution: 2 fXa (m) : e—(w—XV/(Zax) Note that Also X+0x / fX,a—$(ar:)dar: 2 0.68 :> 68% X—ax X+20$ / fX,a—$(m)dm 2 0.954 :s 95.4% X—ZG'E Let’s suppose that we made N measurements of a: and obtained the values m1,ar:2, - - - ,mN. Let’s compute a7: and 0'35. From the discussion above we can conclude that If our measurements are normally distributed and if we were to continue the measurement of a: many more times (after making N measurements and using the same equipment), then about 68 % of our new measure- ments would lie within a distance 0'3, on either side of 5:; that is, 68 % of our new measurements would lie in the range 5: :: 0'35. We can rephrase the above conclusion as follows: Error Analysis for PHYS375 4 Suppose, as before, that we obtain the values m1,ar:2, - - - ,mN and com- pute a7: and 0'35. If we then make one more measurement (using the same equipment), there is a 68 % probability that the new measurement will be within 0'35 of 5:. Now, if the original number of measurements N was large, then 5: should be a very reliable estimate for the actual value of 1:. Therefore, we can say that there is a 68 % probability that a single measurement (using the same equipment) will be within 0'35 of the actual value. 5 The Standard Deviation of the Mean If m1,ar:2, - - - ,mN are the results of N measurements of the same quantity 1:, then, as we have discussed earlier, our best estimate for the quantity 1: is their mean, 5:. We have also discussed that the standard deviation 0'35 characterizes the aver- age uncertainty of the separate measurements m1,ar:2, - - - ,mN. However, our answer mbest : 5: represents a judicious combination of all N measurements, and there is every reason to believe that 5: will be more reliable than any one of the measurements (mi) considered separately. As we will show later, the uncertainty in the final answer mbest : 5: turns out to be the standard deviation 0'35 divided by \/N. This quantity is called the standard deviation of the mean, and is denoted by 0'55: We can now state our final answer for the value of 1: (based on the N measurements of m) as 0x W (Value of m) = a7: :: 6 More on Standard Deviation and Standard Deviation of the Mean Let’s assume we have made many sets of N measurements of a: with the same equipment: m1,m2,uuu,mN, m1,m2,uuu,mN,-uu h\/_’_\/_’ In other words, we have made many determinations of the average of N measurements. Each set of the N -measurements will be normally distributed about the true value X with width 0'35, shown as dashed curve below. The average of each set of the N-measurements will also be normally distributed about X, but with width 0'55 : ax/x/N, shown as solid curve below. Error Analysis for PHYS375 5 7 Weighted Averages It often happens that a physical quantity is measured several times, perhaps in several separate laboratories (or by different students), and the question arises how these measurements can be combined to give a single best estimate. Suppose, for example, that two students, A and B, measure a quantity 1: and obtain these results: Student A: a: = mA :: 0'A Student B: a: : m3 :: 03 Each of these results will probably itself be the results of several measurements, in which case, am will be the mean of all A’s measurements and 0A the standard deviation of that mean (and similarly for $3 and 0'3). The question is how best to combine mA and 1:3 as a single best estimate for m. The answer to this question is to use the principle of maximum likelihood as follows: Let’s assume that both measurements are ”correct” (more discussion on this later) and they are governed by the Gaussian distribution. Let’s further assume that the unknown true value of a: is X. Then the probability of A obtaining the particular value of am is: PM“) QC ie—(wA—XV/(zai) 0A Similarly, for B: 12,4303)“ ie—oB—XV/(2fi3) 0'B The probability that A finds the value mA and B the value an; is just the product of the two probabilities: PX($A,$B) = PX($A)PX($B) O( 1 6 20'?4 2033 (TAO-B 1 _ : e 2 (TAO-B where 2 2 2 (mA—X) (mB—X) X _ 2 + 2 0 0 A B 5—962 _ 0_w+2(mB—X)(—1) (9X — — 0'31 0%; —X —X : “THE—2:0 0A OB :> szbestzalA 0'13 Error Analysis for PHYS375 6 Defining the weights as: we obtain This analysis can be generalized to combine several measurements of the same quantity. Suppose we have N separate measurements of a quantity 1:, 11:1 :01, m2::0'2, ---, mNZZO'N Then N ZZZ-:1 cum mbest : N 21'21 wl where 1 w,- — — 0-.2 It is obvious to note that the larger the error 0',- the smaller the contribution of m,- to the mean. It can be shown that N —1/2 Umbest : (2 ml) i=1 A special case: If all 07’s are equal, i.e., 01:02:”‘ZO'NZO' then N 1 N _(,—2:i=1mi_ 1 __ mbESt—l—N—szi—m 0—2 i=1 and 1 —1/2 Umbest : (—2N) : L 0' W This implies that if a quantity is measured N times, the error will be improved over the error of a single measurement by a factor of j—N. This is what we learn when we discussed the error of the mean earlier. 8 Consistency of the Data After calculated the weighted average and the error, we can then calculate the 2 X i N N 2 m' — m X2 : E :wi(mi _ mbest)2 : E ( I 0.255“) and compare it with N — 1, which is the expectation value of X2 if the measurements are from a Gaussian distribution. We have the following three cases: Error Analysis for PHYS375 7 o If XZ/(N — 1) is less than or equal to 1 and there are no known problems with the data, we say that the data are consistent and should accept the results. 0 If XZ/(N — 1) is greater than 1, but not greatly so, we can still accept the weighted average, but then we need to increase the error awbm by a scale factor defined as 1/2 s = [XZ/(N — 1)] . o If XZ/(N — 1) is very large, we say that the data are inconsistent and should suspect that something has gone wrong in at least one of the measurements. In this case, we should examine all the measurements carefully to see whether some (or all) of the measurements might be subject to unnoticed systematic errors (this could result in larger total errors than quoted). We may choose not to use the weighted average at all. Alternatively, we may quote the weighted average, but then make an educated guess of the error. For example, one could use the standard deviation 1 N al‘bes : 2(ml — mbest)2 t N — 1 i=1 as the error for each measurement instead of the original individual 0",, and use ambest/VN as the error in the weighted mean, instead of the weighted error N —1/2 Umbest : (2 ml) i=1 9 Propagation of Errors Suppose that, in order to find a value for the function q(ar:,y), we measure the two quantities a: and y several times, obtaining N pairs of data, (why/1), (m2,y2), ---, (mmyN). From the N measurements m1, m2, - - - , am, we can compute the mean 5: and standard deviation 0'3, in the usual way; similarly, from y1, y2, ---, yN, we can compute 3] and 0y. Next, using the N pairs of measurements we can compute N values of the quantity of interest: (11' : Q($i,yi), (i : 1,2, . . . ,N) Given q1, q2, -.., qN, we can now calculate their mean c], which we assume giving our best estimate for q,, and their standard deviation O'q, which is our measure of the random uncertainty in the value of q,. We will assume, as usual, that all our uncertainties are small, and hence that all the numbers m1, m2, ---, mN are close to E: and that all the y1, y2, ---, yN are close to 3]. We can then use Taylor series expansion to make the approximation: qr = (Mum) 2 Q($ay)+ a (mi—i)+ — Error Analysis for PHYS375 8 And we have 1 N q I Nng' 1 N (9g (9g = — qi,‘+— 001—5: +— 111—3? NZ ( > my ) 8y; )] 1N__ 1Naq _ 1Naq _ — NngyHN;amwz—mHfiga—ymwi—y) _ _ 1 N aq 1 N _ aq 1 N _ — q(m’y)figl+$mfil§(mi—m)+8—ngNZ1(yi—y) v 7 _V—’ b\/—’ N 0 0 : q(a_0,3]) This means, to find the mean q, we have only to calculate the function q(a:,y) at the point a: = E: and y = 3]. The standard deviation of the N values q1, q2, ---, qN is given by 1 N 02 = — (Ch—(1)2 q Ni=1 2 1 N (9g _ (9g _ — EH[auwz—fldra—yuwz—W] 1— any any 89:21 _28q21N _2 — ($>__F§(mz— )+ 8—3; "figfill—y) any 1— any 1— 8q (9g 1 N 2— — — i—_ 1—— + ammammNgw My 1/) The first two terms are just 0': and 0'13. Let’s define the so-called the covariance of a: and y as follows: 8g 2 (9g 2 (9g 89 2_ _ 2 _ 2 _ _ ”‘1‘ (a) “(6:1) ”9+2 602 6y “W (Note that we have dropped the subscripts E: and 3].) then we have If a: and y are independent, then aw : 0. This is because for a given value of y, the quantity (m, — 27:) is just as likely to be negative as it is to be positive. (This is true for any given value of 3],.) Thus, after many measurements, the positive and negative terms in aw should nearly balance. We then have (9g 2 (9g 2 2 _ _ 2 _ 2 ”a - (am) (ay) 2 Error Analysis for PHYS375 a 2 a 2 <a—:> <a—2> If there are more than two variables, then we have a 2 a 2 <a—:> we) 01' 10 Specific Formulas 10.1 Sums and Differences Let q = 0m :: by, then 861 861 — = — = :5 (9m 0, (9y 0'q : (0)2032” —|— (::b)20'§ : 020'; —|— (flag <— added in quadrature 10.2 Products and Quotients o For products: Let q = may, then 861 fl : ay 861 8—3] I am 0q = (ay)20§+(am)205 _ q2 2 2 2 2 2 2 _ 02m2y2(a y 0'35 —|— a a: 0g) 0' 0' 2 0' 2 : _q (—22) + (—9) q m y => Added in quadrature for the fractional errors. 0 For quotients: Let q = 0%, then @ (91: y Error Analysis for PHYS375 a 2 J: 2 “q = (a) “rt—r) a 22 22 222 : qy aaw+am0y 23:2 2 4 3/ 3/ 2 (2% 2% = (2H?) => Added in quadrature for the fractional errors. Q | | Q 10.3 Powers Let q : of“, then (9g __n_1 nq — = a: :1: —__— (9m ( ) a: 2 n => O'q = (::—q) 0'3 1: n2q2 2 2 2 q 112(2) a: 0' 0'3” 2» —q = n<—> q a: => The fractional error in a: is increased by a factor n. 10.4 Exponentials Let q : aerx, then % = “(:b)e--bx — __bq O'q : w/(::bq)20'g : bqam :> 0" — b _ _ 0'35 q 10.5 Logarithm Error Analysis for PHYS375 11 Let q : aln(::bar:), then (9g _ ::b _ a a _ a E _ E a 2 O'q : — 0'3; : —0'w a: 10.6 Three Examples 1. Measurement of g, the acceleration of gravity, using a simple pendulum. We have I T : 27r\/: g 471'2l =>g= T2 Using the formulae given above, we have (2)1002 g l T2 But, 0']? 01‘ T2 Z 2— 0'9 0'1 2 ( 0T)2 _ : _ 2— :> g ( l) + T Suppose I = 92.95 :: 0.1 cm T = 1.936 :: 0.004 sec 471'2 X (92.95) 2 => gbest = W = 979 cm/sec We have 0'] 0.1 — = — = 0.1 l 92.95 % 07 0.004 — = — = .2 T 1.936 0 % g = (0.1%)2 + (2 x 0.2%)2 = 0.4% :> 0'g : 0.004 X 979 : 4cm/sec2 => 9 = (979 :: 4)cm/.sec2 Note: There is no need to improve the accuracy in the measurement of l since final error is dominated by the error in T. Error Analysis for PHYS375 2. Acceleration of a cart down a slope. We have “M. I. I r Hill-Rafi t.- 3%. Iii-‘5 .H . ‘- ""-~-. I II _""--.. . E42322. But We assume of —|— 2as 2 2 '02 _'U1 2s l = 5.0 ::0.05cm (1%) .s = 100.0 : t1 = 0.054 : t2 = 0.031 : :0.2cm (0.2%) :0.001sec (2%) To calculate on, let’s do it in steps. (a) First, let’s find 0(E): 25 : 2 Note that the uncertainty in 5 makes no appreciable contribution. (b) Next, let’s find 0(é) and 0(é): : 0.001 sec (3%) (2 x 1%)2 + (0.2%)2 % C(35) = 20(L) :2Ut122X2%:4% t1 t1 0(1 2 t2 V II [\9 q A [file )22at222x3%:6% 12 Error Analysis for PHYS375 13 1 => —2 = (343 :: 14)sec_2 t1 if 1 => —2 = (1041 :: 62 )sec—2 t2 E“ c ext, ets n 0' N l ’ fi d (1 1) ?T_?7 2 1 : 64566—2 1 1 _ :> —2 — —2 — (698 :: 64) sec (9%) t2 t1 (d) Lastly, 0a _ a _ = (2%)2 + (9%)2 = 9% 5 x 5 :5 a _ (2X 100) x698::9% = 87.3 cm/sec2 :: 9% = (87.3 :: 8) cm/sec2 3. Refractive index using Snell’s law. We have sin 61 n = _ s1n 62 where 61 (the angle of incidence) and 62 (the angle of refraction) are measured. 0'" _ (Usin01)2 + (Usin02)2 n sin 61 sin 62 To find em let’s use 5111 04 (9 2 O'q : (i) 0'2 <2 use original formula 1‘ asin 04 => 2 cos oz (904 Error Analysis for PHYS375 14 And asin 91 = cos 61 0'91 asin 91 cos 61 _ = _ 0'91 : cot 01091 s1n 61 s1n 61 Similarly, we have 0Isinfi _ 2 : cot 62092 s1n 62 it | || 0' n w/cot2 01031 —|— cot2 02032 71 Suppose we now measure the angle 62 for a couple of values of 01, and get the results shown in the first two columns of the table below. Let’s assume that all angles are measured to an accuracy of :10, or 0.0175 radians. And the results are: * cot(20) x 0.0175 2 0.05 = 5% T cot(13) x 0.0175 2 0.08 = 8% And 1.52 __ 1.61 fl : (0.114) (0.38) : 1.59 (0.14)2 __ (0.08)2 1 1/2 Unbest : fl 2 0'07 (0.14)2 + (0.08)2 Thus, (Value of n) = 1.59 :: 0.07 11 Least-Aquare Fit to a Straight Line It often happens that we wish to determine one characteristic of an experiment y as a function of some other quantity 1:. That is, we wish to find the function f such that y : f(ar:). Instead of making a number of measurements of the quantity y for one particular value of a: (of course, one does this so that one can determine 0y), we make a series of N measurements yi, one for each of several values of the quantity 1: : m, where i is an index that runs from 1 to N to identify the measurements. Error Analysis for PHYS375 15 Probably the most important experiments of this type are those where the expected relation (function) is linear, and this is the case we consider here. In other words, our data consist of pairs of measurements (ml-,yi). We wish to fit the data with an equation of the form: yza—l—bm by determining the values of the coefficients a and I) such that the discrepancy is minimized between the values of our measured y,- and the corresponding values y,- = f (0%)- The problem is to establish and to optimize the estimates of the coefficients. We will again use the principle of maximum likelihood for this problem. If we knew the constants a and b, then, for any given value 1:,- (which we assume to have no uncertainty), we could compute the true value of the corresponding yi: (True value for y,) = a —|— baa,- Assuming that the measurement of y,- is governed by a normal (Gaussian) distribution centered on this true value, with a width aw. Therefore, the probability of obtaining the observed value y,- is p. b(y.-) oc ie-wi-a-W/aazn O'y. ’L where the subscripts a and b indicate that this probability depends on the unknown values of a and b. The probability of obtaining our complete set of measurements ($1,311), (mZayZ)? ' ' 'a (mNayN) is the PIOduCt: Pa,b(y1ay23 ' ' ' ayN) OC Pa,b(y1)Pa,b(y2) ' I ' Pa,b(yN) 1 {_§ 2 (yr-(149002} cc —6 Z a” 0-141 0-142 ' ' ' UyN 1 _fi : e 2 0-141 0-142 ' ' ' UyN where y-—a—bar:-2) M2 i=1 As before, the best estimate for the unknown constants a and I), based on the given measurements, are those values of a and b for which the probability Payb(y1, yg, - - - ,yN) is maximum or for which X2 (the sum of squares) is a minimum (that is why this method is known as the least-squares fitting). To find the values of a and b which yield the minimum value of X2, we differentiate X2 with respect to a and b and set the derivatives to zero: 8X2 N yli—a—bm) — = —2 —=0 (9a 2 0y, 8X2 N (yi_a_bmi)mi W — ”20—2—0 Error Analysis for PHYS375 These equations can be rearranged to yield a pair of simultaneous equations: N 3/1 N 1 N 301 Z 2 = “2—2 “52 2 1:1 0-141 1:1 0-91 1:1 0-91" N N N 2 301311 $1 391 Z 2 = “Z 2 -- 52 2 1:1 0-141 1:1 0-91 1:1 0-91" The solutions are: N y' N a: 1 21:1 Tl 21:1 21 a — — a“ ”3'21 — A N 901141 N w_1 1:1 0.2 1:1 0.2. 1 y’L N 2 N N N _ i 2 m1' 2 3/1 _ 2 m1' m1y1 _ 2 2 2 2 A ':1 0-111 1:1 0-111 1:1 0-111 1:1 0-111 N 1 N 111 b 1 21:1 ”—5, 1:1 UT _ ’L ’L — — N & N 331141 A 21:1 0-51. 21:1 0.21 N N N N — 1 2—1 2% Z 211 yl _ _ 2 2 _ 2 2 A 1:1 0-111 1:1 0-111 1:1 0-111 1:1 0-111 N 1 N an 2121—, 21:1 —. A Z N 12-1 N 1121 ’L 1 21:1 ”—5. 21:1 0—5, ’L ’L 2 N N 2 — 2—1 Z 211. Z 211 _ 2 2 _ 2 1:1 0111 1:1 0111 1:1 0111- We state without proof that the errors in a and b are given as: then where NZ ”2 01? 'ZO'yNZO'y N N N NZMM _ EJ512291 1:1 1:1 1:1 N N 2 16 Error Analysis for PHYS375 17 and q H l>|H l>|H q 0“ | | 12 Uncertainty in the Measurement of y If each of the y,- for a given 1:,- is measured many times, one can certainly get an idea of am by examining the spread in their values, 2'....
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