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# Advanced Lab I PHYSICS 441

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This 25 page Class Notes was uploaded by Lamont Barton on Thursday October 29, 2015. The Class Notes belongs to PHYSICS 441 at University of Michigan taught by Carl Akerlof in Fall. Since its upload, it has received 15 views. For similar materials see /class/231642/physics-441-university-of-michigan in Physics 2 at University of Michigan.

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Date Created: 10/29/15

Notes on Statistics Lecture I Carl W Akerlof quotThere are three kinds of lies lies damned lies and statistics iMark Twain Autobiography 7 1924 Why Statistics 1 Many important discoveries are made at the edge of technical feasibility Every little bit of improvement in the signaltonoise ratio helps Example Nobel Prize in 2006 to John C Mather and George F Smoot for measurements of the cosmic microwave background anisotropy 2 Often the results of experiments are numerical values used to make further predictions One would like to determine such numbers as accurately as possible My own occasional efforts in this area 1 Application of cubic splines to the spectral analysis of unequally spaced data Astrophysical Journal 436 787794 1994 method for finding periodically variable stars A 39 39 image 39 quot by r 39 quot A 39 39 Journal 677 808872 2008 method for finding optical transients such as supernovae 2 Notation for expectation values f x E06 Measurements Either discrete like counting events or people or continuous like measuring a length For discrete associate a probability pl for the i th value 2p 1 11 For continua associate a probability distribution function PDF f x so that the probability of observing a value on the interval x x dx is given by f xdx j fxdx 1 Probability distributions come in all shapes and avors although the most important is the Gaussian 7 riff 2 e 2039 fx V 7T0 where f is the mean and 0392 is the variance The Gaussian has some very nice mathematical properties it is de ned everywhere from 00 to 00 and it is differentiable to all orders Unfortunately its integral is a transcendental function that must be evaluated by suitable approximations In Excel use ERF or ERFC but divide the argument byxE The mean of a continuous distribution is defined by x x j gfxdx The variance is defined by 1Cmax 039 x2gt ltxgt2 J ye ff fxdx 1Cmm These two moments are the most important quantities for characterizing a probability distribution Two higher order moments are sometimes computed skewness and kurtosis but they have little use in physics or astronomy For discrete variables the mean and variance are defined by the analogous summations f Z X p of Z X f2 p One of the simplest probability distribution functions is the uniform distribution 1 OSXSl 0 elsewhere f This distribution is particularly easy to synthesize with a digital computer and is the root of almost every computer simulation of probabilistic phenomena Problem 1 Find the mean and variance for the uniform probability distribution defined above We now ask what is the distribution of the sum of the two quantities each uniformly distributed x2 xa xb The result is obtained by convolution Igdtx OSxSl f22x Llildt2 x leSZ 0 otherwise Figure 1 The geometry corresponding to the addition of two independent uniformly distributed variables In the region bounded by the two dotted red lines the sum xa xb is approximately constant The sum can range from 0 to 2 with amax1mum ofl The results for larger sums of uniformly distributed numbers can be obtained by iteration Distribution for sum of 3 unifonnly distributed values fmoc fawnu och lx20gxg1 2 7 736x72x2 1 x 2 976xx2 2 x 3 0 otherwise Distribution for sum of 4 unifonnly distributed values fzx Mommalt lx30gcg1 6 4712x12x2 73x3 1 x S 2 74460x724x2 315 2 g x g3 47x3 3 x 4 0 otherwise These four distributions for the sum of 1 to 4 uniformly distributed values are plotted in gure 2 below In addition the Gaussian distributions with identical mean and variance for E3 and 24 are shown as dotted lines I hope the trend is clear the sum of a set of uniformly distributed numbers approaches a Gaussian distribution ever more closely as the number of items increases In fact a good computer algorithm for generating Gaussian random numbers simply adds twelve uniformly distributed numbers on the interval 01 and subtracts the mean of 6 12 415 00 05 10 15 20 25 30 35 40 Figure 2 Plots of the distribution mction for the sum of the one to four uniformly distributed numbers Magenta 1 blue 2 green 3 orange 4 The dotted lines show the Gaussian distributions corresponding to the mean and variance for 3 and 4 summed values Problem 2 Compute the mean and variance for the f f23 and f24 distributions Suggestion Use Mathematica or a similar program to perform the integrations symbolically The results shown above are quite general Under very broad limits any sum of randomly distributed variables approach Gaussian distributions as the number of variables as increases This result is called the Central Limit Theorem and gives some idea of why the Gaussian distribution plays such a signi cant role in statistics More generally the following relationships for the mean and variance are independent of the underlying distribution functions mab Ha rfltrmltrdrdr arfrdr 61 b b2gt abgt2 a t lta ltb2 mtg3700515515 Calm a l39 llta2gtltaz ltbgt2 215 Himmm a2gtltagt2ltb2gt ltbgtzaa These relations can be extended to an arbitrary number of variables independent of the distribution function for each element Scaling the mean and variance 39 Z Z Z If gx cx where c is a constant then g cx 0g c oquot If gx is a more complicated function reasonable approximations are given by 2 2 0x xx as long as higherorder derivatives are not important over the range x 0 x 0 If this becomes d g E gx 039 E Ci i questionable it is best to estimate the statistical parameters by Monte Carlo simulation Note that I have carefully avoided discussing the standard deviation which is defined as the square root of the variance Although the standard deviation is important it is the variance that is the more fundamental statistical quantity 4 z z 4 n z 2 z n 3 a 1 u i r z A Figllre 3 TWO dlSUibmiOHS W131 differfem means Figure 4 Two distributions with identical means but Idemlcal variances Standard devmuor sl but different variances standard deviations Binomial distribution There are two common discrete distributions that one usually encounters in science One of these is the binomial distribution that describes the outcome of such activities as coin tossing Suppose for the sake of argument that coins have probability p of landing heads and a probability q of landing tails q I p Ifthe coins are fair p q 05 but that may not always be true Suppose we toss N coins The probability of all possible outcomes is given by N p r 1 The expression on the le side of this equation can be rewritten as P N gm qm Each individual term in this sum represents a different outcome of the total number of heads and tails with m denoting the number of heads and N m the number of tails Therefore the binomial distribution function for the probability of observing each value of our random variable m is Nl N P m m w m N7m zgt 4 Note that the total number of heads or tails is strictly bound between the limits of 0 and N This is distinctly different from the Poisson distribution that will be discussed shortly where no such upper bound exists It is fairly straightforward to show that the mean and variance are given by Mean 5 p N Variance a p q N Proof N l m m nltngtpNZ p1qN m lN m N71 JV 1 m errm N N p gmN 1 mpq p 2 2 N N2 mrz NrM n gt npNN 1 m 2N mp q 2 N72 Iv 2 m N727m 2 pNN I mN2mpq pNN 1 thus 039 n2gt ltngt2 pZNN lltn n2 qu Note that both the mean and the variance scale directly with the number of events N Thus the precision of an observation given by the ratio of the standard deviation to the mean decreases at a rate proportional to 1 J17 Making N as large as possible is good but you improve more and more slowly for the extra time and resources required probability a a ca number of heads Figure 5 Probability distribution for the number of heads observed when tossing 50 coins The points show the exact probability computed from the binomial distribution The smooth curve is the Gaussian distribution with W 25 and 039 125 At the peak of the distributions the two probabilities differ by 05 7 Figure 5 shows a plot of these terms for N 50 for fair coins As expected from the Central Limit Theorem the points follow very closely a Gaussian distribution with the mean and variance computed from the formulas above with p q 05 Poisson distribution The most important statistical distribution function for Physics 441442 is the Poisson 7 it describes the probability of nding 71 discrete events in a specific time interval assuming uniform arrival probability Unlike the binomial distribution the number of events is not fixed In principle a very large number of events can occur randomly within a short interval although the probability becomes vanishingly small as 71 gets large Assume the average interval between events is given by t TNT where T is the length of an appropriately long time interval and NT is the number of corresponding events The probability of zero events within a time interval of length t is pm 6 For very short time intervals p0 is close to one but decreases rapidly as t gets appreciably larger than t For 71 events the corresponding probability is M AWquot w n It is trivial to show that pn reaches a maximum when I nlttgt Figure 6 below shows such behavior graphically One can also see that ipn1 independent of the duration t The mean number of events can be computed from n71 x inpnxim e39x xttgt 1 Problem 3 Using similar techniques find the variance of n 02 n2 r02 1 0 mm pm pm 0 3 pm m 06 E n 04 02 00 0 1 2 3 4 5 6 7 8 may Figure 6 The probability of detecting exactly 0 1 2 3 or 4 events as a mction of the mtio of t divided by the avemge time between events rm Note that each cmve peaks When t n t Appendix A Distribution functions for sums of three and four uniformly distributed variables 3 variables 3x tf1tdt xjx tdt Ostl zzle UWH 1 xtdt Istz x71 1 f3xj2 x tdt 23x33 x72 4 variables Jf3x tf1tdt f4x Iix tzalt 0 S x S1 f4xxJ 36x t 2x t2 3x tfdt lSXSZ f4xxj29 6x tx t2dt l 36x t 2x r2dt 23x33 f4x 39 6x tx tzdt 3S xS 4 x73 Notes on Statistics Lecture 11 Carl W Akerlof Sample statistics with n observations 5 Sample Mean f lzyq 7 11 1 n 1 n 2 Sample Variance 039 lez Zq J quot 1 11 7 11 Why the factor of ln 1 instead of 111 L quot 2l quot 2 J39s n4 n glxzx 1 ixlullexj 1 11 1 71 H n l j Since there are 7 terms in the rst sum and n n 1 terms in the second the average value for an ensemble of samples coincides with 039 Sample Standard Deviation as 039 With these de nitions the sample statistics cluster around the population statistics Mean of Sample Means f x Mean of Sample Variances 0392 x The next question is how accurately do the sample statistics is and 039 compare with the population statistics f and 039 Beware that the statistical quantities described below are population statistics based on the ensemble of all possible similar experiments usually an infinite set l Variance of Sample Means 039 039 n Proof 1 quot 2 1 2 n2ltl1xlgt glycli x 1 2 1 2 2 1 2 1 2 ltx gtltxgt ltx x gt x 2 039 7 214 43 3 2242 3 2 222 3 4 a Z ltx gtx Xx nib 1 n 1ltx Xx 507 1906 With the presence of the third and fourth moments the variance of the sample variance depends on the functional form of the probability distribution function For the important case of the Gaussian distribution 1 7 Hf e 202 VZIZ39O39Z fx we can choose a coordinate system so that f is zero x4 30394 leading to 20394 Variance of sample variances 02 1 1 n and the related equations 2 Standard dev1ation of sample var1ance 039r2 1 0392 1 n Z Variance of sample standard deviations 039 2n l 039 Standard dev1ation of sample standard dev1ations 039 3 2n 1 For the binomial distribution 20394 l 6 o2 Variance of sample var1ances 0 l m and for the Poisson distribution 20394 0392 Variance of sample var1ances 03922 S m l m where m is the number of observations Note the similarities in form to the corresponding expression for the Gaussian distribution For many applications statistical distributions are plagued by outliers that can badly contaminate the mean and variance Statistical parameters based on the ordering of values are much more stable against the effects of poorly understood background effects The median of a set of values is easily de ned but it does require that the set first be ordered l l Variance of median o2 F E rm 4nf2x12 xlZ 2 For Gaussian distributions 0392 7r 2 1C12 2n 039 There are two disadvantages incurred by using the median instead of the mean to represent the average of a sample Ordering takes a signi cant amount of computational effort especially if the set is large Secondly for a fixed value of n the variance of the median is n39 2 greater than for the mean This ratio is called the Asymptotic Relative Efficiency ARE The choice between the two will depend on the application There is no free lunch The variance is even more susceptible to contamination than the mean As an example a 1 admixture of values that are 10 times larger than the 039 for the rest of the sample will double the variance Understanding background outliers at the 1 level is usually an extremely daunting task An order statistic that will robustly estimate dispersion is the InterQuartile Difference IQD defined by Assume Fx14Ei Fx34E Then IQD x34 x14 For a Gaussian distribution Q 039 where q 13489798 l H H edex q 2 27239 9 a q 2 2 27239 82 4 OIQDqz 7 qgtk2 a 4 272092i Note that the variance of the IQD estimated dispersion is a factor of 272 larger than for the normal variance again an example of the cost in precision that must be paid for a more reliable statistical measure Statistical signi cance A principle goal of statistics is to determine the probability that two measured values are consistent in view of the estimated errors In the first lecture several cases were shown in which the effective probability distribution functions were to a very good approximation Gaussian In such case we are interested in the cumulative distribution function Fx39 I fxdx which for a Gaussian distribution is HF 1639 Z J 6 2quot dx foo Fx l 1272390392 In Excel this can be easily computed from the normalized integral 1 x i NORMSDISTx e 2 du V2 J where u x x Similar functions are available on other mathematical evaluation packages Most of the time we are interested in fairly simple questions such as whether a particular measurement is 39 with a p vi 4 l w 11 39 quot 39 A value Suppose that you found a value for Planck s constant of 415gtlt103934 Js with a standard deviation of ll6gtlt103934 for the measurement error The presently accepted value is 6626069X10734 Your result is certainly numerically different Should you call the science editor of the New York Times and inform him that a fundamental constant of physics is changing with time or perhaps location The question that statistics can answer is the probability that such a measurement could occur by chance alone 14 The normalized error in this ease is 4150 6626116 2 13 standard deviations The probability that a measurement would have such a departure from the true value is 13 3 F 213 j e 2 dx00164 Given that there are 35 students in Physics 441 this term there is greater than 50 probability that somebody will get such a large discrepancy during the next few months We have also lowballed the likelihood of chance error you might be equally willing to call the New York Times if the error were correspondingly higher ie 910 gtlt103934 In this case the chance probability is px gt 213aj13e2dxj w edx 00328 112 213 Note that statistics has no way of estimating the effects of systematic problems with your results Problem 4 You have made two measurements of the counting rate produced by a radioactive isotope For the first observation you detected 820 counts in 1000 seconds in the second you detected 1625 counts in 1700 seconds Find the twosided probability that the apparent difference in rates is due to chance alone Problem 5 You are given a single penny How many times would you have to ip it to determine with 99 confidence that probability of heads lies in the range 0495 S pH S 0505 assuming that for your specific experiment exactly half of the tosses were heads There has recently arisen a new catechism for corporations called 6 o Originating with some folks at Motorola who wanted to reduce the number of defective products the idea is to make sure that your production techniques are good enough so that only a 6 6 error would lead to failure The true doctrine is a bit of a fake it s actually 45 6 errors that are bad news and the probability is calculated as one sided not two For many cases you are likely to deal with you may need to use the exact distribution function if the deviation from Gaussianity is large or instead rely on numerical simulations The more anomalous or significant your result the more care you need to take to properly calculate the statistical significance Weighted means In earlier discussion a formula for the sample mean was presented that assumed that each of the n observations is associated with identical standard deviation monoscedastic That situation is frequently violated leading to the need to find a way of weighting the formula to provide a value for the mean with the least error Assume a set ofweights wi so that We must find a set of values for w to minimize 039 2 n H n 3 607 2wp f 2w ZZwlzo2 JZwlj 0 6W I 11 11 I 11 This can be solved by setting w izfor all 139 039 1 The variance of the mean is thus caliii 11 0 The formula given above for weighting observations should be followed whenever considering data with different measurement errors Problem 6 Find the weighted mean of the two counting rates described in Problem 4 Notes on Statistics Lecture 1H Carl W Akerlof Parameter estimation In the last lecture a procedure was developed for nding the mean of a set of observations with different errors It was shown that the most accurate estimate was obtained by a weighted sum f Z m where w l I ql and thus 1 03 We will now take a slightly different point of view to derive the same result This new method can be extended to much more complex descriptions of data than just the mean For the moment make the assumption that every observation if not perturbed by a variety of unwanted phenomena would yield exactly the same value xm We would like to nd a value for xm so that the difference between observed values q and xm are as small as possible The simplest function suitable for this purpose is the sum of squares of deviations divided by the variance for each point This can be easily recognized as the sum of the squares of the departure of each observation from xm in units of the standard deviation This expression can also be derived more formally by the maximum likelihood principle Minimizing A by varying xm yields the following equation 01 1 ngz x l 20 xm as derived previously For reasons that will soon become clear it is useful to think of this as a simple matrix equation of the form 3 IQ Now assume a slightly more complicated problem depicted in Figure 7 below Suppose we have a total of n observations half clustered around xa and half around x17 We would like to estimate the error of the slope of the line that best ts the entire set of data assuming an average error of o for the y 2 values The mean value for each of the two clusters Will be known With a variance of 0392 and thus n O2 xb xayi 4 the variance of the slope Will be determined to an accuracy of n Figure 7 Least square t of a straight line to two clusters of data points This problem can be framed more generally by the assumption that the data is described by a linear relationship y A Bx Our task is to nd the best values ofA and B and their variances 03931 and Z 039 5 Using the previous idea of minimizing the square error we have kiwi1me Minimizing A by varyingA andB leads to III 1 1 1 r1 r1 B 7 1 1 B 20 12 Eff 12 In this case the inverse matrix H 1 is 1 1 1 201m 20 21 1 Z 1 2 l l 2727 27 Z a quotI 20 Restricting ourselves to the example shown in Figure 7 we nd that H Z n x xa 2 I More generally H 2 H and o jw g2 H111 Z ZS IZH1 3 gg ZH 3 12 H If the data must be modeled by the linear sum of several arbitrary foundations y a1f1Xaz ampan amp then the coef cients a1 can be determined by solving the matrix equation ZWJTZW walwmw Zw 99fn99 a1 walxy szfzvmw waffbe NZszszM 322W1f2f yx zm mmzm qmm zw2ltxgt 4 zw my The variances for each coefficient a is given by 039 H 111 and the variance for functions of a1 is J given by 6g 6g 2 1 OLJMJZ 0 Zjlaa 6a HI I 1 Note that the regression matrix de ned above is by construction symmetric It can also be shown to be positivede nite These criteria are sufficient to satisfy the conditions for using the Cholesky decomposition for nding the inverse matrix see Numerical Recipes Numerical computation issues are often critical if the set of model functions f x is large or the functions are similar in shape Use of double precision oating point representations is essential It may also be helpful to select model functions that are orthogonal in the sense that I f x f xdx 5U Occasionally you may also need to include constraints on the tted coef cients for example 2 a 1 These can be incorporated via the method of LaGrangian multipliers as additional linear equations 20 The expanded regression matrix no longer satis es the requirements for the Cholesky decomposition and you must turn to the Singular Value Decomposition SVD instead see Numerical Recipes As long as the model function can be represented by linear sums of welldefined functions unique solutions can be obtained by matrix inversion with estimates for parameter uncertainty extracted from the appropriate elements of the inverse of the regression matrix Unfortunately a large class of problems exist in which the basic model functions depend on parameters in highly nonlinear ways An example is 0pr y Ae quot2 The B and C parameters cannot be computed by any single pass algorithm For such problems one must provide initial guesses followed by a search based on trialanderror that minimizes the square error In Excel this is generally accomplished with the Solver addin On other platforms algorithms such as the downhill simplex method AMOEBA in the Numerical Recipes library are robust procedures that will usually find the desired optimal values However one should always be aware that unlike the situation with linear parameters the sum of square errors can exhibit local minima in parameter space which are not necessarily the global minima Thus the numerical search may find a solution but not the best possible solution Estimating the error of nonlinear parameters is also much messier In general the most reliable technique is performed by simulation of the original data set followed by the identical procedures for estimation of the nonlinear parameters Problem 7 In Physics 141 an experiment is performed with a dropping ball to determine the local gravitational acceleration g Assume the ball is dropped at I 0 and the ycoordinate is measured in the downward direction The standard deviation for the vertical measurements is uniformly 0012 meters 12 mm From the following measurements determine g vertical measurements in meters t I005 I010 I015 I020 I025 I030 I035 I040 I045 I050 y I0463 0483 0578 0652 0747 0879 1051 1224 1422 1671 Also determine the probable error in g from the information provided Fitting smooth curves to data often requires a certain amount of judgment particularly if there is no strong physical reason to choose one set of functions over another One choice when all that is required is smoothness is the cubic spline an example of which is shown in Figure 2 the orange curve Tests for goodness of fit If the data is well described by the chosen model function one would expect that the average deviation between each data point and the fitted curve would be approximately one standard deviation If there are 71 data points this would lead to the expectation that the total square error 21 1 201 fxx2 would be approximately 71 where fq alf1 q azf2 akfk 9g In fact we know that if the number of tting parameters k is exactly equal to the number of data points n the tted curve will go exactly through each observation and the sum of square errors will be zero Thus what really matters is the number of degrees of freedom defined by V n k and we expect that the total square error is on average v instead of n The discussion above suggests how we can assess whether a t to data is good bad or indifferent If we have v independent normally distributed quantities x1 each with unit variance and zero mean the sum Q 2 x12 is distributed according to the 1 distribution The probability peaks at 11 Q V 2for V Z 2 and approaches a Gaussian as v becomes large another example of the Central Limit Theorem A graph of the differential probability distribution function for I is shown below in Figure 8 Note the similarity of the curves to the Poisson distribution shown in Figure 6 The functional forms are closely related The ratio of square error to number of degrees of freedom is often quoted as a measure of goodness of t If this is markedly different from unity your data may have signi cant problems or your choice of tting curve was inappropriate Signi cance levels for the 12 statistic are usually available on most computing platforms 22 gure a The arraehm pmbabdny ruhenuh fur x fur n n m k 5 Graph ub39amed hm hm Nehwuq emeu wkChrs uared dasmbuum mhmx Cumpuleths 1 pa degree uf eed m runhe sued curve hr yrrmpmh1ah7 Tesung the smulanty ufstausuml msmhuums hm w n H V m the me urdlffermt Thatprublem Bubs gamhzed m the quesum ufwhethertvm prutabxhty Ms useful m khumhenheuare wellheahed pmcedures u make sunhtests The bass fursuch cumpansunsxs the cumulativepmbabdny msmhuum runehuh de ned by Slppusethalwe wAshm enmparewm sueh cumulauve hemhuuuhs momma The T chameevm Mses lest upmtes m the mlegal unhe sqmred dlffa mce FunhEr mrmheuuh En he mm m swam Memnds m Expanmental Physici39 T Edmun Fredenek James and Fmbabdny and Stancs m Expmmm39zl Physici39 aymh F Rue m Fm Figure 9 Histogram of two distributions taken Figure 10 The cumulative distributions for from a aus ian population with the two samples shown in Figure 9 The 6 10 blue and 6 18 red vertical green line indicates the maximum separation between F1x and Fzx The graphs shown in Figures 9 and 10 above depict the distribution of 100 observations drawn from Gaussian parent populations with mean of zero and two different values of the variance The right hand plot illustrates the application of the KolmogorovSmimov test In this case the probability of seeing such a large deviation between the two cumulative distributions is approximately 24 Signal detection Although outside the scope of the experiments performed in Physics 441442 there are a variety of techniques used in signal analysis and detection that rely on statistical methods similar to those described previously One of the most interesting of these is the matched lter In this case it is assumed that one is looking for a timedependent signal with a wellde ned functional form It is a fairly straightforward task to devise an optimal weighting scheme that will maximize the signalto noise ratio an important problem for sonar or radar These techniques easily map to similar problems for twodimensional image detection and recognition Finally the Fast Fourier Transform FFT should be mentioned because of its importance for detecting harmonic signals in the face of a variety of noisy backgrounds Notes on Statistics Lecture IV Carl W Akerlof The abuse of the coef cient of distribution R2 A persistent misconception of folks who use Microso Excel or similar applications is the signi cance of the coef cient of distribution R2 or its square root R the coef cient of correlation These two statistics are appropriate for rat psychologists who wish to demonstrate that there is some connection between two observables although the underlying source of correlation is illde ned or unknowable These quantities are incapable of answering the question of whether numerical data are consistent with a particular theoretical prediction To understand this problem we start with the de nition of R2 R1 1M E yx 72 where y is the dependent observable associated with the independent variable x and x is the model function that is postulated to describe the correlation of x and y 7 is the arithmetic mean of the data set 2 This de nition of R2 is unambiguous if the functional relation de ned by x is linear For more complex mathematical forms this convention is not unique For Excel a cionados dealing with powerlaw exponential or logarithmic behaviors the computed value for R2 is obtained from the linearization obtained following the appropriate transformation of variables Note that the de nition for R2 is independent of the statistical errors for the dependent variable y A consequence is shown below for two identical data sets with the errors in Figure llb ve times smaller than in Figure 11a The t of the data in Fig 11a is quite good 7 the 2 statistic shows a high probability that the data is adequately described by a straight line For Fig llb the situation is just the opposite 7 the 2 statistic indicates a vanishingly small probability that the data is described by such a linear relationship The R2 u a 04 u an m m Figure 11a The straight line ts the data Figure 11b The straight line ts the data with ax of 9485 for 9 degrees of freedom with a X2 of 237126 for 9 degrees of The probability of this value or larger for X2 freedom The probability of this value or occurring by chance alone is 039 The larger for X2 occurring by chance alone is coef cient ofdistributionRZ is 08778 52 x 10 The coef cient of distribution R2 is 08778

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