620 Chapter 14.Statistical Description of Data 14.3 Are Two Distributions Different? Given two sets of data,we can generalize the questions asked in the previous section and ask the single question:Are the two sets drawn from the same distribution function,or from different distribution functions?Equivalently,in proper statistical language,"Can we disprove,to a certain required level of significance,the null hypothesis that two data sets are drawn from the same population distribution function?"Disproving the null hypothesis in effect proves that the data sets are from different distributions.Failing to disprove the null hypothesis,on the other hand, 三 only shows that the data sets can be consistent with a single distribution function. 81 One can never prove that two data sets come from a single distribution,since (e.g.) no practical amount of data can distinguish between two distributions which differ only by one part in 1010 Proving that two distributions are different,or showing that they are consistent, is a task that comes up all the time in many areas of research:Are the visible stars distributed uniformly in the sky?(That is,is the distribution of stars as a function % of declination-position in the sky-the same as the distribution of sky area as a function of declination?)Are educational patterns the same in Brooklyn as in the 9 Bronx?(That is,are the distributions of people as a function of last-grade-attended the same?)Do two brands of fluorescent lights have the same distribution of burn-out times?Is the incidence of chicken pox the same for first-born,second-born, third-born children,etc.? These four examples illustrate the four combinations arising from two different a之w 9 9 dichotomies:(1)The data are either continuous or binned.(2)Either we wish to compare one data set to a known distribution,or we wish to compare two equally unknown data sets.The data sets on fluorescent lights and on stars are continuous, since we can be given lists of individual burnout times or of stellar positions.The data sets on chicken pox and educational level are binned,since we are given tables of numbers of events in discrete categories:first-born,second-born,etc.;or 6th Grade,7th Grade,etc.Stars and chicken pox,on the other hand,share the property that the null hypothesis is a known distribution(distribution of area in the sky,or incidence of chicken pox in the general population).Fluorescent lights and 10621 educational level involve the comparison of two equally unknown data sets(the two brands,or Brooklyn and the Bronx). One can always turn continuous data into binned data,by grouping the events 43106 into specified ranges of the continuous variable(s):declinations between 0 and 10 degrees,10 and 20,20 and 30,etc.Binning involves a loss of information,however. 腿 Also,there is often considerable arbitrariness as to how the bins should be chosen. North Along with many other investigators,we prefer to avoid unnecessary binning of data. The accepted test for differences between binned distributions is the chi-square test.For continuous data as a function of a single variable,the most generally accepted test is the Kolmogorov-Smirnov test.We consider each in turn. Chi-Square Test Suppose that Ni is the number of events observed in the ith bin,and that ni is the number expected according to some known distribution.Note that the Ni's are
620 Chapter 14. Statistical Description of Data Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). 14.3 Are Two Distributions Different? Given two sets of data, we can generalize the questions asked in the previous section and ask the single question: Are the two sets drawn from the same distribution function, or from different distribution functions? Equivalently, in proper statistical language, “Can we disprove, to a certain required level of significance, the null hypothesis that two data sets are drawn from the same population distribution function?” Disproving the null hypothesis in effect proves that the data sets are from different distributions. Failing to disprove the null hypothesis, on the other hand, only shows that the data sets can be consistent with a single distribution function. One can never prove that two data sets come from a single distribution, since (e.g.) no practical amount of data can distinguish between two distributions which differ only by one part in 1010. Proving that two distributions are different, or showing that they are consistent, is a task that comes up all the time in many areas of research: Are the visible stars distributed uniformly in the sky? (That is, is the distribution of stars as a function of declination — position in the sky — the same as the distribution of sky area as a function of declination?) Are educational patterns the same in Brooklyn as in the Bronx? (That is, are the distributions of people as a function of last-grade-attended the same?) Do two brands of fluorescent lights have the same distribution of burn-out times? Is the incidence of chicken pox the same for first-born, second-born, third-born children, etc.? These four examples illustrate the four combinations arising from two different dichotomies: (1) The data are either continuous or binned. (2) Either we wish to compare one data set to a known distribution, or we wish to compare two equally unknown data sets. The data sets on fluorescent lights and on stars are continuous, since we can be given lists of individual burnout times or of stellar positions. The data sets on chicken pox and educational level are binned, since we are given tables of numbers of events in discrete categories: first-born, second-born, etc.; or 6th Grade, 7th Grade, etc. Stars and chicken pox, on the other hand, share the property that the null hypothesis is a known distribution (distribution of area in the sky, or incidence of chicken pox in the general population). Fluorescent lights and educational level involve the comparison of two equally unknown data sets (the two brands, or Brooklyn and the Bronx). One can always turn continuous data into binned data, by grouping the events into specified ranges of the continuous variable(s): declinations between 0 and 10 degrees, 10 and 20, 20 and 30, etc. Binning involves a loss of information, however. Also, there is often considerable arbitrariness as to how the bins should be chosen. Along with many other investigators, we prefer to avoid unnecessary binning of data. The accepted test for differences between binned distributions is the chi-square test. For continuous data as a function of a single variable, the most generally accepted test is the Kolmogorov-Smirnov test. We consider each in turn. Chi-Square Test Suppose that Ni is the number of events observed in the ith bin, and that ni is the number expected according to some known distribution. Note that the N i’s are
14.3 Are Two Distributions Different? 621 integers,while the ni's may not be.Then the chi-square statistic is x2=∑-n2 (14.3.1) ni 2 where the sum is over all bins.A large value of x2 indicates that the null hypothesis (that the Ni's are drawn from the population represented by the ni's)is rather unlikely. Any term j in (14.3.1)with 0=nj =Ni should be omitted from the sum.A term with n;=0,N;0 gives an infinite x2,as it should,since in this case the Ni's cannot possibly be drawn from the ni's! 8 The chi-square probability function Q(xv)is an incomplete gamma function, and was already discussed in $6.2 (see equation 6.2.18).Strictly speaking (x2) 18881892 nted for is the probability that the sum of the squares of v random normal variables of unit variance (and zero mean)will be greater than x2.The terms in the sum (14.3.1) are not individually normal.However,if either the number of bins is large (1), or the number of events in each bin is large (1),then the chi-square probability from NUMERICAL RECIPES I function is a good approximation to the distribution of(14.3.1)in the case of the null hypothesis.Its use to estimate the significance of the chi-square test is standard. The appropriate value of v,the number of degrees of freedom,bears some additional discussion.If the data are collected with the model ni's fixed-that is,not later renormalized to fit the total observed number of events XN;-then v 邑免乡 equals the number of bins NB.(Note that this is not the total number of events! Much more commonly.the ni's are normalized after the fact so that their sum equals the sum of the Ni's.In this case the correct value for v is NB-1,and the model is said to have one constraint(knstrn=1 in the program below).If the model that OF SCIENTIFIC gives the ni's has additional free parameters that were adjusted after the fact to agree with the data,then each of these additional "fitted"parameters decreases v(and increases knstrn)by one additional unit. We have,then,the following program: void chsone(float bins[],float ebins,int nbins,int knstrn,float *df, COMPUTING (ISBN 18881292 float *chsq,float *prob) Given the array bins[1..nbins]containing the observed numbers of events,and an array ebins[1..nbins]containing the expected numbers of events,and given the number of con- 10621 straints knstrn (normally one),this routine returns (trivially)the number of degrees of freedom df,and (nontrivially)the chi-square chsq and the significance prob.A small value of prob Fuunrgroirioh Numerical Recipes 43106 indicates a significant difference between the distributions bins and ebins.Note that bins and ebins are both float arrays,although bins will normally contain integer values. (outside float gammq(float a,float x); void nrerror(char error_text[]); Software. int j; float temp; ying of *df=nbins-knstrn; *chsq=0.0; for (j=1;j<=nbins;j++) if (ebins[i]<0.0)nrerror("Bad expected number in chsone"); temp=bins[j]-ebins[j]; *chsq +temp*temp/ebins[j]; *prob=gammq(0.5*(*df),0.5*(*chsq)); Chi-square probability function.See $6.2
14.3 Are Two Distributions Different? 621 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). integers, while the ni’s may not be. Then the chi-square statistic is χ2 = i (Ni − ni)2 ni (14.3.1) where the sum is over all bins. A large value of χ2 indicates that the null hypothesis (that the Ni’s are drawn from the population represented by the ni’s) is rather unlikely. Any term j in (14.3.1) with 0 = nj = Nj should be omitted from the sum. A term with nj = 0, Nj = 0 gives an infinite χ2, as it should, since in this case the Ni’s cannot possibly be drawn from the ni’s! The chi-square probability function Q(χ2|ν) is an incomplete gamma function, and was already discussed in §6.2 (see equation 6.2.18). Strictly speaking Q(χ2|ν) is the probability that the sum of the squares of ν random normal variables of unit variance (and zero mean) will be greater than χ2. The terms in the sum (14.3.1) are not individually normal. However, if either the number of bins is large ( 1), or the number of events in each bin is large ( 1), then the chi-square probability function is a good approximation to the distribution of (14.3.1) in the case of the null hypothesis. Its use to estimate the significance of the chi-square test is standard. The appropriate value of ν, the number of degrees of freedom, bears some additional discussion. If the data are collected with the model ni’s fixed — that is, not later renormalized to fit the total observed number of events ΣN i — then ν equals the number of bins NB. (Note that this is not the total number of events!) Much more commonly, the ni’s are normalized after the fact so that their sum equals the sum of the Ni’s. In this case the correct value for ν is NB − 1, and the model is said to have one constraint (knstrn=1 in the program below). If the model that gives the ni’s has additional free parameters that were adjusted after the fact to agree with the data, then each of these additional “fitted” parameters decreases ν (and increases knstrn) by one additional unit. We have, then, the following program: void chsone(float bins[], float ebins[], int nbins, int knstrn, float *df, float *chsq, float *prob) Given the array bins[1..nbins] containing the observed numbers of events, and an array ebins[1..nbins] containing the expected numbers of events, and given the number of constraints knstrn (normally one), this routine returns (trivially) the number of degrees of freedom df, and (nontrivially) the chi-square chsq and the significance prob. A small value of prob indicates a significant difference between the distributions bins and ebins. Note that bins and ebins are both float arrays, although bins will normally contain integer values. { float gammq(float a, float x); void nrerror(char error_text[]); int j; float temp; *df=nbins-knstrn; *chsq=0.0; for (j=1;j<=nbins;j++) { if (ebins[j] <= 0.0) nrerror("Bad expected number in chsone"); temp=bins[j]-ebins[j]; *chsq += temp*temp/ebins[j]; } *prob=gammq(0.5*(*df),0.5*(*chsq)); Chi-square probability function. See §6.2. }
622 Chapter 14.Statistical Description of Data Next we consider the case of comparing two binned data sets.Let Ri be the number of events in bin i for the first data set,S;the number of events in the same bin i for the second data set.Then the chi-square statistic is x2=∑ (R-S)2 (14.3.2) Ri+Si Comparing (14.3.2)to (14.3.1),you should note that the denominator of (14.3.2)is not just the average of Ri and Si(which would be an estimator of n;in 14.3.1) Rather,it is twice the average,the sum.The reason is that each term in a chi-square 81 sum is supposed to approximate the square of a normally distributed quantity with unit variance.The variance of the difference of two normal quantities is the sum of their individual variances,not the average. 茶 If the data were collected in such a way that the sum of the Ri's is necessarily equal to the sum of Si's,then the number of degrees of freedom is equal to one less than the number of bins,NB-1 (that is,knstrn =1),the usual case.If this requirement were absent,then the number of degrees of freedom would be N B. Example:A birdwatcher wants to know whether the distribution of sighted birds as a function of species is the same this year as last.Each bin corresponds to one species.If the birdwatcher takes his data to be the first 1000 birds that he saw in 33 Press. each year,then the number of degrees of freedom is NB-1.If he takes his data to be all the birds he saw on a random sample of days,the same days in each year,then the number of degrees of freedom is NB(knstrn =0).In this latter case,note that he is also testing whether the birds were more numerous overall in one year or the other:That is the extra degree of freedom.Of course,any additional constraints on the data set lower the number of degrees of freedom (i.e.,increase knstrn to more 61 positive values)in accordance with their number. The program is void chstwo(float binsi[],float bins2[],int nbins,int knstrn,float *df, float *chsq,float *prob) Given the arrays bins1[1..nbins]and bins2[1..nbins],containing two sets of binned data,and given the number of constraints knstrn (normally 1 or 0),this routine returns the Fuunrggoirioh Numerical Recipes 10621 number of degrees of freedom df,the chi-square chsq,and the significance prob.A small value of prob indicates a significant difference between the distributions bins1 and bins2.Note that 43106 bins1 and bins2 are both float arrays,although they will normally contain integer values. float gammq(float a,float x); int j; (outside float temp; Software. *df=nbins-knstrn; *chsq=0.0; for (j=1;j<=nbins;j++) if(b1ns1[j]=0.0&&bins2[j]==0.0) --(*df); No data means one less degree of free- else dom. temp=bins1[j]-bins2[j]; *chsq +temp*temp/(bins1[j]+bins2[j]); *prob=gammq(0.5*(*df),0.5*(*chsq)); Chi-square probability function.See 86.2
622 Chapter 14. Statistical Description of Data Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). Next we consider the case of comparing two binned data sets. Let Ri be the number of events in bin i for the first data set, Si the number of events in the same bin i for the second data set. Then the chi-square statistic is χ2 = i (Ri − Si)2 Ri + Si (14.3.2) Comparing (14.3.2) to (14.3.1), you should note that the denominator of (14.3.2) is not just the average of Ri and Si (which would be an estimator of ni in 14.3.1). Rather, it is twice the average, the sum. The reason is that each term in a chi-square sum is supposed to approximate the square of a normally distributed quantity with unit variance. The variance of the difference of two normal quantities is the sum of their individual variances, not the average. If the data were collected in such a way that the sum of the Ri’s is necessarily equal to the sum of Si’s, then the number of degrees of freedom is equal to one less than the number of bins, NB − 1 (that is, knstrn = 1), the usual case. If this requirement were absent, then the number of degrees of freedom would be N B. Example: A birdwatcher wants to know whether the distribution of sighted birds as a function of species is the same this year as last. Each bin corresponds to one species. If the birdwatcher takes his data to be the first 1000 birds that he saw in each year, then the number of degrees of freedom is N B − 1. If he takes his data to be all the birds he saw on a random sample of days, the same days in each year, then the number of degrees of freedom is N B (knstrn = 0). In this latter case, note that he is also testing whether the birds were more numerous overall in one year or the other: That is the extra degree of freedom. Of course, any additional constraints on the data set lower the number of degrees of freedom (i.e., increase knstrn to more positive values) in accordance with their number. The program is void chstwo(float bins1[], float bins2[], int nbins, int knstrn, float *df, float *chsq, float *prob) Given the arrays bins1[1..nbins] and bins2[1..nbins], containing two sets of binned data, and given the number of constraints knstrn (normally 1 or 0), this routine returns the number of degrees of freedom df, the chi-square chsq, and the significance prob. A small value of prob indicates a significant difference between the distributions bins1 and bins2. Note that bins1 and bins2 are both float arrays, although they will normally contain integer values. { float gammq(float a, float x); int j; float temp; *df=nbins-knstrn; *chsq=0.0; for (j=1;j<=nbins;j++) if (bins1[j] == 0.0 && bins2[j] == 0.0) --(*df); No data means one less degree of freeelse { dom. temp=bins1[j]-bins2[j]; *chsq += temp*temp/(bins1[j]+bins2[j]); } *prob=gammq(0.5*(*df),0.5*(*chsq)); Chi-square probability function. See §6.2. }
14.3 Are Two Distributions Different? 623 Equation (14.3.2)and the routine chstwo both apply to the case where the total number of data points is the same in the two binned sets.For unequal numbers of data points,the formula analogous to (14.3.2)is X2= (VS/RR-VR/SS) (14.3.3) R1+S: where 三 R≡∑R: S=∑S (14.3.4) are the respective numbers of data points.It is straightforward to make the 公 corresponding change in chstwo. Kolmogorov-Smirnov Test 令 2 The Kolmogorov-Smirnov (or K-S)test is applicable to unbinned distributions that are functions of a single independent variable,that is,to data sets where each data point can be associated with a single number(lifetime of each lightbulb when Press. it burns out,or declination of each star).In such cases,the list of data points can be easily converted to an unbiased estimator S()of the cumulative distribution function of the probability distribution from which it was drawn:If the N events are located at values i,i=1,...,N,then SN()is the function giving the fraction of data points to the left of a given value z.This function is obviously constant SCIENTIFIC( between consecutive(i.e.,sorted into ascending order)zi's,and jumps by the same constant 1/N at each xi.(See Figure 14.3.1.) 61 Different distribution functions,or sets of data,give different cumulative distribution function estimates by the above procedure.However,all cumulative distribution functions agree at the smallest allowable value of z(where they are zero),and at the largest allowable value of(where they are unity).(The smallest and largest values might of course be too.)So it is the behavior between the largest 10.621 and smallest values that distinguishes distributions. Numerica One can think of any number of statistics to measure the overall difference 431 between two cumulative distribution functions:the absolute value ofthe area between (outside Recipes them,for example.Or their integrated mean square difference.The Kolmogorov- Smirnov D is a particularly simple measure:It is defined as the maximum value of the absolute difference between two cumulative distribution functions.Thus, North for comparing one data set's SN()to a known cumulative distribution function P(x),the K-S statistic is D=-aISN()-P川 (14.3.5) while for comparing two different cumulative distribution functions SN,(x)and SNa(x),the K-S statistic is D=-xISN,(e)-SN(z川 (14.3.6)
14.3 Are Two Distributions Different? 623 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). Equation (14.3.2) and the routine chstwo both apply to the case where the total number of data points is the same in the two binned sets. For unequal numbers of data points, the formula analogous to (14.3.2) is χ2 = i ( S/RRi − R/SSi)2 Ri + Si (14.3.3) where R ≡ i Ri S ≡ i Si (14.3.4) are the respective numbers of data points. It is straightforward to make the corresponding change in chstwo. Kolmogorov-Smirnov Test The Kolmogorov-Smirnov (or K–S) test is applicable to unbinned distributions that are functions of a single independent variable, that is, to data sets where each data point can be associated with a single number (lifetime of each lightbulb when it burns out, or declination of each star). In such cases, the list of data points can be easily converted to an unbiased estimator SN (x) of the cumulative distribution function of the probability distribution from which it was drawn: If the N events are located at values xi, i = 1,...,N, then SN (x) is the function giving the fraction of data points to the left of a given value x. This function is obviously constant between consecutive (i.e., sorted into ascending order) xi’s, and jumps by the same constant 1/N at each xi. (See Figure 14.3.1.) Different distribution functions, or sets of data, give different cumulative distribution function estimates by the above procedure. However, all cumulative distribution functions agree at the smallest allowable value of x (where they are zero), and at the largest allowable value of x (where they are unity). (The smallest and largest values might of course be ±∞.) So it is the behavior between the largest and smallest values that distinguishes distributions. One can think of any number of statistics to measure the overall difference between two cumulative distribution functions: the absolute value of the area between them, for example. Or their integrated mean square difference. The KolmogorovSmirnov D is a particularly simple measure: It is defined as the maximum value of the absolute difference between two cumulative distribution functions. Thus, for comparing one data set’s SN (x) to a known cumulative distribution function P(x), the K–S statistic is D = max −∞<x<∞ |SN (x) − P(x)| (14.3.5) while for comparing two different cumulative distribution functions S N1 (x) and SN2 (x), the K–S statistic is D = max −∞<x<∞ |SN1 (x) − SN2 (x)| (14.3.6)
624 Chapter 14.Statistical Description of Data Sv(x) read able files 人 P(x) http://www.nr.com or call 1-800-872- (including this one) to any server computer, -7423 (North America tusers to make one paper 1988-1992 by Cambridge University Press. from NUMERICAL RECIPES IN C: THE Figure 14.3.1.Kolmogorov-Smirnov statistic D.A measured distribution of values in x (shown only). as N dots on the lower abscissa)is to be compared with a theoretical distribution whose cumulative ictly proh Programs probability distribution is plotted as P().A step-function cumulative probability distribution ()is constructed,one that rises an equal amount at each measured point.D is the greatest distance between the two cumulative distributions. What makes the K-S statistic useful is that its distribution in the case of the null to dir hypothesis(data sets drawn from the same distribution)can be calculated,at least to useful approximation,thus giving the significance of any observed nonzero value of ART OF SCIENTIFIC COMPUTING (ISBN D.A central feature of the K-S test is that it is invariant under reparametrization 1988-19920 of r;in other words,you can locally slide or stretch the z axis in Figure 14.3.1, and the maximum distance D remains unchanged.For example,you will get the same significance using x as using log r. 10-521 The function that enters into the calculation of the significance can be written as the following sum: Numerical Recipes 43198-5 Qks(A)=2 -1-1e-2212 (14.3.7) (outside j=1 North Software. which is a monotonic function with the limiting values QKs(0)=1QKs(∞)=0 (14.3.8) visit website machine In terms of this function,the significance level of an observed value of D(as a disproof of the null hypothesis that the distributions are the same)is given approximately [1]by the formula Probability(D>observed)=Qks(N+.12+0.11/N D 14.3.9
624 Chapter 14. Statistical Description of Data Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). x x D P(x) SN (x) cumulative probability distribution Figure 14.3.1. Kolmogorov-Smirnov statistic D. A measured distribution of values in x (shown as N dots on the lower abscissa) is to be compared with a theoretical distribution whose cumulative probability distribution is plotted as P(x). A step-function cumulative probability distribution SN (x) is constructed, one that rises an equal amount at each measured point. D is the greatest distance between the two cumulative distributions. What makes the K–S statistic useful is that its distribution in the case of the null hypothesis (data sets drawn from the same distribution) can be calculated, at least to useful approximation, thus giving the significance of any observed nonzero value of D. A central feature of the K–S test is that it is invariant under reparametrization of x; in other words, you can locally slide or stretch the x axis in Figure 14.3.1, and the maximum distance D remains unchanged. For example, you will get the same significance using x as using log x. The function that enters into the calculation of the significance can be written as the following sum: QKS(λ)=2∞ j=1 (−1)j−1 e−2j2λ2 (14.3.7) which is a monotonic function with the limiting values QKS(0) = 1 QKS(∞)=0 (14.3.8) In terms of this function, the significance level of an observed value of D (as a disproof of the null hypothesis that the distributions are the same) is given approximately [1] by the formula Probability (D > observed ) = QKSNe + 0.12 + 0.11/ Ne D (14.3.9)