1 CHI-SQUARE: TESTING FOR GOODNESS OF FIT In the previous chapter we discussed procedures for fitting a hypothesized function to a set of experimental data points. Such procedures involve minimizing a quantity we called Φ in order to determine best estimates for certain function parameters, such as (for a straight line) a slope and an intercept. Φ is proportional to (or in some cases equal to) a statistical measure called χ 2, or chi-square, a quantity commonly used to test whether any given data are well described by some hypothesized function. Such a determination is called a chi-square test for goodness of fit. In the following, we discuss χ 2 and its statistical distribution, and show how it can be used as a test for goodness of fit. 1 The definition of χ 2 If ν independent variables x i are each normally distributed with mean µ i and variance σ 2 i, then the quantity known as chi-square 2 is defined by χ 2 (x 1 µ 1 ) 2 σ (x 2 µ 2 ) 2 σ (x ν µ ν ) 2 σ 2 ν = ν (x i µ i ) 2 σ 2 i=1 i (1) Note that ideally, given the random fluctuations of the values of x i about their mean values µ i, each term in the sum will be of order unity. Hence, if we have chosen the µ i and the σ i correctly, we may expect that a calculated value of χ 2 will be approximately equal to ν. If it is, then we may conclude that the data are well described by the values we have chosen for the µ i, that is, by the hypothesized function. If a calculated value of χ 2 turns out to be much larger than ν, and we have correctly estimated the values for the σ i, we may possibly conclude that our data are not welldescribed by our hypothesized set of the µ i. This is the general idea of the χ 2 test. In what follows we spell out the details of the procedure. 1 The chi-square distribution and some examples of its use as a statistical test are also described in the references listed at the end of this chapter. 2 The notation of χ 2 is traditional and possibly misleading. It is a single statistical variable, and not the square of some quantity χ. It is therefore not chi squared, but chi-square. The notation is merely suggestive of its construction as the sum of squares of terms. Perhaps it would have been better, historically, to have called it ξ or ζ. 4 1
2 4 2 Chi-square: Testing for goodness of fit The χ 2 distribution The quantity χ 2 defined in Eq. 1 has the probability distribution given by f(χ 2 ) = 1 2 ν/2 Γ(ν/2) e χ2 /2 (χ 2 ) (ν/2) 1 (2) This is known as the χ 2 -distribution with ν degrees of freedom. ν is a positive integer. 3 Sometimes we write it as f(χ 2 ν) when we wish to specify the value of ν. f(χ 2 ) d(χ 2 ) is the probability that a particular value of χ 2 falls between χ 2 and χ 2 + d(χ 2 ). Here are graphs of f(χ 2 ) versus χ 2 for three values of ν: Figure 1 The chi-square distribution for ν = 2, 4, and 10. Note that χ 2 ranges only over positive values: 0 < χ 2 <. The mean value of χ 2 ν is equal to ν, and the variance of χ 2 ν is equal to 2ν. The distribution is highly skewed for small values of ν, and becomes more symmetric as ν increases, approaching a Gaussian distribution for large ν, just as predicted by the Central Limit Theorem. 3 Γ(p) is the Gamma function, defined by Γ(p + 1) 0 x p e x dx. It is a generalization of the factorial function to non-integer values of p. If p is an integer, Γ(p+1) = p!. In general, Γ(p+1) = pγ(p), and Γ(1/2) = π.
3 Chi-square: Testing for goodness of fit 4 3 How to use χ 2 to test for goodness of fit Suppose we have a set of N experimentally measured quantities x i. We want to test whether they are well-described by some set of hypothesized values µ i. We form a sum like that shown in Eq. 1. It will contain N terms, constituting a sample value for χ 2. In forming the sum, we must use estimates for the σ i that are independently obtained for each x i. 4 Now imagine, for a moment, that we could repeat our experiment many times. Each time, we would obtain a data sample, and each time, a sample value for χ 2. If our data were well-described by our hypothesis, we would expect our sample values of χ 2 to be distributed according to Eq. 2, and illustrated by example in Fig. 1. However, we must be a little careful. The expected distribution of our samples of χ 2 will not be one of N degrees of freedom, even though there are N terms in the sum, because our sample variables x i will invariably not constitute a set of N independent variables. There will, typically, be at least one, and often as many as three or four, relations connecting the x i. Such relations are needed in order to make estimates of hypothesized parameters such as the µ i, and their presence will reduce the number of degrees of freedom. With r such relations, or constraints, the number of degrees of freedom becomes ν = N r, and the resulting χ 2 sample will be one having ν (rather than N) degrees of freedom. As we repeat our experiment and collect values of χ 2, we expect, if our model is a valid one, that they will be clustered about the median value of χ 2 ν, with about half of these collected values being greater than the median value, and about half being less than the median value. This median value, which we denote by χ 2 ν,0.5, is determined by f(χ 2 ) dχ 2 = 0.5 χ 2 ν,0.5 Note that because of the skewed nature of the distribution function, the median value of χ 2 ν will be somewhat less than the mean (or average) value of χ 2 ν, which as we have noted, is equal to ν. For example, for ν = 10 degrees of freedom, χ 2 10, , a number slightly less than 10. Put another way, we expect that a single measured value of χ 2 will have a probability of 0.5 of being greater than χ 2 ν, In the previous chapter, we showed how a hypothesized function may be fit to a set of data points. There we noted that it may be either impossible or inconvenient to make independent estimates of the σ i, in which case estimates of the σ i can be made only by assuming an ideal fit of the function to the data. That is, we assumed χ 2 to be equal to its mean value, and from that, estimated uncertainties, or confidence intervals, for the values of the determined parameters. Such a procedure precludes the use of the χ 2 test.
4 4 4 Chi-square: Testing for goodness of fit We can generalize from the above discussion, to say that we expect a single measured value of χ 2 will have a probability α ( alpha ) of being greater than χ 2 ν,α, where χ 2 ν,α is defined by χ 2 ν,α f(χ 2 ) dχ 2 = α This definition is illustrated by the inset in Fig. 2 on page 4 9. Here is how the χ 2 test works: (a) We hypothesize that our data are appropriately described by our chosen function, or set of µ i. This is the hypothesis we are going to test. (b) From our data sample we calculate a sample value of χ 2 (chi-square), along with ν (the number of degrees of freedom), and so determine χ 2 /ν (the normalized chi-square, or the chi-square per degree of freedom) for our data sample. (c) We choose a value of the significance level α (a common value is.05, or 5 per cent), and from an appropriate table or graph (e.g., Fig. 2), determine the corresponding value of χ 2 ν,α/ν. We then compare this with our sample value of χ 2 /ν. (d) If we find that χ 2 /ν > χ 2 ν,α/ν, we may conclude that either (i) the model represented by the µ i is a valid one but that a statistically improbable excursion of χ 2 has occurred, or (ii) that our model is so poorly chosen that an unacceptably large value of χ 2 has resulted. (i) will happen with a probability α, so if we are satisfied that (i) and (ii) are the only possibilities, (ii) will happen with a probability 1 α. Thus if we find that χ 2 /ν > χ 2 ν,α/ν, we are 100 (1 α) per cent confident in rejecting our model. Note that this reasoning breaks down if there is a possibility (iii), for example if our data are not normally distributed. The theory of the chi-square test relies on the assumption that chi-square is the sum of the squares of random normal deviates, that is, that each x i is normally distributed about its mean value µ i. However for some experiments, there may be occasional non-normal data points that are too far from the mean to be real. A truck passing by, or a glitch in the electrical power could be the cause. Such points, sometimes called outliers, can unexpectedly increase the sample value of chi-square. It is appropriate to discard data points that are clearly outliers. (e) If we find that χ 2 is too small, that is, if χ 2 /ν < χ 2 ν,1 α /ν, we may conclude only that either (i) our model is valid but that a statistically improbable excursion of χ 2 has occurred, or (ii) we have, too conservatively, over-estimated the values of σ i, or (iii) someone has given us fraudulent data, that is, data too good to be true. A too-small value of χ 2 cannot be indicative of a poor model. A poor model can only increase χ 2.
5 Chi-square: Testing for goodness of fit 4 5 Generally speaking, we should be pleased to find a sample value of χ 2 /ν that is near 1, its mean value for a good fit. In the final analysis, we must be guided by our own intuition and judgment. The chi-square test, being of a statistical nature, serves only as an indicator, and cannot be iron clad. An example The field of particle physics provides numerous situations where the χ 2 test can be applied. A particularly simple example 5 involves measurements of the mass M Z of the Z 0 boson by experimental groups at CERN. The results of measurements of M Z made by four different detectors (L3, OPAL, Aleph and Delphi) are as follows: Detector Mass in GeV/c 2 L ± OPAL ± Aleph ± Delphi ± The listed uncertainties are estimates of the σ i, the standard deviations for each of the measurements. The figure below shows these measurements plotted on a horizontal mass scale (vertically displaced for clarity). Measurements of the Z 0 boson. The question arises: Can these data be well described by a single number, namely an estimate of M Z made by determining the weighted mean of the four measurements? 5 This example is provided by Pat Burchat.
6 4 6 Chi-square: Testing for goodness of fit to find We find the weighted mean M Z, and its standard deviation σ M Z Mi /σi 2 M Z = 1/σ 2 and σ 2 = 1 MZ i 1/σ 2 i like this:6 M Z ± σ M Z = ± Then we form χ 2 : χ 2 = 4 (M i M Z ) 2 σ 2 i=1 i 2.78 We expect this value of χ 2 to be drawn from a chi-square distribution with 3 degrees of freedom. The number is 3 (not 4) because we have used the mean of the four measurements to estimate the value of µ, the true mass of the Z 0 boson, and this uses up one degree of freedom. Hence χ 2 /ν = 2.78/ Now from the graph of α versus χ 2 /ν shown in Fig. 2, we find that for 3 degrees of freedom, α is about 0.42, meaning that if we were to repeat the experiments we would have about a 42 per cent chance of finding a χ 2 for the new measurement set larger than 2.78, assuming our hypothesis is correct. We have therefore no good reason to reject the hypothesis, and conclude that the four measurements of the Z 0 boson mass are consistent with each other. We would have had to have found χ 2 in the vicinity of 8.0 (leading to an α of about 0.05) to have been justified in suspecting the consistency of the measurements. The fact that our sample value of χ 2 /3 is close to 1 is reassuring. Using χ 2 to test hypotheses regarding statistical distributions The χ 2 test is used most commonly to test the nature of a statistical distribution from which some random sample is drawn. It is this kind of application that is described by Evans in his text, and is the kind of application for which the χ 2 test was first formulated. Situations frequently arise where data can be classified into one of k classes, with probabilities p 1, p 2,..., p k of falling into each class. If all the data are accounted for, pi = 1. Now suppose we take data by classifying it: We count the number of observations falling into each of the k classes. We ll have n 1 in the first class, n 2 in the second, and so on, up to n k in the k th class. We suppose there are a total of N observations, so n i = N. 6 These expressions are derived on page 2 12.
7 Chi-square: Testing for goodness of fit 4 7 It can be shown by non-trivial methods that the quantity (n 1 Np 1 ) 2 Np 1 + (n 2 Np 2 ) 2 Np (n k Np k ) 2 Np k = k (n i Np i ) 2 (3) Np i has approximately the χ 2 distribution with k r degrees of freedom, where r is the number of constraints, or relations used to estimate the p i from the data. r will always be at least 1, since it must be that n i = Np i = N p i = N. Since Np i is the mean, or expected value of n i, the form of χ 2 given by Eq. 3 corresponds to summing, over all classes, the squares of the deviations of the observed n i from their mean values divided by their mean values. At first glance, this special form looks different from that shown in Eq. 1, since the variance for each point is replaced by the mean value of n i for each point. Such an estimate for the variance makes sense in situations involving counting, where the counted numbers are distributed according to the Poisson distribution, for which the mean is equal to the variance: µ = σ 2. Equation 3 forms the basis of what is sometimes called Pearson s Chi-square Test. Unfortunately some authors (incorrectly) use this equation to define χ 2. However, this form is not the most general form, in that it applies only to situations involving counting, where the data variables are dimensionless. Another example Here is an example in which the chi-square test is used to test whether a data sample consisting of the heights of 66 women can be assumed to be drawn from a Gaussian distribution. 7 We first arrange the data in the form of a frequency distribution, listing for each height h, the value of n(h), the number of women in the sample whose height is h (h is in inches): h n(h) We make the hypothesis that the heights are distributed according to the Gaussian distribution (see page 2 6 of this manual), namely that the probability p(h) dh that a height falls between h and h + dh is given by p(h) dh = 1 σ 2 /2σ 2π e (h µ) 2 dh i=1 7 This sample was collected by Intermediate Laboratory students in 1991.
8 4 8 Chi-square: Testing for goodness of fit This expression, if multiplied by N, will give, for a sample of N women, the number of women n th (h) dh theoretically expected to have a height between h and h + dh: n th (h) dh = N σ 2 /2σ 2π e (h µ) 2 dh (4) In our example, N = 66. Note that we have, in our table of data above, grouped the data into bins (we ll label them with the index j), each of size 1 inch. A useful approximation to Eq. 4, in which dh is taken to be 1 inch, gives the expected number of women n th (j) having a height h j : n th (j) = N σ 2 2π e (h j µ) /2σ 2 (5) Now the sample mean h and the sample standard deviation s are our best estimates of µ and σ. We find, calculating from the data: h = 64.9 inches, and s = 2.7 inches Using these values we may calculate, from Eq. 5, the number expected in each bin, with the following results: h n th (j) In applying the chi-square test to a situation of this type, it is advisable to re-group the data into new bins (classes) such that the expected number occurring in each bin is greater than 4 or 5; otherwise the theoretical distributions within each bin become too highly skewed for meaningful results. Thus in this situation we shall put all the heights of 61 inches or less into a single bin, and all the heights of 69 inches or more into a single bin. This groups the data into a total of 9 bins (or classes), with actual numbers and expected numbers in each bin being given as follows (note the bin sizes need not be equal): h n(h) n th (j) Now we calculate the value of χ 2 using these data, finding χ 2 = (6 6.5) (6 5.5) (6 5.8)2 5.8 = 6.96 Since we have grouped our data into 9 classes, and since we have used up three degrees of freedom by demanding (a) that the sum of the n j be equal to N, (b) that
9 Chi-square: Testing for goodness of fit 4 9 the mean of the hypothesized distribution be equal to the sample mean, and (c) that the variance of the hypothesized distribution be equal to the sample variance, there are 6 degrees of freedom left. 8 Hence χ 2 /ν = 6.96/6 1.16, leading to an α of about Therefore we have no good reason to reject our hypothesis that our data are drawn from a Gaussian distribution function. Figure 2 α versus the normalized chi-square: χ 2 ν/ν. α is the probability that a sample chi-square will be larger than χ 2 ν, as shown in the inset. Each curve is labeled by ν, the number of degrees of freedom. References 1. Press, William H. et. al., Numerical Recipes in C The Art of Scientific Computing, 2nd Ed. (Cambridge University Press, New York, 1992). Press devotes considerable discussion to the subject of fitting parameters to data, including the use of the chi-square test. Noteworthy are his words of advice, appearing on pages : To be genuinely useful, a fitting procedure should provide (i) parameters, (ii) error estimates on the parameters, and (iii) a statistical measure of goodness-of-fit. When the third item suggests that the model is an unlikely match to the data, then items (i) and (ii) are probably worthless. Unfortunately, many practitioners of parameter estimation never proceed beyond item (i). They deem a fit acceptable 8 Note that if we were hypothesizing a Poisson distribution (as in a counting experiment), there would be 7 degrees of freedom (only 2 less than the number of classes). For a Poisson distribution the variance is equal to the mean, so there is only 1 parameter to be determined, not 2.
10 4 10 Chi-square: Testing for goodness of fit if a graph of data and model looks good. This approach is known as chi-by-eye. Luckily, its practitioners get what they deserve. 2. Bennett, Carl A., and Franklin, Norman L., Statistical Analysis in Chemistry and the Chemical Industry (Wiley, 1954). An excellent discussion of the chi-square distribution function, with good examples illustrating its use, may be found on pages 96 and Evans, Robley D., The Atomic Nucleus (McGraw-Hill, 1969). Chapter 27 of this advanced text contains a description of Pearson s Chi-square Test.
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17. SIMPLE LINEAR REGRESSION II The Model In linear regression analysis, we assume that the relationship between X and Y is linear. This does not mean, however, that Y can be perfectly predicted from X.
5 POWER-LAW FITTING AND LOG-LOG GRAPHS She had taken up the idea, she supposed, and made everything bend to it. --- Emma 5.1 DEALING WITH POWER LAWS Although many relationships in nature are linear, some
University of California, Davis Department of Statistics Summer Session II Statistics 13 August 20, 2012 Date of latest update: August 20 Lecture 5: Estimation with Confidence intervals 5.1 Identifying
FACT SHEET Using Excel for inferential statistics Introduction When you collect data, you expect a certain amount of variation, just caused by chance. A wide variety of statistical tests can be applied
PholC60 September 001 DATA INTERPRETATION AND STATISTICS Books A easy and systematic introductory text is Essentials of Medical Statistics by Betty Kirkwood, published by Blackwell at about 14. DESCRIPTIVE
Age to Age Factor Selection under Changing Development Chris G. Gross, ACAS, MAAA Introduction A common question faced by many actuaries when selecting loss development factors is whether to base the selected
The Analysis of Research Data The design of any project will determine what sort of statistical tests you should perform on your data and how successful the data analysis will be. For example if you decide
Chapter 45 Two-Sample T-Tests Allowing Unequal Variance (Enter Difference) Introduction This procedure provides sample size and power calculations for one- or two-sided two-sample t-tests when no assumption