Topic 19: Goodness of Fit

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1 Topic 19: November 24, 2009 A goodness of fit test examine the case of a sequence if independent experiments each of which can have 1 of k possible outcomes. In terms of hypothesis testing, let π = (π 1,..., π k ) be postulated values of the probability P π {experiment takes on the i-th outcome} = π i and let p = (p 1,..., p m ) denote the actual state of nature. Then, the parameter space is Θ = {p = (p 1,..., p m ); p i 0 for all i = 1,..., k, p i = 1}. The hypothesis test is H 0 : p i = π i, for all i = 1,..., m versus H 1 : p i π i, for some i = 1,..., m, The data x = (x 1,..., x n ) is the outcome of the n experiments. Set n i = #{j; x j = i} to be the number of times the outcome i occurs in the data. The likelihood function L(p n) = p n1 1 pnm m. Its logarithm ln L(p n) = n i ln p i. We maximize this using the method of Lagrange multipliers with constraint s(p) = Thus, at the maximum likelihood estimator (ˆp 1,..., ˆp m ), So, n i /ˆp i = λ, n i = λˆp i. Now sum on i to obtain p i = 1. p ln L(ˆp n) = λ ˆp s(p). ( n1,..., n ) m = λ(1,..., 1) ˆp 1 ˆp m n i = λ ˆp i and n = λ. 119

2 Consequently, The likelihood ratio test Λ(n) = L(n π) L(n ˆp) = Recall that as the number of experiments n, n 1 ˆp i = n and ˆp i = n i n. ( ) n1 ( ) nk nπ1 nπk. n 1 2 ln Λ n (N) = 2 n k ln nπ i converges to a χ 2 k 1 random variable. Here N = (N 1,..., N k ) is the observed number of occurrences of outcome i. The traditional method was introduced between 1895 and 1900 by Karl Pearson and consequenttly has been in use for longer that the idea of likelihood ratio tests. To show the connection between the two tests, recall that ln a (a 1) 1 (a 1)2 2 is the quadratic Taylor polynomial approximation of ln a. Apply this to the logarithm of the likelihood ratio, we find that 2 ln Λ n (N) = 2 = 2 = 0 + ( (nπi ) (nπ i ) + (nπ i ) 2 ( ) ) 2 nπi 1 ( ) 2 nπi 1 The is generally rewritten by writing O i = to be the number of observed occurrences of i and E i = nπ i to be the number of expected occurrences of i as given by H 0. The data can be stored in a table Then, (nπ i ) 2 i 1 2 k observed O 1 O 2 O k expected E 1 E 2 E k (nπ i ) 2 nπ i (O i E i ) 2 E i. Example 1 (Hardy-Weinberg equilibrium). The two allele Hardy Weinberg principle states that after one generation of random mating the genotypic frequencies can be represented by a binomial distribution. So, if a population is segregating for two alleles A 1 and A 2 at an autosomal locus with frequencies p 1 and p 2. Then, random mating would give a proportion p 11 = p 2 1 for the A 1 A 1 genotype, p 12 = 2p 1 p 2 for the A 1 A 2 genotype, and p 22 = p 2 2 for the A 2 A 2 genotype. Then, p 1 = p p 12 p 2 = p p 12. This give us 5 parameters: p 1, p 2, p 11, p 12, p 22, plus the fact p 1 + p 2 = p 11 + p 12 + p 22 = 1. Thus n = 3. The two restrictions from the Handy-Weinberg equilibrium above give k =

3 McDonald et al. (1996) examined variation at the CVJ5 locus in the American oyster, Crassostrea virginica. There were two alleles, L and S, and the genotype frequencies in Panacea, Florida were 14 LL, 21 LS, and 25 SS. So, ˆp 11 = 14 60, ˆp 12 = 21 60, ˆp 22 = So, the estimate of p 1 and p 2 are ˆp 1 = , ˆp 2 = So, the expected number of observations is E 11 = 60ˆp 2 1 = , E 12 = 60 2ˆp 1 ˆp 2 = , E 22 = 60ˆp 2 2 = The chi-square statistic The p-value (14 10)2 10 > 1-pchisq(4.569,1) [1] (21 29) (25 21)2 21 = = Contingency tables For an r c contingency table, we consider two classifications for an experiment. Thus, we can partition the outcome of each experiment into two groups: A 1,... A c and B 1,... B r. Here, we write O ij to denote the number of occurences of the outcome A i B j and organize the results in a two-way table. A 1 A 2 A c total B 1 O 11 O 12 O 1c O 1 B 2 O 21 O 22 O 2c O B r O r1 O r2 O rc O r total O 1 O 2 O c n The null hypothesis is that the classifications A and B are independent. To set the parameter space for this model, we have Θ = {p = p ij, 1 i r, 1 j c); p ij 0 for all i, j = 1, p ij=1 }. Write The hypothesis test is p i = p ij and p j = j=1 p ij. j=1 H 0 : p ij = p i p j, for all i, j versus H 1 : p ij p i p j, for some i, j. Follow the procedure as before for the goodness of fit test to end with the test statistic 2 j=1 O ij ln E ij O ij j=1 (O ij E ij ) 2 E ij. 121

4 or The null hypothesis p ij = p i p j can be written in terms of observed and expected observations as E ij n = O i O j n n. E ij = O i O j /n. Example 2. We have the following data on malaria in three regions of the world. The expected table is To compute the chi-square statistic South Asia America Africa Total Malaria Type A Malaria Type B Malaria Type C Total South Asia America Africa Total Malaria Type A Malaria Type B Malaria Type C Total ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = The degrees of freedom are (c 1)(r 1) = 2 2 = 4. In R, we have > malaria<-matrix(c(31,2,53,45,53,2,14,5,45),nrow=3,ncol=3) > malaria [,1] [,2] [,3] [1,] [2,] [3,] > chisq.test(malaria) Pearson s Chi-squared test data: malaria X-squared = , df = 4, p-value < 2.2e

5 2 Applicability of Chi-squared Tests The chi-square test uses the central limit theorem and so is based on the ability to us a normal approximation. On criterion, the Cochran conditions requires no cell has count zero, and more than 80% of the cells have counts at least 5. If this does not hold, then Fisher s exact test uses the hypergeometric distribution directly rather than normal approximation. 123

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