Chapter 7 Pearson’s chi-square test

Chapter 7 Pearson s chi-square Null hypothesis asymptoticsLetX1,X2, be independent from a multinomial(1,p) distribution, wherepis ak-vectorwith nonnegative entries that sum to one. That is,P(Xij= 1) = 1 P(Xij= 0) =pjfor all 1 j k( )and eachXiconsists of exactlyk 1 zeros and a single one, where the one is in the componentof the success category at triali. Note that the multinomial distribution is a generalizationof the binomial distribution to the case in which there arekcategories of outcome insteadof only purpose of this section is to derive the asymptotic distribution of the Pearson chi-squarestatistic 2=k j=1(nj npj)2npj,( )wherenjis the random variablenXj, the number of successes in thejth category for trials1, .. , n. In a real application, the true value ofpis not known, but instead we assumethatp=p0for some null valuep0. We will show that 2converges in distribution to thechi-square distribution onk 1 degrees of freedom, which yields to the familiar chi-squaretest of goodness of fit for a multinomial ( ) implies that VarXij=pj(1 pj).

sample is multivariate normal, then [(n−k)/(nk−k)]T2 is distributed as F k,n−k. A Pearson chi square statistic may be shown to be a special case of Hotelling’s T2. ] (a) You may assume that S−1 n →P Σ−1 (this follows from the weak law of large numbers since P(S n is nonsingular) → 1). Prove that under the null hypothesis, T2 ...

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  Normal, Multivariate, Multivariate normal

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