Question

1. Derive the exact distribution to test the independence for the 2 x 2 contingency table Response 1 n1 n12 n1+ 2 n21 n22 n2+ +1 m+2 The hypergeometric distribution can be used to find the probability of observing a particular 2x2 table under independence . In the independent binomial model, we observe two random variables, Nu and N21 . We use the observed values n1u and n21 to compare the respective probabilities of success, π1 and π2 . The total number of successes, n+1, tells us nothing about how different π1 and π2 might be, and hence it is not important to the comparison. Thus, there is no harm in "conditioning on" n+ .e., assuming it is known instead of random) In fact, there are mathematical benefits to making this assumption . When H0 : π1-π2 is true, then it can be shown that conditioning on n+1 in the independent binomial model leads to the hypergeometric model The hypergeometric distribution for a 2 × 2 table can be found by starting from two indepen- dent binomial distributions and applying the following steps: Write out the pdf for Ni1 and N21 with binomial distributions involving nn and n21 trials, respectively, and a common probability of success π. Note that n11 and n21 are known constants Find the joint pdf for N1 and N21 Discuss why the pdf for N+1 of success T N1+N21 is binomial with ni1+n21 trials and probability . Calculate the conditional pdf for Nu given N+ as P(N11N+ n n P(NI1 and n+1 11, N2121)/P(N+1+) forw-0,...,n+1 subject to n w11 n2+. Relate the pdf found to a hypergeometric distribution

Answer 1

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