hw7sol.pdf - STAT244 | HW7 solutions | Due Total 100 pts[5...

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STAT244 | HW7 solutions | Due 2/3/2017 Total: 100 pts. Notation : 1. If n 2 N , i.e., n is an integer, [ n ] denotes 1 , 2 , . . . , n - 1 , n (sometimes denoted as 1 , n ); 2. l ( | x ) will denote the log-likelihood for parameter and data x ; 3. ˆ MLE stands for the MLE of ; 4. df stands for degrees of freedom. Problem 1 [20 pts] Call “This, it, thus, and” class I words; class II is “everything else”. For each of N = 215 groups of n = 5 of James Mill’s sentences, the number of class I words was counted. # of class I words 0 1 2 3 4 5 # of groups 87 11 51 42 20 4 Test whether a Bin ( n, ) fits these data. Solution : first we find ˆ MLE . Let x = ( x 1 , . . . , x N ) denote the counts of class I words within given groups. l ( | x ) = N X i =1 log ( 5 x i ) + log · N X i =1 x i + log(1 - ) N X i =1 ( n - x i ) @ l ( | x ) = N ¯ x - n - ¯ x 1 - = 0 when = ˆ MLE , ¯ x n @ 2 ✓✓ l ( | x ) = - N ¯ x 2 + n - ¯ x (1 - ) 2 0 8 We have 87 0 + 11 1 + 2 51 + 3 42 + 4 20 + 5 4 = 339 sentences with class I words in total. Thus, ˆ MLE = 339 215 / 5 . 315 . [5 pts ] # of sentences with class I words, i 0 1 2 3 4 5 Observed, O i 87 11 51 42 20 4 Expected, E i 32.35 74.5 68.61 31.59 7.27 .67 [5 pts] Under the null hypothesis, the test statistics is X = 5 X i =0 ( E i - O i ) 2 E i 193 . 238 with df = 6 - 1 - 1 = 4 [5 pts] The p -value is nearly 0 , so we have strong evidence to reject the null hypothesis, i.e., we conclude that the count of sentences with class I words does not fit a binomial distribution. [5 pts] Problem 2 [15 pts] The members of a community are classified by Blood type: 0 A B AB Total 121 120 79 33 353 Theory has it that the probabilities of those types depend on gene frequency param- eters r, p, q , where r + p + q = 1 , P { ”0” } = r 2 , P { A } = p 2 +2 pr , P { B } = q 2 +2 qr and P { AB } = 2 pq . Using numerical methods (see Chapter 5), we can get ˆ r MLE = . 580 , ˆ p MLE = . 246 and ˆ q MLE = . 173 .
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