sol5 - PSTAT 120C: Solutions to Assignment # 5 May 23, 2006...

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Unformatted text preview: PSTAT 120C: Solutions to Assignment # 5 May 23, 2006 1. The distribution of the committee compositions is multinomial with probabilities 0.4, 0.4, and 0.2. (a) The probability is P { X 1 = 2 ,X 2 = 2 ,X 3 = 1 } = 5 2 2 1 ( . 4) 2 ( . 4) 2 ( . 2) = 30( . 4) 4 ( . 2) = . 1536 (b) This event is the union of the disjoint events, { 4, 5, or 6 Independents } or { 3 Independents and 2, 1, or 0 Democrats } or { 2 Independents and 1 or 0 Democrats } or { 1 Independent and no Democrats } . One way to do this calculation is to condition on the number of Independents. First, the number of Independents is marginally binomial so P { I = k } = 6 k . 2 k . 8 6- k Independents 6 5 4 3 2 1 Probability 6 . 4 10- 5 0.0015 0.015 0.082 0.246 0.393 If there are 3 Independents, then the distribution of the Democrats is Bin(3 ,. 5). ( p = 1 / 2 because the probability of one of the other three being a Republican or Democrat is the same.) Therefore, P { D = 0 , 1 , 2 | I = 3 } = 1 8 + 3 8 + 3 8 = 7 8 Likewise, if there are two independents then D Bin(4 , 1 / 2) P { D = 0 , 1 | I = 2 } = 1 16 + 1 4 = 5 16 If there is one independent then D Bin(5 , 1 / 2) P { D = 0 | I = 1 } = 1 32 Now we need to add all these things together, multiplying by the marginal probabilities, P { I > D } = P { I = 4 , 5 , 6 } + P { I = 3 } P { D = 0 , 1 , 2 | I = 3 } + + P { I = 2 } P { D = 0 , 1 | I = 2 } + P { I = 1 } P { D = 0 | I = 1 } = 0 . 000064 + 0 . 0015 + 0 . 015 + +( . 082)(7 / 8)+ + ( . 246)(5 / 16) + ( . 393)(1 / 32) = . 017 + . 0718 + . 0769 + . 0123 = 0 . 178 1 2. Here is a table with the expected values and 2 statistics Number of Games 4 5 6 7 Count 11 8 12 24 Expected 6.875 13.75 17.1875 17.1875 ( O- E ) 2 /E 2.475 2.405 1.566 2.70 The X 2 = 9 . 145. Comparing this with the 2 with three degrees of freedom critical value = 7.81. Therefore, there is a significant deviation from proposed model....
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sol5 - PSTAT 120C: Solutions to Assignment # 5 May 23, 2006...

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