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# hw8sol - 36-226 Summer 2010 Homework 8 Solutions 1(a For...

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36-226 Summer 2010 Homework 8 Solutions 1. (a) For the candidate to have more than 18500 valid signatures, it is necessary that a fraction of 18500 19850 = 0 . 932 of the signatures are valid. Thus, we set p = true proportion of signatures on the petition that are valid H 0 : p 0 . 932 H 1 : p > 0 . 932 Following the discussion we had in class, we can test this equivalently as H 0 : p = 0 . 932 H 1 : p > 0 . 932 NOTE: I don’t care which way you set up the hypotheses as long as you do the rest correctly. This is the way I chose to do it because the problem says that we are interested in testing whether more than 18500 of the signatures are correct. Thus, I feel that the the generally accepted value is that that is not the case. (b) Type I error: does not have enough signatures, but remains on the ballot. Type II error: the candidate has enough signatures but he is removed from the ballot. (c) Since n = 18500 is very large, we can invoke the CLT and use the sample for the large- sample test for a population proportion. Test statistic: Z = b p - 0 . 932 q b p (1 - b p ) n . Since the alternative is that the true proportion is greater than 0.932, we reject for large values of Z . To find the test of size 0 . 05, we need to solve the following equation: α = 0 . 05 = P H 0 (reject H 0 ) = P p =0 . 932 ( Z > k ) where Z N(0 , 1) under H 0 By the definition of the quantiles, we have k = z

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