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Unformatted text preview: value. From Table A we find 8106 . ) 88 . ( = ≤ Z P . The probability in each tail is 10.8106 = 0.1894, and the Pvalue is 38 . 1894 . 2 = × = P . Since Pvalue is larger then 05 . = α , we do not reject 5 . : = p H at the level 05 . = . Figure 1. The Pvalue for Example 1. Example 2. A coin was tossed n=4040 times and we observed X=1992 tails. We want to test the null hypothesis that the coin is fair that is, that the probability of tails is 0.5. So p is the probability that the coin comes up tails and we test 5 . : = p H 5 . : ≠ p H a The test statistic is 88 . 4040 ) 5 . 1 ( 5 . 5 . 4931 . 4040 ) 5 . 1 ( 5 . 5 . ˆ=== p z Using Table A, we find that 38 . 1894 . 2 = × = P . Since Pvalue is larger then 05 . = α , we do not reject 5 . : = p H at the level 05 . = ....
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 Spring '09
 Statistics, Statistical hypothesis testing, sample proportion, Count Buffon, Population Proportion Draw

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