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Unformatted text preview: would have been v u u t n1 2 1,n1 s = s 99 69 . 2 . 025 = 0 . 030 , and it would have been safe. 1 3. a) If its known that p does not exceed 0.2, we can nd the required sample size as n = z 2 / 2 E 2 p (1p ) = 1 . 96 2 . 02 2 . 2 . 8 = 1537 . b) If nothing is known about the true proportion, we have to use the worst case ( p = 0 . 5): n = z 2 / 2 E 2 . 25 = 1 . 96 2 . 02 2 . 25 = 2401 . 4. Since the CI is symmetric with respect to x , we can nd the latter as the middle of the CI: x = 5228 + 5051 2 = 5139 . 5 . In order to nd s , note that the upper limit of the interval has the form u = x + z . 025 15 , and therefore can be found as s = u x z . 025 15 = 52515139 . 5 1 . 96 15 = 220 . 2...
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 Spring '08
 Perevalov

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