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Unformatted text preview: would have been σ ≤ v u u t n1 χ 2 1α,n1 s = s 99 69 . 2 · . 025 = 0 . 030 , and it would have been safe. 1 3. a) If it’s known that p does not exceed 0.2, we can ﬁnd the required sample size as n = z 2 α/ 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 α/ 2 E 2 · . 25 = 1 . 96 2 . 02 2 · . 25 = 2401 . 4. Since the CI is symmetric with respect to ¯ x , we can ﬁnd the latter as the middle of the CI: ¯ x = 5228 + 5051 2 = 5139 . 5 . In order to ﬁnd s , note that the upper limit of the interval has the form u = ¯ x + z . 025 σ √ 15 , and therefore σ can be found as s = u¯ x z . 025 · √ 15 = 52515139 . 5 1 . 96 · √ 15 = 220 . 2...
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 Spring '08
 Perevalov
 Null hypothesis, Statistical hypothesis testing, upper bound, appropriate hypothesis, alternative H1

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