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Unformatted text preview: when p = 0 : 5. Then n & (0 : 25) z 2 &= 2 B 2 : With 1 ± & = 0 : 95 or & = 0 : 05, we have z &= 2 = z : 025 = 1 : 96. Hence, n & (0 : 25) (1 : 96) 2 (0 : 01) 2 = 9604 or n = 9604 : Exercise 8.75 . Assume n 1 = n 2 = n and B = 5. Thus, n & z 2 &= 2 ( ± 2 1 + ± 2 2 ) B 2 : With 1 ± & = 0 : 99 or & = 0 : 01, we have z &= 2 = z : 005 = 2 : 575. The standard deviations can be estimated as ± 1 = ± 2 ² Range 4 = 104 4 = 26 : Then n & (2 : 575) 2 (26 2 + 26 2 ) 5 2 = 358 : 58 Hence, n 1 = n 2 = 359 : Exercise 8.77 . For estimating ² , the required sample size n must satisfy the following inequality: n & z 2 &= 2 ± 2 B 2 : With 1 ± & = 0 : 95 or & = 0 : 05, we have z &= 2 = z : 025 = 1 : 96. Assume that ± ² 10 and B = 2. Hence, n & (1 : 96) 2 (10) 2 2 2 = 96 : 04 or n = 97 : 2...
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 Spring '08
 Cheng
 Statistics, Simple random sample, largest possible variation

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