stats 363 HW-Section-8-9

# stats 363 HW-Section-8-9 - when p = 0 5 Then n&(0 25 z...

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Section 8.9 Exercise 8.68 . The parameter on interest is ° . Then n ° z 2 °= 2 ± 2 B 2 : It is given that ± = 12 : 7 and B = 1 : 6. With 1 ± ² = 0 : 95 or ² = 0 : 05, we have z °= 2 = z 0 : 025 = 1 : 96. Hence, n ° (1 : 96) 2 (12 : 7) 2 (1 : 6) 2 = 242 : 036 or n = 243 : Exercise 8.69 . The parameter on interest is p . Then n ° z 2 °= 2 pq B 2 : It is given that 0 : 1 < p < 0 : 3 and B = 0 : 04. We assume the largest possible variation for 0 : 1 < p < 0 : 3, which occurs when p = 0 : 3. With 1 ± ² = 0 : 95 or ² = 0 : 05, we have z °= 2 = z 0 : 025 = 1 : 96. Hence, n ° (1 : 96) 2 (0 : 3) (0 : 7) (0 : 04) 2 = 504 : 21 or n = 505 : Exercise 8.71 . The parameter on interest is p 1 ± p 2 . Assume that n 1 = n 2 and B = 0 : 05. Then n ° 2 (0 : 25) z 2 °= 2 B 2 : Since we have no prior knowledge about p 1 and p 2 , we assume the largest possible variation, which occurs when p 1 = p 2 = 0 : 5. With 1 ± ² = 0 : 90 or ² = 0 : 10, we have z °= 2 = z 0 : 05 ² 2 : 33. Hence, n ° 2 (0 : 25) (2 : 33) 2 (0 : 05) 2 = 1085 : 78 or n 1 = n 2 = 1086 :

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Exercise 8.73 ( a ) We would use a simple random sample. It may be di°cult because the number of people needed to be polled could become overwhelming and problem- atic. ( b ) The parameter on interest is p . Assume that B = 0 : 01. Since we have no prior knowledge about p , we assume the largest possible variation, which occurs
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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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