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Mth562-HW8-7

# Mth562-HW8-7 - MTH/STA 562 q p(1 p n Exercise 8.58 We need...

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MTH/STA 562 Exercise 8.58 . We need to solve for n in the equation 1 : 96 q p (1 ° p ) n = B . ( a ) If p = 0 : 9 and B = 0 : 05, then 1 : 96 s (0 : 9) (0 : 1) n = 0 : 05 or n = 139 : ( b ) If p is unknown, we use p = 0 : 5 and 1 : 96 s (0 : 5) (0 : 5) n = 0 : 05 or n = 385 : Exercise 8.59 . We need to solve for n in the equation 2 ° p n = B . If B = 2 and ° = 10 ; we have 2 10 p n = 2 or n = 100 : Exercise 8.60 . ( a ) We assume p 1 = p 2 = 0 : 75 and n 1 = n 2 = 1500. Then B = 2 s (0 : 75) (0 : 25) 1500 + (0 : 75) (0 : 25) 1500 = 0 : 0316 : ( b ) For a 90% con°dence level, z 0 : 05 = 1 : 645. Assuming equal sample sizes, that is, n 1 = n 2 = n , we consider 1 : 645 s (0 : 75) (0 : 25) n + (0 : 75) (0 : 25) n = 0 : 02 or n = 2537 : Hence, n 1 = n 2 = 2537. Exercise 8.61 . For a 99% con°dence level, z 0 : 005 = 2 : 575. Then we need to solve for n in the equation 2 : 575 q p (1 ° p ) n = B . It is given that b p = 2 3 and B = 0 : 02. Thus, 2 : 575 v u u t ° 2 3 ± ° 1 3 ± n = 0 : 02 or n = 3687 : Exercise 8.62 . For a 95% con°dence level, z 0 : 025 = 1 : 96. Then we need to solve for n in the equation 1 : 96 ° p n = B . It is given that ° = 0 : 5 and B = 0 : 1. Thus, 1 : 96 0 : 5 p n = 0 : 1 or n = 97 :

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Exercise 8.63 . For a 90% con°dence level, z 0 : 05 = 1 : 645. Assuming n 1 = n 2 = n . Then we need to solve for n in the equation 1 : 645 q ° 2 1
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