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Mth562-HW10-10

# Mth562-HW10-10 - MTH/STA 562 Exercise 10.76 Refer to...

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Unformatted text preview: MTH/STA 562 Exercise 10.76 . Refer to Exercise 10.2 . Since the rejection region is Y & 12, the power is 1 Â¡ & = P f Y & 12 j p g . To calculate power, use Table 1 . ( a ) 1 Â¡ & = P f Y & 12 j p = 0 : 4 g = 0 : 979. ( b ) 1 Â¡ & = P f Y & 12 j p = 0 : 5 g = 0 : 868. ( c ) 1 Â¡ & = P f Y & 12 j p = 0 : 6 g = 0 : 584. ( d ) 1 Â¡ & = P f Y & 12 j p = 0 : 7 g = 0 : 228. Exercise 10.77 . Refer to Exercise 10.5 . Since the rejection region is Y 1 > : 95, the power is 1 Â¡ & = P f Y 1 > : 95 j Â¡ g . ( a ) 1 Â¡ & = P f Y 1 > : 95 j Â¡ = 0 : 1 g = 1 : 1 Z : 95 dy = 0 : 15. ( b ) 1 Â¡ & = P f Y 1 > : 95 j Â¡ = 0 : 4 g = 1 : 4 Z : 95 dy = 0 : 45. ( c ) 1 Â¡ & = P f Y 1 > : 95 j Â¡ = 0 : 7 g = 1 : 7 Z : 95 dy = 0 : 75. ( d ) 1 Â¡ & = P f Y 1 > : 95 j Â¡ = 1 g = 2 Z : 95 dy = 1 : 05. Exercise 10.79 . ( a ) Refer to Example 10.23 in the text. The uniformly most powerful test is found to be the z test of Section 10.3 . That is, reject H if Z = Y Â¡ 7 Â¢ / p n = Y Â¡ 7 p 5 . p 20 Â¢ 1 : 645 or Y Â¢ 7 + (1 : 645) s 5 20 = 7 : 82 : ( b ) The power is 1 Â¡ & = P n Y Â¢ 7 : 82 & & & Â£ o . 1 Â¡ & = P n Y Â¢ 7 : 82 & & & Â£ = 7 : 5 o = P Â¡ Z Â¢ 7 : 82 Â¡ 7 : 5 : 5 Â¢ = P f Z Â¢ : 64 g = 0 : 2611 1 Â¡ & = P n Y Â¢ 7 : 82 & & & Â£ = 8 : o = P Â¡ Z Â¢ 7 : 82 Â¡ 8 : : 5 Â¢ = P f Z Â¢ Â¡ : 36 g = 0 : 6406 1 Â¡ & = P n Y Â¢ 7 : 82 & & & Â£ = 8 : 5 o = P Â¡ Z Â¢ 7 : 82 Â¡ 8 : 5 : 5 Â¢ = P f Z Â¢ Â¡ 1 : 36 g = 0 : 9131 1 Â¡ & = P n Y Â¢ 7 : 82 & & & Â£ = 9 : o = P Â¡ Z Â¢ 7 : 82 Â¡ 9 : : 5 Â¢ = P f Z Â¢ Â¡ 2...
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Mth562-HW10-10 - MTH/STA 562 Exercise 10.76 Refer to...

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