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Mth562-HW8-5

# Mth562-HW8-5 - MTH/STA 562 Exercise 8.35 Note that Y has a...

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Unformatted text preview: MTH/STA 562 Exercise 8.35. Note that Y has a gamma distribution with parameters & = 2 and an unknown Â¡ . Then m Y ( t ) = (1 & Â¡t ) & 2 . Now m 2 Y=& ( t ) = E & e t ( 2 Y & ) Â¡ = E & e ( 2 t & ) Y Â¡ = m Y 2 t Â¡ ! = 1 & Â¡ 2 t Â¡ ! & 2 = (1 & 2 t ) & 2 : This implies that 2 Y=Â¡ has a chi-square distribution with 4 degrees of freedom. Using Table 6 with Â¢ = 4 degrees of freedom, we have P ( 2 Y Â¡ > 9 : 48773 ) = 0 : 05 and P ( 2 Y Â¡ > : 710721 ) = 0 : 95 : Thus, P ( : 710721 Â¡ 2 Y Â¡ Â¡ 9 : 48773 ) = 0 : 90 or P Â¢ 2 Y 9 : 48773 Â¡ Â¡ Â¡ 2 Y : 710721 Â£ = 0 : 90 : or P Â¢ Y 4 : 743865 Â¡ Â¡ Â¡ Y : 3553605 Â£ = 0 : 90 : Hence, the interval h Y 4 : 743865 ; Y : 3553605 i forms a 90% con&dence interval for Â¡ . Exercise 8.36 . Since Y Â¢ N ( Â£; 1), Y & Â£ Â¢ N (0 ; 1). ( a ) Since P f& 1 : 96 Â¡ Y & Â£ Â¡ 1 : 96 g = 0 : 95, we obtain P f Y & 1 : 96 Â¡ Â£ Â¡ Y + 1 : 96 g = 0 : 95 : Hence, the interval [ Y & 1 : 96 ; Y + 1 : 96] forms a 95% con&dence interval for Â£ . ( b ) Since P f Y & Â£ Â£ & 1 : 645 g = 0 : 95, we obtain P f Â£ Â¡ Y + 1 : 645 g = 0 : 95. Hence, the interval ( &1 ; Y + 1 : 645] forms a 95% upper con&dence interval for Â£ . ( c ) Since P f Y & Â£ Â¡ 1 : 645 g = 0 : 95, we obtain P f Y & 1 : 645 Â¡ Â£ g = 0 : 95....
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