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 . Then m Y ( t ) = (1 & t ) & 2 . Now m 2 Y=& ( t ) = E & e t ( 2 Y & ) = E & e ( 2 t & ) Y = m Y 2 t ! = 1 & 2 t ! & 2 = (1 & 2 t ) & 2 : This implies that 2 Y= has a chi-square distribution with 4 degrees of freedom. Using Table 6 with = 4 degrees of freedom, we have P ( 2 Y > 9 : 48773 ) = 0 : 05 and P ( 2 Y > : 710721 ) = 0 : 95 : Thus, P ( : 710721 2 Y 9 : 48773 ) = 0 : 90 or P 2 Y 9 : 48773 2 Y : 710721 = 0 : 90 : or P Y 4 : 743865 Y : 3553605 = 0 : 90 : Hence, the interval h Y 4 : 743865 ; Y : 3553605 i forms a 90% con&dence interval for . Exercise 8.36 . Since Y N ( ; 1), Y & N (0 ; 1). ( a ) Since P f& 1 : 96 Y & 1 : 96 g = 0 : 95, we obtain P f Y & 1 : 96 Y + 1 : 96 g = 0 : 95 : Hence, the interval [ Y & 1 : 96 ; Y + 1 : 96] forms a 95% con&dence interval for . ( b ) Since P f Y & & 1 : 645 g = 0 : 95, we obtain P f Y + 1 : 645 g = 0 : 95. Hence, the interval ( &1 ; Y + 1 : 645] forms a 95% upper con&dence interval for . ( c ) Since P f Y & 1 : 645 g = 0 : 95, we obtain P f Y & 1 : 645 g = 0 : 95....
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This note was uploaded on 04/20/2008 for the course MTH 562 taught by Professor Cheng during the Spring '08 term at Creighton.

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

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