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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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