HW8 - Stat 415 HW 8 10.87 Y1 , , Yn Poisson(1 ) H1 : = a yi...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Stat 415 HW 8 10.87 Y 1 , ··· ,Y n Poisson ( λ 1 ) a H 0 : λ = λ 0 H 1 : λ = λ a L ( λ ) = Q f ( y i λ ) = λ y i exp( ) Q y i ! By Neyman-Pearson Lemma, the test that maximizes the power at θ a has a rejection region determined by L ( λ 0 ) λ 1 < k λ y i exp( 0 ) 0 Q y i ! λ y i exp( nλa ) a Q y i ! < k ± λ 0 λ a ² y i exp( n ( λ a - λ )) < k X y i ln ± λ 0 λ a ² + n ( λ a - λ ) < ln k Since λ 0 < λ a X y i > k 0 b Y i Poisson ( ) . Given α , the constant k 0 is the value such that P ( Y i > k 0 | λ = λ 0 ) = α c The rejection region does not depend on the particular value assigned to λ a . Therefore the test derived in part a is the uniformly most powerful for the composite hypothesis. d Start with L ( λ 0 ) λ 1 < k Since λ a < λ 0 X y i ln ± λ 0 λ a ² < ln k - n ( λ a - λ ) X Y i < k 0 11.5 a i
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Stat 415 HW 8 b ˆ β 1 = S xx S xy = - 0 . 0837 , ˆ β 0 = ¯ y - ˆ β 1 ¯ x = 34 . 0092 c d The estimate of y when x=125 is ˆ y = 34 . 0092 - 0 . 0837 × 125 = 23 . 5467 11.6 Minimize SSE = [ y i - ˆ β 1 x i ] 2 d SSE d ˆ β 1 = - 2[ y i - ˆ β 1 x i ] x i = 0 = - 2[ ( y i x i - ˆ β 1 x 2 i )] = 0 y i x i - ˆ β 1 x 2 i ) = 0 ˆ β 1 = y i x i x 2 i 11.12 a ˆ β 1 = S xx S xy = - . 31667 , ˆ β 0 = ¯ y - ˆ β 1 ¯ x = 46 ˆ y = 46 - . 317 x b c SSE= S yy - ˆ β 1 S xy = 190 . 333 , S 2 = SSE n - 2 = 19 .
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

HW8 - Stat 415 HW 8 10.87 Y1 , , Yn Poisson(1 ) H1 : = a yi...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online