chapter 1

Chapter 6: Point Estimation9.a. E( X ) = µ = E(X ) = λ,so X is an unbiased estimator for the Poisson parameter∑ iλ ; = ( 0)(18) + (1)(37) + ... + (7)(1) = 317,x since n = 150,ˆ 317= x = 2.11150=λ .b.σ λλˆ2.11σx= = , so the estimated standard error is = = . 119n nn 15010.a.E222 σ 2( X ) Var(X ) + [ E(X )] = + µthus=n2X tends to overestimate2µ ., so the bias of the estimator2X is2σ ;nb.Eσn22 222 22( X − kS ) = E(X ) − kE(S ) = µ + − kσ2 2 2E ( X − kS ) = µ ., so withk1= ,n11.⎛ XE⎜⎝ n1 2− = E X − E X = ( n p ) − ( n p ) = p − p .a. ( ) ( )121 12 2 1 21Xn2⎞⎟⎠1n11n21n11n2⎛ X⎜⎝ n⎛ 1 ⎞⎜n⎟⎝ 1 ⎠⎛ 1n ⎟ ⎞⎜⎝ 2 ⎠1 212b. Var − = Var Var = Var(X ) + Var(X )12n1Xn2⎞⎟⎠12n⎛ X⎜⎝ n1⎞ ⎛ X ⎞⎟ +⎜n⎟⎠ ⎝ 2 ⎠p1q1p2q= +n2( n p q ) ( n p q ) ,+ and the standard error is the square1 1 12 2 2n1 21 2root of this quantity.2122c. Withx1p ˆ1= , qˆ11−pˆ1n11qˆˆ ˆ1p2q2pˆn1+ .n2= ,x2p ˆ2= , qˆ21−pˆ2n2= , the estimated standard error isˆ ˆ2127176d. ( p − p ) = − = .635 − .880 = . 2451−200200208