chapter 1

Chapter 6: Point Estimation19.λ , so p = 2λ− . 3 and ˆ ⎛Y⎞pˆ= 2λ− .3 = 2⎜⎟ − .3;⎝ n ⎠⎛ 20 ⎞the estimate is 2 ⎜ ⎟ − .3 = . 2 .⎝ 80 ⎠a. = . 5 p + .15 ⇒ 2λ= p + . 3ˆb. E ( pˆ)= E( 2λ− .3) = 2E( λ ) − .3 = 2λ− . 3 = pˆ, as desired.c. Here λ = . 7 p + (.3)(.3),so10 9= λ −7 70p and10 ⎛ Y ⎞ 9ˆ = ⎜ ⎟ −7 ⎝ n ⎠ 70p .Section 6.220.⎡⎛n⎞x−x ⎤a. We wish to take the derivative of ⎜ ⎟ ( ) n⎥⎦d ⎡⎢dp ⎣⎛n⎞⎝ x⎠for p. ln ⎜ ⎟ + x ln ( p) + ( n − x) ln ( 1−p)zero and solving for p yieldsln ⎢ p 1−p⎣⎝x⎠⎤ x n − x⎥ = −⎦ p 1−px3p ˆ = . For n = 20 and x = 3, p ˆ = = . 15n20, set it equal to zero and solve; setting this equal to⎛ X ⎞ 1 1ˆ ⎜ ⎟; thus pˆ is an unbiased estimator of p.⎝ n ⎠ n nb. E ( p) = E = E( X ) = ( np) = p5=c. ( 1 − .15) . 4437211