chapter 1

Chapter 6: Point Estimation(.635)(.365) (.880)(.120)e. += . 041200 20012.22( n − ) S + ( n −1) S ⎤ ( n −1) ( n −1)⎡1E⎢⎣⎦n1−12n2−12= σ + σ = σn + n − 2 n + n − 2111 2 21222= E(S1) + E(S2n1+ n2− 2⎥n1+ n2− 2 n1+ n2− 2( ) ( )2212.)2 31x θx11E ∫ 2E( X ) = θ− 14 6 3313. ( X ) = x⋅1 ( 1+θx) dx = + = θ1−11E ( X ) = θ θ ˆ⎛ 1 ⎞= 3X⇒ E(θˆ)= E(3X) = 3E(X ) = 3⎜⎟θ= θ3⎝ 3 ⎠14.a. min(x i ) = 202 and max(x i ) = 525, so the estimate of the number of planes manufactured ismax(x i ) - min(x i ) + 1 = 525 – 202 + 1 = 324.b. The estimate will equal the true number of planes manufactured iff min(x i ) = α andmax(x i ) = β, i.e., iff the smallest serial number in the population and the largest serialnumber in the population both appear in the sample. The estimator is not unbiased. Thisis because max(x i ) never overestimates β and will usually underestimate it ( unlessmax(x i ) = β) , so that E[max(x i )] < β. Similarly, E[min(x i )] > α ,so E[max(x i ) - min(x i )]