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exam00s

# exam00s - Z 2 are independent 2 Not covered by the midterm...

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THE UNIVERSITY OF CHICAGO Graduate School of Business Business 424-01, Spring Quarter 2000, Mr. Ruey S. Tsay Solutions to Midterm 1. By re-arranging the ordering of the components based on the covariance matrix, we have Z * = ( Z 1 , Z 3 , Z 2 ) 0 with mean μ = ( - 3 , 2 , 4) 0 and covariance matrix Σ * = 1 - 1 0 - 1 2 0 0 0 2 . Using Result 4.5(b), ( Z 1 , Z 3 ) 0 and Z 2 are independent. (a) True, because Z 1 + Z 3 normal and cov( Z 1 + Z 3 , Z 2 ) = 0. (b) True, as stated above. (c) Taking expectation of the model, E ( Z 1 ) = β 0 + β 1 E ( Z 3 ). Thus, β 0 = - 3 - 2 β 1 . Next, cov( Z 1 , Z 3 ) = cov( β 0 + β 1 Z 3 + ², Z 3 ) = β 1 Var( Z 3 ). Therefore, β 1 = cov( Z 1 , Z 3 )/Var( Z 3 ) = - 1 / 2 = - 0 . 5. Consequently, β 0 = - 3 + 0 . 5 × 2 = - 2. (d) Use Result 4.6. Z 1 | Z 3 = 3 is normal with mean - 3 . 5 and variance 0.5. (e) Yes, because Z 1 and
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Unformatted text preview: Z 2 are independent. 2. Not covered by the midterm of 2002. 3. Answer: (a) S pooled = n 1-1 n 1 + n 2-2 S 1 + n 2-1 n 1 + n 2-2 S 2 . Simple calculation shows S pooled = " 1 . 6-1 . 398-1 . 398 1 . 998 # . (b) Use the Hotelling T 2 test of Result 6.2. Simple calculation gives T 2 = 3 . 87, which is distributed as 5 × 2 4 F 2 , 4 . The p value is 0.318. Thus, one cannot reject H o at the 5% level. 4. Answer (a) See the two-way MANOVA table on page 310 with n = 1, i.e. no interaction. (b) There is no interaction between the two factors....
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