Dr. Hackney STA Solutions pg 192

# Dr. Hackney STA Solutions pg 192 - 12-2 Solutions Manual...

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Unformatted text preview: 12-2 Solutions Manual for Statistical Inference ^ ^ b. We have from part a), > 0 Sxx > Sxy / and 2 > 0 Syy > Sxy . Furthermore, ^2 ^ 2 ^ ^ ^ > 0 implies that Sxy and have the same sign. Thus Sxx > |Sxy |/|| and Syy > |||Sxy |. ^ Combining yields |Sxy | Syy ^ < < . Sxx |Sxy | 12.11 a. Cov(aY +bX, cY +dX) = E(aY + bX)(cY + dX) - E(aY + bX)E(cY + dX) = E acY 2 +(bc + ad)XY +bdX 2 - E(aY + bX)E(cY + dX) = acVarY + ac(EY )2 + (bc + ad)Cov(X, Y ) +(bc + ad)EXEY + bdVarX + bd(EX)2 - E(aY + bX)E(cY + dX) = acVarY + (bc + ad)Cov(X, Y ) + bdVarX. b. Identify a = , b = 1, c = 1, d = -, and using (12.3.19) Cov(Yi +Xi , Yi -Xi ) = VarY + (1 - 2 )Cov(X, Y ) - VarX 2 2 2 2 = 2 + 2 + (1 - 2 ) - + 2 = 2 - = 0 2 if 2 = . (Note that we did not need the normality assumption, just the moments.) c. Let Wi = Yi + Xi , Vi = Yi + Xi . Exercise 11.33 shows that if Cov(Wi , Vi ) = 0, 2 then n - 2r/ 1 - r2 has a tn-2 distribution. Thus n - 2r ()/ 1 - r () has a tn-2 distribution for all values of , by part (b). Also 2 (n - 2)r() 2 (n - 2)r () F1,n-2, 2 () 1 - r P : 1- 2 r() F1,n-2, =P (X, Y ) : = 1 - . 12.13 a. Rewrite (12.2.22) to get ^ : - t^ t^ ^ + n-2 n-2 2 ^ (-) = : F . 2 (n - 2) ^ b. For of (12.2.16), the numerator of r () in (12.2.22) can be written ^ Syy +(1- 2 )S xy -Sxy = 2 (Sxy ) + (Sxx - Syy ) + Sxy = Sxy ( - ) + Again from (12.2.22), we have 2 r () 2 1 - r () 1 . ^ = Syy +(1- 2 )Sxy -Sxy 2 2, ( 2 2 Syy +2Sxy +Sxx ) (Syy -2Sxy + 2 Sxx ) - (Syy +(1- 2 )Sxy -Sxx ) and a great deal of straightforward (but tedious) algebra will show that the denominator of this expression is equal to 2 (1 + 2 )2 Syy Sxx - Sxy . ...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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