Stat 511 Solutions to Exam I Spring 2000 There may be more than one way to correctly answer a question, or several ways to describe the same answer. Not all of the possible correct answers may be listed here. 1. (a) (6 points) Since the statement of the problem implies that the conditional means of the responses are linear functions of the parameters ,) ,( 21Tβββ == to obtain a Gauss-Markov model we simply need to add that .)(2IYVar σ== To obtain a normal theory Gauss-Markov model we must have ). ,(~2IXNY σβ (b) (4 points) 22β is the difference between the mean amount of escaped vapor when the device is installed in the gas tank and the mean amount of escaped vapor when no device is installed, and gasoline temperature kept the same for both cases. (c) (6 points) )( 1 = 1 22YPIYSSEXT−−σσ has a central chi-squared distribution with 2 degrees of freedom, because 2. = 2-4 = ) rank(XPI−− The non-centrality parameter is 0 = )( = 2ββδ XPIXXTT−− . (d) (6 points) Since the model matrix has full column rank, T) ,( 21βββ == is estimable and the unique solution to the normal equations is .)( TYXXXbT== For the normal theory Gauss-Markov model in part (a), bis the unique best linear unbiased estimator for T) ,( 21βββ == . Also, ).)( ,(~12 −−XXNbTσβ (e) (14 points) There are several ways to approach this problem. You could use the fact that ) ,(~2IXNY σβ to argue that , ) ,N( ~ Y- Y- Y Y1) 1, 1,- (-1, 22143ccXc Y YcTTTσβ++==== where .4 and 4 222σσββ ==== ccXcTT Then, )1 , 2( ~ 2 ) Y- Y- Y (Y22143σβσN++ and by the definition of the non-central chi-squared distribution . 4 ~ 4 ) Y- Y- Y (Y22221222143++σβχσ Next show that )( 1 = 1 22YPIYSSEXT−−σσ= )()( 1 2YPIPIYXTXT−−−−σhas a central chi- square distribution with 2 degrees of freedom. This follows from Result 4.7 in the course notes because )( 1 = A 2XPI −−σ is a symmetric matrix with rank(A) = 2 and )( )( 1 = A 22XXPIPII −−==−−σσΣ is an idempotent matrix. To show that 4 ) Y- Y- Y (Y222143σ++ is stochastically independent of 1 2SSEσ, note that4 ) Y- Y- Y (Y222143σ++is a function only of Y- Y- Y Y1) 1, 1,- (-1, 2143++== Y and 1 2SSEσ= )()( 1 2YPIPIYXTXT−−−−σis a function only of )( YPIX−− . Then, 4 ) Y- Y- Y (Y222143σ++is stochastically independent of 1 2SSEσ if Y 1) 1, 1,- (-1, is independent of the residual vector )( YPIX−− . Since Yhas a multivariate normal distribution, it follows that YX P- I 1 1 1- 1- has a multivariate normal distribution. Then, to show that Y 1) 1, 1,- (-1, is independent of ,)( YPIX−− we only need to show that the quantities are uncorrelated. Here, (0,0,0,0) )1)( 1, 1,- (-1, ))(1)( 1, 1,- (-1, ))( ,1) 1, 1,- -1,((22==−−==−−==−−XXXPIPIIYPIYCov σσ because T1) 1, 1,- (-1, is a column of X. Consequently, 2SSE)Y-Y-Y(Y 2SSE4)Y-Y-Y(Y F221432222143++==++==σσ has an F-distribution with (1,2) degrees of freedom and non-centrality parameter 222/4 σβ because it is the ratio of a non-central chi-square random variable, divided by its degrees of freedom, over an independent central chi-square random variable divided by its degrees of freedom. This is the definition of a non-central F random variable. Alternatively, you could note that ,)()())(( 4 ) Y- Y- Y (Y122222143YAYYccccYccYcYcTTTTTTT======++σσσ and apply Result 4.7 to this quadratic form and use Result 4.8 to show that this quadratic form is independent of SSE. (f) (4 points) Since the non-centrality parameter for the F-statistic in part (e) is 222/4 σβ , the null hypothesis is 0 :H20==β and the alternative is 0. :H2a≠≠β (g) (10 points) Since any linear function of a normally distributed random vector has a normal distribution, it follows that ). ,N( ~ 2aaXaYaTTTσβ Then, 1) , aN( ~ a2T2TaaXaaYTTσβσ and it follows that .)X( ~ )(22T2122aaaaaYaTTTσβχσ In part (e) we also showed that )( 1 = 1 22YPIYSSEXT−−σσhas a central chi- square distribution with 2 degrees of freedom. Then, the statistic shown in this part has an F-distribution if )(22aaYaTTσis independent of )( 1 = 1 22YPIYSSEXT−−σσ. Using the argument from part (e), thisis true if (0,0,0,0) )( ))(( ))( ,(22==−−==−−==−−XTXTXTPIaPIIaYPIYaCov σσ . Hence, the statistic has an F-distribution if a is in the space spanned by the columns of X, that is, if ++==++==bab b-a b- 1 1 1-1-b 1010a a for some constants a and b where either 0.bor 0a≠≠≠≠ 2. (a) (6 points) 1γ is the mean corn yield for variety A grown with nitrogen applied at 150lb/a. 100 1δ is the difference between mean yields for nitrogen applied at 200lb/a and 100lb/a for either variety A or variety B. (b) (6 points) Since the model matrix has full column rank, any linear combination of the parameters is estimable. Hence, both (i) and (ii) are estimable. (c) (4 points) 1αµ ++ is estimable because E[ (Y1 + Y2 + Y3)/3) ] = 1αµ ++ . (d) (6 points) Delete the third row and the third column of XXT. Invert the remaining 4x4 matrix. Fill in zeros for the third row and third column of the generalized inverse. Alternatively you could use the spectral or singular value decomposition of XXTto obtain a generalized inverse. (e) (8 points) First show that 21 - αα is estimable. Then, b0) 0, 1,- 1, (0, is the unique best linear unbiased estimator for 21 0) 0, 1,- 1, (0, ααβ −−== . Furthermore, since the random errors are independent ) ,0(2σN random variables, ),)(