hw2s05a - Solutions and Comments on Assignment 2 Stat 501...

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Stat 501 Solutions and Comments on Assignment 2 Spring 2005 1. (a) Σ 96 . 4 27.80 92 . 14 80 . 27 275.05 70 . 92 92 . 14 92.70 84 . 97 = 8 . 5 5 . 62 6 . 37 = _ X (b) Σ 51 . 5 89 . 30 58 . 16 89 . 30 61 . 305 103 58 . 16 103 71 . 108 = 9 10 = S (c) You can compute the maximum likelihood estimate for the correlation coefficient from either Σ or S. Then 1.9373 = (.5651) - 1 2 - 10 (0.5651) = t and 0.5651 = 5.05) (97.84)(27 92.70 = r 2 8 = 2 - 10 = df with (d) 0.6403 = .5651 - 1 .5651 + 1 log 2 1 = r - 1 r + 1 log 2 1 = z 1.3811 = 3 - 10 1 (1.96) + 0.6403 = 0.1005 - = 3 - 10 1 (1.96) - 0.6403 = z z u L and an approximate 95% confidence interval for the correlation coefficient is 2(-.1005) 2(1.3811) 2(-.1005) 2(1.3811) e - 1 e - 1 , = (-0.1002, 0.8812) e + 1 e + 1 ⎛⎞ ⎜⎟ ⎝⎠ (e) An estimate of the generalized variance is 42403.59 = S , Answers may vary due to rounding off numbers in the computation of S or using the maximum likelihood estimator instead of S. . (f) An estimate of total variance is
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trace(S) = 108.71 + 305.61+ 5.51 = 419.83 (g) An approximate 95% confidence interval for the correlation coefficient is (0.0831, 0.9162) (h) 0.4638 = ) - )(1 - (1 - = r r r r 2 23 2 12 23 12 13 13.2 r r 0.5021 = .4638 - 1 .4638 + 1 log 2 1 = r - 1 r + 1 log 2 1 = z and an approximate 95% confidence interval for the partial correlation coefficient is (-0.2894, 0.8623) (i) The partial correlation for and given could be zero since the 95% CI includes 0. This partial correlation is the correlation between levels of aspartate aminotransferase and glutamate dehydrogenase for any subpopulation of patients defined by a particular level of alanine aminotransferase. 1 X X 3 X 2 (j) 1.384 = (.4638) - 1 3 - 10 (0.4638) = t 2 7 = 3 - 10 = df with and p-value = 0.104 There is insufficient evidence to reject the null hypothesis 0 2 . 13 = ρ 2. (a) Write the joint likelihood function in the form L( , ) = 1 (2 ) np/2 )' -1 ) µ π µµ Σ Σ Σ Σ n tr A XX ee / () ( _ ( _ 2 1 2 1 1 2 −− n jj j=1 __ where A = (X -X)(X -X) .To get the log-likelihood corresponding to the null hypothesis, substitute ( ) np/ 2 22 2 - 1 2 1 = I, I = , and tr( A) = tr(A), Σσ Σ=σ σ Σ σ Then, the log-likelihood for the null hypothesis is 2 np 1 n ( , ) = - log(2 ) - log( ) - tr(A) - (X )'(X ) µΣ π σ −µ σσ A
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(b) Regardless of the value of , under the null hypothesis the log-likelihood is maximized when σ 2 _ X = µ . Then, the last term in the log-likelihood is zero, and the mle for can be derived by maximizing 2 σ (, ) = - np 2 log(2 ) - np 2 log( ) - 1 2 tr(A) 2 2 A µπ σ σ Σ Setting the first partial derivative equal to zero yields the equation = - np 2 + 1 2 tr(A) = 0 24 ∂µ ∂σ σ σ A Σ 2 The solution to this equation yields the maximum likelihood estimate = 1 np tr(A) = 1 p tr( ) = n-1 np tr(S) = 1 np (X - X _ )(X -X _ ) jj j=1 n σ ∧∧ 2 Σ ' which is simply the average of the estimated variances for the p responses obtained from the units in the sample. .
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hw2s05a - Solutions and Comments on Assignment 2 Stat 501...

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