{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw2s05a - Solutions and Comments on Assignment 2 Stat 501 1...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon