{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw2s05a

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

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

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

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

View Full Document