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Unformatted text preview: STAT 3022, April 4, 2009 TA: Jinghan Meng Homework 6 Solutions EX 11.13 (a) We see from the table of coefficients that there were p = 5 explanatory variables. The F statistic has df 5 and 524, so n p 1 = 524, meaning n = 530. (b) We were given R = 0 . 44, so the regression explains R 2 = 19 . 36% of the variation in percent fat mass. (c) Based on the positive coefficients, predicted fat mass is higher for females, those who take in higher percents of energy at dinner, and children of parents with higher BMIs. The one negative coefficient tells us that predicted fat mass is higher for those with underreported intake (low values of EI/predicted BMR) and lower for those who overreported intake. (d) With df = 524, the appropriate critical value for a 95% confidence interval is t * = 1 . 965, so a 95% CI for 2 is 2 t * 2 = 0 . 08 1 . 965(0 . 02) = [0 . 04 , . 12] Therefore, when that explanatory variable changes by 5%, percent fat mass changes by 5 [0 . 04 , . 12] = [0 . 2 , . 6]. EX 11.15 (a) Hypotheses t P Conclusion H : 1 = 0 vs H a : 1 6 = 0 4.55 p < . 001 Reject H ; GPA is significant. H : 2 = 0 vs H a : 2 6 = 0 2.69 p < . 01 Reject H ; popularity is significant. H : 3 = 0 vs H a : 3 6 = 0 2.69 p < . 01 Reject H ; depression is significant. (b) b 1 < 0, so marijuana use decreases with increasing GPA; b 2 and b 3 are positive, so marijuana use increases with popularity and depression. (c) The numbers 3 and 85 are the degrees of freedom of the F statistic ( p = 3 explanatory variables and n p 1 = 85 degrees of freedom left over after estimating the four regression coefficients). (d) H : 1 = 2 = 3 = 0 vs H a : at least one i 6 = 0. EX 11.20 (a) The regression equation is Score = 3 . 96 + 0 . 86 Unfav + 0 . 66 Fav (b) Because p < . 01, we reject H : 1 = 2 = 0 in favor of H a : at least one 1 , 2 is nonzero....
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This note was uploaded on 02/24/2011 for the course STAT 3022 taught by Professor Staff during the Spring '08 term at Minnesota.
 Spring '08
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