STAT 426 HW1 Solutions (provided by TA for your information) 1.3 Let X be the number of defects, = 1. -1 0 (a) P (X = 0) = e 0!1 = e-1 = 0.3679 (b) P (X = 1) =
e-1 11 1!
= e-1 = 0.3679
(c) P (X 2) = 1 - P (X = 0) - P (X = 1) = 1 - 2e-1 = 0.2642 1.8 a. Let
STAT 426 HW1 Solutions (provided by TA for your information) 1.3 Let X be the number of defects, = 1. -1 0 (a) P (X = 0) = e 0!1 = e-1 = 0.3679 (b) P (X = 1) =
e-1 11 1!
= e-1 = 0.3679
(c) P (X 2) = 1 - P (X = 0) - P (X = 1) = 1 - 2e-1 = 0.2642 1.8 a. Let
STAT 426 HW7. 1. (5.24) logit() = + 1 A + 2 S + 3 R + 4 R S, where A=average number of alcoholic drink drinks consumed per day. 1, at least one pack per day 1, black 1, yes S= R= Y = 0, o.w. 0, white 0, no a. Prediction equation: ^ ^ ^ when R = 1, logit(^
STAT424 Spring 2008
Solution to Homework 3
1) Let X = [X1 X2 ] where X1 is n p and X2 is n q, so X is n (p + q). Let M be the projection matrix for C(X) and let M1 be the projection matrix for C(X1 ). X1 and X2 are not necessarily of full rank, and the co
STAT 426 HW2. 2.8 Let p1 be the probability that a subject would be referred for heart catheterization for blacks, and p2 be the probability that a subject would be referred for heart catheterization for whites. p1 = 0.847, p2 = 0.906 a The odds of referr
Homework #11 (Optional) If you choose to turn in your work, you must leave your work in the TA's mailbox in Room 101, Illini Hall by 4 pm, May 1. 1. Refer to the multicenter study discussed in class on April 22 (also see the same data in http:/www.stat.ui
STAT 426 HW7. 1. (5.24) logit() = + 1 A + 2 S + 3 R + 4 R S, where A=average number of alcoholic drink drinks consumed per day. 1, at least one pack per day 1, black 1, yes S= R= Y = 0, o.w. 0, white 0, no a. Prediction equation: ^ ^ ^ when R = 1, logit(^
STAT 426 HW4. 1. (2.21) Let u = (1, 2, 3) for women's attitudes (not likely, somewhat likely, very likely), and v = (0, 2, 1) for mammography experience (never, over one year ago, within the past year). From hw3, we know that 22% cells have expected count
Solutions to STAT 426 HW3 (provided by TA) 1. Computer usage Age Yes No Percentage of Yes young (20-30) 130 100 56.52 middle (30-50) 30 70 30 senior (> 50) 45 125 26.47
(1) 2 = 42.745, p-value< 0.0001. We reject the hypothesis of independence between age
Solutions to STAT 426 HW2. 2.8 Let p1 be the estimated probability that people who suffer fatal injuries in car accident without wearing their seat belts, and p2 be the estimated probability that people who suffer fatal injuries in car accident with weari
STAT 426 HW8. 6.4 a. Summary of Goodness of fit tests: Only (GH, GI) model (lack of HI term) is a poor fit (p-value is very small). b. Model G2 df (GH,GI) 11.6657 2 (GH, HI) 4.1267 2 (GI,HI) 2.3831 2 (GH,GI,HI) 0.3007 1 p-value 0.0029 0.127 0.3030 0.5834