Answer Happy

8. A bank operates a drive-up facility and a walk-up window. Let X = the proportion of time that the drive-up facility is in use and Y = the proportion of time that the walk-up window is in use. The joint probability density function (pdf) of (X,Y) is given by f(x,y) = 1.2 (x + y²), 0 < x < 1, 0<y<1. = == = You are given that the marginal pdf of X is fx(x) = 1.2x + 0.4, 0 < x < 1, and the marginal pdf of Y is fy(y) = 1.2y2 +0.6, 0 <y< 1. Then regression function of Y on x is (a) E(Y ) Soy 1.2(x+y) 1.2x+0.4 dy +y) (b) E(Y | 2) = Só 97.29270.6 dy (c) E(Y | x) = Soy (1.242 +0.6) dy (d) E(Y | 2) = So 27:29,270.6 (+y) = Y = = dx (e) E(Y | x) = Só 2 1:2 (0+22) dy 2 +0.4

9. A study was carried out to compare the job satisfaction of elementary school teachers with that in past years. A survey was made of 395 elementary school teachers. Of the elementary school teachers, 224 were very satisfied with their jobs. The 95% confidence interval for the proportion of elementary school teach- ers that were very satisfied with their jobs was found to be (0.518, 0.616), and the 90% confidence interval for the proportion of elementary school teachers that were very satisfied with their jobs was found to be (0.526,0.608). Using this information to test whether the proportion of elementary school teachers who are very satisfied with their jobs differs from 0.6, the proportion that was reported for the previous year, the appropriate conclusion is to (a) reject Ho at both level a = 0.05 and a = 0.10. (b) reject Ho at level a = 0.10 but not level a = 0.05. (c) reject H, at level a 0.05 but not level a 0.10. (d) do not reject Ho at either level a = 0.05 or a = 0.10. (e) conclude that there is not enough information to carry out a test.

10. For the geometric distribution with probability mass function (pmf) f(x; 0) = 0(1 - 0)*, x = 0,1,2, ..., 0<B<1, 2 X Ꮎ An Fisher's information for 8 based on a random sample X1, ..., Xn of size n is given by (a) E [. (on (1 - 0)SP-, *:)] (b) Elog (on (1 - 0)2?-.*)] X; (c) E [ do log (on (1 – 0)X?=1 X:)]? (d) – E [ S: (0" (1 – 0)*?-, X:)] (e) – E [des log (or (1 – 0)3%-1 X:)] a2 ) 202