Question

Attention! Once you have submitted an answer, you will not be able to change it later. You will not be able to view the previous question. Question 10 2 pts Correctional Services Canada uses a rating system to classify prison inmates according to security risk level. A higher security risk level classification has consequences such as withdrawal of certain privileges and increased likelihood of serving out a longer sentence. In a study published in the Canadian Journal of Criminology and Criminal Justice it was observed that Aboriginal women are often classified at a higher security risk level compared with other female prisoners. They found that 41 of 68 Aboriginal women inmates were classified as medium security risk, while 112 of 266 non-Aboriginal women inmates were similarly classified, though fewer infractions had actually been committed by the Aboriginal women while incarcerated. Let TA be the proportion of Aboriginal women inmates classified as medium security risk. Let an be the proportion of non-Aboriginal women inmates classified similarly. Using a uniform prior for A and In.find a 99% Bayesian credible interval for 74 - N. (Use Normal approximation). (Note. Round answer to 4 decimal places. Thanks!) . 0.0132 to 0.3465 0.0705 to 0.2863 0.4507 to 0.7493 0.0502 to 0.3065

Answer 1

x denote whether or not an Aboriginal wormen is at classified at a higher security risk lend. and Y denote whether or not an non aboriginal woman is classified at a higher security risk lend. . &N Bernoulli (TA) and you Berrno alli (0) also TA and to has uniform prius We are given a sample of size an andne from the population of X and Y respectively. let the random sample le X, xr, –, AM Ben (TA) and Y., Y2.., Vn u Bor (a) likelihood The point density of x. xr... ton is given by, XAT LLA, TA) 2 TA Zxi (TA) 7-Ex; and foA)2 I, OCTALI and the likelihood of Y., t. - Yon is given byr un from pois LCY. B). (1 ) 2 - 24: and feap 1, OLTO C ZY 2x The posterios, dists of TA giren X., X. .. , xn, is given by $ CTA IX.-.xna) . TAR-TA)"-! SABRI (MA) ? ! - xi 1. DTA ASK (OFW" =>* OCITACI Beta (Exit, ni-exi+1) Similarly, and þcool XX-Xn2) B (hel, nyiti) 0 LTOL 1

TAX ~ Betalexit1, n. - 2x + 1) and ly ~ Beta (27; +1, N2-27; 41). X ~ Blah) F(x) a otb V(X) 2 ab (16) "latb+1) . TAIX - Exit1 3+1+n,-zAit! No malizing [ {xixl) (n. 2x + 15 WN(0, 1) kit 1 +91-9 + 1)?(3X1 +1 +; -781+1+1) independent Similarly, TBIY - 27 + 1 Xi + 1 + m2- NNCO, 1 (2+1) (92-2 yi+1) V (74 +1 +227 + 1) (3/11 +2 -1 + 1) I. TAIX - 41+1. 68+2 ~ N(0, 1) (41+ 1) (68-41+1) V (68+2)2 (68+3) indep TBIY - 112+1 206+2 ~ N(0) (112+1) (266–112+1) V (26612) 2 (266+2) TAIX - 0.6 ~ NCOI) 0058 TBIY – 0.42 V No 1 0.03

" BAIXA N 10.6, 0.0582 Bly ~ (0.42, 0.082) > indept ITA - 1B NN (0.18 0.0582 +0.032) . (ITA-ITB) - 0.18 8 ~ N(0, 1) Jo.00430 : PL02 .0043 100 P[- & L TA-TIB - 0 18 LTop] = 1-2 Ta upper too (1-22)% point of Here we are required 120.01 a standard normal variate 997. credible introval for A - TB are J 0.18 0 Jo 0048 Toy, 0.18 + $0.0043 Tool t fromR & from R 0.18 – 0.0656x2.5758, 0.18 +0.0656x2 157 58 ] ~ [0 0132, 0.3465] [process is as described alone ]