Solution 10 1. U ~Binom(15,.75) 90% CI for Q3=[X(8), X(14)] with exact coverage (90.25%) (based on the data) which is [63.3, 73.3). U~Binom(20, .5), 90% CI for Q2 =[X(6), X(14)) or [X(7), X(15)) with exact coverage 92.16%. Based on the data, they are [103, 137) or [117, 142) U~ Binom(100, .95), use normal approximation for large sample: r=np+.5-1.96*(np(1-p))^(1/2)=91.23 s= np+.5+1.96*(np(1-p))^(1/2)= 99.77 The 95% CI is then [X(91), X(100))=[70.5, 76.0) See Lecture notes. 5/12/2019 SA3202, Solution 10

(b) U~Binom(10,.25) P(1<=U<6)=P(U>=1)-P(U>=6)=.9437-.0197=.924 (Exact coverage) 90% CI for Q1 is [X(1), X(6)) P(1<=U<10)=P(U>=1)-P(U>=10)=.9437-.0000=.9437 (Exact coverage) 95% CI for Q1 is [X(1),X(10)) (c ) 90% CI for Q3 is [X(5), X(10)) (exact coverage 92.4%) 95% CI for Q3 is [X(1), X(10)) ( exact coverage 94.37%) Use large sample theory, n=50 r=np+.5- table* std, s=np+.5+table* std (a) r=50*.5+.5-1.645*(50*.5*.5)^(1/2)=19.6819 s=50*.5+.5+1.645*(50*.5*.5)^(1/2)=31.3232 so 90% CI for the median (Q2) is [X(19), X(32)) Similarly 95% CI for the median (Q2) is [X(18), X(33)) (r=18.57, s=32.43) (b) 90% CI for Q1 is [X(7), X(19)) (r=7.96, s=18.04) Similarly 95% CI for Q1 is [X(6), X(19)) (r=6.99, s=19.00) (c) 90% CI for the Q3 is [X(32), X(44)) (r=32.96, s=43.03) Similarly 95% CI for Q3 is [X(31), X(45)) (r=31.99, s=44.001) 5/12/2019 SA3202, Solution 10