1. 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 %. Based on the data, they are [103, 137) or [117, 142)
U~ Binom(100, .95), use normal approximation for large sample:
r=np *(np(1-p))^(1/2)=91.23
s= np *(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

2. (b) U~Binom(10,.25)
P(1<=U<6)=P(U>=1)-P(U>=6)= =.924 (Exact coverage)
90% CI for Q1 is [X(1), X(6))
P(1<=U<10)=P(U>=1)-P(U>=10)= = (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* *(50*.5*.5)^(1/2)=19.6819
s=50* *(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