Solution 8 1. (a). H0: x(.5)=63 vs H1: x(.5) not equal to 63 (x(.5) is the median of X) U=#{I|Xi<63} ~ Binom(n,p), n=15,p= P(X<63). The observed U=6. The test is equivalent to H0: p=1/2 vs H1: p is not equal to ½ Reject H0 when U is too large or too small. Under H0, U~Binom(15, .5). Thus, p-value=2P(U<=6)=.6072. Don’t reject H0. That is, the median can be 63. (b). H0: x(.5)>=63 vs H1: x(.5) <63 The observed U=6. The test is equivalent to H0: p<=1/2 vs H1: p>1/2 Reject H0 when U is too large. Under H0, U~Binom(15, .5). Thus, p-value=P(U>=6)=.8491. Don’t reject H0. That is, the median can be at least 63. H0: Q3=193 vs H1: Q3 is not equal to 193 U=#{I|Xi<193} ~ Binom(n,p), n=15, p= P(X<193). There is an observation X12=193. We exclude this X12 so that now n=14. The observed U=6. The test is equivalent to H0: p=3/4 vs H1: p is not equal to 3/4 Reject H0 when U is too large or too small. Under H0, U~Binom(14, 3/4). Thus, p-value=2P(U<=6)=.0206. Reject H0. There is strong evidence against H0. That is, the upper quartile may be not equal to 193. 6/13/2019 SA3202, Solution 8

(i). H0: Q2=103 vs Q2 is not equal to 103. Two ties were thrown away so that n=18 now. U=#{I|Xi<103} ~ Binom(n,p), n=18,p= P(X<103). The observed U=4. The test is equivalent to H0: p=1/2 vs H1: p is not equal to ½ Reject H0 when U is too large or too small (compared to np=18*.5=9). Under H0, U~Binom(18, .5). Thus, p-value=2P(U<=4)=..0308. There is some evidence against H0. (ii) . H0: Q3>=150 vs H1: Q3<150 U=#{I|Xi<150} ~ Binom(n,p), n=20, ,p= P(X<150). The observed U=16. The test is equivalent to H0: p<=3/4 vs H1: p >3/4 Reject H0 when U is too large. Under H0, U~Binom(20, 3/4). Thus, p-value=P(U>=16)=.4148. Don’t reject H0. (iii). H0: x(.30)<=100 vs H1: x(.30)>100 The observed U=4. The test is equivalent to H0: p>=.3 vs H1: p <.3 Reject H0 when U is too small. Under H0, U~Binom(20, .3). Thus, p-value=P(U<=4)=.2375. Don’t reject H0. 6/13/2019 SA3202, Solution 8

H0: x(.95)=70.3 vs x(.95) >70.3. Some remarks are in list: (i) The sample size is n=100, but only the heights of the 12 tallest are given, so that the ordered data are ………………. 70.0, 70.1, 70.5, …. 72.8 76.0 (ii) The test should be one-sided because there will be a problem only when people are taller than we thought. (iii) n=100 is large so we can use large sample theory: normal approximation. U=#{i|Xi<70.3} ~ Binom(n,p), n=100, p= P(X<70.3). The observed U=(100-12)+2=90. The test is equivalent to H0: p=.95 vs H1: p<.95 Reject H0 when U is too small (compared to np=100*.95=95). Under H0, U~Binom(100,.95). Use CLT, we have z=(90+. 5-100*.95)/sqrt(100*.95*.05)=-2.065. p-value=P(z<-2.065)=..0197. There is some evidence against H0. 6/13/2019 SA3202, Solution 8