1. Solution 8
1. (a). H0: x(.5)= vs H1: x(.5) not equal to (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)= Don’t reject H0. That is, the median can be 63.
(b). H0: x(.5)>= 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)= 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)= 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

2. (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)= 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)= 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)= Don’t reject H0.
6/13/2019
SA3202, Solution 8

3. 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
……………… , 70.1, 70.5, …
(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=( *.95)/sqrt(100*.95*.05)=
p-value=P(z<-2.065)= There is some evidence against H0.
6/13/2019
SA3202, Solution 8