Exam 2 Solution
#1(a)
P(X-bar>3400)=P(Z> -1.41 )=0.5+0.4207=0.9207
#1(b)
#(b)
Since sd is known, use z-cv
95% CI for m is
se=
70.710678
LL
3600-1.96*500/sqrt(50)
3461.4071
UL
3600+1.96*500/sqrt(50)
3738.5929
#2
(a) False
#3 (a) B
LL
UL
#3(b) A
Since sd is estimated, use t-cv
t(0.10,8)=1.860
x_2bar-t(.10,8)*s_2/sqrt(n_2)
165.2
x_2bar+t(.10,8)*s_2/sqrt(n_2)
214.8
t-score of 150 = (150-140)/25*sqrt(13)
1.4422205 between 1.397 and 1.860
p-value= P(t_12>1.44) is between 0.05 and 0.1
Let p=P(feeling improved)
1-p=P(feeling worse)
H_0: p=0.5 vs. H_1: p is not equal 0.5
Let X=the number of subjects reports feeling worse among the 10 subjects reported with changed feeling
Then X is binomial distributed with p=0.5 under null hypothesis
p(x)=P(X=x)
p-value=2[p(0)+p(1)+p(2)]= 2*(1+10+45)*(0.5)^10= 0.109375
Conclusion:
There is not strong enough evidence to conclude the drug changes the sense of the subjects (p-value=0.1094)
#4
#5
(a)
(b)
(b) False
© False
(d) True
(e ) True
(f) False
ID
1
2
3
4
5
6
Resting
100
60
70
55
75
90
Post-ex
120
90
80
95
100
130
mean
sd se
Diff
20
30
10
40
25
40
28
12 4.787136
H_0:
mu_1=mu_2
vs
H_1:
mu_1 is not equal to mu_2
t-test
the sample mean of the difference/the sample standard error
28/(12/sqrt(6))
5.715476
Notice that at df=5, the computed t-statistic is between 4.032 and 6.869
p-value= 2P(t_5>5.72)= is between 0.001 and 0.01
Conclusion:
There is significant difference between the post and pre-exercising pulse rates.
Notice all differences are positive so the min of the rank sum T is 0.
From Table A.5, we have T=0<=0 at n=6, thus we reject the equal medians at alpha=0.05.