Download - CHAPTER 11
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CHAPTER 11Two-Sample Tests of Hypothesis
SOLVED PROBLEMS
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Problem 11-3
H
H
Hnσ
nσ
H
HH
0
0
0
22
2
2
2
1
2
1
21
0
21
11
210
rejecttofail:Decision
.05.1736.3264.5000valuep
rejecttofail,αvaluep
valueponbasedRuleDecision
rejecttoFail:Decision6
.94
55
2.9
40
2.3
8.17.6xxz5.
1.645zifRFT:RuleDecision4.
Z xxStatisticTest3.
tail)(one.05α2.
μμ:
μμ:1.
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Problem 11-3 (Cont.)
0 ) - ( 21 21 X- X
p- value = .5000 - .3264 = .1736 .1736 > .05
.3264
=.05a
p - value=.1736
-1.645 -.94 0 Z
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Problem 11-4
H
HH
σσxx
H
HH
0
0
22
2
2
2
1
2
1
21
21
211
210
rejecttofail.05;.1260
.12602x.0630
2.4730).5000(valuep
rejecttofailαvaluepvalue,ponbasedruleDecision
rejecttofail:Decision6.
1.53
40
26
35
30
380370
nn
z5.
1.96z1.96ifrejecttoFail:RuleDecision4.
z xx :StatisticTest3.
.05α2.
μμ:
μμ:1.
0
0
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Problem 11-4 (Cont.)
0252
1.
.0630 value -p 2
1
21 X- X 0 ) - ( 21
0H FTR .05;.1260 .1260 2 x .0630 value-p
.0630 .4370-.5000 value-p 2
1
.4370
.025
-1.96 -1.53 0 +1.96 Z
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Problem 11-5
H
σσxx
H
HH
0
22
2
2
2
1
2
1
21
0
21
211
210
reject:Decision6.
2.66
49
6.7
32
5.1
34.931.4
nn
z5.
rejecttofail2.58,z2.58if:D.R4.
Z )xx( :StatisticTest3.
.01α2.
μμ:
μμ :1.
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Problem 11-5 (Cont.)
.0039 value-)p 2
1(
005.)2
1(
0 ) - ( 21 21 X- X
.5000 - .4961 =.0039p-value = .0039 x 2 = .0078 , .0078<.01; HReject
0
.4961
-2.66 -2.58 + 2.58 Z
005.
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Problem 11-6
.0183.4817.5000valuep
H.Reject 6.Stat.Dec
2.09
45
90.1
40
25.2
19.80-20.75 Z5.
XXZ
2.05 Z
whenH FTR :D.R 4.
Z)XX(-T.S. 3.
.02α 2.
μμ:H
μμ:H 1.
45 1.90 19.80 Union -Non
40 2.25 20.75 Union
n x
0
22
22
Nu
0
Nu
Nu1
Nu0
σ
n
n
u
u
nn
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Problem 11-8
o
2
cc
1
ca
21
21
21c
o
21
211
21o
HReject:Decision6.
2.70
150.20.80
200.20.80
.73.85Z
)(1)(1Z
.80150200
110170
nn
XXp5.
1.96Z1.96ifHFTR:DR4.
Zpp:TS3.
.05α2.
ππ:H
ππ:H1.
npp
npp
pp
.7333150
110p.85
200
170p
110X170X
150n200n
21
21
21
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Problem 11-11
HFTR
.02.3015
.3015.1985.5000valuep
H FTRDec.Stat.6.
.52
800956)(.2044)(.7
1000956)(.2044)(.7
.21-.20
n
)ρ(1ρ
n
)ρ(1ρ
ρρ Z5.
.20448001000
168200ρ
.21800
168
n
xρ .20
1000
200
n
xρ
2.05 Z whenH FTR :D.R 4.
Z)ρ(ρ-T.S. 3.
.02α 2.
:H
:H 1.
168 200 favor(x)In
800 1000 Number
Democrats sRepublican
0
0
D
CC
R
CC
DR
c
DR
0
21
0R1
0R0
ππππ
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Problem 11-12
0
m
CC
S
CC
MS
c
MS
0
21
MS1
MS0
H FTRDec.6.Stat.
1.74
600(.27)(.73)
400(.27)(.73)
.25-.30
n)ρ(1ρ
n)ρ(1ρ
ρρ Z5.
.27600400
150120ρ
.25600
150
n
xρ .30
400
120
n
x5.ρ
1.96Z1.96%-
whenH FTR :D.R. 4.
.05α
Z)ρ(ρ-T.S. 3.
:H
:H 1.
150 120 accident 1least at
600 400 n
Married Single
ππππ
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Problem 11-14
20.187
21715
1511712115
2
11.5
697.1697.1:...4
:...3
10..2
:
:.1
1512
342350
1715
222
21
2
22
2
112
21
211
21
21
21
21
p
p
o
o
S
nn
SnSnS
tifHFTRRD
tXXST
H
H
SS
XX
nn
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Problem 11-14(contd.)
o
p
HFTRDecision
t
nnS
XXt
:.6
651.1
171
151
20.187
342350
11
21
2
12
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Problem 11-15
749.71
297
49.91988.617
2
11.5
624.2:...4
:...3
01..2
:
:.1
97
49.988.6
7879
222
22
2
1
p
mF
mmFFp
o
mF
mF
mFo
mF
mF
mF
S
nn
SnSnS
tifHFTRRD
tXXST
H
H
nn
SS
XX
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Problem 11-15(contd.)
o
mFp
mF
HFTRDecision
t
nnS
XXt
:.6
234.0
91
71
749.71
7879
112
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Problem 11-16
69.278
21215
1.181125.15115
2
11.5
485.2:...4
:...3
01..2
:
:.1
1215
1.185.15
4.4861
222
22
2
1
p
ds
ddssp
o
ds
ds
dso
ds
ds
ss
S
nn
SnSnS
tifHFTRRD
tXXST
H
H
nn
SS
XX
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Problem 11-16(contd.)
o
dsp
ds
HFTRDecision
t
nnS
XXt
:.6
949.1
121
151
69.278
4.4861
112
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Problem 11-19
1957.10'y
.9596(20)-29.3877 b.y
9596.3877.29y
bxay
.9596-
5.874
6.446.874-
3877.298
146)9591.(
8
95 a ;
n
xb
n
y a
S
Srb
'
'
'
x
y
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Problem 11-25 cont’d..
349.36
49.68S
6
32.144183.27911419S
28
)1502)(9596(.)95(3877.291419S
2n
xybyay S
yx
yx
yx
2
yx
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Problem 11-25(a)'y
0613.13)E(y7.3301CI
2.865610.1957
.3717.722810.1957
.137687.722810.1957
.01268.1257.722810.1957
50.241
25.1820
8
19)2.306(3.3410.1957
xx
xx
n
1tSyIC
'
2
2
2yx
'
• Find the 95% CI for when x=20
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Problem 11-25 (b)
433.18y9524.1PI
2373.81957.10PI
01268.125.17228.71957.10PI
50.241
25.1820
8
11)349.3(306.2yPI
xx
xx
n
11tSyPI
2
'
2
2
yx
'
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Problem 11-28
o
n
N
d
d
ND
o
ND
ND
NDo
nd
nD
nD
HFTRDecision
n
S
n
S
XXZ
ZifHFTRRD
ZXXST
H
H
nn
SS
XX
:.6
30.1
60)28(
54)21(
351345
)()(.5
65.1,:...4
:...3
05..2
:
:.1
6054
2821
351345
2222
1
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Problem 11-28 contd.
• DR: FTR Ho when p-value > • P-value = .5000 - .4032 = .0968
.0968 > .05
p-value >
Therefore, FTR Ho .
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Problem 11-32
H
npp
npp
pp
p
H
HH
rejecttofailDecision
z
zifrejecttofailDecision
ppzStasticTest
cccc
c
0
21
21
0
21
211
210
:.6
614.1
50079.21.
100079.21.
234.198.
11
21.5001000
117198.5
96.196.1:.4
:.3
05..2
:
:.1
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Problem 11-32 (Cont.)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
-1.96 -1.614 +1.96 z
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Problem 11-33
H
npp
npp
pp
P
H
HH
rejecttofailDecision
Z
Z
zifjectRtoFailRD
ZppStasticTest
tailone
cccc
c
0
21
21
0
21
211
210
:.6
019.1
300118.882.
200118.882.
87.90.
11
882.300200
261180.5
645.1 e:..4
:.3
)(05..2
:
:.1
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Problem 11-33 contd..
DR: FTR H0 if p-value >
.5000.
.3461.
p-value .1539 > .05.
There fore FTR H0.
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Problem 11-33 (Cont.)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
-1.96 1.019 1.96 z
Accept
Reject .025Reject .025
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Additional Problem (AP # 1)
• The air force trains enlisted computer personnel at two bases—Cass AFB and Kingston AFB. A common final examination is administered. As part of an ongoing study of the training program, a comparison of the final test scores is to be made. Is there any significant difference in the final results of the two educational programs? Use the .04 significance level. Determine the p-value. Explain your decision to the committee studying the program.
Class AFB Kingston AFB
Number sampled 40 50
Mean score 114.6 117.9
Sample standard deviation 9.1 10.4
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Answer AP # 1
H
ns
nsxx
H
HH
rejecttofailDecision
z
zrejecttofailRD
ZxxstatisticTest
taileachtailtwo
if
0
22
2
2
2
1
2
1
21
21
211
210
:.6
60.1
50
4.10
40
1.9
9.1176.114.5
05.205.2:..4
:.3
)02(.,04..2
:
:.1
0
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Answer AP #1(cont.)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
-2.05 -1.60 2.05
.02 .02
z
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Additional problem (AP#2)• Corrigan industries has been awarded a large contract to supply pipeline
parts to Angus Oil, a company drilling in the Scotland-Ireland area. In the past, two subcontractors specializing in steel products have provided Corrigan industries with high quality supplies such as nuts, bolts, steel bars and casing. One of the concerns of Corrigan is the delivery time of the two subcontractors, Jackson steel and Alabama Distributors. The question to be explored is whether there is a difference in the delivery time of the two subcontractors.
Random samples from the files of Corrigan Industries revealed these statistics about delivery time:
Jackson Steel Alabama Distributors
Number in Sample 45 50
Mean Delivery time (days) 20 21
Sample standard deviation (days) 4 3
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Answer AP# 2
0
22
2
22
1
21
21
0
21
21
210
:.6
37.1
503
454
2120.5
96.196.1:..4
.3
05..2
:.1
HrejectDecision
ns
ns
xxZ
ZifHrejecttoFailRD
xxZstatisticTest
H
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Answer AP# 2(contd.)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
.4147
-1.96 -1.37
p- value = 2(5000 - .4147) = .1706
1.96