CHAPTER 11

CHAPTER 11Two-Sample Tests of Hypothesis

SOLVED PROBLEMS

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.

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

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

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

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.

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.

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

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

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

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

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

Problem 11-14(contd.)

o

p

HFTRDecision

t

nnS

XXt

:.6

651.1

171

151

20.187

342350

11

21

2

12

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

Problem 11-15(contd.)

o

mFp

mF

HFTRDecision

t

nnS

XXt

:.6

234.0

91

71

749.71

7879

112

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

Problem 11-16(contd.)

o

dsp

ds

HFTRDecision

t

nnS

XXt

:.6

949.1

121

151

69.278

4.4861

112

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

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

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

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

'

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

Problem 11-28 contd.

• DR: FTR Ho when p-value > • P-value = .5000 - .4032 = .0968

.0968 > .05

p-value >

Therefore, FTR Ho .

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

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

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

Problem 11-33 contd..

DR: FTR H0 if p-value >

.5000.

.3461.

p-value .1539 > .05.

There fore FTR H0.

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

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

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

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

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

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

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