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Business Statistics,4e, by Ken Black. 2003 John Wiley & Sons.

10-1

Business Statistics, 4e

by Ken Black

Chapter 10

StatisticalI nferences aboutTwo Populations

Discrete Distributions

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10-2

Learning Objectives

Test hypotheses and construct confidenceintervals about the difference in two

population means using the Zstatistic. Test hypotheses and construct confidence

intervals about the difference in twopopulation means using the tstatistic.

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10-3

Learning Objectives

Test hypotheses and construct confidenceintervals about the difference in tworelated populations.

Test hypotheses and construct confidenceintervals about the differences in twopopulation proportions.

Test hypotheses and construct confidence

intervals about two population variances.

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10-4

Sampling Distribution of the

Difference Between Two Sample

Means

nx

x

1

1

Population 1

Population 2

nx

x

2

2

1 2X X

1X

2X

1 2X X

1x

1x

1x 2x

2x

2x

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10-5

Sampling Distribution of the

Difference between Two Sample

Means

1 2X X

1 2X X

1 2

1

2

1

2

2

2X X n n

1 2

1 2X X

2121 xx

2

2

2

1

2

1

21

nn

xx

21 xx 21 xx

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10-6

Z Formula for the Difference

in Two Sample Means

nn

xxz

2

2

2

1

2

1

2121

When 12 and2

2are known andIndependent Samples

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10-7

Hypothesis Testing for Differences Between

Means: The Wage Example (part 1)

1 2X X

Rejection

Region

Non Rejection Region

Critical Values

Rejection

Region

1 2X X

025.2

025.2

H

H

o

a

:

:

1 2

1 2

0

0

21 xx

21 xx

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10-8

Hypothesis Testing for Differences Between


.Hz

.Hzz

o

o

rejectnotdo1.96,1.96-If

reject1.96,>or1.96-

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10-9

Hypothesis Testing for Differences

Between Means: The Wage Example

(part 3)Advertising Managers74.256 57.791 71.115

96.234 65.145 67.574

89.807 96.767 59.621

93.261 77.242 62.483

103.030 67.056 69.319

74.195 64.276 35.394

75.932 74.194 86.741

80.742 65.360 57.351

39.672 73.904

45.652 54.270

93.083 59.045

63.384 68.508

164.264253.16

700.70

32

2

1

1

1

1

xn

411.166

900.12

187.62

34

2

2

2

2

2

x

n

Auditing Managers

69.962 77.136 43.649

55.052 66.035 63.369

57.828 54.335 59.676

63.362 42.494 54.449

37.194 83.849 46.394

99.198 67.160 71.804

61.254 37.386 72.401

73.065 59.505 56.470

48.036 72.790 67.814

60.053 71.351 71.492

66.359 58.653

61.261 63.508

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10-10

Hypothesis Testing for Differences between


35.2

34411.166

32253.256

0187.62700.70

2

2

2

1

2

1

2121

nS

nS

XXZ

.Hrejectnotdo1.96,Z1.96-If

.Hreject1.96,>or Z1.96-2.35=ZSince o

Rejection

Region


Critical Values

Rejection

Region

cZ 2 33.

025.2

0 cZ 2 33.

025.2

.rejectnotdo,96.196.1If

.reject,96.1or96.1If

0

0

Hz

Hzz

35.2

34411.166

32253.256

(0)-62.187)-(70.700

()(

2

22

1

21

)2121

nn

xxz

.reject,96.135.2Since 0Hz

33.2c

z 33.2cz

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10-11

Difference Between Means: Using Excel

z-Test: Two Sample for Means

Adv Mgr Auditing Mgr

Mean 70.7001 62.187

Known Variance 264.164 166.411

Observations 32 34

Hypothesized Mean Difference 0

z 2.35

P(Z

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10-12

Demonstration Problem 10.1 (part 1)

H

H

o

a

:

:

1 2

1 2

0

0


Critical Value

Rejection

Region

.001

cZ 3 08. 008.3cz

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10-13

Demonstration Problem 10.1 (part 2)


Critical Value

Rejection

Region

.001

cZ 3 08. 0

.H

.H

o

o

rejectnotdo,08.3zIf

reject3.08,-

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10-14

Confidence Interval to Estimate 1- 2When

1,

2are known

nn

zxxnn

zxx2

2

2

1

2

1

21212

2

2

1

2

1

21

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10-15

Demonstration Problem 10.2

88.142.4

50

99.2 2

50

46.396.16.2445.21

5050

96.16.2445.21

21

2

21

22

2

2

2

1

2

1

21212

2

2

1

2

1

21

99.246.3

nnxx

nnxx zz

46.3

45.21

50

Re

1

1

1

xn

gular

99.2

6.24

50

Pr

2

2

2

xn

emium

1.96=Confidence%95 z

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10-16

The tTest for Differences

in Population Means

Each of the two populations is normallydistributed.

The two samples are independent. The values of the population variances are

unknown.

The variances of the two populations are equal.12= 22

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10-17

tFormula to Test the Difference in

Means Assuming 12= 22

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

xxt

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10-18

Hernandez Manufacturing Company

(part 1)

H

H

o

a

:

:

1 2

1 2

0

0

If t < -2.060 or t > 2.060, reject H .

If -2.060 t 2.060, do not reject H .

o

o

060.2

25212152

025.2

05.

2

25,25.0

21

t

nndf

Rejection

Region


Critical Values

Rejection

Region

2025.

0 . , .025 25 2 060t

2025.

. , .

025 25 2 060t

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10-19


(part 2)

Training Method A

56 51 45

47 52 43

42 53 52

50 42 48

47 44 44

Training Method B

59

52

53

54

57

56

55

64

53

65

53

57

495.19

73.47

15

2

1

1

1

s

x

n

273.18

5.56

12

2

2

2

2

s

x

n

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10-20


(part 3)

.Ht oreject-2.060,or2.060-

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10-21

MINITAB Output for Hernandez

New-Employee Training Problem

Twosample T for method A vs method B

N Mean StDev SE Mean

method A 15 47.73 4.42 1.1

method B 12 56.60 4.27 1.2

95% C.I. for mu method A - mu method B: (-12.2, -5.3)

T-Test mu method A = mu method B (vs not =): T = -5.20

P=0.0000 DF = 25

Both use Pooled StDev = 4.35

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10-22

EXCEL Output for Hernandez

New-Employee Training Problem

t-Test: Two-Sample Assuming Equal Variances

Variable 1 Variable 2

Mean 4 7.73 56.5Variance 19.495 18.27

Observations 15 12

Pooled Variance 18.957


df 25

t Stat - 5.20P(T

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10-23

Confidence Interval to Estimate 1-

2when 12

and

2

2

are unknown and12= 2

2

2where

11

2

)1()1()(

21

2121

2221

21

21

nndf

nnnn

nsnstxx

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10-24

Dependent Samples

Before and aftermeasurements onthe sameindividual

Studies of twins

Studies of spouses

Individual

1

2

3

4

5

6

7

Before

32

11

21

17

30

38

14

After

39

15

35

13

41

39

22

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10-25

Formulas for Dependent Samples

differencesamplemean=

differencesampleofdeviationstandard=

differencepopulationmean=

pairsindifferencesample=

pairsofnumber

1

d

s

D

d

n

ndf

n

s

Ddt

t

d

1

)(1

)(

22

2

n

n

dd

n

dds

n

dd

d

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10-26

P/E Ratios for Nine Randomly Selected

Companies

Company 2001 P/E Ratio 2002 P/E Ratio

1 8.9 12.7

2 38.1 45.4

3 43.0 10.0

4 34.0 27.2

5 34.5 22.8

6 15.2 24.1

7 20.3 32.3

8 19.9 40.1

9 61.9 106.5

H th i T ti ith D d t

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10-27

Hypothesis Testing with Dependent

Samples: P/E Ratios for Nine Companies

0:

0:

DH

DH

a

o

.Hrejectnotdo3.355,3.355-If

.Hreject3.355,>or3.355-

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10-28



Company

2001 P/E

Ratio

2002 P/E

Ratio d

1 8.9 12.7 -3.8

2 38.1 45.4 -7.3

3 43.0 10.0 33.0

4 34.0 27.2 6.8

5 34.5 22.8 11.7

6 15.2 24.1 -8.97 20.3 32.3 -12.0

8 19.9 40.1 -20.2

9 61.9 106.5 -44.6

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10-29



70.0

9

599.21

0033.5

599.21

033.5

t

s

d

d

oHrejectnotdo,355370.03553 .t = -.-Since

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10-30


Samples: P/E Ratios for Nine Companiest-Test: Paired Two Sample for Means

2001 P/E

Ratio

2002 P/E

Ratio

Mean 30.64 35.68

Variance 268.1 837.5

Observations 9 9

Pearson Correlation 0.674


df 8

t Stat -0.7

P(T

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10-31


Samples: Demonstration Problem 10.5

Individual

1

2

3

4

5

6

7

Before

32

11

21

17

30

38

14

After

39

15

35

13

41

39

22

d

-7

-4

-14

4

-11

-1

-8

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10-32



H D

H D

o

a

:

:

0

0

.rejectnotdo-1.943,If

.reject1.943,-If

o

o

Ht

Ht

943.1

6171

05.

6,05.

t

ndf

Rejection

Region


Critical Value

0943.16,05. t

05.


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10-33



54.2

7

0945.6

0857.5

0945.6

857.5

t

s

d

d

.reject1.943,-2.54-= 0HttSince c

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10-34

Confidence Intervals for Mean Difference

for Related Samples

1

ndf

ntdD

ntd ss dd

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10-35

Difference in Number of New-House SalesRealtor May 2001 May 2002 d

1 8 11 -3

2 19 30 -11

3 5 6 -1

4 9 13 -4

5 3 5 -2

6 0 4 -4

7 13 15 -2

8 11 17 -6

9 9 12 -3

10 5 12 -7

11 8 6 2

12 2 5 -3

13 11 10 1

14 14 22 -8

15 7 8 -1

16 12 15 -3

17 6 12 -6

18 10 10 0

27.3

39.3

ds

d

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10-36

Confidence Interval for Mean Difference

in Number of New-House Sales

16.162.523.239.323.239.3

18

27.3898.239.3

18

27.3898.239.3

898.2

171181

17,005.

DD

D

ntdD

ntd

t

ndf

ss dd

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10-37

Sampling Distribution of Differences

in Sample Proportions

n

qp

n

qp

pp

qnpn

qn

pn

pp

pp

pq

2

22

1

11

21

22

22

11

11

and

withddistributenormallyissproportionsampleindifferencethe

-1= where54.

and,53.

,52.

,51.

sampleslargeFor

21

21

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10-38

Z Formula for the Difference

in Two Population Proportions

pq

pq

p

pnnp

p

n

qp

n

qpppppZ

22

11

2

1

2

1

2

1

2

22

1

11

2121

-1

-1

2populationfromproportion

1populationfromproportion

2sampleofsize

1sampleofsize

2samplefromproportion

1samplefromproportion

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10-39

Z Formula to Test the Difference

in Population Proportions

pq

P

qp

Z

nn pnpn

nnxx nn

pppp

1

11

21

2211

21

21

21

2121

T ti th Diff i P l ti

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10-40

Testing the Difference in Population

Proportions (Demonstration Problem 10.6)

0:

0:

21

21

pp

pp

a

o

H

H

.rejectnotdo2.575,2.575-If

.reject2.575,>or2.575-

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10-41

Testing the Difference in Population

Proportions (Demonstration Problem 10.6)

24.100

24

24

100

1

1

1

p

xn

41.95

39

39

95

2

2

2

p

xn

323.95100

3924

21

21

nnxxP

54.2

067.

17.

95

1

100

1677.323.

041.24.

11

21

2121

nn

pppp

qp

z

.Horejectnotdo2.575,2.54-=z2.575-Since

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10-42

Confidence Interval to Estimate p1- p2

n

qp

n

qppppp

n

qp

n

qppp zz

2

22

1

11

21212

22

1

11

21

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10-43

Example Problem:

When do men shop

for groceries?

88.1

12.400

48

48

400

11

1

1

1

pq

p

xn

61.1

39.480

187

187

480

22

2

2

2

pq

p

xn

206.334.

064.27.064.27.

480

61.39.

400

88.12.33.239.12.

480

61.39.

400

88.12.33.239.12.

21

21

21

2

22

1

11

21212

22

1

11

21

pp

pp

pp

n

qp

n

qppppp

n

qp

n

qppp ZZ

2.33.=z,confidenceoflevel98%aFor

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Business Statistics,4e, by Ken Black. 2003 John Wiley & Sons. 10-44

F Test for Two Population Variances

1

1

22min

11

2

2

2

1

ndf

ndfs

sF

atordeno

numerator

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F Distribution with 1= 10 and 2= 8

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.00 1.00 2.00 3.00 4.00 5.00 6.00

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A Portion of the F Distribution Table

for = 0.025

Numerator Degrees of Freedom

Denominator

Degrees of Freedom

. , ,025 9 11F

1 2 3 4 5 6 7 8 9

1 647.79 799.48 864.15 899.60 921.83 937.11 948.20 956.64 963.28

2 38.51 39.00 39.17 39.25 39.30 39.33 39.36 39.37 39.39

3 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.47

4 12.22 10.65 9.98 9.60 9.36 9.20 9.07 8.98 8.90

5 10.01 8.43 7.76 7.39 7.15 6.98 6.85 6.76 6.68

6 8.81 7.26 6.60 6.23 5.99 5.82 5.70 5.60 5.52

7 8.07 6.54 5.89 5.52 5.29 5.12 4.99 4.90 4.82

8 7.57 6.06 5.42 5.05 4.82 4.65 4.53 4.43 4.36

9 7.21 5.71 5.08 4.72 4.48 4.32 4.20 4.10 4.03

10 6.94 5.46 4.83 4.47 4.24 4.07 3.95 3.85 3.78

11 6.72 5.26 4.63 4.28 4.04 3.88 3.76 3.66 3.59

12 6.55 5.10 4.47 4.12 3.89 3.73 3.61 3.51 3.44

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Sheet Metal Example: Hypothesis Test for

Equality of Two Population Variances (Part 1)

22

21

22

21

:

:

a

o

H

H 59.311,9,025. F

.HFIf.HFFIf

o

o

rejectdo,59.30.28reject,3.59>or0.28

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Sheet metal Manufacturer (Part 2)

Rejection Regions

Critical Values

. , , .025 9 11 359F

Non Rejection

Region

. , , .975 11 9 0 28F

.rejectdo,59.30.28

.reject,3.59>or0.28

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Sheet Metal Example (Part 3)

Machine 1

22.3 21.8 22.2

21.8 21.9 21.6

22.3 22.4

21.6 22.5

Machine 222.0

22.1

21.8

21.9

22.2

22.0

21.7

21.9

22.0

22.1

21.9

22.1

1138.0

10

2

1

1

s

n

0202.0

12

2

2

2

s

n63.5

0202.0

1138.02

2

2

1

ssF

.HFF oc reject3.59,=>5.63=Since

ch10

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