business statistics, 4e, by ken black. © 2003 john wiley & sons. 10-1 business statistics, 4e...

Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons. 10-1

Business Statistics, 4eby Ken Black

Chapter 10

StatisticalInferences about Two Populations

Discrete Distributions


Learning Objectives

• Test hypotheses and construct confidence intervals about the difference in two population means using the Z statistic.

• Test hypotheses and construct confidence intervals about the difference in two population means using the t statistic.


Learning Objectives

• Test hypotheses and construct confidence intervals about the difference in two related populations.

• Test hypotheses and construct confidence intervals about the differences in two population proportions.

• Test hypotheses and construct confidence intervals about two population variances.


Sampling Distribution of the Difference Between Two Sample

Means

nxx

11

Population 1

Population 2

nxx

22

1 2X X

1X

2X

1 2X X

1x

1x1x

2x

2x

2x


Sampling Distribution of the Difference between Two Sample

Means

1 2X X1 2X X

1 2

1

2

1

2

2

2X X n n

1 21 2X X

2121

xx2

22

1

21

21 nnxx

21 xx 21 xx


Z Formula for the Difference in Two Sample Means

nn

xxz

2

2

2

1

2

1

2121

When 12 and2

2 are known and Independent Samples


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


Hypothesis Testing for Differences Between Means: The Wage Example (part 2)

.Hz

.Hzz

o

o

reject not do 1.96, 1.96- If

reject 1.96, > or 1.96- < If

Rejection Region


Critical Values

Rejection Region

96.1Z c 0 96.1Z c

025.2

025.

2

96.1cz 96.1cz


Hypothesis Testing for Differences Between Means: The Wage Example

(part 3)Advertising Managers

74.256 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

164.264

253.16

700.70

32

2

1

1

1

1

xn

411.166

900.12

187.62

34

2

2

2

2

2

xn

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


Hypothesis Testing for Differences between Means: The Wage Example (part 4)

35.2

34411.166

32253.256

0187.62700.702

2

2

1

2

1

2121

nS

nS

XXZ

.Hreject not do 1.96, Z 1.96- If

.Hreject 1.96, > or Z 1.96- < ZIf

o

o

.Hreject 1.96, > 2.35 = ZSince o

Rejection Region


Critical Values

Rejection Region

cZ 2 33.

025.2

0 cZ 2 33.

025.2

.reject not do ,96.196.1 If

.reject ,96.1or 96.1 If

0

0

Hz

Hzz

35.2

34411.166

32253.256

(0)-62.187)-(70.700

()(

2

22

1

21

)2121

nn

xxz

.reject ,96.135.2 Since 0Hz

33.2cz 33.2cz


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<=z) one-tail 0.0094

z Critical one-tail 1.64

P(Z<=z) two-tail 0.0189

z Critical two-tail 1.960


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


Demonstration Problem 10.1 (part 2)


Critical Value

Rejection Region

.001

cZ 3 08. 0

.H

.H

o

o

reject not do ,08.3 z If

reject 3.08,- <z If

42.10

761700

871100

05727335222

2

2

2

1

2

1

2121

nn

xxz

.Horeject 3.08,- < 10.42- = z Since

87

100,1$

352,3$

1

1

1

n

x

Women

76

700,1$

727,5$

2

2

2

n

x

Men

08.3cz


Confidence Interval to Estimate 1 - 2 When 1, 2 are known

nn

zxxnnzxx

2

2

2

1

2

12121

2

2

2

1

2

121


Demonstration Problem 10.2

88.142.4

5099.2 2

5046.396.16.2445.21

505096.16.2445.21

21

2

21

22

2

2

2

1

2

12121

2

2

2

1

2

121

99.246.3

nnxxnnxx 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


The t Test for Differencesin Population Means

• Each of the two populations is normally distributed.

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

unknown.• The variances of the two populations are equal.

12 = 2

2


t Formula to Test the Difference in Means Assuming 1

2 = 22

2121

2221

21

2121

112

)1()1(

)()(

nnnnnsns

xxt


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

2

025.

0 . , .025 25 2 060t

2

025.

. , .025 25 2 060t



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

21

1

1

s

x

n

273.18

5.56

12

22

2

2

s

x

n



.Ht oreject -2.060,<-5.20= Since

20.5121

151

2121511273.1814495.19

050.5673.47

112

)1()1(

)()(

2121

2221

21

2121

nnnnnsns

xxt

.Ht

.Htt

o

o

reject not do 2.060, 2.060- If

reject 2.060, > or 2.060- < If


MINITAB Output for HernandezNew-Employee Training Problem

Twosample T for method A vs method B

N Mean StDev SE Mean

method A15 47.73 4.42 1.1

method B12 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


EXCEL Output for HernandezNew-Employee Training Problem

t-Test: Two-Sample Assuming Equal Variances

Variable 1 Variable 2Mean 4 7.73 56.5Variance 19.495 18.27Observations 15 12Pooled Variance 18.957Hypothesized Mean Difference 0df 25t Stat - 5.20P(T<=t) one-tail 1.12E-05t Critical one-tail 1.71P(T<=t) two-tail 2.23E-05t Critical two-tail 2.06


Confidence Interval to Estimate 1 - 2 when 1

2 and 22 are unknown and

12 = 2

2

2 where

11

2

)1()1()(

21

2121

2221

21

21

nndf

nnnn

nsnstxx


Dependent Samples

• Before and after measurements on the same individual

• 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


Formulas for Dependent Samples

difference samplemean =

difference sample ofdeviation standard =

difference populationmean =

pairsin difference sample =

pairs ofnumber

1

d

s

D

d

n

ndfn

sDd

t

t

d

1

)(

1

)(

22

2

nnd

d

n

dds

n

dd

d


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


Hypothesis Testing with Dependent Samples: P/E Ratios for Nine Companies

0:

0:

DH

DH

a

o

.Hreject not do 3.355, 3.355- If

.Hreject 3.355, > or 3.355- < If

o

o

t

tt

355.3

8191

01.

6,005.

t

ndf

Rejection Region


Critical Value

0 355.311,01.

t

2

005.

355.311,01.

t

Rejection Region

2

005.



Company2001 P/E

Ratio2002 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 -8.9

7 20.3 32.3 -12.0

8 19.9 40.1 -20.2

9 61.9 106.5 -44.6



70.0

9

599.210033.5

599.21

033.5

t

s

d

d

oHreject not do ,355370.03553 . t = -.-Since



t-Test: Paired Two Sample for Means

2001 P/E

Ratio2002 P/E

Ratio

Mean 30.64 35.68

Variance 268.1 837.5

Observations 9 9

Pearson Correlation 0.674

Hypothesized Mean Difference 0

df 8

t Stat -0.7

P(T<=t) one-tail 0.252

t Critical one-tail 1.86

P(T<=t) two-tail 0.504

t Critical two-tail 2.306


Hypothesis Testing with Dependent 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



H D

H D

o

a

:

:

0

0

.reject not do -1.943, If

.reject 1.943,- If

o

o

Ht

Ht

943.1

6171

05.

6,05.

t

ndf

Rejection Region


Critical Value

0943.1

6,05.t

05.



54.2

7

0945.60857.5

0945.6

857.5

t

s

d

d

.reject 1.943,- 2.54- = 0HttSince c


Confidence Intervals for Mean Difference for Related Samples

1

ndfn

tdDn

td ss dd


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


Confidence Interval for Mean Difference in Number of New-House Sales

16.162.5

23.239.323.239.318

27.3898.239.3

18

27.3898.239.3

898.2

171181

17,005.

D

D

D

ntdD

ntd

t

ndf

ss dd


Sampling Distribution of Differences in Sample Proportions

nqp

nqp

σ

pp

qn

pn

qn

pn

pp

pp

pq

2

22

1

11

21

22

22

11

11

ˆˆ

and ˆˆ

withddistributenormally is sproportion samplein difference the

ˆ - 1 = ˆ where5 4.

and ,5 3.

,5 2.

,5 1.

samples largeFor

21

21

ˆˆˆˆ


Z Formula for the Difference in Two Population Proportions

pqpq

ppnnpp

nqp

nqp

ppppZ

22

11

2

1

2

1

2

1

2

22

1

11

2121

- 1

- 1

2 population from proportion

1 population from proportion

2 sample of size

1 sample of size

2 sample from proportion

1 sample from proportion

ˆˆ

ˆˆ


Z Formula to Test the Difference in Population Proportions

pq

P

qp

Z

nnpnpn

nnxx

nn

pppp

1

11

21

2211

21

21

21

2121

ˆˆ

ˆˆ


Testing the Difference in Population Proportions (Demonstration Problem 10.6)

0:

0:

21

21

pppp

a

o

H

H

.reject not do 2.575, 2.575- If

.reject 2.575, > or 2.575- < If

o

o

Hz

Hzz

575.2

005.2

01.

2

005.

z

Rejection Region


Critical Values

Rejection Region

2

005.

0cZ 2 575.

2

005.

cZ 2 575.czcz


Testing the Difference in Population Proportions (Demonstration Problem 10.6)

24.100

24

24

100

ˆ1

1

1

p

xn

41.95

39

39

95

ˆ2

2

2

p

xn

323.95100

392421

21

nnxxP

54.2067.

17.

951

1001

677.323.

041.24.

11

21

2121ˆˆ

nn

pppp

qp

z

.Horeject not do 2.575, 2.54- = z 2.575- Since


Confidence Interval to Estimate p1 - p2

nqp

nqp

ppppnqp

nqp

pp zz2

22

1

11

21212

22

1

11

21

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ


Example Problem: When do men shop

for groceries?88.1

12.400

48

48

400

ˆˆ

ˆ

11

1

1

1

pq

p

xn

61.1

39.480

187

187

480

ˆˆ

ˆ

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

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

pppp

pp

nqp

nqp

ppppnqp

nqp

pp ZZ

2.33. = z ,confidence of level 98% aFor


F Test for Two Population Variances

1

1

22min

11

22

21

ndf

ndf

s

sF

atordeno

numerator


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


A Portion of the F Distribution Table for = 0.025

Numerator Degrees of Freedom

DenominatorDegrees of Freedom

. , ,025 9 11F

1 2 3 4 5 6 7 8 91 647.79 799.48 864.15 899.60 921.83 937.11 948.20 956.64 963.282 38.51 39.00 39.17 39.25 39.30 39.33 39.36 39.37 39.393 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.474 12.22 10.65 9.98 9.60 9.36 9.20 9.07 8.98 8.905 10.01 8.43 7.76 7.39 7.15 6.98 6.85 6.76 6.686 8.81 7.26 6.60 6.23 5.99 5.82 5.70 5.60 5.527 8.07 6.54 5.89 5.52 5.29 5.12 4.99 4.90 4.828 7.57 6.06 5.42 5.05 4.82 4.65 4.53 4.43 4.369 7.21 5.71 5.08 4.72 4.48 4.32 4.20 4.10 4.0310 6.94 5.46 4.83 4.47 4.24 4.07 3.95 3.85 3.7811 6.72 5.26 4.63 4.28 4.04 3.88 3.76 3.66 3.5912 6.55 5.10 4.47 4.12 3.89 3.73 3.61 3.51 3.44


Sheet Metal Example: Hypothesis Test for Equality of Two Population Variances (Part 1)

22

21

22

21

:

:

a

o

H

H 59.3 11,9,025. F

.HFIf

.HFFIf

o

o

reject do ,59.3 0.28

reject ,3.59 > or 0.28<

28.059.3

1

1 =

11,9,05.11,9,05.

FF

1

1

22min

11

22

21

ndf

ndf

s

sF

atordeno

numerator

12

10

05.0

2

1

n

n


Sheet metal Manufacturer (Part 2)

Rejection Regions

Critical Values

. , , .025 9 11 359F

Non RejectionRegion

. , , .975 11 9 0 28F

.reject do ,59.3 0.28

.reject ,3.59 > or 0.28<

o

o

HFIf

HFFIf


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 2

22.0

22.1

21.8

21.9

22.2

22.0

21.7

21.9

22.0

22.1

21.9

22.1

1138.0

1021

1

s

n

0202.0

1222

2

s

n63.5

0202.0

1138.02

2

2

1 ssF

.HFF oc reject 3.59, = > 5.63 = Since

business statistics, 4e, by ken black. © 2003 john wiley & sons. 10-1 business statistics, 4e...

Documents

mean difference

t test

construct confidence

ttest mu method

dependent samplespe

dependent samplesbefore

population variances

t formula