Chapter 10 Ken Black

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

Business Statistics, 4e by Ken Black

Chapter 10

Statistical Inferences 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

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


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 2 1 2X X

2121

xx2

2

2

1

2

1

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


.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 c0

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

x

n

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


Hypothesis Testing for Differences between


35.2

34

411.166

32

253.256

0187.62700.70

2

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

34

411.166

32

253.256

(0)-62.187)-(70.700

()(

2

2

2

1

2

1

)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


Demonstration Problem 10.1 (part 1)

H

H

o

a

:

:

1 2

1 2

0

0


Critical Value

Rejection

Region

.001

cZ 3 08. 0 08.3cz


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,-


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

nn

zxxnn

zxx2

2

2

1

2

1

21212

2

2

1

2

1

21


Demonstration Problem 10.2

88.142.4

50

99.2 2

50

46.396.16.2445.21

505096.16.2445.21

21

2

21

22

2

2

2

1

2

1

21212

2

2

1

2

1

21

99.246.3

nn

xxnn

xx zz

46.3

45.21

50

Re

1

1

1

x

n

gular

99.2

6.24

50

Pr

2

2

2

x

n

emium

1.96 = Confidence %95 z


The t Test for Differences

in 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. 1

2 = 22


t Formula to Test the Difference in

Means Assuming 12 = 2

2

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

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

2025.

0 . , .025 25 2 060t

2025.

. ,.

025 252 060t



(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



(part 3)

.Ht oreject -2.060, or 2.060- < If


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


EXCEL Output for Hernandez

New-Employee Training Problem

t-Test: Two-Sample Assuming Equal Variances

Variable 1 Variable 2

Mean 4 7.73 56.5

Variance 19.495 18.27

Observations 15 12

Pooled Variance 18.957


df 25

t Stat - 5.20

P(T


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

2

2

21

2

121

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

ndf

n

s

Ddt

t

d

1

)(

1

)(

22

2

n

n

dd

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

2005.

355.311,01.

t

Rejection

Region

2005.




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 -8.9

7 20.3 32.3 -12.0

8 19.9 40.1 -20.2

9 61.9 106.5 -44.6




70.0

9

599.21

0033.5

599.21

033.5

t

s

d

d

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



Samples: P/E Ratios for Nine Companies t-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



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

0 943.1

6,05.t

05.




54.2

7

0945.6

0857.5

0945.6

857.5

t

s

d

d

.reject 1.943,- 2.54- = 0HttSince c


Confidence Intervals for Mean Difference

for Related Samples

1

ndf

ntdD

ntd ss dd


Difference in Number of New-House Sales Realtor 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

23.239.323.239.3

18

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

n

qp

n

qp

pp

qn

pn

qn

pn

pp

pp

pq

2

22

1

11

21

22

22

11

11

and

withddistributenormally is sproportion samplein difference the

- 1 = where5 4.

and ,5 3.

,5 2.

,5 1.

samples largeFor

21

21


Z Formula for the Difference

in Two Population Proportions

pq

pq

p

p

n

n

p

p

n

qp

n

qp

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


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


Testing the Difference in Population

Proportions (Demonstration Problem 10.6)

0:

0:

21

21

pp

pp

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

2005.

0 cZ 2 575.

2005.

cZ 2 575.czcz


Testing the Difference in Population

Proportions (Demonstration Problem 10.6)

24.100

24

24

100

1

1

1

p

x

n

41.95

39

39

95

2

2

2

p

x

n

323.

95100

3924

21

21

nnxxP

54.2

067.

17.

95

1

100

1677.323.

041.24.

11

21

2121

nn

pppp

qp

z

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


Confidence Interval to Estimate p1 - p2

n

qp

n

qppppp

n

qp

n

qppp zz

2

22

1

11

21212

22

1

11

21


Example Problem:

When do men shop

for groceries?

88.1

12.400

48

48

400

11

1

1

1

pq

p

x

n

61.1

39.480

187

187

480

22

2

2

2

pq

p

x

n

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 ,confidence of level 98% aFor


F Test for Two Population Variances

1

1

22min

11

2

2

2

1

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

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


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<

28.0

59.3

1

1 =

11,9,05.

11,9,05.

FF

1

1

22min

11

2

2

2

1

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 11359F

Non Rejection

Region

. , ,.

975 11 90 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

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 reject 3.59, = > 5.63 = Since

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Chapter 10 Ken Black