Ken Black QA ch10

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Business Statistics, 5th ed.

by Ken Black

Chapter 10

Statistical Inferences about

Two Populations

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh, Morehead State University



Learning Objectives

• Test hypotheses and construct confidenceintervals about the difference in two

population means using the Z statistic.• Test hypotheses and construct confidence

intervals about the difference in twopopulation means using the t statistic.



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.



Sampling Distribution of the

Difference Between Two Sample

Means

n x

x

1

1

Population 1

Population 2

n x

x

2

2

1 X

2 X

1 x

2 x

21x x

21x x



Sampling Distribution of the

Difference between Two Sample

Means

1 2 X X

1 2 X X

1 2

1

2

1

2

2

2 X X n n

1 2

1 2 X X

2121

x x

2

2

2

1

2

1

21 nn x x

21 x x 21 x x



Z Formula for the Difference

in Two Sample Means

nn

x x z

2

2

2

1

2

1

2121

When 12 and2

2 are known andIndependent Samples



Hypothesis Testing for Differences

Between Means: The Wage Example

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





21

21

:

:0

a H

H

=0.05, /2 = 0.025, z0.025 = 1.96





35.2

34411.166

32160.264

)0()187.62700.70(

z

Since the observed value of 2.35 is greater than 1.96,

reject the null hypothesis. That is, there is a significantdifference between the average annual wage of advertising

managers and the average annual wage of an auditing manager.



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



Confidence Interval to Estimate 1 - 2 When

1,

2are known

nn

z x xnn

z x x2

22

1

21

21212

22

1

21

21



Demonstration Problem 10.2

88.142.450

99.2 2

50

46.396.16.2445.21

5050

96.16.2445.21

21

2

21

22

2

2

2

1

2

1

2121

2

2

2

1

2

1

21

99.246.3

nn x x

nn x x z z

46.3

45.21

50

1

1

1

x

n

Regular

99.2

6.24

50

2

2

2

x

n

Premium

1.96=Confidence%95 z



The t Test 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



t Formula to Test the Difference in

Means Assuming 12 = 22

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x xt



Hernandez Manufacturing Company

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



Hernandez Manufacturing Company

(part 3)

. H t oreject-2.060,<-5.20=Since

20.5

12

1

15

1

21215

11273.1814495.19050.5673.47

11

2

)1()1(

)()(

2121

2

2

21

2

1

2121

nnnn

nsns

x xt

. H t

. H t t

o

o

rejectnotdo2.060,2.060-If

reject2.060,>or2.060-<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 56.5Variance 19.495 18.27Observations 15 12Pooled Variance 18.957

Hypothesized Mean Difference 0df 25

t 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

2

2

are unknown and1

2 = 22

2where

11

2

)1()1()(

21

2121

2221

21

21

nndf

nnnn

nsnst x x




056.2 ,05.0

42.1 ,20.184.6 ,35.4

15 ,13

26,025.0

21

21

21

t

ss x x

nn




46.1-52.3

03.149.2

15

1

13

1

21513

)14()42.1()12()20.1(056.2)84.635.4(

21

22

The researcher is 95% confident that the difference in population

average daily consumption of cups of coffee between regular- and

decaffeinated-coffee drinkers is between 1.46 cups and 3.52 cups.



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



Formulas for Dependent Samples

differencesamplemean=

differencesampleof deviationstandard=

differencepopulationmean=

pairsindifferencesample=

pairsof number

1

d

s

D

d

n

ndf

n

s

Dd t

t

d

1

)(1

)(

22

2

n

n

d d

n

d d s

n

d d

d



P/E Ratios for Nine Randomly Selected

Companies

Company Year1 P/E Ratio Year2 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

Company

Year1 P/E

Ratio

Year2 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





70.0

9

599.21

0033.5

599.21

033.5

t

s

d

d

oHrejectnotdo,355370.03553 .t = -.-Since

H0: D = 0

H1: D 0




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

Year1 P/E

Ratio

Year2 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





– MINITAB Output




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 pothesis Testing ith Dependent




Samples: Demonstration Problem 10.5

54.2

7

0945.6 0857.5

0945.6

857.5

t

s

d

d

.reject1.943,-2.54-= 0 H t t Since c

H0: D = 0

H1: D < 0




Samples: Demonstration Problem

10.5 –

MINITAB output



Confidence Intervals

1

ndf

n

st d d



INSERT TABLE 10.8





16.162.5 23.239.323.239.3 18

27.3898.239.3

18

27.3898.239.3

D D

D

n

st d D

n

st d d d

The analyst estimates with a 99% level of confidence that theaverage difference in new-house sales for a real estate company in

Indianapolis between 2005 and 2006 is between -5.62 and -1.16

houses.



Confidence Intervals-MINITAB Solution



Sampling Distribution of Differences

in Sample Proportions

n

q p

n

q pσ

p p

qn pn

qn

pn

p p

p p

pq

2

22

1

11

21

22

22

11

11

ˆˆ

and ˆˆ

withddistributenormallyissproportionsampleindifferencethe

ˆ-1=ˆwhere5 4.

and,5 3.

,5 2.

,5 1.

sampleslargeFor

21

21

ˆ

ˆ

ˆ

ˆ



Z Formula for the Difference

in Two Population Proportions

pq

pq

p

p

n

n

p

p

n

q p

n

q p p p p p Z

22

11

2

1

2

1

2

1

2

22

1

11

2121

-1

-1

2populationfromproportion

1populationfromproportion

2sampleof size

1sampleof size

2samplefromproportion

1samplefromproportion

ˆ

ˆ

ˆˆ



Z Formula to Test the Difference

in Population Proportions

pq

P

q p

Z

nn pn pn

nn x x nn

p p p p

1

11

21

2211

21

21

21

2121

ˆˆ

ˆˆ

T ti th Diff i P l ti



Testing the Difference in Population

Proportions (Demonstration Problem 10.6)

24.100

24

24

100

ˆ1

1

1

p

x

n

41.95

39

39

95

ˆ2

2

2

p

x

n

323.

95100

3924

21

21

nn x x

P

54.2

067.

17.

95

1

100

1677.323.

041.24.

11

21

2121ˆˆ

nn

p p p p

q p

z

. H orejectnotdo2.575,2.54-=z2.575-Since

H0: p1 – p2 = 0

Ha: p1 – p2 0



Confidence Interval to Estimate p1 - p2

n

q p

n

q p p p p p

n

q p

n

q p p p z z

2

22

1

11

21212

22

1

11

21

ˆˆˆˆ

ˆˆ

ˆˆˆˆ

ˆˆ



Example Problem:

When do men shop

for groceries?

88.1

12.400

48

48

400

ˆˆ

ˆ

11

1

1

1

pq

p

x

n

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

ˆˆˆˆ

ˆˆ

ˆˆˆˆ

ˆˆ

p p

p p

p p

n

q p

n

q p p p p p

n

q p

n

q p p p Z Z

For a 98% level of confidence

z = 2.33.



F Test for Two Population Variances

1

1

22min

11

2

2

2

1

ndf

ndf s

sF

ator deno

numerator



Sheet Metal Example: Hypothesis Test for

Equality of Two Population Variances

22

21

22

21

:

:

a

o

H

H 59.311,9,025.

F

. H F If

. H F F If

o

o

rejectdo,59.30.28

reject,3.59>or0.28<

28.0

59.3

1

1 =

11,9,05.

11,9,05.

F F

11

22min

11

2

2

2

1

ndf ndf

s

sF

ator deno

numerator

12

10

05.0

2

1

n

n



Sheet Metal Example

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

s

sF

. H F F oc reject3.59,=>5.63=Since



Sheet Metal Example-MINITAB Solution



Sheet Metal Example-EXCEL Solution



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Ken Black QA ch10

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