Chap 11-1

Chapter 10

Statistical Inferences About Meansand Proportions with Two Populations

Statistics for Business and Economics

9th Edition

Chap 10-2

Two Sample Tests

Two Sample Tests

Population Means,

Independent Samples

Population Means,

Matched Pairs

Group 1 vs. independent Group 2

Same group before vs. after treatment

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

IntervalInterval

EstimationEstimation

Hypothesis Hypothesis TestsTests

Chap 10-3

Difference Between Two Means

Population means, independent

samples

Test statistic is a z value

Test statistic is a a value from the Student’s t distribution

σx2 and σy

2 assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

小样本

大样本

Chap 10-4

Difference Between Two Means

Population means, independent

samples

Goal: Form a confidence interval for the difference between two population means, μx – μy

Different data sources Unrelated Independent

Sample selected from one population has no effect on the sample selected from the other population

Chap 10-5

Population means, independent

samples

σx2 and σy

2 Known

Assumptions:

Samples are randomly and independently drawn

both population distributions are normal

Population variances are known

*σx2 and σy

2 known

σx2 and σy

2 unknown

Chap 10-6

Estimating the Difference BetweenEstimating the Difference BetweenTwo Population MeansTwo Population Means

Let Let 11 equal the mean of population 1 and equal the mean of population 1 and 22 equalequal

the mean of population 2.the mean of population 2. The difference between the two population The difference between the two population means ismeans is 11 - - 22.. To estimate To estimate 11 - - 22, we will select a simple , we will select a simple randomrandom

sample of size sample of size nn11 from population 1 and a from population 1 and a simplesimple

random sample of size random sample of size nn22 from population 2. from population 2. Let equal the mean of sample 1 and Let equal the mean of sample 1 and

equal theequal the

mean of sample 2.mean of sample 2.

x1x1 x2x2

The point estimator of the difference between The point estimator of the difference between thethe

means of the populations 1 and 2 is .means of the populations 1 and 2 is .x x1 2x x1 2

Chap 10-7

Expected ValueExpected Value

Sampling Distribution of Sampling Distribution of x x1 2x x1 2

E x x( )1 2 1 2 E x x( )1 2 1 2

Standard Deviation (Standard Error)Standard Deviation (Standard Error)

x x n n1 2

12

1

22

2

x x n n1 2

12

1

22

2

where: where: 1 1 = standard deviation of population 1 = standard deviation of population 1

2 2 = standard deviation of population 2 = standard deviation of population 2

nn1 1 = sample size from population 1= sample size from population 1

nn22 = sample size from population 2 = sample size from population 2

Chap 10-8

Interval Estimate

Interval Estimation of 1 - 2: 1 and 2 Known

2 21 2

1 2 / 21 2

x x zn n

2 21 2

1 2 / 21 2

x x zn n

where:where:

1 - 1 - is the confidence coefficient is the confidence coefficient

Chap 10-9

Difference Between Two Means

Population means, independent

samples

Test statistic is a z value

Test statistic is a a value from the Student’s t distribution

σx2 and σy

2 assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

(continued)

小样本

大样本

Chap 10-10

Population means, independent

samples

σx2 and σy

2 Known

Assumptions:

Samples are randomly and independently drawn

both population distributions are normal

Population variances are known

*σx2 and σy

2 known

σx2 and σy

2 unknown

Chap 10-11

Population means, independent

samples

…and the random variable

has a standard normal distribution

When σx2 and σy

2 are known and

both populations are normal, the variance of X – Y is

y

2y

x

2x2

YX n

σ

n

σσ

(continued)

*

Y

2y

X

2x

YX

n

σ

nσ

)μ(μ)yx(Z

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 Known

Chap 10-12

Population means, independent

samples

Test Statistic, σx

2 and σy2 Known

*σx2 and σy

2 known

σx2 and σy

2 unknown

y

2y

x

2x

0

n

σ

nσ

Dyxz

The test statistic for

μx – μy is:

Chap 10-13

Interval Estimate

Interval Estimation of 1 - 2: 1 and 2 Known

2 21 2

1 2 /21 2

x x zn n

2 21 2

1 2 /21 2

x x zn n

where:where:

1 - 1 - is the confidence coefficient is the confidence coefficient

Chap 10-14

Hypothesis Tests forTwo Population Means

Lower-tail test:

H0: μx μy

H1: μx < μy

i.e.,

H0: μx – μy 0H1: μx – μy < 0

Upper-tail test:

H0: μx ≤ μy

H1: μx > μy

i.e.,

H0: μx – μy ≤ 0H1: μx – μy > 0

Two-tail test:

H0: μx = μy

H1: μx ≠ μy

i.e.,

H0: μx – μy = 0H1: μx – μy ≠ 0

Two Population Means, Independent Samples

Chap 10-15

Two Population Means, Independent Samples, Variances Known

Lower-tail test:

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 – μ2= 0H1: μ1 – μy ≠ 0

/2 /2

-z -z/2z z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z/2

or z > z/2

Decision Rules

Chap 10-16

Interval Estimation of Interval Estimation of 11 - - 22:: 1 1 and and 2 2 Known Known

In a test of driving distance using a In a test of driving distance using a mechanicalmechanical

driving device, a sample of Par golf balls wasdriving device, a sample of Par golf balls was

compared with a sample of golf balls made by compared with a sample of golf balls made by Rap,Rap,

Ltd., a competitor. The sample statistics appear Ltd., a competitor. The sample statistics appear on theon the

next slide.next slide.

Par, Inc. is a manufacturerPar, Inc. is a manufacturer

of golf equipment and hasof golf equipment and has

developed a new golf balldeveloped a new golf ball

that has been designed tothat has been designed to

provide “extra distance.”provide “extra distance.”

Example: Par, Inc.

Chap 10-17

Example: Par, Inc.

Interval Estimation of Interval Estimation of 11 - - 22:: 1 1 and and 2 2 Known Known

Sample SizeSample Size

Sample MeanSample Mean

Sample #1Sample #1Par, Inc.Par, Inc.

Sample #2Sample #2Rap, Ltd.Rap, Ltd.

120 balls120 balls 80 balls80 balls

275 yards 258 yards275 yards 258 yards

Based on data from previous driving distanceBased on data from previous driving distancetests, the two population standard deviations aretests, the two population standard deviations areknown with known with 1 1 = 15 yards and = 15 yards and 2 2 = 20 yards. = 20 yards.

Chap 10-18

Interval Estimation of Interval Estimation of 11 - - 22:: 1 1 and and 2 2 Known Known

Example: Par, Inc.

Let us develop a 95% confidence interval Let us develop a 95% confidence interval estimateestimate

of the difference between the mean driving of the difference between the mean driving distances ofdistances of

the two brands of golf ball.the two brands of golf ball.

Chap 10-19

Estimating the Difference BetweenEstimating the Difference BetweenTwo Population MeansTwo Population Means

mm11 – – 22 = difference between= difference between the mean distancesthe mean distances

xx11 - - xx22 = Point Estimate of = Point Estimate of mm11 – – 22

Population 1Population 1Par, Inc. Golf BallsPar, Inc. Golf Balls

11 = mean driving = mean driving distance of Pardistance of Par

golf ballsgolf balls

Population 1Population 1Par, Inc. Golf BallsPar, Inc. Golf Balls

11 = mean driving = mean driving distance of Pardistance of Par

golf ballsgolf balls

Population 2Population 2Rap, Ltd. Golf BallsRap, Ltd. Golf Balls

22 = mean driving = mean driving distance of Rapdistance of Rap

golf ballsgolf balls

Population 2Population 2Rap, Ltd. Golf BallsRap, Ltd. Golf Balls

22 = mean driving = mean driving distance of Rapdistance of Rap

golf ballsgolf balls

Simple random sampleSimple random sample of of nn22 Rap golf balls Rap golf balls

xx22 = sample mean distance = sample mean distance for the Rap golf ballsfor the Rap golf balls

Simple random sampleSimple random sample of of nn22 Rap golf balls Rap golf balls

xx22 = sample mean distance = sample mean distance for the Rap golf ballsfor the Rap golf balls

Simple random sampleSimple random sample of of nn11 Par golf balls Par golf balls

xx11 = sample mean distance = sample mean distance for the Par golf ballsfor the Par golf balls

Simple random sampleSimple random sample of of nn11 Par golf balls Par golf balls

xx11 = sample mean distance = sample mean distance for the Par golf ballsfor the Par golf balls

Chap 10-20

Point Estimate of Point Estimate of 11 - - 22

Point estimate of Point estimate of 11 2 2 ==x x1 2x x1 2

where:where:

11 = mean distance for the population = mean distance for the population of Par, Inc. golf ballsof Par, Inc. golf balls

22 = mean distance for the population = mean distance for the population of Rap, Ltd. golf ballsof Rap, Ltd. golf balls

= 275 = 275 258 258

= 17 yards= 17 yards

Chap 10-21

x x zn n1 2 212

1

22

2

2 2

17 1 9615120

2080

/ .( ) ( )

x x zn n1 2 212

1

22

2

2 2

17 1 9615120

2080

/ .( ) ( )

Interval Estimation of 1 - 2:1 and 2 Known

We are 95% confident that the difference betweenWe are 95% confident that the difference betweenthe mean driving distances of Par, Inc. balls and Rap,the mean driving distances of Par, Inc. balls and Rap,Ltd. balls is 11.86 to 22.14 yards.Ltd. balls is 11.86 to 22.14 yards.

17 17 ++ 5.14 or 11.86 yards to 22.14 yards 5.14 or 11.86 yards to 22.14 yards

Chap 10-22

Example: Par, Inc.Example: Par, Inc.

Hypothesis Tests About 1 2: 1 and 2 Known

Can we conclude, usingCan we conclude, using

= .01, that the mean driving= .01, that the mean driving

distance of Par, Inc. golf ballsdistance of Par, Inc. golf balls

is greater than the mean drivingis greater than the mean driving

distance of Rap, Ltd. golf balls?distance of Rap, Ltd. golf balls?

Chap 10-23

HH00: : 1 1 - - 22 << 0 0

HHaa: : 1 1 - - 22 > 0 > 0where: where: 11 = mean distance for the population = mean distance for the population of Par, Inc. golf ballsof Par, Inc. golf balls22 = mean distance for the population = mean distance for the population of Rap, Ltd. golf ballsof Rap, Ltd. golf balls

1. Develop the hypotheses.1. Develop the hypotheses.

pp –Value and Critical Value Approaches –Value and Critical Value Approaches

Hypothesis Tests About 1 2: 1 and 2 Known

2. Specify the level of significance.2. Specify the level of significance. = .01= .01

Chap 10-24

3. Compute the value of the test statistic.3. Compute the value of the test statistic.

Hypothesis Tests About 1 2: 1 and 2 Known

pp –Value and Critical Value Approaches –Value and Critical Value Approaches

1 2 0

2 21 2

1 2

( )x x Dz

n n

1 2 0

2 21 2

1 2

( )x x Dz

n n

2 2

(235 218) 0 17 6.49

2.62(15) (20)120 80

z

2 2

(235 218) 0 17 6.49

2.62(15) (20)120 80

z

Chap 10-25

p p –Value Approach–Value Approach

4. Compute the 4. Compute the pp–value.–value.

For For zz = 6.49, the = 6.49, the pp –value < .0001. –value < .0001.

Hypothesis Tests About 1 2: 1 and 2 Known

5. Determine whether to reject 5. Determine whether to reject HH00..

Because Because pp–value –value << = .01, we reject = .01, we reject HH00..

At the .01 level of significance, the sample At the .01 level of significance, the sample evidenceevidenceindicates the mean driving distance of Par, Inc. indicates the mean driving distance of Par, Inc. golfgolfballs is greater than the mean driving distance balls is greater than the mean driving distance of Rap,of Rap,Ltd. golf balls.Ltd. golf balls.

Chap 10-26

Hypothesis Tests About 1 2: 1 and 2 Known

5. Determine whether to reject 5. Determine whether to reject HH00..

Because Because zz = 6.49 = 6.49 >> 2.33, we reject 2.33, we reject HH00..

Critical Value ApproachCritical Value Approach

For For = .01, = .01, zz.01.01 = 2.33 = 2.33

4. Determine the critical value and rejection rule.4. Determine the critical value and rejection rule.

Reject Reject HH00 if if zz >> 2.33 2.33

The sample evidence indicates the mean The sample evidence indicates the mean drivingdrivingdistance of Par, Inc. golf balls is greater than distance of Par, Inc. golf balls is greater than the meanthe meandriving distance of Rap, Ltd. golf balls.driving distance of Rap, Ltd. golf balls.

Chap 10-27

Population means, independent

samples

σx2 and σy

2 Unknown,

*大样本

σx2 and σy

2 known

σx2 and σy

2 unknown

小样本

Chap 10-28

*

Test Statistic, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

大样本

（ nx>=30 且 ny>=30 ）

小样本（ nx<30 或 ny<30 ）

22

/21 2

yxx y

ssx x Z

n n 22

/21 2

yxx y

ssx x Z

n n

Chap 10-29

*

Test Statistic, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

Assumptions: Samples are randomly and independently drawn

Populations are normally distributed

Population variances are unknown but assumed equal

Chap 10-30

*

Test Statistic, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

Forming interval estimates:

The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ

use a t value with (nx + ny – 2) degrees of freedom

Chap 10-31

*

Test Statistic, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

σ 2 的合并估计量：

Chap 10-32

*

Test Statistic, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

Where t has (n1 + n2 – 2) d.f.,

and

yx

2p

yx

n1

n1

S

μμt

yx

The test statistic for

μx – μy is:

Chap 10-33

*

Interval Estimation, σx

2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

Where t has (n1 + n2 – 2) d.f.,

and

Chap 10-34

*

σx2 and σy

2 Unknown,Assumed Unequal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

Assumptions: Samples are randomly and independently drawn

Populations are normally distributed

Population variances are unknown and assumed unequal

Chap 10-35

*

σx2 and σy

2 Unknown,Assumed Unequal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

Forming interval estimates:

The population variances are assumed unequal, so a pooled variance is not appropriate

use a t value with degrees of freedom, where

1)/(nn

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

v

Chap 10-36

*

Test Statistic, σx

2 and σy2 Unknown, Unequal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

1)/(nn

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

vWhere t has degrees of freedom:

The test statistic for

μx – μy is:

Y

2y

X

2x

0

n

σ

nσ

D)yx(t

Chap 10-37

*

Interval Estimation, σx

2 and σy2 Unknown, Unequal

σx2 and σy

2 assumed equal

且 小样本σx

2 and σy2 unknown

σx2 and σy

2 assumed unequal

1)/(nn

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

vWhere t has degrees of freedom:

Y

2y

X

2x

0

n

σ

nσ

D)yx(t

Chap 10-38

Lower-tail test:

H0: μx – μy 0H1: μx – μy < 0

Upper-tail test:

H0: μx – μy ≤ 0H1: μx – μy > 0

Two-tail test:

H0: μx – μy = 0H1: μx – μy ≠ 0

Decision Rules

/2 /2

-t -t/2t t/2

Reject H0 if t < -tn-1, Reject H0 if t > tn-1, Reject H0 if t < -tn-1 ,

or t > tn-1 , Where t has n - 1 d.f.

Two Population Means, Independent Samples, Variances Unknown

Chap 10-39

Pooled Variance t Test: Example

You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:

NYSE NASDAQNumber 21 25Sample mean 3.27 2.53Sample std dev 1.30 1.16

Assuming both populations are approximately normal with equal variances, isthere a difference in average yield ( = 0.05)?

Chap 10-40

Calculating the Test Statistic

1.5021

1)25(1)-(21

1.161251.30121

1)n()1(n

S1nS1nS

22

21

222

2112

p

2.040

251

211

5021.1

02.533.27

n1

n1

S

μμXXt

21

2p

2121

The test statistic is:

Chap 10-41

Solution

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

= 0.05

df = 21 + 25 - 2 = 44Critical Values: t = ± 2.0154

Test Statistic: Decision:

Conclusion:

Reject H0 at = 0.05

There is evidence of a difference in means.

t0 2.0154-2.0154

.025

Reject H0 Reject H0

.025

2.040

2.040

251

211

5021.1

2.533.27t

Chap 10-42

Two Sample Tests

Two Sample Tests

Population Means,

Independent Samples

Population Means,

Matched Pairs

Group 1 vs. independent Group 2

Same group before vs. after treatment

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

IntervalInterval

EstimationEstimation

Hypothesis Hypothesis TestsTests

Chap 10-43

Matched Pairs

Tests Means of 2 Related Populations Paired or matched samples Repeated measures (before/after) Use difference between paired values:

Assumptions: Both Populations Are Normally Distributed

Matched Pairs

di = xi - yi

Chap 10-44

The test statistic for the mean difference is a t value, with n – 1 degrees of freedom:

n

sDd

td

0

Test Statistic: Matched Pairs

WhereD0 = hypothesized mean differencesd = sample standard dev. of differencesn = the sample size (number of pairs)

Matched Pairs

Chap 10-45

Lower-tail test:

H0: μx – μy 0H1: μx – μy < 0

Upper-tail test:

H0: μx – μy ≤ 0H1: μx – μy > 0

Two-tail test:

H0: μx – μy = 0H1: μx – μy ≠ 0

Paired Samples

Decision Rules: Matched Pairs

/2 /2

-t -t/2t t/2

Reject H0 if t < -tn-1, Reject H0 if t > tn-1, Reject H0 if t < -tn-1 ,

or t > tn-1 ,

Wheren

sDd

td

0

has n - 1 d.f.

Chap 10-46

Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:

Matched Pairs Example

Number of Complaints: (2) - (1)Salesperson Before (1) After (2) Difference, di

C.B. 6 4 - 2 T.F. 20 6 -14 M.H. 3 2 - 1 R.K. 0 0 0 M.O. 4 0 - 4 -21

d = di

n

5.67

1n

)d(dS

2i

d

= - 4.2

Chap 10-47

Has the training made a difference in the number of complaints (at the = 0.01 level)?

- 4.2d =

1.6655.67/

04.2

n/s

Ddt

d

0

H0: μx – μy = 0H1: μx – μy 0

Test Statistic:

Critical Value = ± 4.604 d.f. = n - 1 = 4

Reject

/2

- 4.604 4.604

Decision: Do not reject H0

(t stat is not in the reject region)

Conclusion: There is not a significant change in the number of complaints.

Matched Pairs: Solution

Reject

/2

- 1.66 = .01

Chap 10-48

Two Sample Tests

Two Sample Tests

Population Means,

Independent Samples

Population Means,

Matched Pairs

Group 1 vs. independent Group 2

Same group before vs. after treatment

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

IntervalInterval

EstimationEstimation

Hypothesis Hypothesis TestsTests

Chap 10-49

Two Population Proportions

Goal: Test hypotheses for the difference between two population proportions, P1 – P2

Population proportions

Assumptions: Both sample sizes are large,

Chap 10-50

Two Population Proportions

Population proportions

Chap 10-51

Interval Estimation forTwo Population Proportions

Population proportions

Chap 10-52

Test Statistic forTwo Population Proportions

Population proportions

The test statistic for

H0: P1 – P2 = 0 is a z value:

1 2

0 0 0 0

1 2

ˆ ˆp pz

ˆ ˆ ˆ ˆp (1 p ) p (1 p )

n n

1 1 2 20

1 2

ˆ ˆn p n pp̂

n n

Where

Chap 10-53

Decision Rules: Proportions

Population proportions

Lower-tail test:

H0: p1 – p2 0H1: p1 – p2 < 0

Upper-tail test:

H0: p1 – p2 ≤ 0H1: p1 – p2 > 0

Two-tail test:

H0: p1– p2 = 0H1: p1 – p2 ≠ 0

/2 /2

-z -z/2z z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z or z > z

Chap 10-54

Example: Two Population Proportions

Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?

In a random sample, 36 of 72 men and 31 of 50 women indicated they would vote Yes

Test at the .05 level of significance

Chap 10-55

The hypothesis test is:H0: PM – PW = 0 (the two proportions are equal)

H1: PM – PW ≠ 0 (there is a significant difference between

proportions) The sample proportions are:

Men: = 36/72 = .50

Women: = 31/50 = .62

.549122

67

5072

50(31/50)72(36/72)

nn

pnpnp

yx

yyxx0

ˆˆ

ˆ

The estimate for the common overall proportion is:

Example: Two Population Proportions

(continued)

Mp̂

Wp̂

Chap 10-56

The test statistic for PM – PW = 0 is:

Example: Two Population Proportions

(continued)

.025

-1.96 1.96

.025

-1.31

Decision: Do not reject H0

Conclusion: There is not significant evidence of a difference between men and women in proportions who will vote yes.

1.31

50.549)(1.549

72.549)(1.549

.62.50

n)p(1p

n)p(1p

ppz

2

00

1

00

WM

ˆˆˆˆ

ˆˆ

Reject H0 Reject H0

Critical Values = ±1.96For = .05

Chap 10-57

Chapter Summary

Compared two dependent samples (paired samples) Performed paired sample t test for the mean

difference Compared two independent samples

Performed z test for the differences in two means Performed pooled variance t test for the differences

in two means Compared two population proportions

Performed z-test for two population proportions