1 1 Slide © 2005 Thomson/South-Western Chapter 10 Statistical Inference About Means and Proportions...

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© 2005 Thomson/South-Western© 2005 Thomson/South-Western

Chapter 10Chapter 10 Statistical Inference About Means and Statistical Inference About Means and

Proportions With Two PopulationsProportions With Two Populations

Inferences About the Difference BetweenInferences About the Difference Between

Two Population Means: Two Population Means: 11 and and 22 Known Known

Inferences About the Difference BetweenInferences About the Difference Between Two Population ProportionsTwo Population Proportions

Inferences About the Difference BetweenInferences About the Difference Between Two Population Means: Matched SamplesTwo Population Means: Matched Samples

Inferences About the Difference BetweenInferences About the Difference Between

Two Population Means: Two Population Means: 11 and and 22 Unknown Unknown

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Inferences About the Difference BetweenInferences About the Difference BetweenTwo Population Means: Two Population Means: 1 1 and and 2 2 Known Known

Interval Estimation of Interval Estimation of 11 – – 22

Hypothesis Tests About Hypothesis Tests About 11 – – 22

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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

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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

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Interval EstimateInterval Estimate

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

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


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.”

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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.

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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.

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Estimating the Difference BetweenEstimating the Difference BetweenTwo Population MeansTwo Population Means

11 – – 22 = difference between= difference between the mean distancesthe mean distances

xx11 - - xx22 = Point Estimate of = Point Estimate of 11 –– 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


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

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


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

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


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

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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

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x x zn n1 2 2

12

1

22

2

2 2

17 1 9615120

2080

/ .( ) ( )

x x zn n1 2 2

12

1

22

2

2 2

17 1 9615120

2080

/ .( ) ( )

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

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Hypothesis Tests About Hypothesis Tests About 1 1 2 2:: 1 1 and and 2 2 Known Known

HypothesesHypotheses

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

1 2 0: aH D 1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D 0 1 2 0: H D 0 1 2 0: H D

1 2 0: aH D 1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D 1 2 0: aH D 1 2 0: aH D

Left-tailedLeft-tailed Right-tailedRight-tailed Two-tailedTwo-tailed Test StatisticTest Statistic

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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?

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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


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

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



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

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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.


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.

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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.

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Inferences About the Difference BetweenInferences About the Difference BetweenTwo Population Means: Two Population Means: 1 1 and and 2 2

UnknownUnknown Interval Estimation of Interval Estimation of 11 – – 22

Hypothesis Tests About Hypothesis Tests About 11 – – 22

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

When When 1 1 and and 2 2 are unknown, we will: are unknown, we will:

• replace replace zz/2/2 with with tt/2/2. .

• use the sample standard deviations use the sample standard deviations ss11 and and ss22

as estimates of as estimates of 1 1 and and 2 2 , and , and

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2 21 2

1 2 / 21 2

s sx x t

n n 2 21 2

1 2 / 21 2

s sx x t

n n

Where the degrees of freedom for Where the degrees of freedom for tt/2/2 are: are:

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


22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

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Example: Specific MotorsExample: Specific Motors

Difference Between Two Population Difference Between Two Population Means:Means:

11 and and 2 2 Unknown Unknown

Specific Motors of DetroitSpecific Motors of Detroit

has developed a new automobilehas developed a new automobile

known as the M car. 24 M carsknown as the M car. 24 M cars

and 28 J cars (from Japan) were roadand 28 J cars (from Japan) were road

tested to compare miles-per-gallon (mpg) performance. tested to compare miles-per-gallon (mpg) performance.

The sample statistics are shown on the next slide.The sample statistics are shown on the next slide.

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11 and and 2 2 Unknown Unknown Example: Specific MotorsExample: Specific Motors

Sample SizeSample Size

Sample MeanSample Mean

Sample Std. Dev.Sample Std. Dev.

Sample #1Sample #1M CarsM Cars

Sample #2Sample #2J CarsJ Cars

24 cars24 cars 2 28 cars8 cars

29.8 mpg 27.3 mpg29.8 mpg 27.3 mpg

2.56 mpg 1.81 mpg2.56 mpg 1.81 mpg

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11 and and 2 2 Unknown Unknown

Let us develop a 90% confidenceLet us develop a 90% confidence

interval estimate of the differenceinterval estimate of the difference

between the mpg performances ofbetween the mpg performances of

the two models of automobile.the two models of automobile.


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

Point Estimate of Point Estimate of 1 1 2 2

where:where:

11 = mean miles-per-gallon for the = mean miles-per-gallon for the population of M carspopulation of M cars

22 = mean miles-per-gallon for the = mean miles-per-gallon for the population of J carspopulation of J cars

= 29.8 - 27.3= 29.8 - 27.3

= 2.5 mpg= 2.5 mpg

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Interval Estimation of Interval Estimation of 1 1 2 2:: 1 1 and and 2 2 Unknown Unknown

The degrees of freedom for The degrees of freedom for tt/2/2 are: are:22 2

2 22 2

(2.56) (1.81)24 28

24.07 241 (2.56) 1 (1.81)

24 1 24 28 1 28

df

22 2

2 22 2

(2.56) (1.81)24 28

24.07 241 (2.56) 1 (1.81)

24 1 24 28 1 28

df

With With /2 = .05 and /2 = .05 and dfdf = 24, = 24, tt/2/2 = 1.711 = 1.711

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Interval Estimation of Interval Estimation of 1 1 2 2:: 1 1 and and 2 2 Unknown Unknown

2 2 2 21 2

1 2 / 21 2

(2.56) (1.81) 29.8 27.3 1.711

24 28

s sx x t

n n 2 2 2 21 2

1 2 / 21 2

(2.56) (1.81) 29.8 27.3 1.711

24 28

s sx x t

n n

We are 90% confident that the difference betweenWe are 90% confident that the difference betweenthe miles-per-gallon performances of M cars and J carsthe miles-per-gallon performances of M cars and J carsis 1.431 to 3.569 mpg.is 1.431 to 3.569 mpg.

2.5 2.5 ++ 1.069 or 1.431 to 3.569 mpg 1.069 or 1.431 to 3.569 mpg

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Hypothesis Tests About Hypothesis Tests About 1 1 2 2:: 1 1 and and 2 2 Unknown Unknown


1 2 0

2 21 2

1 2

( )x x Dt

s sn n

1 2 0

2 21 2

1 2

( )x x Dt

s sn n

1 2 0: aH D 1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D 0 1 2 0: H D 0 1 2 0: H D

1 2 0: aH D 1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D 1 2 0: aH D 1 2 0: aH D

Left-tailedLeft-tailed Right-tailedRight-tailed Two-tailedTwo-tailed Test StatisticTest Statistic

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Can we conclude, using aCan we conclude, using a

.05 level of significance, that the.05 level of significance, that the

miles-per-gallon (miles-per-gallon (mpgmpg) performance) performance

of M cars is greater than the miles-per-of M cars is greater than the miles-per-

gallon performance of J cars?gallon performance of J cars?

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

HHaa: : 1 1 - - 22 > 0 > 0where: where: 11 = mean = mean mpgmpg for the population of M cars for the population of M cars22 = mean = mean mpgmpg for the population of J cars for the population of J cars




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2. Specify the level of significance.2. Specify the level of significance.


= .05= .05



1 2 0

2 2 2 21 2

1 2

( ) (29.8 27.3) 0 4.003

(2.56) (1.81)24 28

x x Dt

s sn n

1 2 0

2 2 2 21 2

1 2

( ) (29.8 27.3) 0 4.003

(2.56) (1.81)24 28

x x Dt

s sn n

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pp –Value Approach –Value Approach

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

The degrees of freedom for The degrees of freedom for tt are: are:22 2

2 22 2

(2.56) (1.81)

24 2824.07 24

1 (2.56) 1 (1.81)

24 1 24 28 1 28

df

22 2

2 22 2

(2.56) (1.81)

24 2824.07 24

1 (2.56) 1 (1.81)

24 1 24 28 1 28

df

Because Because tt = 4.003 > = 4.003 > tt.005.005 = 2.797, the = 2.797, the pp–value < .005.–value < .005.

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We are at least 95% confident that the We are at least 95% confident that the miles-per-gallon (miles-per-gallon (mpgmpg) performance of M ) performance of M cars is greater than the miles-per-gallon cars is greater than the miles-per-gallon performance of J cars?.performance of J cars?.




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For For = .05 and = .05 and dfdf = 24, = 24, tt.05.05 = 1.711 = 1.711

Reject Reject HH00 if if tt >> 1.711 1.711


Because 4.003 Because 4.003 >> 1.711, we reject 1.711, we reject HH00..

We are at least 95% confident that the We are at least 95% confident that the miles-per-gallon (miles-per-gallon (mpgmpg) performance of M ) performance of M cars is greater than the miles-per-gallon cars is greater than the miles-per-gallon performance of J cars?.performance of J cars?.

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With a With a matched-sample designmatched-sample design each sampled item each sampled item provides a pair of data values.provides a pair of data values.

This design often leads to a smaller sampling This design often leads to a smaller sampling errorerror than the independent-sample design than the independent-sample design becausebecause variation between sampled items is variation between sampled items is eliminated as aeliminated as a source of sampling error.source of sampling error.

Inferences About the Difference BetweenInferences About the Difference BetweenTwo Population Means: Matched SamplesTwo Population Means: Matched Samples

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Example: Express DeliveriesExample: Express Deliveries


A Chicago-based firm hasA Chicago-based firm has

documents that must be quicklydocuments that must be quickly

distributed to district officesdistributed to district offices

throughout the U.S. The firmthroughout the U.S. The firm

must decide between two deliverymust decide between two delivery

services, UPX (United Parcel Express) and INTEXservices, UPX (United Parcel Express) and INTEX

(International Express), to transport its documents.(International Express), to transport its documents.

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Example: Express DeliveriesExample: Express Deliveries


In testing the delivery timesIn testing the delivery times

of the two services, the firm sentof the two services, the firm sent

two reports to a random sampletwo reports to a random sample

of its district offices with oneof its district offices with one

report carried by UPX and thereport carried by UPX and the

other report carried by INTEX. Do the data on theother report carried by INTEX. Do the data on the

next slide indicate a difference in mean deliverynext slide indicate a difference in mean delivery

times for the two services? Use a .05 level oftimes for the two services? Use a .05 level of

significance.significance.

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32323030191916161515181814141010 771616

2525242415151515131315151515 88 991111

UPXUPX INTEXINTEX DifferenceDifferenceDistrict OfficeDistrict OfficeSeattleSeattleLos AngelesLos AngelesBostonBostonClevelandClevelandNew YorkNew YorkHoustonHoustonAtlantaAtlantaSt. LouisSt. LouisMilwaukeeMilwaukeeDenverDenver

Delivery Time (Hours)Delivery Time (Hours)

7 7 6 6 4 4 1 1 2 2 3 3 -1 -1 2 2 -2 -2 55


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HH00: : d d = 0= 0

HHaa: : dd Let Let d d = the mean of the = the mean of the differencedifference values for the values for the two delivery services for the populationtwo delivery services for the population of district officesof district offices




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ddni ( ... )

.7 6 5

102 7d

dni ( ... )

.7 6 5

102 7

sd dndi

( ) ..

2

176 1

92 9s

d dndi

( ) ..

2

176 1

92 9

2.7 0 2.94

2.9 10d

d

dt

s n

2.7 0

2.942.9 10

d

d

dt

s n

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We are at least 95% confident that We are at least 95% confident that there is a difference in mean delivery there is a difference in mean delivery times for the two services?times for the two services?


For For tt = 2.94 and = 2.94 and dfdf = 9, the = 9, the pp–value is –value is betweenbetween.02 and .01. (This is a two-tailed test, so we .02 and .01. (This is a two-tailed test, so we double the upper-tail areas of .01 and .005.)double the upper-tail areas of .01 and .005.)




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For For = .05 and = .05 and dfdf = 9, = 9, tt.025.025 = 2.262. = 2.262.

Reject Reject HH00 if if tt >> 2.262 2.262


Because Because tt = 2.94 = 2.94 >> 2.262, we reject 2.262, we reject HH00..

We are at least 95% confident that there We are at least 95% confident that there is a difference in mean delivery times for is a difference in mean delivery times for the two services?the two services?

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Inferences About the Difference BetweenInferences About the Difference BetweenTwo Population ProportionsTwo Population Proportions

Interval Estimation of Interval Estimation of pp11 - - pp22

Hypothesis Tests About Hypothesis Tests About pp11 - - pp22

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Expected ValueExpected Value

Sampling Distribution of Sampling Distribution of p p1 2p p1 2

E p p p p( )1 2 1 2 E p p p p( )1 2 1 2

p pp pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( ) p p

p pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( )

where: where: nn11 = size of sample taken from population 1 = size of sample taken from population 1

nn22 = size of sample taken from population 2 = size of sample taken from population 2

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

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If the sample sizes are large, the sampling distributionIf the sample sizes are large, the sampling distribution of can be approximated by a normal probabilityof can be approximated by a normal probability distribution. distribution.

If the sample sizes are large, the sampling distributionIf the sample sizes are large, the sampling distribution of can be approximated by a normal probabilityof can be approximated by a normal probability distribution. distribution.

p p1 2p p1 2

The sample sizes are sufficiently large if The sample sizes are sufficiently large if allall of these of these conditions are met:conditions are met: The sample sizes are sufficiently large if The sample sizes are sufficiently large if allall of these of these conditions are met:conditions are met:

nn11pp11 >> 5 5 nn11(1 - (1 - pp11) ) >> 5 5

nn22pp22 >> 5 5 nn22(1 - (1 - pp22) ) >> 5 5


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pp11 – – pp22pp11 – – pp22

p pp pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( ) p p

p pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( )

p p1 2p p1 2

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1 1 2 21 2 / 2

1 2

(1 ) (1 )p p p pp p z

n n

1 1 2 21 2 / 2

1 2

(1 ) (1 )p p p pp p z

n n

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Market Research Associates isMarket Research Associates isconducting research to evaluate theconducting research to evaluate theeffectiveness of a client’s new adver-effectiveness of a client’s new adver-tising campaign. Before the newtising campaign. Before the newcampaign began, a telephone surveycampaign began, a telephone surveyof 150 households in the test marketof 150 households in the test marketarea showed 60 households “aware” ofarea showed 60 households “aware” ofthe client’s product. the client’s product.


Example: Market Research AssociatesExample: Market Research Associates

The new campaign has been initiated with TV andThe new campaign has been initiated with TV andnewspaper advertisements running for three weeks.newspaper advertisements running for three weeks.

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A survey conducted immediatelyA survey conducted immediatelyafter the new campaign showed 120after the new campaign showed 120of 250 households “aware” of theof 250 households “aware” of theclient’s product.client’s product.



Does the data support the positionDoes the data support the positionthat the advertising campaign has that the advertising campaign has provided an increased awareness ofprovided an increased awareness ofthe client’s product?the client’s product?

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Point Estimator of the Difference BetweenPoint Estimator of the Difference BetweenTwo Population ProportionsTwo Population Proportions

= sample proportion of households “aware” of the= sample proportion of households “aware” of the product product afterafter the new campaign the new campaign = sample proportion of households “aware” of the= sample proportion of households “aware” of the product product beforebefore the new campaign the new campaign

1p1p

2p2p

pp11 = proportion of the population of households = proportion of the population of households “ “aware” of the product aware” of the product afterafter the new campaign the new campaign pp22 = proportion of the population of households = proportion of the population of households “ “aware” of the product aware” of the product beforebefore the new campaign the new campaign

1 2

120 60.48 .40 .08

250 150p p 1 2

120 60.48 .40 .08

250 150p p

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.08 .08 ++ 1.96(.0510) 1.96(.0510)

.08 .08 ++ .10 .10

.48(.52) .40(.60).48 .40 1.96

250 150

.48(.52) .40(.60).48 .40 1.96

250 150


Hence, the 95% confidence interval for the differenceHence, the 95% confidence interval for the differencein before and after awareness of the product isin before and after awareness of the product is-.02 to +.18.-.02 to +.18.

For For = .05, = .05, zz.025.025 = 1.96: = 1.96:

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Hypothesis Tests about Hypothesis Tests about pp11 - - pp22


HH00: : pp11 - - pp22 << 0 0

HHaa: : pp11 - - pp22 > 0 > 0 1 2: 0aH p p 1 2: 0aH p p 0 1 2: 0H p p 0 1 2: 0H p p 0 1 2: 0H p p 0 1 2: 0H p p

1 2: 0aH p p 1 2: 0aH p p 0 1 2: 0H p p 0 1 2: 0H p p 1 2: 0aH p p 1 2: 0aH p p

Left-tailedLeft-tailed Right-tailedRight-tailed Two-tailedTwo-tailed

We focus on tests involving no difference We focus on tests involving no difference betweenbetweenthe two population proportions (i.e. the two population proportions (i.e. pp11 = = pp22))

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1 2p p1 2p p Pooled Estimate of Standard Error of Pooled Estimate of Standard Error of

1 2

1 2

1 1(1 )p p p p

n n

1 2

1 2

1 1(1 )p p p p

n n

1 1 2 2

1 2

n p n pp

n n

1 1 2 2

1 2

n p n pp

n n

where:where:

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1 2

1 2

( )

1 1(1 )

p pz

p pn n

1 2

1 2

( )

1 1(1 )

p pz

p pn n

Test StatisticTest Statistic

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Can we conclude, using a .05 levelCan we conclude, using a .05 level

of significance, that the proportion ofof significance, that the proportion of

households aware of the client’s producthouseholds aware of the client’s product

increased after the new advertisingincreased after the new advertising

campaign?campaign?



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pp -Value and Critical Value Approaches -Value and Critical Value Approaches

HH00: : pp11 - - pp22 << 0 0

HHaa: : pp11 - - pp22 > 0 > 0

pp11 = proportion of the population of households = proportion of the population of households “ “aware” of the product aware” of the product afterafter the new campaign the new campaign pp22 = proportion of the population of households = proportion of the population of households “ “aware” of the product aware” of the product beforebefore the new campaign the new campaign

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pp -Value and Critical Value Approaches -Value and Critical Value Approaches

p

250 48 150 40250 150

180400

45(. ) (. )

.p

250 48 150 40250 150

180400

45(. ) (. )

.

sp p1 245 55 1

2501150 0514 . (. )( ) .sp p1 2

45 55 1250

1150 0514 . (. )( ) .

(.48 .40) 0 .08 1.56

.0514 .0514z

(.48 .40) 0 .08 1.56

.0514 .0514z

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We We cannotcannot conclude that the proportion of conclude that the proportion of householdshouseholdsaware of the client’s product increased after aware of the client’s product increased after the newthe newcampaign.campaign.


For For zz = 1.56, the = 1.56, the pp–value = .0594–value = .0594

Because Because pp–value > –value > = .05, we = .05, we cannotcannot reject reject HH00..


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Because 1.56 < 1.645, we cannot reject Because 1.56 < 1.645, we cannot reject HH00..

For For = .05, = .05, zz.05.05 = 1.645 = 1.645


Reject Reject HH00 if if zz >> 1.645 1.645

We We cannotcannot conclude that the proportion of conclude that the proportion of householdshouseholdsaware of the client’s product increased after aware of the client’s product increased after the newthe newcampaign.campaign.

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