1 1 slide ma4704 problem solving 4 statistical inference about means and proportions with two...

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Problem solving 4Problem solving 4 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

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

Interval Estimate of Interval Estimate 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.”

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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.

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


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

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.


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

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3. Compute the Alarm Signal : Noise3. Compute the Alarm Signal : Noise


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

Because Alarm Signal: Noise= 6.49 Because Alarm Signal: Noise= 6.49 >> 2.33, we 2.33, we reject reject HH00..

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 Alarm Signal: Noise if Alarm Signal: Noise >> 2.332.33

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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Example: Express DeliveriesExample: Express Deliveries

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

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

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


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

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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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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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.

Interval Estimation of Interval Estimation of pp11 - - pp22

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

HypothesesHypotheses

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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1. Develop the hypotheses.1. Develop the hypotheses.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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3. Compute the value of the test statistic.3. Compute the value of the test statistic.

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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Because 1.56 < 1.645, we Fail to Reject Because 1.56 < 1.645, we Fail to 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.

1 1 slide ma4704 problem solving 4 statistical inference about means and proportions with two...

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