Chapter 10b

Chapter 10b

•Hypothesis Tests About the DifferenceHypothesis Tests About the DifferenceBetween the Means of Two Populations:Between the Means of Two Populations: Independent Samples, Small-Sample Case Independent Samples, Small-Sample Case

•Using Excel to Conduct aUsing Excel to Conduct aHypothesis Test about Hypothesis Test about μμ1 – 1 – μμ2: Small Sample2: Small Sample

•Inference About the Difference between the Means of Two Populations: Matched Samples

Hypothesis Tests About the DifferenceHypothesis Tests About the DifferenceBetween the Means of Two Populations:Between the Means of Two Populations: Independent Samples, Small-Sample CaseIndependent Samples, Small-Sample Case

Example: Specific MotorsExample: Specific Motors

Recall that Specific Motors ofRecall that Specific Motors of

Detroit has developed a newDetroit has developed a new

automobile known as the M car.automobile known as the M car.

12 M cars and 8 J cars (from Japan)12 M cars and 8 J cars (from Japan)

were road tested to compare miles-per-gallon were road tested to compare miles-per-gallon

(mpg)(mpg)

performance. The sample statistics are shown performance. The sample statistics are shown

on theon the

next slide.next slide.

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

12 cars12 cars 8 cars8 cars

29.8 mpg 27.3 mpg29.8 mpg 27.3 mpg

2.56 mpg 1.81 mpg2.56 mpg 1.81 mpg

Can we conclude, using a .05 level of Can we conclude, using a .05 level of significance, that the miles-per-gallon (significance, that the miles-per-gallon (mpgmpg) ) performance of M cars isperformance of M cars isgreater than the miles-per-gallon performance of greater than the miles-per-gallon performance of J cars?J cars?


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

1. Determine the hypotheses.1. Determine the hypotheses.


Using the Test StatisticUsing the Test Statistic

HypothesisHypothesis Tests About the DifferenceTests About the DifferenceBetween the Means of Two Populations:Between the Means of Two Populations: Independent Samples, Small-Sample CaseIndependent Samples, Small-Sample Case

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

3. Select the test statistic.3. Select the test statistic.

= .05= .05

4. State the rejection rule.4. State the rejection rule.Reject Reject HH00 if if tt > 1.734 > 1.734

(18 degrees of freedom)(18 degrees of freedom)

tx x

s n n

( ) ( )

( )1 2 1 2

21 21 1

t

x x

s n n

( ) ( )

( )1 2 1 2

21 21 1

where:where:2 2

2 1 1 2 2

1 2

( 1) ( 1)

2

n s n ss

n n

2 22 1 1 2 2

1 2

( 1) ( 1)

2

n s n ss

n n


Hypothesis Tests About the DifferenceBetween the Means of Two Populations:

Independent Samples, Small-Sample Case

5. Compute the value of the test statistic.5. Compute the value of the test statistic.

Pooled Variance Estimator of Pooled Variance Estimator of 22

2 22 1 1 2 2

1 2

( 1) ( 1)

2

n s n ss

n n

2 22 1 1 2 2

1 2

( 1) ( 1)

2

n s n ss

n n

2 2

2 (12 1)(2.56) (8 1)(1.81) 95.02235.27902

12 8 2 18s

2 22 (12 1)(2.56) (8 1)(1.81) 95.0223

5.2790212 8 2 18

s

continuedcontinued



5. Compute the value of the test statistic.5. Compute the value of the test statistic.(continued)(continued)

tt Statistic Statistic

(29.8 27.3) 0 2.5 2.384

1.04871 15.27902( )12 8

t

(29.8 27.3) 0 2.5 2.384

1.04871 15.27902( )12 8

t

tx x

s n n

( ) ( )

( )1 2 1 2

21 21 1

t

x x

s n n

( ) ( )

( )1 2 1 2

21 21 1


6. Determine whether to reject 6. Determine whether to reject HH00..

At the .05 level of significance, the At the .05 level of significance, the samplesampleevidence indicates that the mean evidence indicates that the mean mpgmpg of of M carsM carsis greater than the mean is greater than the mean mpgmpg of J cars. of J cars.

tt = 2.384 > = 2.384 > tt.05.05 = 1.734, so we reject = 1.734, so we reject HH00..



Step 1Step 1 Select the Select the ToolsTools menu menu

Step 2Step 2 Choose the Choose the Data AnalysisData Analysis option option

Step 3Step 3 Choose Choose tt-Test: Two Sample Assuming Equal-Test: Two Sample Assuming Equal

Variances Variances from the list of Analysis Toolsfrom the list of Analysis Tools

… … continuedcontinued

Excel’s “Excel’s “tt-Test: Two Sample Assuming Equal -Test: Two Sample Assuming Equal Variances” ToolVariances” Tool

Using Excel to Conduct aUsing Excel to Conduct aHypothesis Test about Hypothesis Test about 11 – – 22: : Small SampleSmall Sample



Step 4Step 4 When the When the tt-Test: Two Sample Assuming-Test: Two Sample Assuming

Equal Variances dialog box appears:Equal Variances dialog box appears:


Enter A1:A13 in the Enter A1:A13 in the Variable 1 RangeVariable 1 Range box box

Enter B1:B9 in the Enter B1:B9 in the Variable 2 RangeVariable 2 Range box box

Type 0 in the Type 0 in the Hypothesized MeanHypothesized Mean

DifferenceDifference box box



Click Click OKOK

(Any upper left-hand corner cell indicating(Any upper left-hand corner cell indicating

where the output is to begin may be entered)where the output is to begin may be entered)

Enter D1 in the Enter D1 in the Output RangeOutput Range box boxSelect Select Output RangeOutput Range

Type .01 in the Type .01 in the AlphaAlpha box boxSelect Select LabelsLabels

Step 4Step 4 (continued) (continued)

Value WorksheetValue WorksheetA B C D E F

1 M Car J Car t-Test: Two-Sample Assuming Equal Variances

2 25.1 25.63 32.2 28.1 M Car J Car

4 31.7 27.9 Mean 29.78333 27.3

5 27.6 25.3 Variance 6.556061 3.265714

6 28.5 30.1 Observations 12 8

7 33.6 27.5 Pooled Variance 5.276481

8 30.8 25.1 Hypothesized Mean Diff. 0

9 26.2 28.8 df 18

10 29.0 t Stat 2.368555

11 31.0 P(T<=t) one-tail 0.014626

12 31.7 t Critical one-tail 1.734063

13 30.0 P(T<=t) two-tail 0.029251

14 t Critical two-tail 2.100924



Using the Using the p p ValueValue


5. Compute the 5. Compute the pp–value.–value.

The Excel worksheet shows The Excel worksheet shows pp-value = .0146-value = .0146


Because Because pp–value = .0146 < –value = .0146 < = .05, we reject = .05, we reject HH00..

The Excel worksheet shows The Excel worksheet shows tt = 2.369 = 2.369

Inference About the Difference between the Means of Two Populations: Matched

Samples

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.

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 services, UPX (United Parcel Express) and INTEXINTEX

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

Inference About the Difference between Inference About the Difference between the Means of Two Populations: the Means of Two Populations: Matched Matched

SamplesSamples

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 other report carried by INTEX. Do the data on thethe

next slide indicate a difference in mean next slide indicate a difference in mean deliverydelivery

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


SamplesSamples

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

Inference About the Difference between the Means of Two Populations: Matched Samples

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. Determine the hypotheses.1. Determine the hypotheses.



SamplesSamples


SamplesSamples

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

3. Select the test statistic.3. Select the test statistic.

= .05= .05

4. State the rejection rule.4. State the rejection rule.Reject Reject HH00 if | if |t|t| > 2.262 > 2.262

(9 degrees of freedom)(9 degrees of freedom)

where:where:


d

d

dt

s n

d

d

dt

s n

idd

n id

dn

and and 2( )

1i

d

d ds

n

2( )

1i

d

d ds

n




ddni ( ... )

.7 6 5

102 7d

dni ( ... )

.7 6 5

102 7

sd dndi

( ) ..

2

176 19

2 9sd dndi

( ) ..

2

176 19

2 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



At the .05 level of significance, the sample At the .05 level of significance, the sample evidenceevidence

indicates that there is a significant difference indicates that there is a significant difference betweenbetween

the mean delivery times for the two services. the mean delivery times for the two services.

tt = 2.94 > = 2.94 > tt.05/2.05/2 = 2.262, so we reject = 2.262, so we reject HH00..


Using Excel to Conduct aUsing Excel to Conduct aHypothesis Test about Hypothesis Test about 11 – – 22: : Matched Matched

SamplesSamples

Step 1Step 1 Select the Select the ToolsTools menu menu

Step 2Step 2 Choose the Choose the Data AnalysisData Analysis option option

Step 3Step 3 Choose Choose tt-Test: Paired Two Sample for Means-Test: Paired Two Sample for Means

from the list of Analysis Toolsfrom the list of Analysis Tools


Excel’s “Excel’s “tt-Test: Paired Two Sample for Means” -Test: Paired Two Sample for Means” ToolTool


SamplesSamples Excel’s “Excel’s “tt-Test: Paired Two Sample for Means” Tool-Test: Paired Two Sample for Means” Tool

Enter E2 (your choice) in the Enter E2 (your choice) in the Output RangeOutput Range boxbox

Click Click OKOK

Select Select Output RangeOutput RangeType .05 in the Type .05 in the AlphaAlpha box boxSelect Select LabelsLabels

Type 0 in the Type 0 in the Hypothesized Mean DifferenceHypothesized Mean Difference boxbox

Enter C1:C11 in the Enter C1:C11 in the Variable 2 RangeVariable 2 Range box boxEnter B1:B11 in the Enter B1:B11 in the Variable 1 RangeVariable 1 Range box box

Step 4Step 4 When the When the tt-Test: Paired Two Sample for Means-Test: Paired Two Sample for Means dialog box appears:dialog box appears:


SamplesSamples

Value WorksheetValue Worksheet


SamplesSamples

A B C D E F G1 Office UPX INTEX2 Seattle 32 25 t-Test: Paired Two Sample for Means3 L.A. 30 244 Boston 19 15 UPX INTEX5 Cleveland 16 15 Mean 17.7 156 N.Y.C. 15 13 Variance 62.011 31.77787 Houston 18 15 Observations 10 108 Atlanta 14 15 Pearson Correlation 0.96129 St. Louis 10 8 Hypothesized Mean Difference 010 Milwauk. 7 9 df 911 Denver 16 11 t Stat 2.936212 P(T<=t) one-tail 0.008313 t Critical one-tail 1.833114 P(T<=t) two-tail 0.016615 t Critical two-tail 2.2622


SamplesSamples Using the Using the p p ValueValue


5. Compute the 5. Compute the pp–value.–value.

The Excel worksheet shows The Excel worksheet shows pp-value = .0166-value = .0166


Because Because pp–value = .0166 < –value = .0166 < = .05, we reject = .05, we reject HH00..

The Excel worksheet shows The Excel worksheet shows tt = 2.9362 = 2.9362

Chapter 10b

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