1 Chapter 11 – Test for the Equality of k Chapter 11 – Test for the Equality of k Population Means Population Means Rejection Rejection Rule Rule where the value of where the value of F F is based on an is based on an F F distribution with distribution with k k - 1 numerator d.f. and - 1 numerator d.f. and n n T - - k k denominator d.f. denominator d.f. Reject Reject H H 0 if if p p -value -value < < p p -value Approach: -value Approach: Critical Value Approach: Critical Value Approach: Reject Reject H H 0 if if F F > > F F Hypothese Hypothese s s H H 0 : : 1 = = 2 = = 3 = = . . . . . . = = k H H a : Not all population means are equal : Not all population means are equal Test Test Statistic Statistic F F = MSTR/MSE = MSTR/MSE
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Chapter 11 – Test for the Equality of k Chapter 11 – Test for the Equality of k Population MeansPopulation Means
Rejection Rejection RuleRule
where the value of where the value of FF is based on an is based on an FF distribution distribution with with kk - 1 numerator d.f. and - 1 numerator d.f. and nnTT - - kk denominator denominator d.f.d.f.
Reject Reject HH00 if if pp-value -value << pp-value Approach:-value Approach:
Critical Value Approach:Critical Value Approach: Reject Reject HH00 if if FF >> FF
HypotheseHypothesess
HH00: : 11==22==33==. . . . . . = = kk
HHaa: Not all population means are equal: Not all population means are equal
Test Test StatisticStatistic
FF = MSTR/MSE = MSTR/MSE
22
Test for the Equality of k Population Test for the Equality of k Population MeansMeans
33
AutoShine, Inc. is considering marketing a AutoShine, Inc. is considering marketing a long-long-
lasting car wax. Three different waxes (Type 1, lasting car wax. Three different waxes (Type 1, Type 2,Type 2,
and Type 3) have been developed.and Type 3) have been developed.
Example: AutoShine, IncExample: AutoShine, Inc..
In order to test the durability of these waxes, In order to test the durability of these waxes, 5 new5 new
cars were waxed with Type 1, 5 with Type 2, cars were waxed with Type 1, 5 with Type 2, and 5and 5
with Type 3. Each car was then repeatedly runwith Type 3. Each car was then repeatedly run
through an automatic carwash until the wax through an automatic carwash until the wax coatingcoating
showed signs of deterioration.showed signs of deterioration.
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
44
The number of times each car went The number of times each car went through thethrough the
carwash before its wax deteriorated is shown carwash before its wax deteriorated is shown on theon the
next slide. AutoShine, Inc. must decide which next slide. AutoShine, Inc. must decide which waxwax
to market. Are the three waxesto market. Are the three waxes
equally effective?equally effective?
Example: AutoShine, Inc.Example: AutoShine, Inc.
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
Factor . . . Car waxFactor . . . Car waxTreatments . . . Type I, Type 2, Type 3Treatments . . . Type I, Type 2, Type 3Experimental units . . . CarsExperimental units . . . Cars
Response variable . . . Number of washesResponse variable . . . Number of washes
55
1122334455
27273030292928283131
33332828313130303030
29292828303032323131
Sample MeanSample MeanSample VarianceSample Variance
ObservationObservationWaxWax
Type 1Type 1WaxWax
Type 2Type 2WaxWax
Type 3Type 3
2.52.5 3.3 3.3 2.5 2.529.0 30.429.0 30.4 30.0 30.0
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
66
HypothesesHypotheses
where: where:
1 1 = mean number of washes using Type 1 wax= mean number of washes using Type 1 wax
2 2 = mean number of washes using Type 2 wax= mean number of washes using Type 2 wax
3 3 = mean number of washes using Type 3 wax= mean number of washes using Type 3 wax
HH00: : 11==22==33
HHaa: Not all the means are equal: Not all the means are equal
Testing for the Equality of k Testing for the Equality of k Population Means:Population Means:
A Completely Randomized A Completely Randomized Experimental DesignExperimental Design
77
Because the sample sizes are all Because the sample sizes are all equal:equal:
MSE = 33.2/(15 - 3) = 2.77MSE = 33.2/(15 - 3) = 2.77
Within-treatments estimate of σWithin-treatments estimate of σ22 = {(2.5) + (3.3) + = {(2.5) + (3.3) + (2.5)}/3 = 33.2(2.5)}/3 = 33.2
{(29–29.8){(29–29.8)22 + (30.4–29.8) + (30.4–29.8)22 + (30–29.8) + (30–29.8)22 }/2= .52}/2= .52
Mean Square Error of Within Mean Square Error of Within TreatmentsTreatments
Mean Square Between Treatments Mean Square Between Treatments
2xs 2xs
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
1 2 3( )/ 3x x x x 1 2 3( )/ 3x x x x = (29 + 30.4 + 30)/3 = 29.8= (29 + 30.4 + 30)/3 = 29.8
2 2xx
s ns 2 2xx
s ns Mean Square Between Mean Square Between Treatments Treatments
2
2 2 2 5(.52) 2.6xx xx
ss or s ns
n
22 2 2 5(.52) 2.6xx xx
ss or s ns
n
88
Rejection RuleRejection Rule
where where FF.05.05 = 3.89 is based on an = 3.89 is based on an FF distribution distributionwith 2 numerator degrees of freedom and 12with 2 numerator degrees of freedom and 12denominator degrees of freedomdenominator degrees of freedom
pp-Value Approach: Reject -Value Approach: Reject HH00 if if pp-value -value << .05 .05
Critical Value Approach: Reject Critical Value Approach: Reject HH00 if if FF >> 3.89 3.89
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
99
Test StatisticTest Statistic
There is insufficient evidence to conclude thatThere is insufficient evidence to conclude thatthe mean number of washes for the three waxthe mean number of washes for the three waxtypes are not all the same.types are not all the same.
ConclusionConclusion
F F = MSTR/MSE = 2.60/2.77 = .939 = MSTR/MSE = 2.60/2.77 = .939
The The pp-value is greater than .10, where -value is greater than .10, where FF = 2.81. = 2.81. (Excel provides a (Excel provides a pp-value of .42.)-value of .42.) Therefore, we cannot reject Therefore, we cannot reject HH00..
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
1010
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
TreatmentsTreatments
ErrorError
TotalTotal
22
1414
5.25.2
33.233.2
38.438.4
1212
2.602.60
2.772.77
.939.939
ANOVA TableANOVA Table
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
A Completely Randomized Experimental A Completely Randomized Experimental DesignDesign
pp-Value-Value
.42.42
1111
Example: Reed Example: Reed ManufacturingManufacturing
Janet Reed would like to know if there is Janet Reed would like to know if there is anyanysignificant difference in the mean number of significant difference in the mean number of hourshoursworked per week for the department worked per week for the department managers atmanagers ather three manufacturing plants (in Buffalo,her three manufacturing plants (in Buffalo,Pittsburgh, and Detroit). An Pittsburgh, and Detroit). An FF test will be test will be conductedconductedusing using = .05. = .05.
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
An Observational StudyAn Observational Study
1212
Example: Reed Example: Reed ManufacturingManufacturing A simple random sample of five managers A simple random sample of five managers fromfromeach of the three plants was taken and the each of the three plants was taken and the number ofnumber ofhours worked by each manager in the previous hours worked by each manager in the previous weekweekis shown on the next slide.is shown on the next slide.
Testing for the Equality of k Testing for the Equality of k Population Means:Population Means:
An Observational StudyAn Observational Study
Factor . . . Manufacturing plantFactor . . . Manufacturing plantTreatments . . . Buffalo, Pittsburgh, DetroitTreatments . . . Buffalo, Pittsburgh, DetroitExperimental units . . . ManagersExperimental units . . . Managers
Response variable . . . Number of hours workedResponse variable . . . Number of hours worked
1313
1122334455
48485454575754546262
73736363666664647474
51516363616154545656
Plant 1Plant 1BuffaloBuffalo
Plant 2Plant 2PittsburghPittsburgh
Plant 3Plant 3DetroitDetroitObservationObservation
Sample MeanSample MeanSample VarianceSample Variance
5555 68 68 57 5726.026.0 26.5 26.5 24.5 24.5
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
An Observational StudyAn Observational Study
1414
HH00: : 11==22==33
HHaa: Not all the means are equal: Not all the means are equalwhere: where: 1 1 = mean number of hours worked per= mean number of hours worked per
week by the managers at Plant 1week by the managers at Plant 1 2 2 = mean number of hours worked per= mean number of hours worked per week by the managers at Plant 2week by the managers at Plant 23 3 = mean number of hours worked per= mean number of hours worked per week by the managers at Plant 3week by the managers at Plant 3
1. Develop the hypotheses.1. Develop the hypotheses.
pp -Value and Critical Value Approaches -Value and Critical Value Approaches
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
An Observational StudyAn Observational Study
1515
2. Specify the level of significance.2. Specify the level of significance. = .05= .05
pp -Value and Critical Value Approaches -Value and Critical Value Approaches
3. Compute the value of the test statistic.3. Compute the value of the test statistic.
{(55 - 60){(55 - 60)22 + (68 - 60) + (68 - 60)22 + (57 - 60) + (57 - 60)22 }/2= 49 }/2= 49= (55 + 68 + 57)/3 = 60= (55 + 68 + 57)/3 = 60xx
(Sample sizes are all equal.)(Sample sizes are all equal.)Mean Square Due to TreatmentsMean Square Due to Treatments
Testing for the Equality of k Population Testing for the Equality of k Population Means:Means:
An Observational StudyAn Observational Study
2xs 2xs
2 5(49) 245xMSTR ns 2 5(49) 245xMSTR ns
1616
3. Compute the value of the test statistic.3. Compute the value of the test statistic.
MSE = {(26.0) + (26.5) + (24.5)}/3 = 25.667MSE = {(26.0) + (26.5) + (24.5)}/3 = 25.667
Mean Square Due to ErrorMean Square Due to Error
(con’t.)(con’t.)
FF = MSTR/MSE = 245/25.667 = 9.55 = MSTR/MSE = 245/25.667 = 9.55
Testing for the Equality of k Testing for the Equality of k Population Means:Population Means:
An Observational StudyAn Observational Study
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