of 52 /52
TOPIC 11 TOPIC 11 Analysis of Analysis of Variance Variance

Embed Size (px)

### Transcript of TOPIC 11 Analysis of Variance. Draw Sample Populations μ 1 = μ 2 = μ 3 = μ 4 = ….. μ n...

• TOPIC 11Analysis of Variance

• Analysis of Variance Draw SamplePopulations1 = 2 = 3 = 4 = .. n

• Road MapFactorial DesignDecision MakingOne/Two SamplesAnalysis of VarianceCompletely Randomized Design2 TestsRandomized Block Design

• Completely Randomized DesignIn many situations, you need to examine difference among more than two groups (populations). The group involved can be classified according to factor level of interest (treatments). For example, a factor such as baking temperature may have several groups defined by numerical levels such as 300o, 350o, 400o, 450o and a factor such as preferred supplier for a certain manufacturer may have several groups defined by categorical levels such as Supplier 1, Supplier 2, Supplier 3, Supplier 4.When there is a single factor, the experimental design is called a completely randomized design.

• One Factor Design ExperimentSample of tensile strength of synthetic fibers from four different suppliersDo the synthetic fibers from each of four suppliers have equal strength?

Supplier 1Supplier 2Supplier 3Supplier 418.524.017.219.918.026.325.324.021.224.520.625.220.8 24.722.925.419.922.617.520.4Sample Mean19.5224.2622.8421.16Grand Mean21.945Sample Standard Deviation2.691.922.132.98

• One Way ANOVAThe ANOVA procedure used for the completely randomized design is referred to as the One-Way ANOVA It is the extension of the t-test for the difference between two means.Although ANOVA is the acronym for analysis of variance, the term is misleading because the objective is to analyze differences among the group means, not the variances.By analyzing the variation among and within the groups, you can make conclusions about possible differences in group means.

• Partitioning the Total VariationIn One Way ANOVA, the total variation is subdivided into two parts:Variation that is due to differences among the treatmentsVariation that is due to differences within the treatmentsThe symbol k is used to indicate the number of treatmentsPartitioning the Total Variation: SST = SSTr + SSE

• Hypothesis to be TestedAssumptions:k groups represent populationsIts values are randomly and independently selectedFollowing a normal distributionHaving equal variancesRefer back to the table of synthetic fibers from four suppliers. The null hypothesis of no differences in the population meansis tested against the alternative that at least two of the k treatment means differ (or not all j are equal, where j = 1,2,, k)

• Sums of Squares FormulaWe divide the total variation into variation among the treatments and variation within the treatments.The total variation is presented by the sum of squares total (SST)where= Grand meanXTreat. 1Treat. 2Treat. 3Response, X

• Sums of Squares FormulaThe variation among the treatments is presented by the sum of squares treatments (SSTr)XX3X2X1Treat. 1Treat. 2Treat. 3Response, X

• The within-group variation is given by the sum of squares within treatments (SSE)Sums of Squares FormulaX2X1X3Treat. 1Treat. 2Treat. 3Response, X

• Sums of Squares Formula

• To convert the sums of squares to mean squares, we divide SSTr , SSE and SST by degrees of freedom. We have MSTr (mean square treatments), MSE (mean square error), and MST (mean squares total)Mean SquaresTotal degrees of freedom = (k - 1) + (n k) = n - 1

• Source of VariationDegrees of FreedomSum of SquaresMean Square (Variance)FTreatmentk - 1SSTrMSTr = SSTr/(k - 1)MSTrMSEErrorn - kSSEMSE = SSE/(n - k)Totaln - 1SST= SSTr+SSEOne-Way ANOVA Summary Table

• F Test for Differences Among More than Two MeansMSA and MSE provide estimates of the overall variance in the data. To test the null hypothesis:againstyou compute the One-Way ANOVA F test statistic, which is given byF from F-distribution with (k-1) numerator and (n-k) denominator degrees of freedomReject H0 if F > F , Otherwise, do not reject

• As production manager, you want to see if three filling machines have different mean filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in mean filling times? Mach1 Mach2 Mach3 25.4023.4020.00 26.3121.8022.20 24.1023.5019.75 23.7422.7520.60 25.1021.6020.40One-Way ANOVA Example

• Treatmentk 1 =3 - 1 = 2SSTr =47.1640MSTr =23.5820MSTrMSE

= 25.60Errorn k =15 - 3 = 12SSE =11.0532MSE =.9211Totaln 1 =15 - 1 = 14SST =58.2172Source of VariationDegrees of FreedomSum of SquaresMean Square (Variance)FExample Solution

• F0F = 3.89H0: 1 = 2 = 3Ha: Not All Equal = .05 1 = 2 2 = 12 Critical Value(s):

Test Statistic: Decision:

Conclusion:

Reject H0 at = .05There is evidence population means are different = .05Example Solution

• Youre a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods ( =.05)?M1M2M3M4 10111318 916823 59925Use the following values.ExerciseSSTr = 348SSE = 80

• Factorial DesignDecision MakingOne/Two SamplesAnalysis of VarianceCompletely Randomized Design2 TestsRoad Map

• The Randomized Block DesignA method to analyze more than two treatments using repeated measures or matched samples (related population)The items or individuals that have been matched (or from repeated measurements) are called blocks.Experimental situations that used blocks are called randomized block design.The blocks remove as much variability as possible from the random error so that the differences among the treatments are more evident.

• The Randomized Block DesignBRANDRandomized Block DesignCompletely Randomized DesignBlocks

GolferBrandABCDHit3Hit1Hit4Hit2Hit2Hit4Hit3Hit1

Hit4Hit3Hit1Hit2

• Partitioning the Total VariationThen we need to break the within treatment variation into variation due to differences among the blocks (SSB) and variation due to random error (SSE)Partitioning the Total Variation:SST = SSTr + SSB + SSE

• Sums of Squares FormulaTotal variation in randomized block designwhere= Grand meanAmong treatment variation in randomized block designwhere

• Among block variation in randomized block designRandom error in randomized block designSums of Squares Formulawhere

• You divide each of the sums of squares by its associated degrees of freedom, The Mean Squares

• The null hypothesisRandomized Block F Tests

is tested against

F test statistic

You reject the null hypothesis at the level if

F from F distribution with (k-1) numerator and (k-1) (b-1) denominator degrees of freedom

• The null hypothesisF Tests for Block Effects

is tested against

F test statistic

You reject the null hypothesis at the level if

F from F distribution with (b-1) numerator and (k-1) (b-1) denominator degrees of freedom

• A production manager wants to see if three assembly methods have different mean assembly times (in minutes). Five employees were selected at random and assigned to use each assembly method. At the .05 level of significance, is there a difference in mean assembly times?Employee Method 1 Method 2 Method 315.43.64.0 24.13.82.9 36.15.64.3 43.62.32.6 55.34.73.4Randomized Block Design Example

• Treatment (Methods)3 - 1 = 2SSTr=5.43MSTr = 2.71MSTrMSE= 12.9Error15 - 3 - 5 + 1= 8SSE =1.68MSE =.21Total15 - 1 = 14SST =17.8Source of VariationDegrees of FreedomSum of SquaresMean Square (Variance)FBlock (Employee)5 - 1 = 4SSBL=10.69MSB =2.67MSBMSE= 12.7Example Solution

• F0F = 4.46H0: 1 = 2 = 3 Ha: Not all equal = .05 1 = 2 2 = 8 Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject H0 at = .05There is evidence population means are different = .05Example Solution

• A fast-food chain wants to evaluate the service at four restaurant. The customer service director for the cahin hires six investigators with varied experiences in food service to act as raters. To reduce the effect the variability from rater to rater, you use a randomized block design with raters serving as the blocks. The four restaurants are the groups of interest. The six raters evaluate at each of the four restaurants in a random order. A rating scale from 0 (low) to 100 (high) is used.ExerciseUse the 0.05 level of significance to test for differences among the restaurants. Check also the effectiveness of blocking.

• Road MapDecision MakingOne/Two SamplesAnalysis of VarianceCompletely Randomized Design2 Tests

• The Factorial Design When there are two factors simultaneously evaluated, the experimental design is called a two factor factorial design (or just, factorial design)We can explore interaction between variablesData from a two-factor factorial design are analyzed using Two-Way ANOVA (or two-way table)Let the two factors be Factor A and Factor BWe are going to only deal the equal number of replicates for each combination of the level of factor A with those of factor B

• Example of Two Factors Design Tensile strength of parachutes woven by two types of looms, using synthetic fibers from four suppliersWe want to evaluate the different suppliers but also to determine whether parachutes woven on the Jetta looms are as strong as those woven on Turk looms.

Loom(Factor A)Supplier (Factor B)1234 Jetta20.618.019.021.313.222.624.619.623.827.127.718.620.825.117.721.520.021.123.916.0 Turk18.524.17.219.918.026.325.324.021.224.520.625.220.824.722.925.419.922.617.520.4

• Two Way ANOVA Procedure The following definitions are needed to develop two-way ANOVAwhere= grand mean= mean of the i-th level of factor A (where i = 1,2, , a)= mean of the j-th level of factor B (where j = 1,2, , b)= mean of the cell ij, the combination of the i-th level of factor A and the j-th level of factor B= number of levels of factor A and B, respectively= number of replicates for each cell (combination of a particular level of factor A and that of factor B)

• Main and Interaction Effects No A effect; B main effect123Mean responseLevel of factor ALevel 1, factor BLevel 2, factor B123Mean responseLevel of factor ALevel 1, factor BLevel 2, factor BA main effect; insignificant B effectA and B main effects, no interaction123Mean responseLevel of factor ALevel 1, factor BLevel 2, factor BA and B interact123Mean responseLevel of factor ALevel 1, factor BLevel 2, factor B

• Partitioning the Total Variation Then we need to break the group variability into three components plus one random variation or errorPartitioning the Total Variation:SST = SSA + SSB + SSI + SSE

• Sum of Squares FormulaThe computation for total variation:Factor A variation:Factor B variation:

• Sum of Squares FormulaInteraction variation:Random Error:

• The Mean SquaresIf you divide each of the sums of squares by its associated degrees of freedom, you have the four variances or mean square terms.

• There are three distinct tests to performF Test in Two-Way ANOVATest for Main Effect of Factor A

F test statistic

You reject the null hypothesis at the level if

F from F distribution with (a-1) numerator and (n-ab) denominator degrees of freedom

• F Test in Two-Way ANOVA2)Test for Main Effect of Factor B

F test statistic

You reject the null hypothesis at the level if

F from F distribution with (b-1) numerator and (n-ab) denominator degrees of freedom

• F Test in Two-Way ANOVA3)Test for Factor Interaction

F test statistic

You reject the null hypothesis at the level if

F from F distribution with (a-1)(b-1) numerator and (n-ab) denominator degrees of freedom

• Human Resources wants to determine if training time is different based on motivation level and training method. Conduct the appropriate ANOVA tests. Use = .05 for each test (Interaction, Motivation and Training Method).Factorial Design Example

• Source of VariationDegrees of FreedomSum of SquaresMean SquareFA (Row)1546.75546.75B (Column)2531.5265.75AB (Interaction)2123.561.76Error6188.531.42Total11SSTSame as other designs17.408.461.97Example Solution

• H0: The factors do not interactHa: The factors interact = .05 1 = 2 2 = 6 Critical Value(s):

Test Statistic: Decision:

Conclusion:

Do not reject at = .05There is no evidence the factors interactExample Solution

• H0: Ha: = 1 = 2 = Critical Value(s):

Test Statistic: Decision:

Conclusion:

No difference between motivation levelsMotivation levels differ.051 6Reject at = .05There is evidence motivation levels differExample Solution

• H0: Ha: = 1 = 2 = Critical Value(s):

Test Statistic:

Decision:

Conclusion:

No difference between training methodsTraining methods differ.052 6Reject at = .05There is evidence training methods differExample Solution

• ExerciseTensile strength of parachutes woven by two types of looms, using synthetic fibers from four suppliersUsing 0.05 level of significance, determine whether there is evidence of an interaction between the loom and the supplier, a difference between the two looms, and a difference among the suppliers.

Loom(Factor A)Supplier (Factor B)1234 Jetta20.618.019.021.313.222.624.619.623.827.127.718.620.825.117.721.520.021.123.916.0 Turk18.524.17.219.918.026.325.324.021.224.520.625.220.824.722.925.419.922.617.520.4