TOPIC 11 Analysis of Variance

Draw Sample Populations μ 1 = μ 2 = μ 3 = μ 4 = ….. μ n Evidence to accept/reject our claim Sample mean each group, grand mean, ANOVA test of equality of population means

Road Map Factorial Design Decision Making One/Two SamplesAnalysis of Variance Completely Randomized Design χ 2 Tests Randomized Block Design

Completely Randomized Design In 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 300 o, 350 o, 400 o, 450 o 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 Experiment Supplier 1Supplier 2Supplier 3Supplier Sample Mean Grand Mean Sample Standard Deviation Sample of tensile strength of synthetic fibers from four different suppliers Do the synthetic fibers from each of four suppliers have equal strength?

One Way ANOVA The 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 Variation In One Way ANOVA, the total variation is subdivided into two parts: Variation that is due to differences among the treatments Variation that is due to differences within the treatments The symbol k is used to indicate the number of treatments Total Variation (SST) d.f. = n - 1 Total Variation (SST) d.f. = n - 1 Treatment Variation (SSTr) d.f. = k - 1 Treatment Variation (SSTr) d.f. = k - 1 Random Error Variation (SSE) d.f. = n - k Random Error Variation (SSE) d.f. = n - k Partitioning the Total Variation: SST = SSTr + SSE

Hypothesis to be Tested Assumptions: k groups represent populations Its values are randomly and independently selected Following a normal distribution Having equal variances Refer back to the table of synthetic fibers from four suppliers. The null hypothesis of no differences in the population means is tested against the alternative that at least two of the k treatment means differ (or not all μ j are equal, where j = 1,2,…, k) At least one of the k treatment means differ

Sums of Squares Formula We 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 mean X Treat. 1Treat. 2Treat. 3 Response, X

Sums of Squares Formula The variation among the treatments is presented by the sum of squares treatments (SSTr) X X3X3 X2X2 X1X1 Treat. 1Treat. 2Treat. 3 Response, X

The within-group variation is given by the sum of squares within treatments (SSE) Sums of Squares Formula X2X2 X1X1 X3X3 Treat. 1Treat. 2Treat. 3 Response, X

where = sample mean of treatment j = i-th value of treatment j n j = sample size of treatment j n = total number of values in all treatments = n 1 + n 2 + … + n j 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 Squares Total degrees of freedom = (k - 1) + (n – k) = n - 1

Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F Treatmentk - 1SSTr MSTr = SSTr/(k - 1) MSTr MSE Errorn - kSSE MSE = SSE/(n - k) Totaln - 1 SST= SSTr+SSE One-Way ANOVA Summary Table

F Test for Differences Among More than Two Means MSA and MSE provide estimates of the overall variance in the data. To test the null hypothesis: At least one of the k treatment means differ against you compute the One-Way ANOVA F test statistic, which is given by Rejection Region Critical Value F α from F-distribution with (k-1) numerator and (n-k) denominator degrees of freedom Reject H 0 if F > F α, 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 Mach Mach1 Mach2 Mach One-Way ANOVA Example

Treatment k – 1 = = 2 SSTr = MSTr = MSTr MSE = Error n – k = = 12 SSE = MSE =.9211 Total n – 1 = = 14 SST = Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F Example Solution

F 0F α = 3.89 H 0 :  1 =  2 =  3 H a : Not All Equal  =.05 1 = 2 2 = 12 Critical Value(s): Test Statistic: Decision: Conclusion: Reject H 0 at  =.05 There is evidence population means are different  =.05 F MSTr MSE  Example Solution

You’re a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods (  =.05)? M1M2M3M Use the following values. ExerciseExercise SSTr = 348 SSE = 80

Factorial Design Decision Making One/Two SamplesAnalysis of Variance Completely Randomized Design χ 2 Tests Road Map Randomized Block Design

The Randomized Block Design A 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 Design BRAND Golfer Brand ABCD Hit 3 Hit 1 Hit 4 Hit 2 Hit 2 Hit 4 Hit 3 Hit 1 Hit 4 Hit 3 Hit 1 Hit Randomized Block Design Completely Randomized Design Blocks

Partitioning the Total Variation Then we need to break the within treatment variation into variation due to differences among the blocks (SSB) and variation due to random error (SSE) Total Variation (SST) d.f. = n - 1 Total Variation (SST) d.f. = n - 1 Among-Treatment Variation (SSTr) d.f. = k - 1 Among-Treatment Variation (SSTr) d.f. = k - 1 Within-Treatment Variation (SSE) d.f. = n - k Within-Treatment Variation (SSE) d.f. = n - k Among-Block Variation (SSB) d.f. = b - 1 Among-Block Variation (SSB) d.f. = b - 1 Random-Error Variation (SSE) d.f. = (b - 1)(k - 1) Random-Error Variation (SSE) d.f. = (b - 1)(k - 1) Partitioning the Total Variation: SST = SSTr + SSB + SSE

Sums of Squares Formula Total variation in randomized block design where = Grand mean Among treatment variation in randomized block design where

Among block variation in randomized block design Random error in randomized block design Sums of Squares Formula where

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

The null hypothesis Randomized Block F Tests is tested against F test statistic You reject the null hypothesis at the α level if F α from F distribution with (k-1) numerator and (k- 1) (b-1) denominator degrees of freedom At least one of the k treatment means differ

The null hypothesis F Tests for Block Effects is tested against F test statistic You reject the null hypothesis at the α level if F α from F distribution with (b-1) numerator and (k- 1) (b-1) denominator degrees of freedom At least one of the b block means differ

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 Randomized Block Design Example

Treatment (Methods) = 2SSTr= 5.43 MSTr = 2.71 MSTr MSE = 12.9 Error = 8 SSE = 1.68 MSE =.21 Total = 14 SST = 17.8 Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F Block (Employee) = 4SSBL= MSB = 2.67 MSB MSE = 12.7 Example Solution

F 0F α = 4.46 H 0 :  1 =  2 =  3 H a : Not all equal  =.05 1 = 2 2 = 8 Critical Value(s): Test Statistic: Decision: Conclusion: Reject H 0 at  =.05 There is evidence population means are different  =.05 FTFT MSTr MSE  Example 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. ExerciseExercise Use the 0.05 level of significance to test for differences among the restaurants. Check also the effectiveness of blocking.

Road Map Decision Making One/Two SamplesAnalysis of Variance Completely Randomized Design χ 2 Tests Factorial Design

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 variables Data 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 B We 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 suppliers Loom (Factor A) Supplier (Factor B) 1234 Jetta Turk We 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.

Two Way ANOVA Procedure The following definitions are needed to develop two-way ANOVA where = 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 effect Mean response Level of factor A Level 1, factor B Level 2, factor B Mean response Level of factor A Level 1, factor B Level 2, factor B A main effect; insignificant B effect A and B main effects, no interaction Mean response Level of factor A Level 1, factor B Level 2, factor B A and B interact Mean response Level of factor A Level 1, factor B Level 2, factor B

Partitioning the Total Variation Then we need to break the group variability into three components plus one random variation or error Total Sum of Squares (SST) d.f. = n - 1 Total Sum of Squares (SST) d.f. = n - 1 Interaction Sum of Squares Factors A and B (SSI) d.f. = (a – 1) (b – 1) Interaction Sum of Squares Factors A and B (SSI) d.f. = (a – 1) (b – 1) Sum of Squares Random Error (SSE) d.f. = n - ab Sum of Squares Random Error (SSE) d.f. = n - ab Main Effect Sum of Squares Factor A (SSA) d.f. = a - 1 Main Effect Sum of Squares Factor A (SSA) d.f. = a - 1 Main Effect Sum of Squares Factor B (SSB) d.f. = b - 1 Main Effect Sum of Squares Factor B (SSB) d.f. = b - 1 Partitioning the Total Variation: SST = SSA + SSB + SSI + SSE

Sum of Squares Formula The computation for total variation: Factor A variation: Factor B variation:

Sum of Squares Formula Interaction variation: Random Error:

The Mean Squares If 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 perform F Test in Two-Way ANOVA 1)Test for Main Effect of Factor A F test statistic You reject the null hypothesis at the α level if F α from F distribution with (a-1) numerator and (n- ab) denominator degrees of freedom

F Test in Two-Way ANOVA 2)Test for Main Effect of Factor B F test statistic You reject the null hypothesis at the α level if F α from F distribution with (b-1) numerator and (n- ab) denominator degrees of freedom

F Test in Two-Way ANOVA 3)Test for Factor Interaction F test statistic You reject the null hypothesis at the α level if F α 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 α =.05 for each test (Interaction, Motivation and Training Method). Training Method (Factor B) Factor Levels Self– paced ClassroomComputer 15 hr.10 hr.22 hr. Motivation (Factor A) High 11 hr.12 hr.17 hr. 27 hr.15 hr. 31 hr. Low 29 hr. 17 hr. 49 hr. Factorial Design Example

Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) B (Column) AB (Interaction) Error Total11SST Same as other designs Example Solution

H 0 : The factors do not interact H a : The factors interact  =.05 1 = 2 2 = 6 Critical Value(s): Test Statistic: Decision: Conclusion: F 0F α = 5.14  =.05 Do not reject at  =.05 There is no evidence the factors interact Example Solution

H 0 : H a :  = 1 = 2 = Critical Value(s): Test Statistic: Decision: Conclusion: F 0F α = 5.99  =.05 No difference between motivation levels Motivation levels differ.05 1   6 Reject at  =.05 There is evidence motivation levels differ Example Solution

H 0 : H a :  = 1 = 2 = Critical Value(s): Test Statistic: Decision: Conclusion: F 0F α = 5.14  =.05 No difference between training methods Training methods differ.05 2   6 Reject at  =.05 There is evidence training methods differ Example Solution

ExerciseExercise Tensile strength of parachutes woven by two types of looms, using synthetic fibers from four suppliers Loom (Factor A) Supplier (Factor B) 1234 Jetta Turk Using 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.

ExerciseExercise The sums of squares are already given,