Two way ANOVA with replication

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TWO WAY ANOVA WITHOUT REPLICATION
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Presentation transcript:

Two way ANOVA with replication Also called a Factorial Experiment. Replication means an independent repeat of each factor combination. The purpose of factorial experiment is to examine: 1. The effect of factor A on the dependent variable, y. 2. The effect of factor B on the dependent variable, y along with 3. The effects of the interactions between different levels of the factors on the dependent variable, y. Interaction exists when the effect of a level for one factor depends on which level of the other factor is present.

The effect model for a factorial experiment can be written as:

There are three sets of hypothesis: 1. Factor A effect: 2. Factor B effect: 3. Interaction effect:

The results obtained in this analysis are summarized in the following ANOVA table:

Two way Factorial Treatment Structure

where

Example 4.4 The two-way table gives data for a 2x2 factorial experiment with two observations per factor – level combination. Construct the ANOVA table for this experiment and do a complete analysis at a level of significance 0.05. Factor A Factor B Level 1 2 29.6 35.2 47.3 42.1 12.9 17.6 28.4 22.7

Solution: Factor A Factor B Level 1 2 29.6 35.2 64.8 47.3 42.1 89.4 12.9 17.6 30.5 28.4 22.7 51.1 154.2 81.6 95.3 140.5 235.8

Solution: 1. Set up hypothesis Factor A effect: Factor B effect: Interaction effect:

2. Calculation (given the ANOVA table is as follows): 3 2. Calculation (given the ANOVA table is as follows): 3. Critical value: Factor A, Factor B, Interaction AB, Source of Variation SS df MS F A 658.845 1 46.652 B 255.38 18.083 AB 2 0.1416 Error 56.49 4 14.1225 Total 972.715 7

4. With = 0.05 we reject if :

We conclude that the difference level of A effect the response 5. Factor A : since , thus we reject We conclude that the difference level of A effect the response Factor B : since , thus we reject We conclude that the difference level of B effect the response Interaction: since , thus we failed to reject We conclude that no interaction between factor A and factor B.

Exercise 4.5 Each of three operators made two weighing of several silicon wafers. Results are presented in the following table for three wafers. Construct ANOVA table. Determine whether there is a differenced in the measured weights among the operators and also the difference among wafers at . Wafer Operator 1 Operator 2 Operator 3 1 11 , 15 10 , 6 14 , 10 2 210 , 208 205 , 201 208 , 207 3 111 , 113 102 , 105 108 , 111

Exercise 4.6: In a study to determine which are the important source of variation in an industrial process, 3 measurements are taken on yield for 3 operators chosen randomly and 4 batches a raw materials chosen randomly. It was decided that a significance test should be made at the 0.05 level of significance to determine if the variance components due to batches, operators, and interaction are significant. In addition, estimates of variance components are to be computed. The data are as follows, with the response being percent by weight. Batch 1 2 3 4 Operator 66.9 68.1 67.2 68.3 67.4 67.7 69.0 69.8 67.5 69.3 70.9 71.4 66.3 65.4 65.8 67.6 69.7 68.8 69.2 69.4 69.6 70.0 65.6 65.2 66.0 67.3 67.1 66.2 67.9 68.4 68.7

Perform the analysis of variance of this experiment at level of significance 0.05. State your conclusion