Presentation on theme: "i) Two way ANOVA without replication"— Presentation transcript:
1i) Two way ANOVA without replication CHAPTER 4Analysis of VarianceOne-way ANOVATwo-way ANOVAi) Two way ANOVA without replicationii) Two way ANOVA with replication
2Key ConceptsANOVA can be used to analyze the data obtained from experimental or observational studies.A factor is a variable that the experimenter has selected for investigation.A treatment is a level of a factor.Experimental units are the objects of interest in the experiment.Variation between treatment groups captures the effect of the treatment.Variation within treatment groups represents random error not explained by the experimental treatments.
3One-way ANOVAThe one-way analysis of variance specifically allows us to compare several groups of observations whether or not their population mean are equal.One way ANOVA is also known as Completely Randomized Design (CRD).The application of one way ANOVA requires that the following assumptions hold true:(i) The populations from which the samples are drawn are(approximately) normally distributed.(ii) The populations from which the samples are drawn have the samevariance.(iii) The samples drawn from different populations are random andindependent.
4Each observation may be written as: Or alternatively written as: Where
5The is the total of all observations from the treatment, while is the grand total of all N observations.Then the hypothesis can be written as:Note: This hypothesis is use for modelTreatment12…ik.Total
6For model the hypothesis is as follows: The computations for an analysis of variance problem are usually summarized in tabular form as shown in table below. This table is of the referred to as the ANOVA table.Source of VariationSum of SquaresDegree of freedomMean SquareF CalculatedTreatment(Between levels)SSTRk - 1Error(within levels)SSEN - kTotalSSTN - 1
7where We reject if and conclude that some of the data is due to differences in the treatment levels.
8Example 4.1 Three different types of acid can be used in a particular chemical process. The resulting yield (in %) from several batches using the different types of acid are given below: Test whether or not the three populations appear to have equal means using = 0.05.AcidABC9395769777748784
9Solution: 1. Construct the table of calculation: 2 Solution: 1. Construct the table of calculation: 2. Set up the hypothesis:AcidABC9395769777748784
11Treatment (Between levels) Source of VariationSum of SquaresDegrees of FreedomMean SquareF CalculatedTreatment (Between levels)3 – 1 = 2Error (within levels)9 – 3 = 6Total9 – 1 = 84. At = 0.05, from the statistical table for f distribution, we have5. Since , thus we failed to rejectand conclude that there is no difference in the three types of acid atsignificance at = 0.05
12Exercise: Four catalyst that may affect the concentration of one component in a three-component liquid mixture are being investigated. The following concentrations are obtained. Compute a one-way analysis of variance for this experiment and test the hypothesis at 0.05 level of significance and state your conclusion concerning the effect of catalyst on the concentration of one component in three-component liquid mixture.123458.256.350.152.957.254.554.249.958.457.055.450.055.855.351.754.9
13Two-way ANOVA Without Replication Also known as Randomized Block Design (RBD)In RBD there is one factor or variable that isof primary interest. However, there are alsoseveral other nuisance factors.Nuisance factors are those that may affect themeasured result, but are not of primary interest.The way to control nuisance factor is by blockingthem to reduce or eliminate the contribution toexperimental error contributed by nuisance factors
14The model for a randomized block design with one nuisance variable can be written as: : j th observation from i th treatment,:Overall mean,: i th effect of treatment,: j th effect of block: random error.
15We can use the following layout for this kind of two-way classification: BlocksTotalTreatment 1Treatment 2Treatment iTreatment k
16In the two way analysis of variance where each treatment is represented once in each block, the major objective is to test:The effect of treatment:The effect of block:
20Solution; 1. Construct the table of calculation, we have k = 4 and n = 3: 2. Set up the hypothesis: Engine 1Engine 2Engine 3Detergent ADetergent BDetergent CDetergent DTotals182176207139145153128565454351474652485055423749
22Source of VariationSum of SquaresDegree of FreedomMean SquareF CalculatedTreatments1113Blocks1352Error196Total265114. At = 0.01, from the statistical table for f distribution, we have (treatments) and (blocks). 5. Since , thus we reject and conclude that there are differences in the effectiveness of the 4 detergents at = 0.01 and also since , thus we reject and conclude that there are differences among the results obtained for the 3 engines are significant
23Exercise: Consider the hardness testing experiment described below Exercise: Consider the hardness testing experiment described below. There are 4 tips and 4 available metal coupons. Each tip is tested once on each coupon, resulting in a randomized block design. The data obtained are represented in the below. Analyze the data by test the hypothesis at the 0.05 level of significance and draw the appropriate conclusion.Coupon (Block)Type of tip123220.127.116.110.09.89.99.29.59.710.2