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**i) Two way ANOVA without replication **

CHAPTER 4 Analysis of Variance One-way ANOVA Two-way ANOVA i) Two way ANOVA without replication ii) Two way ANOVA with replication

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Key Concepts ANOVA 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.

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One-way ANOVA The 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 same variance. (iii) The samples drawn from different populations are random and independent.

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**Each observation may be written as: Or alternatively written as: Where **

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**The 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 model Treatment 1 2 … i k . Total

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For 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 Variation Sum of Squares Degree of freedom Mean Square F Calculated Treatment (Between levels) SSTR k - 1 Error (within levels) SSE N - k Total SST N - 1

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**where We reject if and conclude that some of the data is due to differences in the treatment levels. **

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Example 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. Acid A B C 93 95 76 97 77 74 87 84

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**Solution: 1. Construct the table of calculation: 2**

Solution: 1. Construct the table of calculation: 2. Set up the hypothesis: Acid A B C 93 95 76 97 77 74 87 84

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**3. Construct ANOVA table:**

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**Treatment (Between levels)**

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Calculated Treatment (Between levels) 3 – 1 = 2 Error (within levels) 9 – 3 = 6 Total 9 – 1 = 8 4. At = 0.05, from the statistical table for f distribution, we have 5. Since , thus we failed to reject and conclude that there is no difference in the three types of acid at significance at = 0.05

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Exercise: 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. 1 2 3 4 58.2 56.3 50.1 52.9 57.2 54.5 54.2 49.9 58.4 57.0 55.4 50.0 55.8 55.3 51.7 54.9

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**Two-way ANOVA Without Replication**

Also known as Randomized Block Design (RBD) In RBD there is one factor or variable that is of primary interest. However, there are also several other nuisance factors. Nuisance factors are those that may affect the measured result, but are not of primary interest. The way to control nuisance factor is by blocking them to reduce or eliminate the contribution to experimental error contributed by nuisance factors

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**The 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.

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**We can use the following layout for this kind of two-way classification:**

Blocks Total Treatment 1 Treatment 2 Treatment i Treatment k

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In 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:

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We reject if: The effect of treatment: 2. The effect of block:

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**Solution; 1. Construct the table of calculation, we have k = 4 and n = 3: 2. Set up the hypothesis: **

Engine 1 Engine 2 Engine 3 Detergent A Detergent B Detergent C Detergent D Totals 182 176 207 139 145 153 128 565 45 43 51 47 46 52 48 50 55 42 37 49

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**3. Construct ANOVA table:**

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Source of Variation Sum of Squares Degree of Freedom Mean Square F Calculated Treatments 111 3 Blocks 135 2 Error 19 6 Total 265 11 4. 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

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**Exercise: 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 tip 1 2 3 4 9.3 9.4 9.6 10.0 9.8 9.9 9.2 9.5 9.7 10.2

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