RANDOMIZED COMPLETE BLOCK DESIGN in .NET

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13-4 RANDOMIZED COMPLETE BLOCK DESIGN
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where is an overall mean, i is the effect of the ith treatment, j is the effect of the jth block, and ij is the random error term, which is assumed to be normally and independently distributed with mean zero and variance 2. Treatments and blocks will initially be considered as xed factors. Furthermore, the treatment and block effects are de ned as deviations a b from the overall mean, so g i 1 i 0 and g j 1 j 0 . We also assume that treatments and blocks do not interact. That is, the effect of treatment i is the same regardless of which block (or blocks) it is tested in. We are interested in testing the equality of the treatment effects. That is H0: H1:
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Testing the hypothesis that all the treatment effects i are equal to zero is equivalent to testing the hypothesis that the treatment means are equal. To see this, note that the mean of the ith treatment is i, de ned as
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0 , we have the mean of the ith treatment de ned as
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Therefore, testing the hypothesis that the a treatment means are equal is equivalent to testing that all the treatment effects i are equal to zero. The analysis of variance can be extended to the randomized complete block design. The procedure uses a sum of squares identity that partitions the total sum of squares into three components.
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The sum of squares identity for the randomized complete block design is a a 1 yij
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or symbolically SST SSTreatments SSBlocks SSE
Furthermore, the degrees of freedom corresponding to these sums of squares are ab 1 1a 12 1b 12 1a 12 1b 12
CHAPTER 13 DESIGN AND ANALYSIS OF SINGLE-FACTOR EXPERIMENTS: THE ANALYSIS OF VARIANCE
For the randomized block design, the relevant mean squares are MSTreatments MSBlocks MSE SSTreatments a 1 SSBlocks b 1 SSE 1a 121b
The expected values of these mean squares can be shown to be as follows:
E1MSTreatments 2 E1MSBlocks 2
2 i 1
2 2 j 1
E1MSE 2
Therefore, if the null hypothesis H0 is true so that all treatment effects i 0, MSTreatments is an unbiased estimator of 2, while if H0 is false, MSTreatments overestimates 2. The mean square for error is always an unbiased estimate of 2. To test the null hypothesis that the treatment effects are all zero, we use the ratio F0 MSTreatments MSE (13-28)
which has an F-distribution with a 1 and (a 1)(b 1) degrees of freedom if the null hypothesis is true. We would reject the null hypothesis at the -level of signi cance if the computed value of the test statistic in Equation 13-28 is f0 f ,a 1,(a 1)(b 1). In practice, we compute SST, SSTreatments and SSBlocks and then obtain the error sum of squares SSE by subtraction. The appropriate computing formulas are as follows. De nition The computing formulas for the sums of squares in the analysis of variance for a randomized complete block design are
SST SSTreatments SSBlocks and SSE SST
2 a a yij i 1 j 1
2 y.. ab 2 y.. ab 2 y.. ab
(13-29) (13-30) (13-31)
1 a 2 y. b ia i 1 1 b 2 a ja y. j 1 SS Treatments
SSBlocks
(13-32)
13-4 RANDOMIZED COMPLETE BLOCK DESIGN
Table 13-11 ANOVA for a Randomized Complete Block Design Source of Variation Treatments Blocks Error Total Sum of Squares SSTreatments SSBlocks SSE (by subtraction) SST (a Degrees of Freedom a b 1 1 1) 1 Mean Square SSTreatments a 1 SSBlocks b 1 1a SSE 121b 12 F0 MSTreatments MSE
1)(b ab
The computations are usually arranged in an ANOVA table, such as is shown in Table 13-11. Generally, a computer software package will be used to perform the analysis of variance for the randomized complete block design. EXAMPLE 13-5 An experiment was performed to determine the effect of four different chemicals on the strength of a fabric. These chemicals are used as part of the permanent press nishing process. Five fabric samples were selected, and a randomized complete block design was run by testing each chemical type once in random order on each fabric sample. The data are shown in Table 13-12. We will test for differences in means using an ANOVA with 0.01. The sums of squares for the analysis of variance are computed as follows: