FRACTIONAL FACTORIAL DESIGN As the number of factors to be tested increases, the complete set of factorial treatments may become to large to be tested simultaneously in a single experiment. A logical alternative is an experimental design that allows testing of only a fraction of the total numbers of treatments. A design uniquely suited for experiments involving a large number of factors is the fractional factorial design (FFD). It provides a systematic way of selecting and testing only a fraction of the complete set of factorial treatment combinations. In exchange, however, there is loss of information on some preselected effects. Although this information loss may be serious in experiments with one or two factors, such a loss becomes more tolerable with a large number of factors. The number of interactions effects increases rapidly with the number of factors involved, which allows flexibility in the choice of the particular effects to be sacrificed. In fact, in cases where some specific effects are known beforehand to be small or unimportant, use of the FFD results in minimal loss of information. In practice, the effects that are most commonly sacrificed by use of the FFD are high-order interactions-the four-factors or five-factor interactions and, at times, even the three factor interaction. In almost all cases, unless the

researcher has prior information to indicate otherwise, he should select a set of treatments to be tested so that all main effects and two-factor interactions can be estimated. in agricultural research, the FFD is the most commonly used in exploratory trials where the main objective is to examine the interactions between factors. For such trials the most appropriate FFD are those that sacrifice only those interactions that involve more than two factors. With the FFD, the number of effects that can be measured decreases rapidly with the reduction in the number of treatments to be tested. Thus, when the number of effects to be measured is large, the number of treatments to be tested, even with the use of FFD, may still be to large. In such cases, further reduction in the size of the experiment can be achieved by reducing the number of replications. Although use of a FFD without replication is uncommon in agricultural experiments, when FFD is applied to an exploratory trial the number of replications required can be reduced. For example, two replications are commonly used in an exploratory field trial in rice whereas four replications are used for a standard field experiment in rice. Another desirable feature of FFD is that it allows reduced block size by not requiring a block to contain all treatments to be tested. In this way, the homogeneity of experimental units within the same block can be improved. A reduction in block size is, however, accompanied by loss of information in

Addition to that already lost through the reduction in number of treatments. Although the FFD can be tailor-made to fit most factorial experiments, the procedure for doing so is complex and beyond the scope of this book. Thus, we describe only a few selected set of FFD that are suited for exploratory trials in agricultural research. The major features of these selected designs are that they:Apply only to 2n factorial experiments where n, the number of factors, ranges from 5 to 7.Involve only one half of the complete set of factorial treatment combinations (I.e., the number of treatments is 1\2 of 2n or 2n-1). Have a block size of 16 plots or less.Allow all main effects and most, if not all, of the two factor interactions to be estimated.

The selected plans are given in appendix M. each plan provides the list of treatments to be tested and the specific effects the can be estimated. In the designation of the various treatment combinations for all plans, the letters a, b, c, Are used to denote the presence (or use of high level) of factors A, B, C,. Thus, the treatment combination ab in a 25 factorial experiment refers to the treatment combination that contains the high level (or presence) of factors A and

B and low level (or absence) of factors C, D, and E, but this combination that contains the high level of factors A and B and low level of factors C, D, E, and F. In all cases, the treatment combination that consist of the low level of all factors is denoted by the symbol (1). We illustrate the procedure for randomization, layout, and analysis of variance of a FFD with a field experiment involving six factors A, B, C, D, E, and F, each at two levels (i.e. 26 factorial experiment). Only 32 treatments from the total of 64 complete factorial treatment combinations are tested in blocks of 16 plots each. With two replications, the total number of experimental plots is 64.

4.5.1 Randomization and layoutThe steps for randomization and layout are:STEP 1. Choose an appropriate basic plan of a FFD in appendix M. The should correspond to number of factors and the number of levels of each factor to be tested. For basic plans that are not given in appendix M, see Cochran and Cox, 1957.* Our example uses plan 3 of appendix M.STEP 2. If there is more than one block per replication, randomly assign the block arrangement in the basic plan to the actual blocks in the field. For example, the experimental area is first divided into two replications

(Rep. I and Rep. II), each consisting of 32 experimental plots. Each replication is further divided into two blocks (Block 1 and Block 2), each consisting of 16 plots. Following one of the randomization schemes of Chapter 2, Section 2.1.1, randomly reassign the block numbers in the basic plan to the block in the field. The result maybe as follows:Block Number in Block Number Assignment in field Basic Plan Rep. I Rep. II ___________________________________________________________ I 2 1 II 1 2 Note all 16 treatments listed in block 1 of the basic plan are assigned to block 2 of replication 1 in the field all 16 treatments listed in block II of the basic plan are assigned to block 1 of replication I in the field and so onSTEP 3. Randomly reassign the treatments in each block of the basic plan to the experimental plots of the reassigned block in the field (from step 2) For this example, follow the same randomization scheme used in step 2 and randomly assign the 16 treatments of a given block (in the basic plan) to the 16 plots of the corresponding block in the field, separately and independently for

Each of the four blocks (i.e., two block per replication and two replications). The result of the four independent randomization processes may be as follows: Plot Number Assignment in fieldTreatment ___________________________________________________ Rep.1 Rep.2 Number in ___________________ ____________________Basic Plan Block 1 Block 2 Block 1 Block 2 _________________________________________________________________ 1 6 5 4 11 2 3 4 14 7 3 15 10 3 6 4 12 6 8 1 5 1 12 7 15 6 5 1 11 4 7 13 3 16 14 8 7 8 12 9

Plot Number Assignment in fieldTreatment __________________________________________________________________ Rep.1 Rep.2 Number in ___________________ _____________________Basic Plan Block 1 Block 2 Block 1 Block 2 __________________________________________________________________ 9 2 16 9 3 10 10 11 10 5 11 11 15 5 8 12 8 2 6 12 13 4 14 1 16 14 9 9 2 13 15 16 13 15 2 16 14 7 13 10 Note that block one of replication I in the field was assigned to receive treatments of block II in the basic plan (step 2); and according to the basic plan used (i.e., plan 3 of appendix M) treatment 1 of block II is ae. Thus, according to the foregoing

assignment of treatments, treatment ae is assigned to plot 6 in block 1 of replication I. In the same manner because treatment 2 of block II in the basic plan is af, treatment af is assigned to plot 3 in block 1 of replication I; and so on. The final layout is shown in Figure 4.8.4.5.2 Analysis of varianceThe analysis of variance procedures of a FFD, without replication and replication, are illustrated. We Yates method for the computation of large fractional factorial experiments. Other alternative procedure are:The application of the standard riles for the computation of sums of squares in the analysis of variance (Chapter 3), by constructing two-way tables of totals for two-factor interactions, three-way table of totals for three-factor interactions, and so on.The application of the single d.f. contrast method (Chapter 5), by specifying a contrast for each of the main effects and interaction effects that are to be estimated.4.5.2.1 Design without replication. For illustration, we use data from replication I of table 4.20 is used. The computational steps in the analysis of variance are:STEP 1. Outlining the analysis of variance, following that given in appendix M, corresponding to the basic plan used. For our example, the basic plan is

Block 1 Block 2 Block 1 Block 2Plot no.

Treatment

1

cd 2

ad 1

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1 abde 2

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Figure 4.8 A sample layout of a fractional factorial design with two replications 1\2 of 25 factorial treatments arranged in blocks of 16 plots each plan 3 appendix M and the outline of the analysis of variance is:_________________________________________________________________ Source Degree Sum of of of Mean Computed Tabular FVariation Freedom Squares Squares F 5% 1%_________________________________________________________________Block 1Main effect 6Two-factor interaction 15Error 9 Total 31STEP 2. Determine the number of real factors (k) each at two levels, whose complete set of factorial treatments is equal to the number of treatments (t) to be tested (i.e., 2k = t). Then select the specific set of k real factors from

Table 4.20 Grain Yield Data from a 26 Factorial Experiment Planted in a1\2 Fractional Factorial Design in Blocks of 16 Experimental Plots each, and with two replications_________________________________________________________________ Grain Yield, Grain Yield t/ha t/ha _______________ _______________ Treatment Rep. I Rep. II Total Treatment Rep. I Rep. II Total __________________________________________________________________ Block 1 Block 2(1) 2.92 2.76 5.68 ad 3.23 3.48 6.71ab 3.45 3.50 6.95 ae 3.10 3.11 6.21ac 3.65 3.50 7.15 af 3.52 3.27 6.79bc 3.16 3.05 6.21 bd 3.29 3.22 6.51de 3.29 3.03 6.32 be 3.06 3.20 6.26df 3.34 3.3 6.71 bf 3.27 3.27 6.54

Grain Yield, Grain Yield t/ha t/ha _______________ _______________ Treatment Rep. I Rep. II Total Treatment Rep. I Rep. II Total __________________________________________________________________ef 3.28 3.23 6.51 cd 3.68 3.52 7.20abde 3.88 3.79 7.67 ce 3.08 3.02 6.10abdf 3.95 4.03 7.98 cf 3.29 3.10 6.39abef 3.85 3.90 7.75 abcd 3.89 3.99 7.88acde 4.05 4.18 8.23 abce 3.71 3.80 7.51acdf 4.37 4.20 8.57 abcf 3.96 3.98 7.94acef 3.77 3.80 7.57 adef 4.27 3.98 8.25bcde 4.04 3.87 7.91 bdef 3.69 3.62 7.31bcdf 4.00 3.76 7.76 cdef 4.29 4.09 8.38bcef 3.63 3.46 7.09 abcdef 4.80 4.78 9.58Total (RB) 58.63 57.43 58.13 57.43__________________________________________________________________

The original set of n factors and designate all (n k) factors not included in the set of k as dummy factors. For our example, the t = 32 treatment combinations correspond to a complete set of 2k factorial treatment combinations, with k = 5. For simplicity, the first five factors A, B, C, D, and E are designated as the real factors and F as the dummy factor.STEP 3. Arrange the t treatments in a system order based on the k real factors:Treatments with fewer number of letters are listed first. For example, ab comes before abc, and abc comes before abcde, and so on. Note that if treatment (1) is present in the set of t treatments, it always appears as the first treatment in the sequence.Among treatments with the same number of letters, those involving letters, corresponding to factors assigned to the lower-order letters come first. For example, ab comes before ac, ad before bc, and so on.All treatment identification letters corresponding to the dummy factors are ignored in the arrangement process. For our example, factor F is the dummy factor and, thus, af is considered simply as a and comes before ab. In this example, the systematic arrangement of the 32 treatments is shown in the first column of table 4.21. Note that:

The treatments are listed systematically regardless of their block allocation.The dummy factor F is place in parenthesis.STEP 4. Compute the t factorial effect totals:Designate the original data of the t treatments as the initial set or the t0values. For our example, the systematically arranged set of 32 t0 values are listed in the second column of table 4.21.Group the t0 values into t/2 successive pairs. For our example, there are 16 successive pairs: the first pair is 2.92 and 3.52, the second pair is 3.27 and 3.45, and the last pair is 4.04 and 4.80.Add the values of the two treatments in each of the t/2 pairs constituted in task 2 to constitute the first half of the second set or the t1 values. For our example, the first half of the t1 values are computed as: 6.44 = 2.92 + 3.52 6.72 = 3.27 + 3.45 . . .

8.34 = 4.29 + 4.05 8.84 = 4.04 + 4.80Table 4.21 Application of Yates method for the computation of sums of squares of a 26 Factorial Experiment Conducted in a1/2 Fractional Factorial Design, without replication, from Rep. I data from table 4.20_________________________________________________________________ Factorial Effect Identification Treatment ___________________Combination t0 t1 t2 t3 t4 t5 Preliminary Final _________________________________________________________________(1) 2.92 6.44 13.16 27.22 56.97 116.76 (G) (G) a(f) 3.52 6.72 14.06 29.75 59.79 6.14 A A b(f) 3.27 6.94 13.81 27.48 3.07 2.50 B Bab 3.45 7.12 15.94 32.31 3.07 0.56 AB ABc(f) 3.29 6.57 13.29 1.94 0.97 5.98 C C

Factorial Effect Identification Treatment ___________________Combination t0 t1 t2 t3 t4 t5 Preliminary Final ___________________________________________________________________ac 3.65 7.24 14.19 1.13 1.53 -0.08 AC ACbc 3.16 8.05 15.13 1.38 -0.01 -0.48 BC BCabc(f) 3.96 7.89 17.18 1.69 0.57 -0.50 ABC ABC(Block)d(f) 3.34 6.38 0.78 0.46 3.03 7.36 D D ad 3.23 6.91 1.16 0.51 2.95 -0.50 AD ADbd 3.29 6.85 0.55 1.02 0.41 -0.46 BD BDabd(f) 3.95 7.34 0.58 0.51 -0.49 -0.20 ABD ABDcd 3.68 7.56 0.61 0.02 -0.93 2.38 CD CD acd(f) 4.37 7.57 0.77 -0.03 0.45 -1.16 ACD ACDbcd(f) 4.00 8.34 1.17 0.36 -0.71 -0.20 BCD BCDabcd 3.89 8.84 0.52 0.21 0.21 0.94 ABCD EF--------------------------------------------------------------------------------------------------------------e(f) 3.28 0.60 0.28 0.90 2.53 2.82 E Eae 3.10 0.18 0.18 2.13 4.83 0.00 AE AEbe 3.06 0.36 0.67 0.90 -0.81 0.56 BE BEabe(f) 3.85 0.80 -0.16 2.05 0.31 0.58 ABE ABEce 3.08 -0.11 0.53 0.38 0.05 -0.08 CE CEace(f) 3.77 0.66 0.49 0.03 -0.51 -0.90 ACE ACE

Factorial Effect Identification Treatment ___________________Combination t0 t1 t2 t3 t4 t5 Preliminary Final ____________________________________________________________________bce 3.63 0.69 0.01 0.16 -0.05 1.38 BCE BCEabce 3.71 -0.11 0.50 -0.65 -0.15 0.92 ABCE DFde 3.29 -0.18 -0.42 -0.10 1.23 2.30 DE DEade(f) 4.27 0.79 0.44 -0.83 1.15 1.12 ADE ADEbde(f) 3.69 0.69 0.77 -0.04 -0.35 -0.56 BDE BDEabde 3.88 0.08 -0.80 0.49 -0.81 -0.10 ABDE CFcde(f) 4.29 0.98 0.97 0.86 -0.73 -0.08 CDE CDEacde 4.05 0.19 -0.61 -1.57 0.53 -0.46 ACDE BFbcde 4.04 -0.24 -0.79 -1.58 -2.43 1.26 BCDE AFabcde(f) 4.80 0.76 1.00 1.79 3.37 5.80 ABCDE F__________________________________________________________________The result of the first 16 t1 values are shown in the top of the third column of table 4.21.D. Subtract the first value from the second in each of the t/2 pairs constitute the bottom half of the t1 values. For our example, the second half of the t1 values are computed as: 0.61 = 3.52 - 2.92 0.18 = 3.45 - 3.27

. . . -0.24 = 4.05 4.29 0.76 = 4.80 4.04The results of the last 16 t1 values are shown in the bottom half of the third column of table 4.21.E. Reapply tasks B to D using the values t1 instead of t0 to derive the third set or the t2 values. For our example, tasks B to D are reapplied to t1 values to arrive at the t2 values shown in the fourth column of table 4.21.F. Repeat task E, (k 2) times. Each time use the newly derived values of t. for our example, task E is repeated three more times to derive t3 values, t4 values, and t5 values as shown in the fifth, sixth, and seventh columns of table 4.21.STEP 5. Identify the specific factorial effect that is represented by each of the values of the last set (commonly referred to as the factorial effect totals) derived in step 4. use the following guidelines:The first value represents the grand total (G)For the remaining (t 1) values assign the preliminary factorial effects according

To the letters of the corresponding treatments. With the dummy factors ignored. For our example, the second t5 value corresponds to treatment combination a (f) and, hence, is assigned to the 4 main effect. The fourth t5 value corresponds to treatment ab and is assign to the A X B interaction effect, and so on. The result for all 32 treatments are shown in the eighth column of table 4.21.C. For treatments involving the dummy factor (or factors) adjust the preliminary factorial effects derived in task B as follows:Based on the conditions stated in the basic plan of appendix M identify all effects involving the dummy factor that are estimable (i. e., that can be estimated). For our example, the estimable effects involving the dummy factor F consist of the main effect of F and all its two-factor interactions AF, BF, DF, and EF.Identify the aliases of all effects listed immediately above. The alias of many effect is defined as its generalized interaction with the defining contrast. The generalized interaction between any two factorial effects is obtained by combining all the letters that appear in the two effects and canceling all letters that enter twice. For our example, the generalized interaction between ABC and AB is AABBC or C. For our example, because the defining contrast is ABCDEF (see plan 3 or appendix M) the aliases of the six effects involving the dummy factor F are: F = ABCDE, AF = BCDE, BF = ACDE, CF = ABDE, DF = ABCE, and EF = ABCD. The two factorial effects involve in each pair of aliases (one to the left and

another to the right of the equal sign) are not separable (i. e., can not be estimated separately). For example, for the first pair, F and ABCDE, the main effect of factor F cannot be separated from the A X B X C X D X E interaction effect and, hence, unless one of the pair is known to be absent there is no way to known which of the pairs is the contributor to the estimate contained.Replace all preliminary factorial effects that are aliases of the estimable effects involving the dummy factors by the latter. For example, because ABCDE (corresponding to the last treatment in table 4.21) is the alias of F, it is replaced by F. in the same manner, BCDE is replaced by AF, ACDE by BF, ABDE by CF, ABCE by DF, and ABCD by EF.When blocking is used, identify the factorial effects that are confounded with blocks. Such effects are stated for each plan of appendix M. for our example, ABC is confounded with block (see plan 3 of appendix M) and the preliminary factorial effect ABC is, therefore replaced by block effect. That means that the estimate of the ABC effect becomes the measure of the block effect. The final results of the factorial effect identification are shown in the last column of table 4.21.STEP 6. For each source of variation in the analysis of variance (step 1) identify the corresponding factorial effects. For our example there is only one factorial effect (i.e., ABC) corresponding to the first source or variation of block. For the second source of variation (main effects) there are six factorial effects corresponding to the

six main effects (A, B, C, D, E, and F). And, for the third source of variation (two-factor interactions) there are 15 factorial effects (i.e., all 15 possible two-factor interaction effects among the six factors). All the remaining nine factorial effects correspond to the last source of variation (error).STEP 7. For each source of variation in the analysis of variance of step 1, compute its SS as the sum of the squares of the factorial effect totals of the corresponding factorial effects (identified in step 6) divided by the total number of treatments tested in the experiment. For our example, the various SS are computed as: Block SS = (ABC)2 32 = (-0.50)2 = 0.007812 32Main effect SS = (A)2 + (B)2 + (C)2 + (D)2 + (E)2 + (F)2 32 = [(6.14)2 + (2.50)2 + (5.98)2 + (7.36)2 + (2.82)2 + (5.80)2]/32 = 5.483500

Two-factorial interaction SS = [(AB)2 + (AC)2 + (BC)2 + + (CF)2 + (BF)2 + (AF)2]/32 = [(0.56)2 + (-0.08)2 + (-0.48)2 + + (-0.10)2 + (- 0.46)2 + (1.26)2]/32 = 0.494550 Error SS = [(ABD)2 + (ACD)2 + (BCD)2 + + (ADE)2 + (BDE)2 + (CDE)2]/32 = [(-0.20)2 + (-1.16)2 + (-0.20)2 + + (-1.12)2 + (-0.56)2 + (-0.08)2]/32 = 0.189088Note that the error SS can also be computed as the difference between the total SS and the sum of all other SS, where the total SS is computed from all factorial effect total. For our example, the total SS and the error SS are: Total SS = (A)2 + (B)2 + (AB)2 + + (BF)2 + (AF)2 + (F)2 32 = [(6.14)2 + (2.50)2 + (0.56)2 + + (-0.46)2 +

+ (1.26)2 + (5.80)2]/32 = 0.189088 STEP 8. Determine the degree of freedom for each SS as the number of factorial effects totals used in its computation. For example, the computation of the block SS involves only one effect, namely ABC; hence, its d.f. is 1. on the other hand, there are six effect totals involved in the computation of the main effect SS; hence, its d.f. is 6. the results are shown in the second column of table 4.22.STEP 9. Compute the mean square for each source of variation by dividing each SS by its d.f: Block MS = Block SS 1 = 0.007812Table 4.22 Analysis of variance of data from a Fractional Factorial Design: of a 26 Factorial Experiment without Replication

___________________________________________________________________Source Degree Sum Of of of Mean Computed Tabular FVariation freedom Squares Squares Fb 5% 1%___________________________________________________________________Block 1 0.007812 0.007812

= 5.483500 = 0.913917 6 Two-factor interaction MS = Two-factor interaction SS 15 = 0.494550 = 0.032970 15 Error MS = Error SS 9 = 0.189088 = 0.021010 9STEP 10. Compute the F value for each effect by dividing its MS by the error MS: F (Block) = Block MS Error MS = 0.007812 < 1 0.021010

F (main effect) = Main effect MS Error MS = 0.913917 = 43.50 0.021010 F (two-factor interaction) = Two-factor interaction MS Error MS = 0.032970 = 1.57 0.021010STEP 11. Compare each computed F value with the corresponding tabular F values, from appendix E, with f1 = d.f. of the numerator MS and f2 = error d.f. The final analysis of variance is shown in table 4.22. The results indicate a highly significant main effect but not the two-factor interaction effect4.5.2.2 Design with Replication. We show the computations involved in the analysis of variance of a FFD with data from both replications in table 4.20.STEP 1. Outline the analysis of variance, following that given in plan 3 of appendix M:

Source Degree Sum Of of of Mean Computed Tabular FVariation freedom Squares Squares F 5% 1%__________________________________________________________________Replication 1Block 1Block X Replication 1Main effect 6Two-factor interaction 15Three-factor interaction 9Error 30 Total 63STEP 2. Compute the replication X block totals (RB) as shown in table 4.20. Then compute the replication total for each of the two replications (R), the block totals for each of the two blocks (B), and the grand total (G) as: R1 = 58.63 + 58.13 = 116.76 R2 = 57.43 + 57.43 = 114.86 B1 = 58.63 + 57.43 = 116.06 B2 = 58.13 + 57.43 = 115.56 G = 116.76 + 114.86 = 116.06 + 115.56 = 231.62

STEP 3. Let r denote the number of replications, p the number of blocks in each replication, and t the total number of treatments tested. Compute the correction factor, total SS, replication SS, block SS, and block X replication SS as: C.F. = G2 rt = (231.62)2 = 838.247256 (2)(32) Total SS = X2 C.F. = [(2.92)2 + + (4.78)2] 838.247256 = 12.419344 Replication SS = R2 C.F. t = (116.76)2 + (114.86)2 838.247256 32 = 0.056406

Blocks SS = B2 C.F. t = (116.06)2 + (115.56)2 838.247256 32 = 0.003906 Blocks X Replication SS = (RB)2 C.F. Replication SS Block SS t/p = (58.63)2 + (57.43)2 + (58.13)2 + (57.43)2 32/2 - 838.247256 0.056406 0.003906 = 0.003907STEP 4. Follow steps 2 to 7 of section 4.5.2.1; with one modification, namely that the grain yield data in the second column of table 4.21 is replaced by the yield totals over two replications as shown in table 4.23. Then compute the various SS as follows:

Main effect SS = (A)2 + (B)2 + (C)2 + (D)2 + (E)2 + (F)2 (r)(2k) = [(13.86)2 + (6.08)2 + (11.32)2 + (14.32)2 + (5.68)2 + (10.62)2] /(2)(32) = 11.051838 Two-factor interaction SS = [(AB)2 + (AC)2 + (BC)2 + + (CF)2 + (BF)2 + (AF)2]/(r)(2k) = [(1.48)2 + (0.92)2 + (-1.50)2 + + (-0.44)2 + (-0.52)2 + (1.62)2]/(2)(32) = 0.787594Three-factor interaction SS = [(ABD)2 + (ACD)2 + (BCD)2 + + (ADE)2 + (BDE)2 + (CDE)2]/(r)(2k) Error SS = Totals SS (the sum of all other SS) = 12.419344 (0.056406 + 0.003906 + 0.003907 + 11.051838 + 0.787594 + 0.238206)

= 0.277487STEP 5. Compute the mean square for each source of variation, by dividing the SS by its d.f. (see step 8 of section 4.5.2.1 for the determination of d.f) as: Replication MS = Replication SS 1 = 0.056406 = 0.056406 1 Block MS = Block SS 1 = 0.003906 = 0.003906 1 Block X Replication MS = Block X Replication SS 1 = 0.003907 = 0.003907 1

Table 4.23 Application of Yates method for the computation of sums of squares of a 26 Factorial Experiment Conducted in Fractional Factorial Design, with two replications; from data table 4.20__________________________________________________________________ Factorial EffectTreatment Identification Combination t0 t1 t2 t3 t4 t5 Preliminary Final__________________________________________________________________(I) 5.68 12.47 25.96 53.65 112.97 231.62 (G) (G)a(f) 6.79 13.49 27.69 59.32 118.65 13.86 A Ab(f) 6.54 13.54 27.91 55.00 6.97 6.08 B Bab 6.95 14.15 31.41 63.65 6.89 1.48 AB ABc(f) 6.39 13.42 26.73 4.01 2.57 11.32 C Cac 7.15 14.49 28.27 2.96 3.51 0.92 AC ACbc 6.21 15.77 29.55 3.08 0.49 -1.50 BC BCabc(f) 7.94 15.64 34.10 3.81 0.99 -0.50 ABC ABC(Block)

__________________________________________________________________ Factorial Effect Treatment ___Identification____ Combination t0 t1 t2 t3 t4 t5 Preliminary Final__________________________________________________________________d(f) 6.71 12.72 1.52 1.63 5.23 14.32 D Dad 6.71 14.01 2.49 0.94 6.09 -0.32 AD ADbd 6.51 13.67 1.47 2.22 0.99 -1.62 BD BDabd(f) 7.98 14.60 1.49 1.29 -0.07 -0.54 ABD ABDcd 7.20 14.57 1.19 0.27 -1.61 4.78 CD CDacd(f) 8.57 14.98 1.89 0.22 0.11 -2.42 ACD ACDbcd(f) 7.76 16.61 2.29 0.74 -1.05 0.04 BCD BCDabcd 7.88 17.49 1.52 0.25 0.55 1.84 ABCD EFe(f) 6.51 1.11 1.02 1.73 5.67 5.68 E Eae 6.21 0.41 0.61 3.50 8.65 -0.08 AE AEbe 6.26 0.76 1.07 1.54 -1.05 0.94 BE BEabe(f) 7.75 1.73 -0.13 4.55 0.73 0.50 ABE ABE

_________________________________________________________________ Factorial EffectTreatment Identification Combination t0 t1 t2 t3 t4 t5 Preliminary Final_________________________________________________________________ce 6.10 0.00 1.29 0.97 -0.69 0.86 CE CEace(f) 7.57 1.47 0.93 0.02 -0.93 -1.06 ACE ACEbcef(f) 7.09 1.37 0.41 0.70 -0.05 1.72 BCE BCEabce 7.51 0.12 0.88 -0.77 -0.49 1.60 ABCE DF de 6.32 -0.30 -0.70 -0.41 1.77 2.98 DE DEade(f) 8.25 1.49 0.97 -1.20 3.01 1.78 ADE ADEbde(f) 7.31 1.47 1.47 -1.36 -0.95 0.24 BDE BDE abde 7.67 0.42 -1.25 0.47 -1.47 0.44 ABDE CE cde(f) 8.38 1.93 1.79 1.67 -0.79 1.24 CDE CDEacde 8.23 0.36 -1.05 -2.72 0.83 0.52 ACDE BE bcde 7.91 -0.15 -1.57 -2.84 -4.39 1.62 BCDE AEabcde(f) 9.58 1.67 1.82 3.39 6.23 10.62 ABCDE E_________________________________________________________________

Main effect MS = Main effect SS 6 = 11.051838 = 1.841973 6 Two-factor interaction MS = Two-factor interaction SS 15 = 0.787594 = 0.052506 15Three-factor interaction MS = Three-factor interaction SS 9 Error MS = Error SS 30 = 0.277487 = 0.009250 30STEP 6. Compute the F value for each effect, by dividing its MS by the error MS as:

F(replication) = Replication MS Error MS = 0.056406 = 6.10 0.009250 F(block) = Block MS Error MS = 0.003906 < 1 0.009250 F(block X replication) = Block X replication MS Error MS = 0.003907 < 1 0.009250 F(main effect) = Main effect MS Error MS = 0.052506 = 2.86 0.009250

STEP 7. Compare its computed F value with the corresponding tabular F values, from appendix E, with f1 = d.f. of the numerator MS and f2 = error d.f. The results indicate that the main effects, the two-factor interactions, and the three-factor interactions are all significant.The final analysis of variance is shown in table 4.24. there are two important points that should be noted in the results of this analysis of variance obtained from two replications as compared to that without replication (Table 4.22):The effect of the three-factor interactions can be estimated only when there is replication.Table 4.24 Analysis of variance of Grain of Grain Yield Data in table 4.20, from a Fractional Factorial Design: of a 26 Factorial Experiment with Two Replications__________________________________________________________________ Source Degree Sum Of of of Mean Computed Tabular FVariation freedom Squares Squares Fa 5% 1%__________________________________________________________________Replication 1 0.056406 0.056406 6.10 4.17 7.56Block 1 0.003906 0.003906 < 1 - -Block X replication 1 0.003907 0.003907 < 1 - -

Source Degree Sum Of of of Mean Computed Tabular FVariation freedom Squares Squares Fa 5% 1%____________________________________________________________________Main effect 6 11.051838 1.841973 199.13** 2.42 3.47Two-factor interaction 15 0.787594 0.052506 5.68** 2.02 2.70 Three-factor interaction 9 0.238206 0.026467 2.86* 2.21 3.06Error 30 0.277487 0.009250 Total 63 12.419344 ____________________________________________________________________a** = significant at 1% level, * = significant at 5% level.