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completely randomized block design

EXAMPLE . 1. Consider the following data for average daily gain (ADG) by 12 pens of cattle fed three treatment diets: Trt 1 Trt 2 Trt 3 3.40 3.32 3.25 3.59 3.49 3.42 3.65 3.52 3.55 3.85 3.70 3.67Trt means 3.4725 3.6225 3.50752. If we do the ANOVA for this CRD, w21

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completely randomized block design

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    1. Completely randomized block design The randomized complete block design Two-way classification A. So far, our study of the ANOVA has involved the simplest of experimental designs, the completely randomized or completely random design (CRD) The only complexity we have introduced at this point is the factorial arrangement of treatments within the CRD B. What if we use a CRD and we find that the variation among experimental units is so large relative to the treatment effects that we are unable to detect a significant difference with the F test?

    2. EXAMPLE 1. Consider the following data for average daily gain (ADG) by 12 pens of cattle fed three treatment diets: Trt 1 Trt 2 Trt 3 3.40 3.32 3.25 3.59 3.49 3.42 3.65 3.52 3.55 3.85 3.70 3.67 Trt means 3.4725 3.6225 3.5075 2. If we do the ANOVA for this CRD, we would have the following results: SOV df SS MS F P-value Trt 2 0.049267 0.024633 0.812014 0.474013 Error 9 0.273025 0.030336 ----- ----- Total 11 0.322292

    3. Thus, we would conclude from this ANOVA that treatment did not significantly affect the response 3. If we look at the ADG data, there seems to be considerable variation in within each treatment in the ADG In fact, the variation within a treatment is greater than the variation among treatments 4. So, what can the researcher do to control this variation? a. With the CRD design, the researcher could increase the sample size, which should help decrease the error variance (Within group or Residual MS) b. But, are there other ways that we can account for sources of variation that might be affecting the results?

    4. 5. What if we find that within each treatment, the lower values of ADG are associated with pens of cattle that have lower initial body weights when the experiment was started, and that higher values of ADG are associated with cattle that have higher initial body weights? a. Suppose we find out that the technicians at the feedlot randomly allotted treatments to pens of cattle on the basis of their initial body weight (a) Thus, they grouped cattle by initial body weight, sorted them into three pens, and randomly applied the treatments to those three pens within an initial body weight group (i) They repeated this same process with three other groups of three pens each (ii) This type of design is referred to as a two-way classification 1. Specifically in this case, this design is referred to as a randomized complete block design (RCBD)

    5. (b) We can define a block as a collection of experimental units put together so that the variation between two experimental units in the same block is less (on average) than two experimental units in a different block (i) Although our example might lead you to believe otherwise, remember that blocking is done BEFORE treatments are applied, not after the fact (c) Lets rearrange our data to reflect those initial body weight groups: Weight block Trt 1 Trt 2 Trt 3 1 3.40 3.32 3.25 2 3.59 3.49 3.42 3 3.65 3.52 3.55 4 3.85 3.70 3.67 Trt means 3.6225 3.5075 3.4725

    6. Now, we have something that looks like a factorial structure, with three treatments and four weight blocks 6. Lets take a look at the sources of variation that we have in this design a. Just like in the CRD, our total variation can be subdivided into sources: b. Variation among treatments c. We also have another source of variation the weight block 7. In addition, the variation among experimental units treated alike is not quite the same as it was in the CRD Now, we have to consider that differences among experimental units treated alike needs to be evaluated as a function of what block they are in. In fact, each experimental unit is cross-classified by its treatment and its block (a) The residual error, then must represent the failure of treatments to be the same in different blocks (b) Thus, the residual error must be equal to block x treatment interaction

    7. 8. How would the ANOVA look for this type of design? Lets use the notation that we have t treatments, and b blocks SOV Calculation of df df Block b - 1 3 Trt t-1 2 Error (b 1)(t 1) 6 Total tb - 1 11 b. We already know how to obtain the Total SS and the Treatment SS from our previous examples with the CRD, so all we need to do in addition to that is calculate the Block SS Then we should be able calculate the Residual SS by difference c. The weigh block sums can be obtained by summing values in the data table: Block 1 = 9.97 (b) Block 2 = 10.5 (c) Block 3 = 10.72 Block 4 = 11.22

    8. Thus, the Block SS would be: (9.972 + + 11.222)/3 (42.412/12) = 0.268558 9. Thus, we can write our final AOV as follows: SOV df SS MS F P-value Block 3 0.268558 0.089519 120.25 0.00001 Trt 2 0.049267 0.024633 33.08955 0.0006 Error 6 0.004467 0.000744 Total 11 0.322292 Based on this ANOVA, our conclusion would be that the treatment effect was highly significant 10. Why is our conclusion different? We have decreased the Error variance (Residual MS) Essentially, the block effect has removed variance that would been part of the Error in the CRD. Notice also, that compared with the CRD for this analysis, we have less df in the Error 6 for the RCBD vs. 9 for the CRD b. Thus, we have lost Residual (error) df, but we have gained power for detecting treatment differences because SS that would have been in the error term are accounted for in Block SS

    9. 11. Can we determine how efficient our blocking procedure is? And what if we blocked, but the blocks dont account for much variation? a. Recall that we analyzed our data both as a CRD and as a RCBD (a) The Error (Residual) MS (EMS) for the CRD = 0.030336 (i) The variance of a treatment mean would equal 0.030336/r, where r is the number of observations per treatment (b) The Error (Residual) MS (EMS) for the RCBD = 0.000744 (i) The variance of a treatment mean would equal - 0.000744/b, where b is the number of blocks (no replication within blocks) (c) To have equal variance of a treatment mean with a CRD as with a RCBD, the number of replications, r, must satisfy the following: (i) EMS( CRD ) / r = EMS(RCBD) / b r / b = EMS (RCBD) / EMS(CRD ) b. Assuming that all effects in the model are fixed, the ratio of the EMS for the CRD and the RCBC is referred to as the relative efficiency of blocking

    10. In our case, this value = 0.030336/0.000744 = 40.774 a. Thus, in the example we conducted, it would take approximately 40 replications per treatment in the CRD vs. four in the RCBD to obtain the same standard error of the treatment means! (i) Recall that we have used a contrived example, and the efficiency of blocking is rarely this good . However, we certainly would want the ratio to exceed 1.0, or we would conclude that blocking was ineffective d. To calculate the relative efficiency of blocking, one does not have to construct the AOV for the CRD (a) The value can be estimated from information in the AOV for the RCBD as follows: (i) RE of blocking = (b-1)Block MS+b(t-1) Error MS/ (tb-1)EMS 1. Where b = the number of blocks and t = number of treatments (ii) In our case: ([3 x 0.089519] + [3 x 2 x 0.000744])/(11 x 0.000744) = 33.4

    11. Note that the relative efficiency calculation is not quite fair to the CRD because the CRD has more error df than the RCBD (a) The formula can be modified to account for the differences in df See Proc GLM; Class trt block; Model ADG=trt block; Run;

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