In our previous lecture, we talked about blocking a replicated two to the K factorial. Now, let's talk about the unreplicated case. Well, clearly our previous discussion doesn't apply because there's only one replicate. So you can't make each replicate a block, you've got to do something different, and what we end up doing is a very clever solution to the problem, and I'm going to illustrate this using that resin plant experiment from example 6.2 that we've used quite a few times. Remember this is a two to the four, and there's only one replica of the design. Here's the layout of the design. Let's suppose that when we run this experiment, we need batches of raw material, and we can only get eight runs out of one batch. So what are we going to do? We're going to have to split this design up into two blocks of eight runs each. How do we do that? Well, the secret for doing this is shown in this table of contrast constants. The Madison plus signs for the two to the four. This is the complete table. We've got all four of the main effects, we've got all six of the two-factor interactions, we've got all three of the 3-factor and even the 4-factor interaction. Now, every one of these columns contains eight pluses and minuses. Furthermore, the columns are all orthogonal. So the technique that's used here to create the two blocks is to let one of these factor effects essentially be linked to the blocks. Literally, take one of the columns out of this table and use the minus and plus signs in that column to assign the treatment combinations to the two blocks. That is pick a column, and put all of the plus treatment combinations in that column into one block, and all of the minus treatment combinations in that column into the other block. Now, when you do this, what happens to the column that you've chosen to do this? Well, that column is no longer estimable. What can be estimated from that column is the block effect. So what you've done is the technical word now is you have confounded that effect with the blocks. So which effect would you want to confound? Which effect would you want to lose information on? Because that's what you're doing, you're sacrificing information on the confounded effect. Well, doesn't it make sense to pick an effect that is least likely to be significant. What's your guess as to the factor that's least likely to be significant here. What do you think? One of the higher order interactions, and my guess would be ABCD. So I would confound ABCD with blocks, just change the label from ABCD to blocks. Now all the runs that are plus here go in Block 1, and all the runs that are minus go in Block 2, and that produces a design that looks like this. The figure on the right shows you the two blocks. Now, I did something I think interesting here, when I actually put the observations into the table, I took all the observations that you originally observed in Block 1 and I reduced them by 20 units. I've simulated a 20 unit block effect. In other words, if you had actually been able to run these observations in block one in Block 2, and you had enough material to be able to do that, the results would have been 20 units higher. So what effect does that have on the analysis? Well, here are the model regression coefficients and the factor effects and the sums of squares for all of these 15 effects, including my block effect. Well, if you return to the original example back in Chapter 6, these regression coefficients and effect estimates are exactly the same as they were when the design was treated as a completely randomized design, except for the effect estimate for blocks The effect estimate for blocks is now equal to the original effect estimate for ABCD minus 20. Why minus 20? Because minus 20 was the block effect that we deliberately introduced into the data. So this confounding scheme enables us to essentially get the same information on all of our factorial effects, except the effect that we choose to confound the blocks in an unreplicated design. Here's what the ANOVA would look like for the largest effects, and just as before, the three main effects of AC and DC, and AC and AD, or significant. There is a single degree of freedom for blocks, because we have two blocks. Instead of having ten degrees of freedom for error, we only have nine because before, we were able to pool the ABCD interaction into error. Well, we can't do that now because the ABCD effect is confounded with blocks, it can't be part of the error. So the rest of the analysis then is completely unchanged from the original block effect did not introduce anything into the experiment unusual. Below we're going to take a look at the jump analysis, assuming blocks are random, and this of course, will allow us to use Remmel for the estimate of the block effect, the block variants component. This analysis considers only the main effects and two factor interactions, but it's basically in agreement with what we just basically saw. It identifies the same factors and interactions as being significant. Here is the analysis. This is the first part of the jump display, and here's the second part of the jump display showing you the tests on these parameter estimates, and then here is the Remmel output, and you notice that we do have a very large variance component now for blocks 169.68, and in fact, the block variance component accounts for almost 85 percent of the total variability in the data. Wow, the block effect is huge. Well, remember I made up a huge block effect, that's not a surprise. But again, notice that the lower and upper confidence limits span zero, is this an indication that the variance component is really zero? Well, maybe I don't really think so. It accounts for 85 percent of the total variability. I think this confidence interval is ridiculously wide because you're estimating a variance with only two observations. So I think the fact that the number of blocks is so small that this is why that confidence interval is so wide, and I wouldn't really worry about that, I would look at the fact that that block variance component accounts for almost 85 percent of the total variability, that's a very important result. So this is how you would go about confounding or blocking an unreplicated factorial by using this technique of confounding.
