PMean: Estimating the efficiency of a completely randomized block design

Steve Simon

2014/03/12

I needed to look up a formula for the estimating the relative efficiency to a completely randomized block design to a design without blocking.

A quick search on Google revealed the formula:

Here’s where I needed the formula. I was running a Monte Carlo simulation of three different statistical estimators. Each replication of the simulation required you to generate a binomial random variable with 100 trials and a probability of success of 0.5. You could just generate these values on the fly, but it makes sense to save the binomials that you used for one estimator and re-use the same ones for the other two estimators. This makes each replication in the Monte Carlo simulation a block, and you can estimate the precision that you have gained by the careful re-use of these binomials by fitting a completely randomized block design.

 > tst.m1 <- lm(bb~trt+blk,data=mcd)
 > anova(tst.m1)
 Analysis of Variance Table
 Response: bb
             Df<U+00A0> Sum Sq Mean Sq<U+00A0><U+00A0> F value<U+00A0><U+00A0><U+00A0> Pr(>F)
 trt<U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0> 2 11.5617<U+00A0> 5.7809 35659.326 < 2.2e-16 ***
 blk<U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0><U+00A0> 999<U+00A0> 2.2959<U+00A0> 0.0023<U+00A0><U+00A0><U+00A0> 14.176 < 2.2e-16 ***
 Residuals 1998<U+00A0> 0.3239<U+00A0> 0.0002
 ---
 Signif. codes:<U+00A0> 0 <U+0091>***<U+0092> 0.001 <U+0091>**<U+0092> 0.01 <U+0091>*<U+0092> 0.05 <U+0091>.<U+0092> 0.1 <U+0091> <U+0092> 1
 > (999*0.0023+2000*0.0002)/(2999*0.0002)
 [1] 4.497666

The re-use of the binomials has resulted in a 4.5 fold improvement in precision.

This formula was published in Experimental Design and Data Analysis for Biologists by Gerald Peter Quinn, Michael J. Keough, (ISBN: 9780521009768) and digitized by Google. There is a huge controversy over the<U+00A0> Google project to digitize a whole bunch of books and make them accessible through their search engine. But I have to admit that I’m glad to have access to all those books for finding a formula like this.

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