Randomized Complete Block Design (RCBD)

Today’s goals

Explore key concepts in RCBD:

  1. Heterogeneous experimental material
  2. Treatment randomization
  3. The effects model
  4. The ANOVA table

Motivational example - Treatment design

  • 2-way factorial
  • N fertilizer rates: 0, 100, 200 kg N/ha
  • K fertilizer rates: 0, 30, 60 kg K/ha
  • 3 x 3 = 9 treatment combinations

Experimental design

Assuming we have heterogeneous experimental material (e.g., different soil type, topography, etc.)

  • Thus, we should use a randomized complete block design (RCBD)

Treatment randomization - RCBD

  • Randomization of a treatment to a EU is restricted.

  • In an RCBD, each treatment appears once in each block.

  • Because of that, randomization needs to be performed for each block individually.

In our motivational example:

  • 4 replicates

  • Total observations: 9 x 4 = 36 EUs

Heterogeneous Experimental Material - RCBD ✅

In the plot layout here, all treatments (1 through 9) were randomly assigned to any experimental unit (plot) within each block.

Heterogeneous Experimental Material - RCBD ✅

  • Treatment 1 and its replicates are highlighted.

  • Note how, due to the restricted randomization, treatment 1 appears only once in every block.

Heterogeneous Experimental Material - RCBD ✅

Since the experimental material is heterogeneous (e.g., different soil texture class), we are safeguarding statistical power when estimating treatment means and performing comparisons. 👍

The effects model

\[ y_{ijk} = \mu + \rho_{k} + \alpha_{i} + \beta_{j} + \alpha\beta_{ij} + e_{ijk} \]

  • \(y_{ijk}\) is the observation on the kth block from ith N rate and jth K rate
  • \(\mu\) is the overall mean 
  • \(\rho_{k}\) is the differential effect of kth block
  • \(\alpha_{i}\) is the differential effect of ith N rate
  • \(\beta_{j}\) is the differential effect of jth K rate
  • \(\alpha\beta_{ij}\) is the differential effect of the combination of the ith N rate and ith K rate
  • \(e_{ijk}\) is the residual corresponding to the block k of N rate i and K rate j.

The ANOVA table

In the following ANOVA table…

  • n is number of levels in N rate = 3
  • k is number of levels in K rate = 3
  • r is number of blocks = 4
  • N is total number of obserevations or EUs = 3 x 3 x 4 = 36

The ANOVA table

Source of variation

df

SS

MS

F ratio

Block

dfb =
r - 1 =
4 - 1 = 3

SSb

N rate

dfn =
n - 1 =
3 - 1 = 2

SSn

MSn =
SSn / dfn

MSn / MSe

K rate

dfk =
k - 1 =
3 - 1 = 2

SSk

MSk =
SSk / dfk

MSk / MSe

N x K

dfnk =
(n - 1) x (k - 1) =
(3-1) x (3-1) = 4

SSnk

MSnk =
SSnk / dfnk

MSnk / MSe

Error

dfe =
(ab - 1) x (r - 1) =
(9 - 1) x (4 - 1) =
8 x 3 = 24

SSe

MSe =
SSe / dfe

TOTAL

dft =
N -1 = 35

SSt

  • Notice how now we consumed extra 3 dfs to calculate the block effect.
  • With that, we have 3 fewer dfs in the error df.

ANOVA table - motivational example

Source of variation

Sum Sq

Df

F value

Pr(>F)

(Intercept)

836,829,184.0

1

2,223.1503963

<0.001

rep

2,747,393.1

3

2.4329410

0.09

nrate_kgha

1,491,078.2

2

1.9806258

0.16

krate_kgha

470,445.5

2

0.6249012

0.544

nrate_kgha:krate_kgha

11,107,585.3

4

7.3772023

<0.001

Residuals

9,033,981.9

24

  • What is significant here (say at \(\alpha = 0.05\))?

  • What can we say about the rep/block effect?

Summary

  • RCBD is randomized with one restriction (each block separately)
  • Each term of the effects model has a row in the ANOVA table
  • By including the block in the model, it consumed dfs from error
  • In a properly conducted RCBD, we would hope that the amount of SS explained by block is relatively greater than the consequence of consuming 3 df from the error to calculate the block effect.