Concept

What it is: Randomized Block Design (RBD) groups experimental units into relatively homogeneous blocks (strata). Each block receives all treatments (or one replicate of each treatment). Randomization occurs within each block so treatments are assigned at random to plots inside the block. RBD reduces experimental error due to known nuisance variation by controlling it through blocking.

Why use RBD?

Explanation: When experimental units are heterogeneous (e.g., fertility gradient in a field), blocking isolates that heterogeneity into block effects; this reduces residual variance and increases precision of treatment comparisons. RBD is simple to implement and analyze.

Example: Five fertilizer treatments measured on a field split into four blocks (north–south strips). Each block contains the five treatments randomized within it. Compare treatment means using RBD ANOVA.

Layout

General layout: For \(t\) treatments and \(r\) blocks (replications), arrange an \(r \times t\) table. Each row is a block and in each row all \(t\) treatments appear once (if possible). Label cell observation as \(y_{ij}\) = response of treatment \(i\) in block \(j\).

Notation used

Let \(y_{ij}\) be the observation for treatment \(i\) in block \(j\). Let

\(T_i=\sum_{j=1}^r y_{ij}\) (treatment total), \quad \(B_j=\sum_{i=1}^t y_{ij}\) (block total),\quad \(G=\sum_{i=1}^t\sum_{j=1}^r y_{ij}\) (grand total).

Practical considerations

Explanation: Blocks should be as homogeneous as possible (within-block variation minimal) and selected based on known gradients (e.g., soil depth, initial weight, machine batch). RBD is less suitable when blocks become very large (hard to keep homogeneous) or when many treatments make block size big.

Example layout (t=4 treatments, r=3 blocks):
Block \ TreatmentT1T2T3T4
Block 1y11y21y31y41
Block 2y12y22y32y42
Block 3y13y23y33y43

Statistical analysis (ANOVA for RBD)

Model (fixed-effects):

\( y_{ij} = \mu + \tau_i + \beta_j + \varepsilon_{ij},\quad i=1,\dots,t;\ j=1,\dots,r \)

Components: \(\mu\) = overall mean; \(\tau_i\) = treatment effect (fixed); \(\beta_j\) = block effect; \(\varepsilon_{ij}\) ~ iid \(N(0,\sigma^2)\).

Sum of squares (corrected)

Total SS (corrected): \( \mathrm{TSS} = \sum_{i=1}^t\sum_{j=1}^r y_{ij}^2 - \dfrac{G^2}{tr} \)

Treatment SS: \( \mathrm{SS}_T = \dfrac{1}{r}\sum_{i=1}^t T_i^2 - \dfrac{G^2}{tr} \)

Block SS: \( \mathrm{SS}_B = \dfrac{1}{t}\sum_{j=1}^r B_j^2 - \dfrac{G^2}{tr} \)

Error SS: \( \mathrm{SS}_E = \mathrm{TSS} - \mathrm{SS}_T - \mathrm{SS}_B \).

ANOVA table

Sourced.f.Sum of SquaresMean SquareF
Treatmentst-1\(\mathrm{SS}_T\)\(\mathrm{MS}_T=\mathrm{SS}_T/(t-1)\)\(\dfrac{\mathrm{MS}_T}{\mathrm{MS}_E}\)
Blocksr-1\(\mathrm{SS}_B\)\(\mathrm{MS}_B=\mathrm{SS}_B/(r-1)\)\(\dfrac{\mathrm{MS}_B}{\mathrm{MS}_E}\)
Error(t-1)(r-1)\(\mathrm{SS}_E\)\(\mathrm{MS}_E=\mathrm{SS}_E/((t-1)(r-1))\)
Totaltr-1\(\mathrm{TSS}\)
Numerical example (short): Suppose 4 treatments, 3 blocks; computed totals give \(G\), \(T_i\) and \(B_j\). Using formulas above compute \(\mathrm{SS}_T,\mathrm{SS}_B,\mathrm{SS}_E\), then test \(H_0:\tau_1=\cdots=\tau_t\) via \(F\) with d.f. \((t-1),(t-1)(r-1)\).

Critical Difference (C.D.) and multiple comparisons

Purpose: If ANOVA rejects the global null (treatment means equal), we need pairwise comparisons to know which treatment pairs differ. The simplest is Critical Difference (a t-based pairwise test). :contentReference[oaicite:13]{index=13}

Critical Difference formula

\( \text{C.D.} = t_{\alpha, \nu} \times \text{S.E.}(d) \)

For comparing means \( \bar y_i, \bar y_j\) with replications \(r_i,r_j\):

\( \text{S.E.}(d) = \sqrt{\mathrm{MS}_E\left(\dfrac{1}{r_i} + \dfrac{1}{r_j}\right)}. \)

When replications are equal \(r_i=r_j=r\):

\( \text{S.E.}(d) = \sqrt{\dfrac{2\mathrm{MS}_E}{r}}. \)

Reject equality of two means at level \(\alpha\) if \(|\bar y_i - \bar y_j| > \text{C.D.}\).

Example: From an RBD ANOVA with \(\mathrm{MS}_E = 25\), \(r=4\), error d.f. = 12 and \(t_{0.05,12}=2.179\):

S.E. = \(\sqrt{2\times 25/4} = \sqrt{12.5} = 3.535\). C.D. = \(2.179\times 3.535 \approx 7.70\). Any difference larger than 7.70 is significant at 5%.

RBD with one missing value — estimation & analysis

Problem: When one observation is missing in an RBD, we can estimate it (by solving normal equations or using formulae) and then proceed with ANOVA. The course notes give a direct formula for the missing value estimate and the adjustment for upward bias.

Notation for one missing observation

Suppose the missing observation is \(x\) for treatment \(i_0\) in block \(j_0\). Let:

\(R'=\) total of the observed block \(j_0\) (without x), \quad \(T'=\) total of observed treatment \(i_0\) (without x), \quad \(G'=\) grand total of observed values (without x).

Estimate for missing value (direct formula)

One commonly used estimator is:

\( \displaystyle \hat x \;=\; \dfrac{r\,R' + t\,T' - G'}{(r-1)(t-1)}. \)

Upward bias correction: After substitution one may compute an upward bias correction term \(B\) and adjust treatment SS as described in the notes. The worked formula and substitution are shown in the PDF.

Worked numeric example (from your PDF)

Data: \(t=5\) treatments, \(r=4\) replications. One missing observation at treatment 2, replication III. The observed totals (without x) are:

  • \(R' =\) replication-III total (observed) = 135.1
  • \(T' =\) treatment-2 total (observed) = 89.5
  • \(G' =\) grand total (observed) = 590.2

Apply the formula:

\( \hat x = \dfrac{rR' + tT' - G'}{(r-1)(t-1)} = \dfrac{4(135.1) + 5(89.5) - 590.2}{3\times 4} = \dfrac{397.7}{12} = 33.1. \)

After substituting \(x=33.1\) the grand total becomes \(623.3\) and treatment 2 total becomes \(122.6\); the adjusted treatment SS is computed and an upward-bias term \(B=0.3645\) is used to correct SS (see the ANOVA table below). Final ANOVA (as in PDF) gave Treatment MS = 130.3659, Error MS = 31.6316 and \(F \approx 4.117\) (d.f. for Error = 11).

Takeaway: The formula gives a quick estimate of a single missing plot; the bias term is small here. Replace the missing value, recompute totals and S.S., then finish the ANOVA. Full worked arithmetic is in the course document. :contentReference[oaicite:18]{index=18}

Advantages, disadvantages & applications

Advantages

  • Increases precision by removing block-to-block variability from the error term.
  • Flexible: any number of treatments/replications (practically at least 2) and easy to include checks or repeated controls. }
  • Analysis is straightforward (simple ANOVA computations).

Disadvantages

  • May give misleading results if blocks are not homogeneous.
  • Not suitable for very large number of treatments (blocks would be large).
  • If many plots are missing analysis becomes tedious (one or two missing observations are manageable using formulas/worked methods).

Applications

Common in agriculture (fertilizers, varieties across field blocks), manufacturing (batches as blocks), medical trials (centres as blocks) and lab experiments where a known nuisance factor can be blocked.

Problems (practice)

Below are two practice problems — work them by hand or plug into a calculator following the steps shown above.

Problem 1 — Basic RBD ANOVA

Five treatments with 4 blocks. Observed treatment totals: 121.8, 122.6, 113.9, 162.8, 102.2; grand total 623.3. Compute treatment SS, block SS, error SS and test if treatments differ (use formulas in section "Statistical analysis").

Problem 2 — One missing observation (use the formula)

Same as Problem 1 but assume the observation for treatment 2 in block 3 is missing and the observed totals are \(R'=135.1\), \(T'=89.5\), \(G'=590.2\). Find \(\hat x\) and complete the ANOVA. (This is the worked example in the notes; answer: \(\hat x\approx 33.1\).)

If you want, I can convert these problems into fillable HTML forms that compute S.S. and F-values automatically in the browser (small JS calculator).

17.3 Latin Square Design

Explanation: Controls two sources of variation.

1. Concept of LSD

(A) Complete 3×3 LSD — concise walk‑through

Suppose the 3×3 table (treatments labelled A,B,C) has observations:

Col1Col2Col3
Row110(A)12(B)11(C)
Row213(B)14(C)15(A)
Row39(C)8(A)16(B)

Compute treatment totals: \(T_A=10+15+8=33,\; T_B=12+13+16=41,\; T_C=11+14+9=34\). Grand total \(G=108\).

Then form SS_{trt}, SS_rows, SS_cols and SSE. (This page keeps algebra concise — full numeric ANOVA can be computed by following standard SS formulas.)

(B) Estimation of one missing value (summary)

If one cell is missing (row \(i\), col \(j\), treatment \(k\)), estimate with:

\[ \widehat{y}_{ij(k)} = \dfrac{t(R_i + C_j + T_k) - 2G}{(t-1)(t-2)}. \]

After replacing the missing value use completed ANOVA but reduce total and error df by 1.

9. Problems & practice questions

  1. Given the 4×4 square in section 1 with numeric observations (provide a table), compute the ANOVA table and test treatments at 5% level.
  2. For a 5×5 Latin square with one missing observation at (row2, col4, treatment C), using the marginal totals below, estimate the missing value with the formula in section 8(B).
  3. Compare CRD, RBD and LSD using hypothetical MSEs: CRD=10, RBD=6, LSD=3. Compute relative efficiencies and interpret.
  4. Find an orthogonal Latin square for t=3 (note: small t has limitations) and discuss if Graeco‑Latin square is possible.

10. Critical differences: LSD vs RBD vs CRD

Summary of key differences:

FeatureCRDRBD (RCBD)LSD
Blocks controllednoneonetwo (rows & columns)
Number of experimental unitsflexibleneeds blocks × treatmentsmust be t×t
Error dfN-t(b-1)(t-1)(t-1)(t-2)
When preferablehomogeneous experimental unitsone nuisance factor importanttwo nuisance factors important & equal levels
Complexitysimplemoderatehigher (layout constraints)

Example (critical differences)

If field variability exists in two directions (rows and columns), LSD will often yield lower MSE than RBD and CRD. But if only one direction matters, RBD may suffice and is simpler.

References & further reading

Standard textbooks: Montgomery Design and Analysis of Experiments, Cochran & Cox Experimental Designs, and class lecture notes on Latin squares. For missing-value derivations see lecture notes on imputation in experimental designs.

Latin Square Design (LSD) — Estimating one missing value & efficiencies (versus RBD, CRD)

You can find derivations of this in standard ANOVA/DOE lecture notes (see references).

Example problem (practice)

Given an order‑3 Latin square with observed table (one missing at row2,col2):

C1C2C3
R1A:10B:12C:11
R2B:13C:?A:14
R3C:9A:8B:15

Compute \(R_2, C_2, T_{\text{treatment at missing}}\) and \(G\) (exclude missing), then use the formula (with \(t=3\)) to estimate the missing value. (Try it — for \(t=3\) the denominator \((t-1)(t-2)=2\times1=2\).)

Efficiency: RBD relative to CRD (intuitive + formula)

Why RBD can be more efficient: Randomized Block Design (RCBD/RBD) accounts for a known nuisance factor (blocks). When blocks explain variation, the residual (error) variance is reduced compared to CRD — so treatment comparisons become more precise.

Relative efficiency (practical definition)

A common definition of relative efficiency of RBD compared to CRD is the ratio of their mean‑squared errors (or the ratio of sample sizes required to achieve the same precision):

\[ \text{RE}_{\text{RBD:CRD}} = \frac{\text{MSE}_{\text{CRD}}}{\text{MSE}_{\text{RBD}}}. \]

If \(\text{RE}>1\) then RBD is more efficient (i.e. a CRD would need roughly \(\text{RE}\) times more observations per treatment to match the precision).

How to estimate \(\text{MSE}_{\text{CRD}}\) when you only have RBD data: common practice is to re‑compute the CRD MSE by "undoing" the block partitioning — for example, use the RCBD sums of squares to form a hypothetical CRD error:

\[ \text{MSE}_{\text{CRD}} \approx \frac{\text{SS}_{\text{blocks}} + \text{SS}_{\text{error (RBD)}}}{\text{df}_{\text{CRD}}}, \]

and then take the ratio with \(\text{MSE}_{\text{RBD}}\). Textbook notes and lecture slides give worked examples; the practical formula used by many instructors is

\[ \text{RE} \approx \frac{\text{MSE}_{\text{CRD (reconstructed)}}}{\text{MSE}_{\text{RBD}}}. \]

Short numeric illustration

Suppose an RBD ANOVA gives \(\text{MSE}_{RBD}=3.7\) and a hypothetical CRD MSE (reconstructed) would be \(\text{MSE}_{CRD}=5.2\). Then

\[ \text{RE} = 5.2/3.7 \approx 1.41. \]

This means CRD would need about 1.41 times as many observations per treatment to achieve the same precision as the RBD.

Efficiency: LSD relative to RBD and CRD

Because LSD blocks in two directions (rows and columns), it can reduce residual variance further than RBD (which blocks in one direction) provided the row and column sources actually contribute meaningful variability. The same relative efficiency idea applies:

\[ \text{RE}_{\text{LSD:RBD}} = \frac{\text{MSE}_{\text{RBD}}}{\text{MSE}_{\text{LSD}}}, \qquad \text{RE}_{\text{LSD:CRD}} = \frac{\text{MSE}_{\text{CRD}}}{\text{MSE}_{\text{LSD}}}. \]

Interpretation: values >1 mean the denominator design (LSD) is more efficient.

Comparative example (toy)

Imagine MSEs from analyses are: \(\text{MSE}_{CRD}=8.0,\; \text{MSE}_{RBD}=4.0,\; \text{MSE}_{LSD}=2.0\). Then

\[ \text{RE}_{RBD:CRD}=8/4=2, \quad \text{RE}_{LSD:RBD}=4/2=2, \quad \text{RE}_{LSD:CRD}=8/2=4. \]

So LSD here is 4× more efficient than CRD, and 2× more efficient than RBD for treatment comparisons.

Practical remarks and when LSD helps

  • LSD is best when you have exactly \(t\) treatments and can block in two directions (e.g., rows and columns) — otherwise it's not applicable.
  • The gain in precision depends on how much variation is explained by rows and columns. If once estimated, row and column effects are negligible, LSD may not beat RBD.
  • Degrees of freedom for error in LSD are limited: \((t-1)(t-2)\). So for small \(t\) there are few error df — be cautious when drawing strong conclusions.
  • When a missing observation is imputed, adjust total and error df by −1 before testing.

More worked problems (suggested exercises)

  1. Take the 4×4 numeric example in section 3, replace the missing value with the estimate, form the ANOVA table and show the adjustment of df. Compute F for treatments.
  2. Given an RBD ANOVA with SSblocks = 40 (df = 3), SSE = 18 (df = 8), compute MSE_RBD and reconstruct MSE_CRD and the relative efficiency RE = MSE_CRD / MSE_RBD.
  3. Design a 5×5 LSD dataset with two strong blocking directions and show numerically how MSE drops from CRD → RBD → LSD.