Randomised Block Design (RBD) — Design & Analysis of Experiments (Unit 3)

Topics Covered

RBD Concept Layout Advantages Statistical Analysis Critical Difference Missing Value

On this page

1. Concept of RBD
2. Layout of RBD
3. Advantages and Disadvantages
4. Statistical Analysis of RBD
5. Estimating One Missing Value in RBD
Key Take-aways

1. Concept of RBD

DEFINITION

Randomised Block Design (RBD) uses all three Fisher principles — replication, randomization and local control. The experimental units are grouped into homogeneous blocks; within each block, treatments are randomly allotted, with each treatment appearing exactly once per block.

RBD is appropriate when there is a known source of variability across the experimental area (gradient of soil fertility, batch of animals, time periods). Blocking removes that variability from the error.

2. Layout of RBD

For \(t\) treatments and \(b\) blocks, total plots \(= t \times b\). Within each block of \(t\) plots, the \(t\) treatments are randomly assigned.

Fig 3.1 — RBD layout: each block contains all treatments in random order

3. Advantages and Disadvantages

Advantages

More precise than CRD when block-to-block variation is significant.
Simple statistical analysis (two-way ANOVA without interaction).
Flexible: any number of treatments and blocks.
Missing observations can still be recovered via formula.

Disadvantages

Requires homogeneity within blocks; if violated, RBD loses efficiency.
If number of treatments is large, blocks become large and homogeneity is harder.
Less flexible than CRD when replicates differ across treatments.

Applications

Agricultural field trials with a fertility gradient.
Industrial experiments where machines or operators cause variation.
Animal feeding trials where animals can be paired by weight/age.

4. Statistical Analysis of RBD

Mathematical Model

\[ y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij}, \]

where \(\alpha_i\) = effect of treatment \(i\), \(\beta_j\) = effect of block \(j\), \(\epsilon_{ij} \sim N(0,\sigma^2)\) and \(\sum \alpha_i = \sum \beta_j = 0\).

Hypotheses

\(H_0^T: \alpha_1 = \cdots = \alpha_t = 0\) (no treatment effect)
\(H_0^B: \beta_1 = \cdots = \beta_b = 0\) (no block effect)

Sums of Squares

Notation: \(T_i\) treatment totals (\(b\) obs each); \(B_j\) block totals (\(t\) obs each); \(G\) grand total; \(N = t b\); \(C = G^2/N\).

\[ \text{SS}_T = \sum y_{ij}^2 - C, \quad \text{SS}_{Tr} = \dfrac{\sum T_i^2}{b} - C, \quad \text{SS}_{Bl} = \dfrac{\sum B_j^2}{t} - C, \] \[ \text{SS}_E = \text{SS}_T - \text{SS}_{Tr} - \text{SS}_{Bl}. \]

ANOVA Table

Source	df	SS	MS	F
Treatments	\(t-1\)	SS_Tr	MS_Tr	MS_Tr/MS_E
Blocks	\(b-1\)	SS_Bl	MS_Bl	MS_Bl/MS_E
Error	\((t-1)(b-1)\)	SS_E	MS_E	—
Total	\(N-1\)	SS_T

Critical Difference

\[ \text{CD} \;=\; t_{\alpha/2,\,(t-1)(b-1)} \,\sqrt{\dfrac{2\, \text{MS}_E}{b}}. \]

Used to compare any pair of treatment means: differ if |\(\bar y_i - \bar y_j\)| > CD.

EXAMPLE 1 (Full RBD analysis)

4 treatments × 3 blocks (yields):

	B1	B2	B3	T_i
T1	10	12	14	36
T2	14	16	18	48
T3	8	10	12	30
T4	12	14	16	42
B_j	44	52	60	G=156

\(N = 12, C = 156^2/12 = 2028\).

\(\sum y^2 = 100+144+196+196+256+324+64+100+144+144+196+256 = 2120\).

\(\text{SS}_T = 2120 - 2028 = 92\).

\(\text{SS}_{Tr} = (36^2 + 48^2 + 30^2 + 42^2)/3 - 2028 = (1296+2304+900+1764)/3 - 2028 = 6264/3 - 2028 = 2088 - 2028 = 60\).

\(\text{SS}_{Bl} = (44^2 + 52^2 + 60^2)/4 - 2028 = (1936+2704+3600)/4 - 2028 = 8240/4 - 2028 = 2060 - 2028 = 32\).

\(\text{SS}_E = 92 - 60 - 32 = 0\). df: 3, 2, 6.

(Constructed data has zero error — illustrates the decomposition.)

EXAMPLE 2

3 treatments × 4 blocks: \(\text{SS}_{Tr} = 60,\; \text{SS}_{Bl} = 24,\; \text{SS}_E = 12,\; \text{MS}_E = 12/6 = 2\).

\(F_T = 30/2 = 15\); \(F_{0.05, 2, 6} = 5.14\) ⇒ reject (treatments differ).

\(F_B = 8/2 = 4\); \(F_{0.05, 3, 6} = 4.76\) ⇒ accept (no block effect).

CD: \(t_{0.025, 6} = 2.447;\; \text{CD} = 2.447 \sqrt{2 \cdot 2/4} = 2.447\). Compare with treatment-mean differences.

5. Estimating One Missing Value in RBD

If one observation \(y_{ij}\) is missing (in row \(i\), column \(j\)), it can be estimated by minimising the error sum of squares.

YATES' FORMULA (Missing value in RBD) \[ \hat y_{ij} \;=\; \dfrac{t\, T'_i + b\, B'_j - G'}{(t - 1)(b - 1)}, \]

where \(T'_i, B'_j\) are the sums of available values in the corresponding treatment row and block column; \(G'\) is the grand total of all available values; \(t\) = number of treatments; \(b\) = number of blocks.

Adjustments after substitution

Insert \(\hat y_{ij}\) and analyse as usual.
Reduce error df by 1 (loss of one observation): df_E = \((t-1)(b-1) - 1\).
Treatment SS is biased upward — apply the bias correction: \[ B = \dfrac{(B'_j - (t-1)\hat y_{ij})^2}{t(t-1)}. \] Reduce SS_Tr by \(B\) before testing treatments.

EXAMPLE 1

RBD with 3 treatments × 3 blocks. Suppose value at T2-B3 is missing.

	B1	B2	B3
T1	10	12	14
T2	14	16	?
T3	8	10	12

\(T'_2 = 14 + 16 = 30; B'_3 = 14 + 12 = 26;\) \(G' = 96\).

\(\hat y_{23} = \dfrac{3(30) + 3(26) - 96}{2 \cdot 2} = \dfrac{90 + 78 - 96}{4} = \dfrac{72}{4} = 18\).

Insert 18 and analyse normally; reduce error df by 1.

EXAMPLE 2

RBD with 4 treatments × 5 blocks. \(T'_i = 80,\; B'_j = 60,\; G' = 380,\; t = 4, b = 5\).

\(\hat y_{ij} = \dfrac{4(80) + 5(60) - 380}{(3)(4)} = \dfrac{320 + 300 - 380}{12} = \dfrac{240}{12} = 20\).

Key Take-aways

RBD applies all three principles — replication, randomization, local control.
Layout: each treatment appears once per block; randomized within block.
Analysis = two-way ANOVA without interaction.
df: treatment \((t-1)\), block \((b-1)\), error \((t-1)(b-1)\).
CD = \(t_{\alpha/2, df_E}\sqrt{2\,\text{MS}_E/b}\).
Missing value: Yates' formula \(\hat y_{ij} = (tT'_i + bB'_j - G')/[(t-1)(b-1)]\); reduce df_E by 1.

Extra Practical Problems

PRACTICE

Additional worked problems with step-by-step procedures to support self-study, matching this unit's topics.

STEP-BY-STEP PROCEDURE (RBD / two-way ANOVA)

RBD uses all three principles — replication, randomization and local control (blocking). With \(k\) treatments in \(r\) blocks, \(N = rk\).
Grand total \(G\); correction factor \(C = G^2/(rk)\).
\(\text{TSS} = \sum\sum x_{ij}^2 - C\).
\(\text{Treatment (Variety) SS} = \dfrac{\sum T_i^2}{r} - C\).
\(\text{Block SS} = \dfrac{\sum B_j^2}{k} - C\).
\(\text{Error SS} = \text{TSS} - \text{Treatment SS} - \text{Block SS}\).
df: Treatment \(= k-1\), Block \(= r-1\), Error \(= (k-1)(r-1)\), Total \(= rk-1\).
Compute \(F = \text{MS}/\text{MSE}\) for treatments and blocks; compare with the table value.
Critical difference \(\text{CD} = t_{\alpha,\text{error df}}\times SE(D)\), where \(SE(D) = \sqrt{\dfrac{2\,\text{MSE}}{r}}\); CV \(= \dfrac{\sqrt{\text{MSE}}}{\bar X}\times 100\).

Problem 1 — RBD Analysis (fodder sorghum)

DATA

Green-matter yield (kg/plot) of 5 sorghum varieties in 4 blocks:

Variety	I	II	III	IV	Total
African Tall	22.9	25.9	39.1	33.9	121.8
Co-11	29.5	30.4	35.3	29.6	124.8
FS-1	28.8	24.4	32.1	28.6	113.9
K-7	47.0	40.9	42.8	32.1	162.8
Co-24	28.9	20.4	21.1	31.8	102.2
Block total	157.1	142.0	170.4	156.0	625.5

\(C = 625.5^2/20 = 19562.51\). \(\text{TSS} = 952.44\); \(\text{Block SS} = \frac{\sum B_j^2}{5} - C = 80.80\); \(\text{Variety SS} = \frac{\sum V_i^2}{4} - C = 520.53\); \(\text{Error SS} = 952.44 - 80.80 - 520.53 = 351.11\).

Source	df	SS	MS	\(F\)	\(F_{0.05}\)
Replication	3	80.80	26.9	<1	3.49
Variety	4	520.53	130	4.45*	3.26
Error	12	351.11	29.3
Total	19	952.44

Varieties differ significantly. \(SE(D) = \sqrt{\frac{2\times29.26}{4}} = 3.83\); \(CD = 2.179\times 3.83 = 8.33\). Variety K-7 (mean 40.7) yields significantly higher than all others, which are on par.

Problem 2 — RBD Analysis (6 crop varieties, 5 blocks)

DATA

Yields (lb) of varieties A–F in 5 blocks; rearranged variety/block table:

Block	A	B	C	D	E	F	Block total
B1	26	12	15	10	26	62	151
B2	30	10	16	20	23	56	155
B3	28	9	14	23	35	64	173
B4	23	7	14	20	30	75	169
B5	20	9	12	17	28	70	156
Variety total	127	47	71	90	142	327	804

\(C = 804^2/30 = 21547.2\). \(\text{Variety SS} = \frac{\sum V^2}{5} - C = 10167.2\); \(\text{Block SS} = \frac{\sum B^2}{6} - C = 61.47\); \(\text{TSS} = 10646.8\); \(\text{Error SS} = 10646.8 - 61.47 - 10167.2 = 418.13\).

Source	df	SS	MS	\(F\)	\(F_{0.05}\)
Blocks	4	61.47	15.37	0.74	2.87
Varieties	5	10167.2	2033.44	97.25*	2.71
Error	20	418.13	20.91
Total	29	10646.8

Highly significant difference between varieties. \(SE_m = \sqrt{20.91/5} = 2.04\), \(SED = 1.414\times 2.04 = 2.88\), \(CD = 2.88\times 2.09 = 6.04\), CV \(= \frac{\sqrt{20.91}}{26.8}\times100 = 17\%\). Variety F gives a significantly higher yield than all others; D, C, B are on par and exceed A.

Unsolved Exercises

PRACTICE

Wheat yields of 4 varieties × 5 replications (V1: 7,9,8,10,10; V2: 12,13,15,11,13; V3: 15,20,15,18,16; V4: 8,10,12,10,8). Analyse and conclude. (Ans: Variety variance 66.13, Error variance 2.59)
Yields of 5 varieties applied to 4, 3, 2, 4, 3 plots (A: 8,8,6,10; B: 10,9,8; C: 8,10; D: 7,10,9,8; E: 12,8,10). Analyse and conclude. (Ans: Variety variance 1.86, Error variance 2.28)
Write short notes on: (a) local control, (b) replication, (c) advantages of CRD.

← Previous: Unit 2 Next: Unit 4 — LSD →

Unit 3 — Randomised Block Design (RBD)

Topics Covered

1. Concept of RBD

2. Layout of RBD

3. Advantages and Disadvantages

Advantages

Disadvantages

Applications

4. Statistical Analysis of RBD

Mathematical Model

Hypotheses

Sums of Squares

ANOVA Table

Critical Difference

5. Estimating One Missing Value in RBD

Adjustments after substitution

Key Take-aways

Extra Practical Problems

Problem 1 — RBD Analysis (fodder sorghum)

Problem 2 — RBD Analysis (6 crop varieties, 5 blocks)

Unsolved Exercises