Standard Experimental Designs & ANOVA

Topic - Experimental designs structure how treatments are allocated to experimental units to isolate factor effects from background error. The four classical designs - Completely Randomized Design (CRD), Randomized Block Design (RBD), Latin Square Design (LSD), and $2^k$ Full Factorial Design - progressively control for zero, one, two, or multiple interactive sources of variation through linear hypothesis testing and Analysis of Variance (ANOVA).

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1. Comparative Architecture of Standard Designs

Selecting an experimental layout depends strictly on the physical homogeneity of the experimental material and the presence of external nuisance factors:

Design	Blocking Structure	ANOVA Type	Sources of Variation	Key Precondition
CRD	No blocks (homogeneous units)	One-Way ANOVA	Treatments, Error	All experimental units are uniform.
RBD	1 blocking factor	Two-Way ANOVA	Treatments, Blocks, Error	One-directional gradient across units.
LSD	2 blocking factors (Rows & Columns)	Three-Way ANOVA	Treatments, Rows, Columns, Error	Exactly $p \times p$ matrix; no interactions between factors.
$2^k$ Full Factorial	All combinations of $k$ factors at 2 levels	Multi-Factor ANOVA	Main Effects, Interactions, Error	Evaluates joint variable interactions.

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2. Completely Randomized Design (CRD)

Theoretical Model & Assumptions

CRD is the basic single-factor design where treatments are assigned completely at random to all units. It is appropriate when units are completely homogeneous (e.g., standard laboratory test tubes, incubator shelves, well-mixed chemicals).

$\text{Statistical Model: } y_{ij} = \mu + \tau_i + \epsilon_{ij}$

Where $\mu$ is the overall mean, $\tau_i$ is the effect of treatment $i$, and $\epsilon_{ij} \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2)$ is the random experimental error.

Mathematical Layout & Formulas

Given $t$ treatments each replicated $r$ times ($N = t \times r$ observations):

• Grand Total ($G$): $G = \sum_{i=1}^t \sum_{j=1}^r y_{ij}$
• Correction Factor ($CF$): $CF = \frac{G^2}{N}$
• Total Sum of Squares ($\text{SS}_{\text{Total}}$): $\text{SS}_{\text{Total}} = \sum_{i=1}^t \sum_{j=1}^r y_{ij}^2 - CF$
• Treatment Sum of Squares ($\text{SS}_{\text{Treat}}$): $\text{SS}_{\text{Treat}} = \sum_{i=1}^t \frac{T_i^2}{r} - CF$
• Error Sum of Squares ($\text{SS}_{\text{Error}}$): $\text{SS}_{\text{Error}} = \text{SS}_{\text{Total}} - \text{SS}_{\text{Treat}}$

Standard CRD ANOVA Table

Source of Variation	Degrees of Freedom ($\text{df}$)	Sum of Squares ($\text{SS}$)	Mean Square ($\text{MS}$)	Calculated $F$ ($F_{\text{cal}}$)
Treatments	$t - 1$	$\text{SS}_{\text{Treat}}$	$\text{MS}_{\text{Treat}} = \frac{\text{SS}_{\text{Treat}}}{t - 1}$	$\frac{\text{MS}_{\text{Treat}}}{\text{MS}_{\text{Error}}}$
Error	$N - t = t(r - 1)$	$\text{SS}_{\text{Error}}$	$\text{MS}_{\text{Error}} = \frac{\text{SS}_{\text{Error}}}{N - t}$	-
Total	$N - 1$	$\text{SS}_{\text{Total}}$	-	-

Decision Rule: Compare $F_{\text{cal}}$ with tabulated $F_{\alpha, (t-1), (N-t)}$. If $F_{\text{cal}} \gt F_{\text{tab}}$, reject $H_0$ (at least one treatment mean differs significantly).

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Solved Problem: Fertilizer Yield Evaluation

Problem Statement: Three fertilizers ($A, B, C$) are tested on a homogeneous field. Each treatment is replicated 4 times ($t=3, r=4, N=12$). Observed crop yields (kg) are:

• Fertilizer A: $18, 20, 19, 22$
• Fertilizer B: $24, 22, 21, 23$
• Fertilizer C: $26, 25, 27, 28$

Test at $\alpha = 0.05$ whether there are significant treatment differences.

Step 1: Compute Totals and Correction Factor

• $T_A = 18 + 20 + 19 + 22 = 79$
• $T_B = 24 + 22 + 21 + 23 = 90$
• $T_C = 26 + 25 + 27 + 28 = 106$
• $G = 79 + 90 + 106 = 275$
• $CF = \frac{G^2}{N} = \frac{275^2}{12} = \frac{75625}{12} \approx 6302.0833$

Step 2: Sum of Squares

• $\sum y_{ij}^2 = 18^2 + 20^2 + 19^2 + 22^2 + 24^2 + 22^2 + 21^2 + 23^2 + 26^2 + 25^2 + 27^2 + 28^2 = 6413$
• $\text{SS}_{\text{Total}} = 6413 - 6302.0833 = 110.9167$
• $\text{SS}_{\text{Treat}} = \frac{79^2 + 90^2 + 106^2}{4} - CF = \frac{6241 + 8100 + 11236}{4} - 6302.0833 = \frac{25577}{4} - 6302.0833 = 6394.25 - 6302.0833 = 92.1667$
• $\text{SS}_{\text{Error}} = \text{SS}_{\text{Total}} - \text{SS}_{\text{Treat}} = 110.9167 - 92.1667 = 18.7500$

Step 3: Mean Squares & F-Ratio

• $\text{df}_{\text{Treat}} = 3 - 1 = 2$
• $\text{df}_{\text{Error}} = 12 - 3 = 9$
• $\text{MS}_{\text{Treat}} = \frac{92.1667}{2} = 46.0833$
• $\text{MS}_{\text{Error}} = \frac{18.7500}{9} = 2.0833$
• $F_{\text{cal}} = \frac{46.0833}{2.0833} \approx 22.12$

Step 4: Decision & Conclusion

• Critical $F_{0.05, 2, 9} \approx 4.26$.
• Since $F_{\text{cal}} = 22.12 \gt 4.26$, we reject the null hypothesis $H_0$ at the $5\%$ level of significance.
• Conclusion: The three fertilizers differ significantly in their mean crop yield.

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3. Randomized Block Design (RBD)

Theoretical Model & Assumptions

RBD is an improvement over CRD when experimental units exhibit a known one-directional gradient (e.g., slope fertility, temperature gradient across an oven, different laboratory technicians). Units are divided into homogeneous groups called blocks. Each treatment appears exactly once in every block in a randomized sequence.

$\text{Statistical Model: } y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}$

Where $\tau_i$ is the $i$-th treatment effect, $\beta_j$ is the $j$-th block effect, and $\epsilon_{ij} \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2)$.

Mathematical Formulas

Given $t$ treatments and $r$ blocks ($N = t \times r$):

• Block Sum of Squares ($\text{SS}_{\text{Block}}$): $\text{SS}_{\text{Block}} = \sum_{j=1}^r \frac{B_j^2}{t} - CF$
• Error Sum of Squares ($\text{SS}_{\text{Error}}$): $\text{SS}_{\text{Error}} = \text{SS}_{\text{Total}} - \text{SS}_{\text{Treat}} - \text{SS}_{\text{Block}}$

Standard RBD ANOVA Table

Source	Degrees of Freedom ($\text{df}$)	Sum of Squares ($\text{SS}$)	Mean Square ($\text{MS}$)	Calculated $F$
Treatments	$t - 1$	$\text{SS}_{\text{Treat}}$	$\text{MS}_{\text{Treat}} = \frac{\text{SS}_{\text{Treat}}}{t - 1}$	$\frac{\text{MS}_{\text{Treat}}}{\text{MS}_{\text{Error}}}$
Blocks	$r - 1$	$\text{SS}_{\text{Block}}$	$\text{MS}_{\text{Block}} = \frac{\text{SS}_{\text{Block}}}{r - 1}$	$\frac{\text{MS}_{\text{Block}}}{\text{MS}_{\text{Error}}}$ (Optional)
Error	$(t - 1)(r - 1)$	$\text{SS}_{\text{Error}}$	$\text{MS}_{\text{Error}} = \frac{\text{SS}_{\text{Error}}}{(t - 1)(r - 1)}$	-
Total	$N - 1$	$\text{SS}_{\text{Total}}$	-	-

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Solved Problem: Cholesterol Analysis Across 4 Diets in 4 Labs

Problem Statement: Measurement of cholesterol content (%) is performed in 4 different laboratories ($A, B, C, D$) across 4 diet foods ($D_1, D_2, D_3, D_4$). Analyze the data using RBD at $\alpha = 0.05$ (Given critical $F_{0.05}(3, 9) = 3.86$).

Laboratory (Block)	$D_1$	$D_2$	$D_3$	$D_4$	Block Total ($B_j$)
A	2	5	4	3	14
B	3	7	4	2	16
C	5	5	6	4	20
D	5	5	8	5	23
Treatment Total ($T_i$)	15	22	22	14	$G = 73$

Step 1: Totals and Correction Factor

• $t = 4, r = 4, N = 16$
• $G = 73$
• $CF = \frac{73^2}{16} = \frac{5329}{16} = 333.0625$

Step 2: Sum of Squares

• $\sum y_{ij}^2 = 2^2 + 5^2 + 4^2 + 3^2 + 3^2 + 7^2 + 4^2 + 2^2 + 5^2 + 5^2 + 6^2 + 4^2 + 5^2 + 5^2 + 8^2 + 5^2 = 373$
• $\text{SS}_{\text{Total}} = 373 - 333.0625 = 39.9375$
• $\text{SS}_{\text{Treat}} = \frac{15^2 + 22^2 + 22^2 + 14^2}{4} - CF = \frac{225 + 484 + 484 + 196}{4} - 333.0625 = \frac{1389}{4} - 333.0625 = 347.25 - 333.0625 = 14.1875$
• $\text{SS}_{\text{Block}} = \frac{14^2 + 16^2 + 20^2 + 23^2}{4} - CF = \frac{196 + 256 + 400 + 529}{4} - 333.0625 = \frac{1381}{4} - 333.0625 = 345.25 - 333.0625 = 12.1875$
• $\text{SS}_{\text{Error}} = 39.9375 - 14.1875 - 12.1875 = 13.5625$

Step 3: Mean Squares & Decision

• $\text{df}_{\text{Treat}} = 4 - 1 = 3$
• $\text{df}_{\text{Block}} = 4 - 1 = 3$
• $\text{df}_{\text{Error}} = (4 - 1)(4 - 1) = 9$
• $\text{MS}_{\text{Treat}} = \frac{14.1875}{3} \approx 4.7292$
• $\text{MS}_{\text{Error}} = \frac{13.5625}{9} \approx 1.5069$
• $F_{\text{cal}} = \frac{4.7292}{1.5069} \approx 3.14$
• Comparison: $F_{\text{cal}} = 3.14 \lt F_{\text{tab}} = 3.86$.
• Conclusion: Fail to reject $H_0$. At the $5\%$ significance level, there is no statistically significant difference in cholesterol content among the four diet foods after accounting for laboratory blocking variations.

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4. Latin Square Design (LSD)

Theoretical Model & Assumptions

When two independent sources of nuisance variability exist simultaneously (e.g., Machine differences along rows and Operator differences along columns), the Latin Square Design (LSD) isolates both.

• A $p \times p$ square arrangement is used where each of $p$ treatments appears exactly once in each row and once in each column.
• The number of treatments must equal the number of rows and columns ($t = \text{rows} = \text{columns} = p$).
• Strict Assumption: There are no interactions between rows, columns, and treatments.

$\text{Statistical Model: } y_{ijk} = \mu + \tau_i + \rho_j + \gamma_k + \epsilon_{ijk}$

Where $\tau_i$ is treatment effect, $\rho_j$ is row effect, and $\gamma_k$ is column effect.

Mathematical Formulas ($N = p^2$)

• $CF = \frac{G^2}{p^2}$
• $\text{SS}_{\text{Total}} = \sum y^2 - CF$
• $\text{SS}_{\text{Treat}} = \frac{\sum T_i^2}{p} - CF, \quad \text{SS}_{\text{Row}} = \frac{\sum R_j^2}{p} - CF, \quad \text{SS}_{\text{Col}} = \frac{\sum C_k^2}{p} - CF$
• $\text{SS}_{\text{Error}} = \text{SS}_{\text{Total}} - \text{SS}_{\text{Treat}} - \text{SS}_{\text{Row}} - \text{SS}_{\text{Col}}$
• $\text{df}_{\text{Treat}} = p - 1, \quad \text{df}_{\text{Row}} = p - 1, \quad \text{df}_{\text{Col}} = p - 1, \quad \text{df}_{\text{Error}} = (p - 1)(p - 2), \quad \text{df}_{\text{Total}} = p^2 - 1$

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Solved Problem: 4x4 Latin Square Evaluation

Problem Statement: Four treatments ($A, B, C, D$) are allocated in a $4 \times 4$ Latin square to control for row and column effects ($p=4, N=16$).

• Row 1: $A(18), B(22), C(25), D(28) \implies R_1 = 93$
• Row 2: $B(21), C(24), D(26), A(20) \implies R_2 = 91$
• Row 3: $C(19), D(23), A(22), B(25) \implies R_3 = 89$
• Row 4: $D(20), A(21), B(23), C(27) \implies R_4 = 91$

Column totals: $C_1 = 78, C_2 = 90, C_3 = 96, C_4 = 100$. Grand Total $G = 364$. Treatment totals: $T_A = 81, T_B = 91, T_C = 95, T_D = 97$. Total sum of squares $\sum y^2 = 8408$.

Test at $\alpha = 0.05$ whether treatment differences are significant ($F_{0.05}(3, 6) = 4.76$).

Step 1: Sum of Squares

• $CF = \frac{364^2}{16} = \frac{132496}{16} = 8281.0$
• $\text{SS}_{\text{Total}} = 8408.0 - 8281.0 = 127.0$
• $\text{SS}_{\text{Treat}} = \frac{81^2 + 91^2 + 95^2 + 97^2}{4} - 8281.0 = \frac{33276}{4} - 8281.0 = 8319.0 - 8281.0 = 38.0$
• $\text{SS}_{\text{Row}} = \frac{93^2 + 91^2 + 89^2 + 91^2}{4} - 8281.0 = \frac{33132}{4} - 8281.0 = 8283.0 - 8281.0 = 2.0$
• $\text{SS}_{\text{Col}} = \frac{78^2 + 90^2 + 96^2 + 100^2}{4} - 8281.0 = \frac{33400}{4} - 8281.0 = 8350.0 - 8281.0 = 69.0$
• $\text{SS}_{\text{Error}} = 127.0 - 38.0 - 2.0 - 69.0 = 18.0$

Step 2: LSD ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	$F_{\text{cal}}$	Critical $F_{0.05}$
Rows	3	2.0	0.6667	-	-
Columns	3	69.0	23.0000	-	-
Treatments	3	38.0	12.6667	$\frac{12.6667}{3.0} \approx 4.222$	4.76
Error	$(4-1)(4-2)=6$	18.0	3.0000	-	-
Total	15	127.0	-	-	-

• Decision: $F_{\text{cal}} = 4.222 \lt 4.76$. Fail to reject $H_0$.
• Conclusion: After adjusting for row and column variations, treatments do not produce statistically significant differences at the $5\%$ level.

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5. Full Factorial Design ($2^2, 2^3, 2^k$)

Orthogonal Coding & Contrast Formulation

In a $2^k$ factorial design, $k$ factors are evaluated at 2 coded levels: Low ($-1$) and High ($+1$).

• Total combinations: $2^k$. With $r$ replicates, total runs $N = r \cdot 2^k$.
• Contrast ($C$): Sum of signed responses for an effect: $C = \sum_{i=1}^{2^k} \text{Sign}_i \cdot (\text{Sum of replicates for run } i)$
• Estimated Effect: $\text{Effect} = \frac{C}{2^{k-1} \cdot r}$
• Sum of Squares for any Effect: $\text{SS} = \frac{C^2}{r \cdot 2^k}$

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Solved Problem: $2^2$ Chemical Reaction Factorial Analysis

Problem Statement: A chemical engineer studies the effect of Temperature ($A$) and Pressure ($B$) on product yield (%). Both factors are tested at two coded levels: Low ($-1$) and High ($+1$) with $r=2$ replicates ($N = 2^2 \times 2 = 8$).

Run	A	B	Interaction ($AB = A \times B$)	Run 1 ($y_{i1}$)	Run 2 ($y_{i2}$)	Treatment Total ($y_{i+}$)
1	$-1$	$-1$	$+1$	42	43	85
2	$+1$	$-1$	$-1$	50	49	99
3	$-1$	$+1$	$-1$	46	45	91
4	$+1$	$+1$	$+1$	62	61	123
Grand Total						$G = 398$

Step 1: Contrasts & Main Effects

• Contrast for A: $C_A = -85 + 99 - 91 + 123 = 46$ $\hat{A} = \frac{C_A}{2^{2-1} \cdot 2} = \frac{46}{4} = 11.5$
• Contrast for B: $C_B = -85 - 99 + 91 + 123 = 30$ $\hat{B} = \frac{C_B}{4} = \frac{30}{4} = 7.5$
• Contrast for AB: $C_{AB} = +85 - 99 - 91 + 123 = 18$ $\widehat{AB} = \frac{C_{AB}}{4} = \frac{18}{4} = 4.5$

Step 2: Sum of Squares ($N = 8$)

• $\text{SS}_A = \frac{C_A^2}{8} = \frac{46^2}{8} = \frac{2116}{8} = 264.5$
• $\text{SS}_B = \frac{C_B^2}{8} = \frac{30^2}{8} = \frac{900}{8} = 112.5$
• $\text{SS}_{AB} = \frac{C_{AB}^2}{8} = \frac{18^2}{8} = \frac{324}{8} = 40.5$
• $\text{SS}_{\text{Treat}} = 264.5 + 112.5 + 40.5 = 417.5$
• $\sum y^2 = 42^2 + 43^2 + 50^2 + 49^2 + 46^2 + 45^2 + 62^2 + 61^2 = 20220$
• $CF = \frac{398^2}{8} = \frac{158404}{8} = 19800.5$
• $\text{SS}_{\text{Total}} = 20220 - 19800.5 = 419.5$
• $\text{SS}_{\text{Error}} = \text{SS}_{\text{Total}} - \text{SS}_{\text{Treat}} = 419.5 - 417.5 = 2.0$

Step 3: $2^2$ Factorial ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	Calculated $F$	Critical $F_{0.05}(1, 4)$	Conclusion
A (Temperature)	1	264.5	264.5	$\frac{264.5}{0.5} = 529.0$	7.708	Reject $H_0$ ($p \lt 0.001$)
B (Pressure)	1	112.5	112.5	$\frac{112.5}{0.5} = 225.0$	7.708	Reject $H_0$ ($p \lt 0.001$)
AB (Interaction)	1	40.5	40.5	$\frac{40.5}{0.5} = 81.0$	7.708	Reject $H_0$ ($p \lt 0.001$)
Error (Residual)	$2^2(2 - 1) = 4$	2.0	$\text{MSE} = 0.5$	-	-	-
Total	7	419.5	-	-	-	-

Engineering Interpretation: Increasing Temperature raises yield by an average of $11.5\%$. Increasing Pressure raises yield by $7.5\%$. The strong positive interaction ($AB = +4.5\%$, $F = 81.0$) demonstrates synergistic behavior: the positive impact of temperature is amplified at high pressure.

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6. Exam Traps & Operational Nuances

1. Degrees of Freedom Trap in LSD

• The Trap: Confusing error degrees of freedom in Latin Square Design.
• The Reality: In a $p \times p$ square, $\text{df}_{\text{Error}} = (p - 1)(p - 2)$.
• Exam Consequence: For a $2 \times 2$ Latin Square, $\text{df}_{\text{Error}} = (2 - 1)(2 - 2) = 0$. For a $3 \times 3$ square, $\text{df}_{\text{Error}} = (2)(1) = 2$ (too small to achieve statistical power). Hence, Latin Squares should practically be at least $4 \times 4$ ($p \ge 4$).

2. Factorial Divisor Confusion

• The Trap: Using $N$ instead of $2^{k-1} \cdot r$ when computing the estimated main effect from a contrast.
• The Reality: The effect is the difference between two averages (High average minus Low average). Each average contains half the observations ($2^{k-1} \cdot r$). In contrast, when computing Sum of Squares, the denominator is the total observations $N = 2^k \cdot r$.

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7. Summary & Cheatsheet

ANOVA Decomposition

• CRD: $\text{SST} = \text{SSTreat} + \text{SSE}$ ($\text{df}: t-1, N-t$).
• RBD: $\text{SST} = \text{SSTreat} + \text{SSBlock} + \text{SSE}$ ($\text{df}_{\text{Error}} = (t-1)(r-1)$).
• LSD: $\text{SST} = \text{SSTreat} + \text{SSRow} + \text{SSCol} + \text{SSE}$ ($\text{df}_{\text{Error}} = (p-1)(p-2)$).

2^k Factorial Rules

• Contrast: $C = \sum (\pm 1) \cdot y_{i+}$.
• Effect: $\text{Effect} = C / (2^{k-1} r)$.
• Sum of Squares: $\text{SS} = C^2 / (2^k r)$ with $\text{df} = 1$ per effect.

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Key Takeaways

• Blocking Reduces MSE: Introducing blocks (RBD) or rows and columns (LSD) partitions nuisance variation out of the error sum of squares, lowering $\text{MSE}$ and boosting the $F$-test sensitivity.
• Synergy in Factorials: Full factorial designs are uniquely capable of discovering non-linear interactions where the effect of one parameter depends on another.
• Exponential Scaling: As $k$ increases, full factorial runs escalate exponentially ($2^k$), motivating fractional factorial designs and Taguchi orthogonal arrays.

Next Section: Fractional Factorials & Taguchi Robust Design - Learn how fractional designs and Taguchi orthogonal arrays ($L_9$) dramatically reduce required runs while optimizing robust performance against noise.