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How to use this manual: In the lab, copy the blank working table (usually the skeleton ANOVA table) at the start of the Calculation into your record book and fill the sums of squares, degrees of freedom, mean squares and F-ratios. The Calculation section shows the completed table with the arithmetic, and the Result states the test decision and any critical differences or efficiencies.

List of Practical Experiments (Official Syllabus)

  1. ANOVA — one-way classification with equal observations.
  2. ANOVA — one-way classification with unequal observations.
  3. ANOVA — two-way classification.
  4. Analysis of CRD and critical differences.
  5. Analysis of RBD, critical differences, and relative efficiency of CRD with RBD.
  6. Estimation of single missing observation in RBD.
  7. Analysis of LSD and efficiency of LSD over CRD and RBD.
  8. Estimation of single missing observation in LSD.

Experiment 1 — One-way ANOVA (Equal Observations)

1. Problem

The yields (kg/plot) of three fertilizers, each applied to four plots, are recorded below. Test whether the three fertilizers differ significantly in mean yield.

F122262428
F230343236
F320222422

2. Aim

To test \(H_0:\mu_1 = \mu_2 = \mu_3\) against \(H_1:\) not all equal, using one-way analysis of variance with equal group sizes.

3. Formula

\[ SS_T = \sum x^2 - \frac{G^2}{N}, \qquad SS_{Tr} = \sum_i \frac{T_i^2}{n_i} - \frac{G^2}{N}, \qquad F = \frac{MS_{Tr}}{MS_E} \]

Applying it:

  1. Find the grand total \(G\), \(N = \sum n_i\) and correction factor \(CF = G^2/N\).
  2. Compute \(SS_T = \sum x^2 - CF\) and \(SS_{Tr} = \sum(T_i^2/n_i) - CF\); then \(SS_E = SS_T - SS_{Tr}\).
  3. Enter the sums of squares, df and mean squares in the ANOVA table and form \(F = MS_{Tr}/MS_E\).
  4. Compare with \(F_{0.05}(k-1,\,N-k)\).

4. Calculation

Blank ANOVA table:

SourceSSdfMSF
Treatments
Error
Total——

Totals: \(T_1 = 100,\; T_2 = 132,\; T_3 = 88\); \(G = 320,\; N = 12,\; CF = 320^2/12 = 8533.33\).

\(\sum x^2 = 8840\Rightarrow SS_T = 306.67\); \(SS_{Tr} = (100^2+132^2+88^2)/4 - CF = 8792 - 8533.33 = 258.67\); \(SS_E = 48\).

SourceSSdfMSF
Treatments258.672129.3324.25
Error48.0095.33—
Total306.6711——

5. Result

\(F = 24.25 > F_{0.05}(2,9) = 4.26\), so \(H_0\) is rejected: the three fertilizers differ significantly in mean yield.

Experiment 2 — One-way ANOVA (Unequal Observations)

1. Problem

Three varieties are grown on different numbers of plots: V1 (3 plots): 20, 24, 22; V2 (4 plots): 28, 30, 32, 34; V3 (2 plots): 18, 20. Test whether the variety means differ.

2. Aim

To perform one-way ANOVA with unequal group sizes and test \(H_0:\mu_1=\mu_2=\mu_3\).

3. Formula

\[ SS_{Tr} = \sum_i \frac{T_i^2}{n_i} - \frac{G^2}{N}, \qquad df_{Tr} = k-1, \quad df_E = N-k \]

Applying it:

  1. Compute \(G\), \(N = \sum n_i\) and \(CF = G^2/N\).
  2. \(SS_T = \sum x^2 - CF\); \(SS_{Tr} = \sum(T_i^2/n_i) - CF\) (each treatment total divided by its own \(n_i\)); \(SS_E = SS_T - SS_{Tr}\).
  3. Complete the ANOVA table and test \(F = MS_{Tr}/MS_E\) against \(F_{0.05}(k-1, N-k)\).

4. Calculation

Blank ANOVA table:

SourceSSdfMSF
Treatments
Error
Total——

Totals: \(T_1 = 66\,(n_1=3),\; T_2 = 124\,(n_2=4),\; T_3 = 38\,(n_3=2)\); \(G = 228,\; N = 9,\; CF = 228^2/9 = 5776\).

\(\sum x^2 = 6048\Rightarrow SS_T = 272\); \(SS_{Tr} = (66^2/3 + 124^2/4 + 38^2/2) - 5776 = 6018 - 5776 = 242\); \(SS_E = 30\).

SourceSSdfMSF
Treatments242212124.2
Error3065—
Total2728——

5. Result

\(F = 24.2 > F_{0.05}(2,6) = 5.14\), so \(H_0\) is rejected: the varieties differ significantly.

Experiment 3 — Two-way ANOVA

1. Problem

Yields are classified by three treatments (rows) and four blocks (columns), one observation per cell. Test whether treatment means and block means differ.

B1B2B3B4
T122262824
T230363432
T320242226

2. Aim

To perform two-way ANOVA (without replication) and test the equality of treatment (row) effects and of block (column) effects.

3. Formula

\[ SS_R = \frac{\sum R_i^2}{c} - CF, \quad SS_C = \frac{\sum C_j^2}{r} - CF, \quad SS_E = SS_T - SS_R - SS_C \]

Applying it:

  1. Find row totals \(R_i\), column totals \(C_j\), grand total \(G\) and \(CF = G^2/N\).
  2. \(SS_T = \sum x^2 - CF\); \(SS_R = \sum R_i^2/c - CF\); \(SS_C = \sum C_j^2/r - CF\); \(SS_E = SS_T - SS_R - SS_C\).
  3. Degrees of freedom: rows \(r-1\), columns \(c-1\), error \((r-1)(c-1)\).
  4. Test \(F_R = MS_R/MS_E\) and \(F_C = MS_C/MS_E\).

4. Calculation

Blank ANOVA table:

SourceSSdfMSF
Treatments (rows)
Blocks (columns)
Error—
Total——

Row totals \(R = 100, 132, 92\); column totals \(C = 72, 86, 84, 82\); \(G = 324,\; N = 12,\; CF = 324^2/12 = 8748\).

\(\sum x^2 = 9032\Rightarrow SS_T = 284\); \(SS_R = 35888/4 - 8748 = 224\); \(SS_C = 26360/3 - 8748 = 38.67\); \(SS_E = 284 - 224 - 38.67 = 21.33\).

SourceSSdfMSF
Treatments (rows)224.002112.0031.50
Blocks (columns)38.67312.893.63
Error21.3363.56—
Total284.0011——

5. Result

Treatments: \(F_R = 31.50 > F_{0.05}(2,6) = 5.14\) ⇒ treatment effects differ significantly. Blocks: \(F_C = 3.63 < F_{0.05}(3,6) = 4.76\) ⇒ block effects are not significant.

Experiment 4 — Analysis of CRD & Critical Differences

1. Problem

Using the completely randomised design of Experiment 1 (3 fertilizers × 4 replicates, \(MS_E = 5.33,\; df_E = 9\)), determine the critical difference and identify which pairs of fertilizer means differ significantly.

2. Aim

To compute the critical difference (CD) for a CRD and compare all pairs of treatment means.

3. Formula

\[ CD = t_{\alpha/2,\,df_E}\sqrt{\frac{2\,MS_E}{r}} \]

Applying it:

  1. Take \(MS_E\) and \(df_E\) from the ANOVA.
  2. Compute \(CD = t_{0.025,\,df_E}\sqrt{2\,MS_E/r}\).
  3. Form all pairwise differences of treatment means; a difference exceeding CD is significant.

4. Calculation

Blank working table:

Pair\(|\bar y_i - \bar y_j|\)vs CDDecision
F1 vs F2
F1 vs F3
F2 vs F3

\(t_{0.025,9} = 2.262\), so \(CD = 2.262\sqrt{2(5.33)/4} = 2.262\times 1.633 = 3.69\).

Means: \(\bar y_1 = 25,\; \bar y_2 = 33,\; \bar y_3 = 22\).

Pair\(|\bar y_i - \bar y_j|\)vs CD (3.69)Decision
F1 vs F28>Differ
F1 vs F33<No difference
F2 vs F311>Differ

5. Result

\(CD = 3.69\). F2 differs from both F1 and F3, but F1 and F3 do not differ significantly. F2 is the best-yielding fertilizer.

Experiment 5 — RBD Analysis & Relative Efficiency vs CRD

1. Problem

A randomised block design has 3 treatments in 4 blocks with \(SS_{Tr} = 60,\; SS_{Bl} = 24,\; SS_E = 12,\; df_E = 6\). Test the treatment effect, obtain the CD, and find the relative efficiency of RBD over CRD.

2. Aim

To analyse an RBD, compute the critical difference, and evaluate the gain from blocking via the relative efficiency of RBD over CRD.

3. Formula

\[ F_{Tr} = \frac{MS_{Tr}}{MS_E}, \qquad RE = \frac{MS_E^{CRD}}{MS_E^{RBD}} = \frac{(SS_{Bl}+SS_E)/(df_{Bl}+df_E)}{MS_E^{RBD}} \]

Applying it:

  1. Form the ANOVA table: \(MS_{Tr} = SS_{Tr}/(t-1)\), \(MS_E = SS_E/df_E\), \(F_{Tr} = MS_{Tr}/MS_E\).
  2. \(CD = t_{0.025,\,df_E}\sqrt{2\,MS_E/b}\).
  3. Relative efficiency: \(MS_E^{CRD} = (SS_{Bl} + SS_E)/(df_{Bl} + df_E)\) and \(RE = MS_E^{CRD}/MS_E^{RBD}\).

4. Calculation

Blank ANOVA table:

SourceSSdfMSF
Treatments
Blocks—
Error—
SourceSSdfMSF
Treatments6023015
Blocks2438—
Error1262—

\(F_{Tr} = 30/2 = 15\). \(CD = t_{0.025,6}\sqrt{2(2)/4} = 2.447\times 1 = 2.447\).

\(MS_E^{CRD} = (24 + 12)/(3 + 6) = 36/9 = 4\), so \(RE = 4/2 = 2.0\).

5. Result

\(F_{Tr} = 15 > F_{0.05}(2,6) = 5.14\) ⇒ treatments differ significantly; \(CD = 2.447\). RBD is twice as efficient as CRD, so blocking was worthwhile.

Experiment 6 — Estimation of a Missing Value in RBD

1. Problem

In a 3 treatments × 3 blocks RBD, the observation at T2–B3 is missing. Estimate it.

B1B2B3
T1101214
T21416?
T381012

2. Aim

To estimate a single missing value in an RBD by minimising the error sum of squares.

3. Formula

\[ \hat y = \frac{t\,T' + b\,B' - G'}{(t-1)(b-1)} \]

Applying it:

  1. Let \(T'\) = total of the treatment containing the missing cell, \(B'\) = total of its block, \(G'\) = grand total, all computed from the available observations.
  2. Apply the missing-value formula with \(t\) treatments and \(b\) blocks.
  3. Insert the estimate, then analyse the RBD reducing the error df by 1.

4. Calculation

Blank working table:

QuantityValue
\(T'\) (treatment total)
\(B'\) (block total)
\(G'\) (grand total)
\(\hat y\)
QuantityValue
\(T'_2 = 14+16\)30
\(B'_3 = 14+12\)26
\(G' \) (all available)96
\(t = b\)3

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

5. Result

The missing value is estimated as \(\hat y = 18\). Insert it and analyse the RBD with the error df reduced by one.

Experiment 7 — LSD Analysis & Efficiency vs CRD/RBD

1. Problem

A 5 × 5 Latin square design gives \(SS_R = 30,\; SS_C = 25,\; SS_{Tr} = 200,\; SS_E = 60,\; df_E = 12\). Test the treatment effect, find the CD, and compute the efficiency of LSD relative to CRD and to RBD.

2. Aim

To analyse a Latin square design and evaluate its relative efficiency against CRD and RBD.

3. Formula

\[ RE_{LSD/CRD} = \frac{MS_E^{CRD}}{MS_E^{LSD}}, \qquad RE_{LSD/RBD} = \frac{MS_E^{RBD}}{MS_E^{LSD}} \]

Applying it:

  1. \(MS_E = SS_E/df_E\); \(F_{Tr} = MS_{Tr}/MS_E\) with \(df_{Tr} = t-1\).
  2. \(CD = t_{0.025,\,df_E}\sqrt{2\,MS_E/t}\).
  3. RE over CRD: rows and columns join the error — \(MS_E^{CRD} = (SS_R + SS_C + SS_E)/(df_R + df_C + df_E)\).
  4. RE over RBD: only one blocking factor joins the error (columns-as-blocks uses \(SS_R\); rows-as-blocks uses \(SS_C\)).

4. Calculation

Blank working table:

Comparison\(MS_E\) of alternativeRE
LSD vs CRD
LSD vs RBD (cols as blocks)
LSD vs RBD (rows as blocks)

\(MS_E^{LSD} = 60/12 = 5\); \(F_{Tr} = (200/4)/5 = 10 > F_{0.05}(4,12) = 3.26\) ⇒ treatments differ.

\(CD = t_{0.025,12}\sqrt{2(5)/5} = 2.179\sqrt{2} = 3.08\).

Comparison\(MS_E\) of alternativeRE
LSD vs CRD\((30+25+60)/20 = 5.75\)1.15 (15 %)
LSD vs RBD (cols as blocks)\((30+60)/16 = 5.625\)1.125 (12.5 %)
LSD vs RBD (rows as blocks)\((25+60)/16 = 5.31\)1.06 (6 %)

5. Result

Treatments differ significantly (\(F = 10\)); \(CD = 3.08\). The LSD is 15 % more efficient than CRD and 6–12.5 % more efficient than RBD, so controlling both row and column variation was beneficial.

Experiment 8 — Estimation of a Missing Value in LSD

1. Problem

In a 4 × 4 Latin square, one observation is missing. From the available data the row total \(R' = 65\), column total \(C' = 70\), treatment total \(T' = 60\) and grand total \(G' = 240\) (\(t = 4\)). Estimate the missing value.

2. Aim

To estimate a single missing value in a Latin square design.

3. Formula

\[ \hat y = \frac{t\,(R' + C' + T') - 2\,G'}{(t-1)(t-2)} \]

Applying it:

  1. Compute the row, column and treatment totals and the grand total of the available observations.
  2. Apply the LSD missing-value formula.
  3. Insert the estimate and analyse the LSD, reducing the error df by 1.

4. Calculation

Blank working table:

QuantityValue
\(R'\)
\(C'\)
\(T'\)
\(G'\)
\(\hat y\)
QuantityValue
\(R'\)65
\(C'\)70
\(T'\)60
\(G'\)240

\(\hat y = \dfrac{4(65 + 70 + 60) - 2(240)}{(3)(2)} = \dfrac{780 - 480}{6} = \dfrac{300}{6} = 50\).

5. Result

The missing value is estimated as \(\hat y = 50\). Insert it and analyse the LSD with the error df reduced by one.

Lab Record Format (to be followed for every experiment)

  1. 1. Problem — the data/design given and what is to be tested or estimated.
  2. 2. Aim — the hypothesis or quantity the experiment addresses.
  3. 3. Formula — the formula, then the numbered steps that apply it.
  4. 4. Calculation — the filled table with the arithmetic worked through.
  5. 5. Result — the test decision, critical differences, or efficiency, with interpretation.
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