TR

Analysis of variance

Enter the replicates of each treatment to get the ANOVA table, the F test, multiple comparisons and the coefficient of variation of the trial.

Completely randomized design

When treatments were assigned to plots completely at random. Treatments may have different numbers of replicates.

Question: Is there a difference among three or more treatments (fertilizer rates, varieties, irrigation intervals)? If so, which ones differ?
Example: Four nitrogen rates (kg/da) were tested with four replicates each in a completely randomized design. Yield (kg/da) was measured. Do the rates differ?

n = 4
n = 4
n = 4
n = 4
H0:μ1=μ2=⋯=μ4H_0: \mu_1 = \mu_2 = \cdots = \mu_{4}H1:At least one mean differs from the othersH_1: \text{At least one mean differs from the others}
Analysis of variance table
Source of variationdfSSMSF
Treatment323053.197684.465.67 **
Error121404.25117.02
Total1524457.44

df: degrees of freedom, SS: sum of squares, MS: mean square. * significant at 5%, ** at 1%, ns: not significant. CV = 2.97%.

Treatment means
TreatmentMean
0 kg N308 c
5 kg N351.25 b
10 kg N396.25 a
15 kg N401.75 a

Duncan (5%) critical ranges: R2 = 16.67, R3 = 17.44, R4 = 17.92. Means sharing a letter are not significantly different.

Pairwise comparisons: Duncan’s multiple range test

Means are ranked; a larger critical difference is used for means far apart in rank and a smaller one for neighbours. Widely used in agricultural field trials.

  1. Rank the means from largest to smallest. The critical range depends on how many means (p) the pair spans.

    Rp=qαp; p; 12MSE4,αp=1−(1−α)p−1R_p = q_{\alpha_p;\,p;\,12} \sqrt{\dfrac{MS_E}{4}}, \qquad \alpha_p = 1 - (1 - \alpha)^{p-1}

  2. Critical ranges:

    R2=16.67,R3=17.44,R4=17.92R_{2} = 16.67,\quad R_{3} = 17.44,\quad R_{4} = 17.92

  3. If a wider range is not significant, the differences between means inside it are also declared not significant.

  4. Means whose difference is below the critical value share a letter. 5 of 6 pairs differ significantly.

ComparisonDifferencepCritical differenceResult
0 kg N – 5 kg N43.25216.67Significant
0 kg N – 10 kg N88.25317.44Significant
0 kg N – 15 kg N93.75417.92Significant
5 kg N – 10 kg N45216.67Significant
5 kg N – 15 kg N50.5317.44Significant
10 kg N – 15 kg N5.5216.67Not significant
012345673.49calculated 65.67 →F (3; 12)

The red area is the rejection region (α = 0.05). The green line is the value calculated from your data.

H₀ rejected: the difference is significant

The difference between treatments is statistically significant: at least one group mean differs from the others. See the comparison table for which groups differ.

p < 0.0001 (significant at 1%, **)

How was it calculated?

FcalcF_{calc} = 65.667

  1. Correction factor (CF): the square of the grand total divided by the number of observations.

    CF=(∑X)2N=5829216=2123577.56CF = \dfrac{(\sum X)^2}{N} = \dfrac{5829^2}{16} = 2123577.56

  2. Total sum of squares: the overall variability of all observations.

    SST=∑X2−CF=2148035−2123577.56=24457.44SS_T = \sum X^2 - CF = 2148035 - 2123577.56 = 24457.44

  3. Treatment sum of squares: variability coming from differences between groups.

    SSTr=12322+14052+15852+160724−CF=23053.19SS_{Tr} = \dfrac{1232^2 + 1405^2 + 1585^2 + 1607^2}{4} - CF = 23053.19

  4. Error sum of squares: variability within groups (due to chance).

    SSE=SST−SSTr=24457.44−23053.19=1404.25SS_E = SS_T - SS_{Tr} = 24457.44 - 23053.19 = 1404.25

  5. Divide the sums of squares by their degrees of freedom to get the mean squares.

    MSTr=SSTrk−1=23053.193=7684.396MS_{Tr} = \dfrac{SS_{Tr}}{k-1} = \dfrac{23053.19}{3} = 7684.396
  6. MSE=SSEN−k=1404.2512=117.021MS_E = \dfrac{SS_E}{N-k} = \dfrac{1404.25}{12} = 117.021

  7. F: the ratio of between-group to within-group variability. If the treatment has no effect it comes out around 1.

    F=MSTrMSE=7684.396117.021=65.667F = \dfrac{MS_{Tr}}{MS_E} = \dfrac{7684.396}{117.021} = 65.667

  8. Compare the calculated value with the table (critical) value.

    Fcalc=65.667>Ftab=3.49F_{calc} = 65.667 > F_{tab} = 3.49

  9. The calculated value is in the rejection region: H₀ is rejected. The p-value (< 0.0001) is smaller than α = 0.05.

  10. The coefficient of variation of the experiment shows how precisely it was carried out.

    CV=MSEXˉ⋅100=10.818364.313⋅100=2.97%CV = \dfrac{\sqrt{MS_E}}{\bar{X}} \cdot 100 = \dfrac{10.818}{364.313} \cdot 100 = 2.97\%

Randomized complete block design

When the field was divided into blocks and every treatment was tried once in each block. The table can be pasted straight from Excel.

Question: When the field is not uniform, do the treatments differ? Once the block effect is removed, which treatments differ?
Example: Five wheat varieties were tested in 4 blocks on a sloping field. Blocks run across the slope: block 1 at the top, block 4 at the bottom. Yield (kg/da) was measured. Do the varieties differ?

Total
2051
2140
1992
2233
2044
Block total236225822677283910460

If a plot was lost (animals, flooding, a measurement error), leave its cell empty; its value will be estimated.

H0:μ1=μ2=⋯=μ5H_0: \mu_1 = \mu_2 = \cdots = \mu_{5}H1:At least one treatment mean differsH_1: \text{At least one treatment mean differs}
Analysis of variance table
Source of variationdfSSMSF
Block323823.67941.264.05 **
Treatment49042.52260.6318.23 **
Error121487.9123.99
Total1934354

df: degrees of freedom, SS: sum of squares, MS: mean square. * significant at 5%, ** at 1%, ns: not significant. CV = 2.13%.

Treatment means
TreatmentMean
Variety 1512.75 c
Variety 2535 b
Variety 3498 c
Variety 4558.25 a
Variety 5511 c

Duncan (5%) critical ranges: R2 = 17.16, R3 = 17.96, R4 = 18.44, R5 = 18.76. Means sharing a letter are not significantly different.

Pairwise comparisons: Duncan’s multiple range test

Means are ranked; a larger critical difference is used for means far apart in rank and a smaller one for neighbours. Widely used in agricultural field trials.

  1. Rank the means from largest to smallest. The critical range depends on how many means (p) the pair spans.

    Rp=qαp; p; 12MSE4,αp=1−(1−α)p−1R_p = q_{\alpha_p;\,p;\,12} \sqrt{\dfrac{MS_E}{4}}, \qquad \alpha_p = 1 - (1 - \alpha)^{p-1}

  2. Critical ranges:

    R2=17.16,R3=17.96,R4=18.44,R5=18.76R_{2} = 17.16,\quad R_{3} = 17.96,\quad R_{4} = 18.44,\quad R_{5} = 18.76

  3. If a wider range is not significant, the differences between means inside it are also declared not significant.

  4. Means whose difference is below the critical value share a letter. 7 of 10 pairs differ significantly.

ComparisonDifferencepCritical differenceResult
Variety 1 – Variety 222.25217.16Significant
Variety 1 – Variety 314.75317.96Not significant
Variety 1 – Variety 445.5317.96Significant
Variety 1 – Variety 51.75217.16Not significant
Variety 2 – Variety 337418.44Significant
Variety 2 – Variety 423.25217.16Significant
Variety 2 – Variety 524317.96Significant
Variety 3 – Variety 460.25518.76Significant
Variety 3 – Variety 513217.16Not significant
Variety 4 – Variety 547.25418.44Significant

Did blocking help?

The blocks differ significantly (F = 64.05, p < 0.0001): the field really had a gradient, and blocking removed that variation from the error.

Relative efficiency: 10.95. A completely randomized design would need about 11 times as many replicates for the same precision.

Had the same data been analysed ignoring blocks (as a completely randomized design): F = 1.34, p = 0.3009, so the treatment difference would not be significant. Block variation would have inflated the error and hidden the real difference.

01234563.259calculated 18.23 →F (4; 12)

The red area is the rejection region (α = 0.05). The green line is the value calculated from your data.

H₀ rejected: the difference is significant

After removing the block effect, the difference between treatments is statistically significant. Letters show which treatments differ.

p < 0.0001 (significant at 1%, **)

How was it calculated?

FcalcF_{calc} = 18.232

  1. Correction factor (CF): the square of the grand total divided by the number of observations.

    CF=(∑X)2t⋅r=1046025⋅4=5470580CF = \dfrac{(\sum X)^2}{t \cdot r} = \dfrac{10460^2}{5 \cdot 4} = 5470580

  2. Total sum of squares: the overall variability of all plots.

    SST=∑X2−CF=34354SS_T = \sum X^2 - CF = 34354

  3. Block sum of squares: squares of the block totals divided by the number of treatments. Variability due to slope, soil or moisture differences in the field.

    SSBl=23622+25822+26772+⋯5−CF=23823.6SS_{Bl} = \dfrac{2362^2 + 2582^2 + 2677^2 + \cdots}{5} - CF = 23823.6

  4. Treatment sum of squares: squares of the treatment totals divided by the number of blocks.

    SSTr=20512+21402+19922+⋯4−CF=9042.5SS_{Tr} = \dfrac{2051^2 + 2140^2 + 1992^2 + \cdots}{4} - CF = 9042.5

  5. Error sum of squares: what remains after removing blocks and treatments, the truly random part.

    SSE=SST−SSBl−SSTrSS_E = SS_T - SS_{Bl} - SS_{Tr}
  6. SSE=34354−23823.6−9042.5=1487.9SS_E = 34354 - 23823.6 - 9042.5 = 1487.9

  7. Degrees of freedom: blocks r − 1, treatments t − 1, error (t − 1)(r − 1).

    dfB=3,dfM=4,dfH=4⋅3=12df_B = 3, \quad df_M = 4, \quad df_H = 4 \cdot 3 = 12

  8. Mean squares:

    MSTr=9042.54=2260.625MS_{Tr} = \dfrac{9042.5}{4} = 2260.625
  9. MSE=1487.912=123.992MS_E = \dfrac{1487.9}{12} = 123.992

  10. F value for treatments:

    F=MSTrMSE=2260.625123.992=18.232F = \dfrac{MS_{Tr}}{MS_E} = \dfrac{2260.625}{123.992} = 18.232

  11. Compare the calculated value with the table (critical) value.

    Fcalc=18.232>Ftab=3.259F_{calc} = 18.232 > F_{tab} = 3.259

  12. The calculated value is in the rejection region: H₀ is rejected. The p-value (< 0.0001) is smaller than α = 0.05.

  13. Coefficient of variation of the experiment:

    CV=MSEXˉ⋅100=11.135523⋅100=2.13%CV = \dfrac{\sqrt{MS_E}}{\bar{X}} \cdot 100 = \dfrac{11.135}{523} \cdot 100 = 2.13\%