APPENDIX 17

ANALYSIS OF VARIANCE FOR A RHIZOBIUM STRAIN SELECTION EXPERIMENT

The data in Table A.12 presents the dry weight (g) of plant tops from a strain selection experiment for soybean (G. max var. Jupiter). The experiment was a Randomized Complete Block Design (RCBD), with 3 blocks and 16 treatments (14 inoculated + 2 controls). Each treatment was replicated once within each block. Each treatmentplot was a Leonard jar unit with two soybean plants. The plant tops were harvested at 32 days and oven dried at 70C. The strains of Bradyrhizobium japonicum have been ranked according to dry weight.

Summary of calculations for the analysis of variance for the strain selection experiment.

No. of treatments = k = 16

No. of blocks = b = 3

No. of replicates per treatment per block = n = 1

Calculate the Grand Total (GT) by adding up all the treatment totals:

GT = T1 + T2 T3 + Tk

= 31.09 + 28.85 + 28.04 + 20.07

= 344.83

Table A.12. Data from a strain selection experiment for soybean.

Dry Weight of Plant Tops (g)
BLOCKS / Treatment / Treatment
TREATMENTS / B1 / B2 / B3 / Total (T) / Means (x)
TAL 102 / 9.66 / 10.60 / 10.83 / 31.09 / 10.36
TAL 379 / 9.36 / 9.00 / 10.49 / 28.85 / 9.62
TAL 206 / 8.41 / 9.44 / 10.19 / 28.04 / 9.35
TAL 435 / 8.61 / 9.23 / 8.22 / 26.06 / 8.69
TAL 411 / 9.20 / 8.19 / 8.46 / 25.85 / 8.62
Allen 527 / 8.11 / 8.82 / 8.62 / 25.55 / 8.52
TAL 211 / 8.83 / 6.32 / 9.14 / 24.29 / 8.10
TAL 487 / 6.27 / 8.67 / 8.35 / 23.29 / 7.76
CB 1795 / 6.79 / 8.17 / 5.70 / 20.66 / 6.89
TAL 650 / 6.95 / 5.83 / 6.83 / 19.61 / 6.54
TAL 649 / 6.55 / 4.82 / 8.10 / 19.47 / 6.49
TAL 860 / 6.00 / 4.83 / 6.54 / 17.37 / 5.79
TAL 183 / 6.11 / 3.46 / 5.51 / 15.08 / 5.03
TAL 378 / 5.39 / 4.46 / 5.07 / 14.92 / 4.97
Control* / 1.53 / 1.30 / 1.80 / 4.63 / 1.54
Control** / 8.41 / 7.83 / 5.83 / 20.07 / 6.36
114.18 / 110.97 / 119.68 / 344.83 / 116.36

* Uninoculated

** 70 ppm N

Calculate the Grand Mean (X) by adding up all the treatment means:

X = x1 + x2 + x3 xk

= 10.36 + 9.62 + 9.35 + 6.36

= 116.13

Calculate the Correction Factor (CF)

CF= (GT)2 = (344.83)2

bkn 3 x 16 x 1

= 2477.2444

Calculate the total sum of Squares (SS)

SS = Σx2 CF

= 9.662 + 10.602 + 10.832 + 5.832 2477.244

= 247.8507

Calculate the Treatment Sum of Squares (SST)

SST = ∑T2 – CF

bn

= 31.092 + 28.852...+ 4.632 + 20.072 – 2477.2444

3 X 1

= 217.4785

Calculate the Block Sum of Squares (SSB)

SSB = ∑B2 – CF

kn

= 114.182 + 110.972 + 119.682 – 2477.2444

16

= 2.4253

Calculate the Error Sum of Squares (SSE)

SSE = SS (SST + SSB)

= 247.8507 (217.4785 + 2.4253)

= 27.9469

Prepare the Analysis of Variance according to Table A.13.

Table A.13. Analysis of Variance

Sources of Sum of Degrees of Mean

Variation Squares Freedom Squares FRatio

Treatments SST k1 SST SSTx(bknkb+1)

(k1) SSE (k1)

Blocks SSB b1 SSB SSBx(bknkb+1)

(b1) SSE (b1)

Error SSE bknkb+1 SSE

(bknkb+1)

Total SS bkn1

Using the above formulations, substitute with actual figures from the calculations and prepare the Table A.14.:

Table A.14. Analysis of Variance

Sources of Variation / Sum of
Squares / Degrees of
Freedom / Mean
Squares / F-Ratio
(calcu-
lated) / F-Ratio
(tabular 5%)
Treatments / 217.4785 / 16-1=15 / 217.4785= 14.4986
15 / 14.4986= 15.5
0.9316 / 2.01
Blocks / 2.4253 / 3-1=2 / 2.4253= 1.2126
2 / 1.2126= 1.30
0.9316 / 3.32
Error / 27.9469 / 48-16-3+1=30 / 27.9469= 0.9316
30
Total / 247.8507 / 48-1=47

Use of the Fdistribution

The statistic F is a ratio of two variances and these variances are the 'mean squares'. To identify the Fdistribution, the degrees of freedom (df) of each variance needs to be specified. The degrees of freedom of two variances may be represented as df1 and df2, where df1 is the number of degrees of freedom in the numerator and df2 is the number of degrees of freedom in the denominator.

From the calculations in the table of analysis of variance for Treatments, F(df1, df2) = F(15,30). From a F distribution table, the critical value for F(15,30) with p = 0.05 is 2.01. Enter this tabular value into the table.

Similarly for Blocks, the critical value for F(2,30) with p = 0.05 is 3.32. Enter this tabular value into the table.

Since the calculated Fratio for treatments is greater than the tabular value of F at the 5% level, the results indicate significant differences between the strains of B. japonicum in their nitrogenfixing effectiveness.

The calculated Fratio for blocks is less than the tabular value indicating the "blocking" of the experiment did not create any significant disuniformity in the aeration, light, or other environmental factors in the greenhouse.

Calculate the Least Significant Different (LSD)

Where t0.05 =The tabular value of t for degrees of freedom for error at the 5% probability level

s2 =Mean square for error

n =Number of replications

= 2.042 x 0.79

= 1.60 g

The LSD is used to compare values of two adjacent means. A pair of means which differ by more than the LSD is considered significantly different at the probability level of t employed. If comparison between means not adjacent to each other in a ranked array are made, the Duncan's Multiple Range test should be used. However, this test requires the computation of the Bayes LSD whose value may differ from the LSD as calculated above. The calculation of the Bayes LSD is not presented here but its use is illustrated in Figure A.17.

Means not joined by the same line differ at p = 0.05 as given by Duncan's New Multiple Range Test.

Figure A.17. Effect of various strains of B. japonicum on the dry weight of shoots of soybean (G. max var. Jupiter)