Question

Five different environments were chosen to compare the seed yield of four strains of a weed species. In each environment, four adjacent plots of approximately same fertility and moisture level were found and each strain was randomly assigned to a plot. At maturity, ten random samples were taken from each plot and the mean number of seeds per plant was recorded. These data are presented below:

Strain (*I*)

Environment (*J*)
A
B
C
D

_______________________________________________________________

1 18 20 17 15

2 16 18 16 18

3 18 21 16 13

4 18 20 17 16

5 19 17 18 20

- Complete the analysis of variance (ANOVA) table for the experiment.

- Was it a reasonable design to use for the experiment? Explain.

Answer #1

a)

Two-way ANOVA: YIELD versus ENVIRONMENT, STRAIN

Source DF SS MS F P

ENVIRONMENT 4 11.20 2.80000 0.92 0.482

STRAIN 3 23.35 7.78333 2.57 0.103

Error 12 36.40 3.03333

Total 19 70.95

S = 1.742 R-Sq = 48.70% R-Sq(adj) = 18.77%

Since p-value>0.05 in both cases for ENVIRONMENT and STRAIN, at 5% level we can conclude that different environments as well as different strains cannot give markedly different yields.

b)Here yield is continuous variable whereas the two factors are dummy variable.So ANOVA is appropriate for analysis.Moreover there are two factors-environment and strain we use two way ANOVA. Each strain is randomly designed in each plot and each plot randomle belongs to any of the five environments,So randomization is present here. Overall the design is exactly what is needed to perform two-way ANOVA.

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