Question

1.Crop rotation is a common strategy used to improve the yields of certain crops in subsequent growing seasons. An experiment was performed to assess the effects of crop rotation plant type and crop rotation plant density levels on the yield of corn, the primary crop of interest. A field was separated into 12 plots and each of the treatments was randomly applied. After 2 months of growth of the rotated crops, the plots were cleared, and corn seeds were applied evenly to each plot. After 5 months of growth of the corn, the yields were assessed. The data, in kg/m2, are shown below. Determine if crop rotation plant type and density affect the yields of corn in this field. What treatment should the farmers use to maximize the yield?

	Density (k/ha)
Rotation Variety	05 k/ha	10 k/ha	15 k/ha	20 k/ha
Pea	7.8	11.2	18.5	15.4
	9.1	12.7	16.7	14.7
	10.6	13.3	15.4	11.3
Soy	7	9.3	13.8	11.3
	6.7	10.9	14.3	12.7
	8.1	11.8	15.4	14.3
Wheat	6.4	4.9	3.6	2.8
	4.5	7.1	3.9	6.1
	5.9	3.2	5.8	4.6

1. What kind of statistical test will you be performing?
2. Will you need to test for equal variance? If so what are your results and how does that influence the next steps in your analysis?
3. What are your null and alternative hypotheses?
4. Discuss the results of your analysis. Will you accept or reject your null hypothesis? Why? What can you specifically say about the data?

Answer 1

The appropriate statistical test - Two way ANOVA

For this testing problem, equal variance assumption is necessary.

Using Excel, (Data -> Data Analysis -> Anova: Two-Factor Without Replication), we get the following output -

Testing the significance of Rotation Variety,

The null hypothesis is

The alternative hypothesis is H₁ : at least one mean is different

The value of test statistic F = 11.39

and P-value = 0.00000161

Since P-value < 0.05, so we reject H₀ at 5% level of significance and we can conclude that there is significant of Rotation Variety on yields.

Testing the significance of Density,

The null hypothesis is

The alternative hypothesis is H₁ : at least one mean is different

The value of test statistic F = 6.66

and P-value = 0.001977

Since P-value < 0.05, so we reject H₀ at 5% level of significance and we can conclude that there is significant of Density on yields.