Question

A farmer is interested in finding a new strain of alfalfa to grow on his land. He finds six new varieties and divides his land into thirty equally sized plots, with five randomly chosen plots for each strain.

Strain   Yield
A   1849
A   1430
A   2389
A   1920
A   2145
B   1943
B   2354
B   1990
B   2063
B   2390
C   1791
C   2125
C   1084
C   1620
C   2033
D   2179
D   1846
D   1938
D   1410
D   1983
E   2012
E   1653
E   1489
E   1696
E   2206

Analyze the data using the appropriate model. Including any relevant graphs or output, report the results of the analysis. Give giving test statistic and p-values. Make a recommendation based on the scenario and your results.

Answer 1

Here we have 5 different strains.

Here we can test five population means.

Hypothesis :

H0 : mu1 = mu2 = mu3 = mu4 = mu5

H1 : Atleast one population mean differ.

Assume alpha = level of significance = 0.05

This is one way anova problem.

We can do one way anova in excel.

Test statistic follows F-distribution.

steps :

ENTER data into excel sheet --> Data --> Data analysis --> Anova : Single factor --> Input range : Select all the data together --> Labels in first row --> Output range : select one empty cell --> ok

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
A	5	9733	1946.6	128047.3
B	5	10740	2148	43803.5
C	5	8653	1730.6	170242.3
D	5	9356	1871.2	81260.7
E	5	9056	1811.2	84574.7
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	505376.2	4	126344.1	1.243719	0.324505	2.866081
Within Groups	2031714	20	101585.7
Total	2537090	24

Test statistic = 1.24

P-value = 0.3245

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that population means differ.