Question

An experiment has 5 treatments; 6 replicates were obtained for each of the treatments. Use the treatments below to solve

Treatment A	Treatment B	Treatment C	Treatment D	Treatment E
0.58	-0.01	-0.64	0.14	-0.56
0.92	-0.04	0.22	0.3	0.29
0.43	-0.73	-0.55	0.36	-0.37
-0.1	0.07	-0.91	0.68	-0.32
0.57	-0.16	0.56	-0.5	-0.34
-0.63	-0.17	-0.25	-0.06	0.29

A) Do the ANOVA computations on this data; what is the P-value?

B) On the basis of this P-value, do you reject the hypothesis that the means of the 5 treatments are the same at an α = 0.05confidence level?

Answer 1

A) The null and alternative hypotheses are:

H0: Means of the 5 treatments are the same.

Ha: Atleast one treatment group has different mean.

By using excel, the mean and standard deviation of these groups are:

Groups	Count	Average	Standard deviation
Treatment A	6	0.295	0.562
Treatment B	6	-0.173	0.288
Treatment C	6	-0.262	0.557
Treatment D	6	0.153	0.404
Treatment E	6	-0.168	0.365

There are groups and total sample size is .

The grand mean is:

By using excel function =FDIST(1.735,4,25), the p-value = 0.174.

The complete ANOVA table is shown below:

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1.395	4	0.349	1.735	0.174
Within Groups	5.026	25	0.201
Total	6.421	29

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B) Since p-value = 0.174 is greater than 0.05, fail to reject null hypothesis. So, answer should be:

No, we do not reject null hypothesis.