Answer:-
Given that:-
Group1 | Group2 | Group3 | Total | |
Count1 | 4 | 4 | 4 | 12 |
Mean,Sum/n | 5.75 | 4.25 | 2.75 | |
Standard deviation,s |
0.96 | 0.96 | 0.50 |
a)Null and Alternative Hypothesis:
At least one mean is different.
--
b)Number of treatement ,k=3
Total sample Size, N=12
df(between)=k-1=2
df(within)=N-k=9
df(total)=N-1=11
Critica value Fc =F.INV.RT(0.05,2,9)=4.256
--
c)SS(between)=
SS(within)
SS(total)=SS(between)+SS(within)=24.25
MS(within)=SS(within)/df(within)=0.6944
F=MS(between)/MS(within)=12.96
Decision:Reject null hypothesis.
ANOVA | ||||
Source of Variation | SS | df | MS | F |
Between Groups | 18 | 2 | 9 | 12.96 |
Within Groups | 6.25 | 9 | 0.69 | |
Total | 24.25 | 11 |
d)Eta,SS(between)/SS(total)=18/24.25=0.7423
As per the HOMEWORKLIB RULES, I have solved the
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Please Re-post for the Rest.
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