Solution:
Part 6-7)
We have:
dftreatment = 3
dfresidual = 80
SStreatment = 187.19
SSresidual =872.8
Thus
MStreatment = SStreatment / dftreatment
MStreatment = 187.19 / 3
MStreatment = 62.40
MSresidual = SSresidual / dfresidual
MSresidual = 872.8 / 80
MSresidual = 10.91
Part 8) F statistic:
F = MStreatment / MSresidual
F = 62.40 / 10.91
F = 5.72
Part 9) F critical value = 2.72
Since F test statistic value = 5.72 > F critical value = 2.72, we reject null hypothesis.
Thus answer is:
Reject null
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