5. Using the following data, perform a oneway analysis of variance using α=0.05. group 1 group...
5. Using the following data, perform a oneway analysis of variance using α=0.05.
group 1 group 2 group 3
51 23 56
45 43 76
33 23 74
45 43 87
67 45 56
a) What are the null and alternative hypotheses for this test?
b) What is your conclusion?
Homework Answers
Overall total = 767
Overall mean Y̅.. = 767 / 15 = 51.1333
Y̅ .. is overall mean
y̅i . is treatment mean
SS total = ΣΣ(Yij - & Y̅..)2 = 4883.7332
SS treatment = ΣΣ(Yij - & y̅i.)2 = 3022.9332
SS error = Σ(y̅i. - & Y̅..)2 = 1860.8
MS treatment = ΣΣ(Yij - & y̅i.)2 / a - 1 = 1511.4666
MS error = Σ(y̅i. - & Y̅..)2 / N - a = 155.0667
| ANOVA table | |||||
| Source | SS | df | MS | F | p-value |
| Treatment | 3,022.93 | 2 | 1,511.467 | 9.75 | 0.0031 |
| Error | 1,860.80 | 12 | 155.067 | ||
| Total | 4,883.73 | 14 |
To Test :-
H0 :- µ1 = µ2 = µ3 = 0 i.e There is no difference in the group
means
H0 :- µ1 = µ2 = µ3 ≠ 0 i.e Some means are different
Test Statistic :-
f = MS treatment / MS error = 9.7472
Test Criteria :-
Reject null hypothesis if f > f(α , a-1 , N-a )
Critical value f(0.05, 2 , 12 ) = 3.8853 (From F table)
Since 9.7472 > 3.8853, we reject H0
conclusion = Group means may differs
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