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Answer:
Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis and conclude all three means are not equal.
Explanation:
The given data is
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12 8 10
10 8 8
10 6 6
12 8 8
<u>10 10 10</u>
The results of excel are:
<u><em>Anova: Single Factor </em></u>
SUMMARY
Groups Count Sum Average Variance
Column 1 5 54 10.8 1.2
Column 2 5 40 8 2
Column 3 5 42 8.4 2.8
1) Let the null and alternate hypotheses be
H0: u1=u2=u3 i.e all the three means are equal and
Ha: Not all three means are equal
2) The significance level is set at ∝ =0.05
3)The test statistic to use is
F= sb²/ sw²
which has F distribution with v1= k-1 →3-1=2 and v2= n-k →15-3=12 degrees of freedom
After calculations the following table is obtained.
<u><em>ANOVA </em></u>
Source SS df MS F P-value F crit
of Variation
B/w Groups 22.93 2 11.467 5.733 0.01788 3.885
<u>Within Groups 24 12 2 </u>
<u>Total 46.93 14 </u>
4) The critical region is F ≥ F(0.05, 2,12) = 3.885
5) Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis and conclude all three means are not equal.
