In: Statistics and Probability
Examine the three samples obtained independently from three populations:
Population Data
Item | Group 1 | Group 2 | Group 3 |
---|---|---|---|
1 | 14 | 17 | 17 |
2 | 13 | 16 | 14 |
3 | 12 | 16 | 15 |
4 | 15 | 18 | 16 |
5 | 16 | 14 | |
6 | 16 |
Conduct a one-way analysis of variance on the data assuming the populations have equal variances and the populations are normally distributed. Use alpha = 0.05. Please be sure to clearly state the null and alternative hypotheses, critical value, test statistic, and conclusion.
Group 1 | Group 2 | Group 3 | Total | |
Sum | 70 | 67 | 92 | 229 |
Count | 5 | 4 | 6 | 15 |
Average, Sum/n | 14 | 16.75 | 15.3333 | |
Sum of square, Ʃ(xᵢ-x̅)² | 10 | 2.75 | 7.3333 |
Number of treatment, k = 3
Total sample Size, N = 15
df(between) = k-1 = 2
df(within) = N-k = 12
df(total) = N-1 = 14
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/
N = 16.85
SS(within) = SS1 + SS2 + SS3 =
20.0833
SS(total) = SS(between) + SS(within) =
36.9333
MS(between) = SS(between)/df(between) =
8.425
MS(within) = SS(within)/df(within) =
1.6736
F = MS(between)/MS(within) =
5.0340
p-value = F.DIST.RT(5.0340, 2,12) = 0.0259
Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: at least one mean is different
Test statistic:
F = 5.0340
Critical value:
At α = 0.05, df1 = 2, df2 = 12, critical value, Fc =
3.885
Decision: As F = 5.0340 > Fc =
3.885, Reject the null hypothesis.