In: Statistics and Probability
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For the given data set
Group 1 | Group 2 | Group 3 |
4.2 | 4.5 | 1.2 |
5.3 | 2.2 | −0.3 |
3.4 | 2.3 | 0.6 |
2.6 |
a) The following Hypotheses are made based on the assumption:
B) H0 : μ1 = μ2 = μ3,
H1 : μi ≠ μj for at least one pair
b) Based on the Hypothesis one way ANOVA hypothesis testing is done at 0.05 level of significance in excel which gives the following result:
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Group1 | 3 | 12.9 | 4.3 | 0.91 | ||
Group2 | 3 | 9 | 3 | 1.69 | ||
Group3 | 4 | 4.1 | 1.025 | 1.4825 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 19.073 | 2 | 9.5363 | 6.9193 | 0.0220 | 4.7374 |
Within Groups | 9.648 | 7 | 1.3782 | |||
Total | 28.72 | 9 |
so, SS (Between Groups/Treatment)=19.073
SS(Within Group/Error)=9.648
c) Based on the ANOVA result since P-value<0.05 and F>F-crit we reject the null hypothesis at 0.05 level of significance and conclude that there is enough evidence to support the claim that at least one pair has a different mean.