Solution:
Here we have data:
Fish | Adler | Bell | Linde | Marston | Morgan | Pearson |
1 | 18 | 23 | 18 | 24 | 18 | 20 |
2 | 20 | 26 | 21 | 26 | 15 | 19 |
3 | 17 | 29 | 22 | 23 | 19 | 20 |
4 | 20 | 22 | 26 | 18 | 23 | 28 |
5 | 24 | 29 | 25 | 23 | 21 | 20 |
6 | 28 | 21 | 26 | 25 | 18 | 19 |
7 | 20 | 25 | 23 | 22 | 19 | 21 |
Here we are using Excel for calculation:
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Column 1 | 7 | 147 | 21 | 14.333 | ||
Column 2 | 7 | 175 | 25 | 10.333 | ||
Column 3 | 7 | 161 | 23 | 8.667 | ||
Column 4 | 7 | 161 | 23 | 6.667 | ||
Column 5 | 7 | 133 | 19 | 6.333 | ||
Column 6 | 7 | 147 | 21 | 10 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 154 | 5 | 30.8 | 3.280 | 0.015 | 2.477 |
Within Groups | 338 | 36 | 9.389 | |||
Total | 492 | 41 |
Hypothesis:
Ho; μ1 = μ2 = μ3 = μ4 = μ5 = μ6
Ha; At least one mean is different from others.
Test statistics:
Fmax = 3.280
Critical Value:
Fmax = 2.477
Reject the null hypotheses.
Here we have sufficient evidence to reject the null hypothesis , Test statistics (3.280) is grater than critical value (2.477).
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