In: Statistics and Probability
We consider an ANOVA to test for difference in mean gill beat rates for fish in water with three different levels of calcium. The data are stored in FishGills3. If the ANOVA table indicates that the mean gill rates differ due to the calcium levels, determine which levels lead to different means. If the ANOVA shows no significant difference, find a confidence interval for the mean gill rate at each level of calcium.
Calcium | GillRate |
Low | 55 |
Low | 63 |
Low | 78 |
Low | 85 |
Low | 65 |
Low | 98 |
Low | 68 |
Low | 84 |
Low | 44 |
Low | 87 |
Low | 48 |
Low | 86 |
Low | 93 |
Low | 64 |
Low | 83 |
Low | 79 |
Low | 85 |
Low | 65 |
Low | 88 |
Low | 47 |
Low | 68 |
Low | 86 |
Low | 57 |
Low | 53 |
Low | 58 |
Low | 47 |
Low | 62 |
Low | 64 |
Low | 50 |
Low | 45 |
Medium | 38 |
Medium | 42 |
Medium | 63 |
Medium | 46 |
Medium | 55 |
Medium | 63 |
Medium | 36 |
Medium | 58 |
Medium | 73 |
Medium | 69 |
Medium | 55 |
Medium | 68 |
Medium | 63 |
Medium | 73 |
Medium | 45 |
Medium | 79 |
Medium | 41 |
Medium | 83 |
Medium | 60 |
Medium | 48 |
Medium | 59 |
Medium | 33 |
Medium | 67 |
Medium | 43 |
Medium | 57 |
Medium | 72 |
Medium | 46 |
Medium | 74 |
Medium | 68 |
Medium | 83 |
High | 59 |
High | 45 |
High | 63 |
High | 52 |
High | 59 |
High | 78 |
High | 72 |
High | 53 |
High | 69 |
High | 68 |
High | 57 |
High | 63 |
High | 68 |
High | 83 |
High | 38 |
High | 85 |
High | 68 |
High | 63 |
High | 58 |
High | 48 |
High | 42 |
High | 42 |
High | 80 |
High | 42 |
High | 52 |
High | 37 |
High | 57 |
High | 62 |
High | 40 |
High | 42 |
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Low | 30 | 2055 | 68.5 | 263.569 | ||
Medium | 30 | 1760 | 58.66667 | 204.023 | ||
High | 30 | 1745 | 58.16667 | 189.7989 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 2037.222 | 2 | 1018.611 | 4.648427 | 0.012077 | 3.101296 |
Within Groups | 19064.33 | 87 | 219.1303 | |||
Total | 21101.56 | 89 |
P VALUE =0.012
P VALUE<0.05, result is significant
so , means are different
.....................
Level of significance | 0.05 |
no. of treatments,k | 3 |
DF error =N-k= | 87 |
MSE | 219.130 |
q-statistic value(α,k,N-k) | 3.4000 |
critical value = q*√(MSE/n) |
confidence interval | ||||||
population mean difference | critical value | lower limit | upper limit | result | ||
µ1-µ2 | 9.83 | 9.19 | 0.64 | 19.02 | means are different | |
µ1-µ3 | 9.83 | 9.19 | 0.64 | 19.02 | means are different | |
µ2-µ3 | 0.00 | 9.19 | -9.19 | 9.19 | means are not different |
...........................
Please revert back in case of any doubt.
Please upvote. Thanks in advance.