data
tire 1 | tire 2 | tire 3 | tire 4 |
9.1 | 17.1 | 20.8 | 11.8 |
13.4 | 20.3 | 28.3 | 16 |
15.6 | 24.6 | 23.7 | 16.2 |
Using Excel
Anova single factor
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
tire 1 | 3 | 38.1 | 12.7 | 10.93 | ||
tire 2 | 3 | 62 | 20.66667 | 14.16333 | ||
tire 3 | 3 | 72.8 | 24.26667 | 14.30333 | ||
tire 4 | 3 | 44 | 14.66667 | 6.173333 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 256.6825 | 3 | 85.56083 | 7.510277 | 0.01031 | 4.066181 |
Within Groups | 91.14 | 8 | 11.3925 | |||
Total | 347.8225 | 11 |
a)
SST = 347.8225
SSM + 256.6825
SSE = 91.14
b)
F = 7.5103
p-value = 0.0103
p-value < alpha(0.05)
hence we reject the null hypothesis
we conclude that there is significant difference of wear between tire types
2. Managers of a transit system want to evaluate four types of tires with respect to wear. Three buses are being used for a test drive with one tire of each type placed randomly on the four wheels of...