A researcher wants to know if the clothes a woman wears is a factor in her GPA. The table below shows data that was collected from a survey.
Shorts | Dress | Jeans | Skirt |
---|---|---|---|
3.9 | 2.6 | 2.7 | 3.4 |
2.1 | 2.6 | 1.8 | 3.5 |
2.4 | 2.3 | 3 | 3.5 |
2 | 2.1 | 4 | 3.8 |
3.7 | 2.6 | 2.3 | 3.2 |
2.5 | 2.6 | 2.8 | 3.5 |
3.6 | 1.9 | 2.9 | 2.9 |
2.2 | 2.3 | 2.2 | 2.1 |
2.7 | 3.5 |
Assume that all distributions are normal, the four population
standard deviations are all the same, and the data was collected
independently and randomly. Use a level of significance of
α=0.1α=0.1.
H0: μ1=μ2=μ3=μ4
H1: At least two of the means differ from each other.
Excel output:
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Shorts | 9 | 25.1 | 2.788889 | 0.551111 | ||
Dress | 8 | 19 | 2.375 | 0.073571 | ||
Jeans | 8 | 21.7 | 2.7125 | 0.435536 | ||
Skirt | 9 | 29.4 | 3.266667 | 0.2525 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 3.467361111 | 3 | 1.155787 | 3.469915 | 0.028253 | 2.922277 |
Within Groups | 9.992638889 | 30 | 0.333088 | |||
Total | 13.46 | 33 |
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