In: Statistics and Probability
In the mall |
Online |
At a stand-alone store |
2 |
3 |
6 |
5 |
3 |
6 |
1 |
4 |
5 |
0 |
5 |
6 |
1 |
3 |
5 |
3 |
5 |
4 |
In the mall | Online | At a stand-alone store |
2 | 3 | 6 |
5 | 3 | 6 |
1 | 4 | 5 |
0 | 5 | 6 |
1 | 3 | 5 |
3 | 5 | 4 |
Excel
data -> data analysis -> Anova Single Factor
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
In the mall | 6 | 12 | 2.0000 | 3.2000 | ||
Online | 6 | 23 | 3.8333 | 0.9667 | ||
At a stand-alone store | 6 | 32 | 5.3333 | 0.6667 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 33.4444 | 2 | 16.7222 | 10.3793 | 0.0015 | 3.6823 |
Within Groups | 24.1667 | 15 | 1.6111 | |||
Total | 57.6111 | 17 |
p-value = 0.0015
We conduct Anova One way to test if there is any difference in groups
we get F = 10.3793 , p-value = 0.0015
since p-value < 0.05 , we reject the null hypothesis. We conclude that there is significant difference in means of different groups