Here we have data:
|
Edward |
Cathy |
Brad |
|
55 |
66 |
55 |
|
59 |
76 |
51 |
|
66 |
67 |
46 |
|
60 |
71 |
48 |
Here we are using excel for calculation:
|
Anova: Single Factor |
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SUMMARY |
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Groups |
Count |
Sum |
Average |
Variance |
|
|
| 55 |
3 |
185 |
61.6667 |
14.3333 |
|
|
| 66 |
3 |
214 |
71.3333 |
20.3333 |
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55 |
3 |
145 |
48.3333 |
6.3333 |
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ANOVA |
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Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
| Between
Groups |
800.2222 |
2 |
400.1111 |
29.2764 |
0.0008 |
10.9248 |
| Within
Groups |
82 |
6 |
13.6667 |
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Total |
882.2222 |
8 |
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Hypothesis:
Ho; μ1 = μ2 =
μ3
Ha; At least one mean is difference from others.
Test statistics:
F = 29.2764
P-value = 0.0008
Critical value:
Fc = 10.9248
Reject the null hypothesis
Here we have sufficient evidence to reject the null hypothesis ,
F-observed value (29.2764) is greater than F-critical value
(10.9248) and p-value (0.0008) is less than alpha value (0.01).
We can say that there is a difference in the number
of customers served.
