The following is the sample information. Test the hypothesis at the 5% level of
ID: 2947592 • Letter: T
Question
The following is the sample information. Test the hypothesis at the 5% level of significance that the treatment means are equal. Present your solution by the p value method
Treatment 1
Treatment 2
Treatment 3
9
13
10
7
20
9
11
14
15
9
13
14
12
15
10
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Treatment 1
6
58
9.66666667
3.066667
Treatment 2
4
60
15
11.33333
Treatment 3
5
63
12.6
8.3
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
70.4
2
35.2
5.117932
0.024704
Within Groups
82.53333333
12
6.87777778
Total
152.9333333
14
Treatment 1
Treatment 2
Treatment 3
9
13
10
7
20
9
11
14
15
9
13
14
12
15
10
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Treatment 1
6
58
9.66666667
3.066667
Treatment 2
4
60
15
11.33333
Treatment 3
5
63
12.6
8.3
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
70.4
2
35.2
5.117932
0.024704
Within Groups
82.53333333
12
6.87777778
Total
152.9333333
14
Explanation / Answer
H0: ?1 = ?2 = ?3 all the means are the same
H1: Two or more means are different from the others
Here, we are trying to test whether the means of the three groups are the same or are different. This is a case of one way ANOVA.
Now as we are given here the level of significance of 0.05, and the p-value here is 0.025 < 0.05, therefore the test is significant and we can reject the null hypothesis here and conclude that there are at least two of the means which are not the same.
Therefore the test is significant and all the means are not the same here.
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