Question 138586: Statistical software is commonly used for problem solving in the workplace. It is useful to look at a computer generated output to understand the format of the output, how the computerized solution treats an ANOVA table, and so on. The ANOVA table is somewhat standardized among various software packages. Formulate the null and alternate hypothesis and decide whether or not to reject it from the ANOVA table data provided below.
Group 1
5
7
4
5
4
6
8
5
6
7
Group 2
6
7
7
5
6
7
8
6
8
6
Group 3
7
8
9
8
9
9
10
11
10
9
One factor ANOVA
Mean n Std. Dev
7.1 5.7 10 1.34 Group 1
7.1 6.6 10 0.97 Group 2
7.1 9.0 10 1.15 Group 3
7.1 30 1.81 Total
ANOVA table
Source SS df MS F p-value
Treatment 58.20 2 29.100 21.53 2.57E-06
Error 36.50 27 1.352
Total 94.70 29
Post hoc analysis
Tukey simultaneous comparison t-values (d.f. = 27)
Group 1 Group 2 Group 3
5.7 6.6 9.0
Group 1 5.7
Group 2 6.6 1.73
Group 3 9.0 6.35 4.62
critical values for experimentwise error rate:
0.05 2.49
0.01 3.18
p-values for pairwise t-tests
Group 1 Group 2 Group 3
5.7 6.6 9.0
Group 1 5.7
Group 2 6.6 .0949
Group 3 9.0 8.54E-07 .0001
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The ANOVA table is somewhat standardized among various software packages. Formulate the null and alternate hypothesis and decide whether or not to reject it from the ANOVA table data provided below.
------------------------------------------
One factor ANOVA
Mean n Std. Dev
7.1 5.7 10 1.34 Group 1
7.1 6.6 10 0.97 Group 2
7.1 9.0 10 1.15 Group 3
7.1 30 1.81 Total
ANOVA table
Source.. SS df MS F p-value
Treatment 58.20 2 29.100 21.53 ...2.57E-06
Error ........36.50 27 1.352
Total ........94.70 29
Post hoc analysis
Tukey simultaneous comparison t-values (d.f. = 27)
Group 1 Group 2 Group 3
5.7 6.6 9.0
Group 1 ........5.7
Group 2 ........6.6 1.73
Group 3 ........9.0 6.35 4.62
critical values for experimentwise error rate:
0.05 2.49
0.01 3.18
p-values for pairwise t-tests
Group 1 Group 2 Group 3
5.7 6.6 9.0
Group 1 ........5.7
Group 2 ........6.6 .0949
Group 3 ........9.0 8.54E-07 .0001
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Ho: The means of the three Groups were all equal
Ha: At least one of the means was different.
Since p-value = 0.00000257, Reject Ho.
Conclusion: At least one of the mean values is different.
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Cheers,
Stan H.
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