SOLUTION: Stan, will you please explain how you got the answers below. Thank you very much
The ANOVA table is somewhat standardized among various software packages. Formulate the null an
Algebra ->
Probability-and-statistics
-> SOLUTION: Stan, will you please explain how you got the answers below. Thank you very much
The ANOVA table is somewhat standardized among various software packages. Formulate the null an
Log On
Question 138706: Stan, will you please explain how you got the answers below. Thank you very much
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
---------------------------------------------------
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.
===============================
Cheers,
Stan H.
(