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Analyze VariabilityAnalysis of Variance Table - P-Values |
Use the p-values (P) in the analysis of variance table to determine which of the effects in the model are statistically significant. Typically, you look at the interaction effects in the model first because a significant interaction will influence how you interpret the main effects. To use the p-value, you need to:
- If the p-value is less than or equal to a, you can conclude that the effect is significant.
- If the p-value is greater than a, you can conclude that the effect is not significant.
Example Output |
Analysis of Variance for Ln(Std)
Source DF Adj SS Adj MS F-Value P-Value Model 10 65.4970 6.5497 21.73 0.002 Linear 4 31.7838 7.9459 26.36 0.001 Material 1 30.0559 30.0559 99.71 0.000 InjPress 1 1.1104 1.1104 3.68 0.113 InjTemp 1 0.1005 0.1005 0.33 0.589 CoolTemp 1 0.5170 0.5170 1.71 0.247 2-Way Interactions 6 33.7132 5.6189 18.64 0.003 Material*InjPress 1 32.0953 32.0953 106.47 0.000 Material*InjTemp 1 1.1466 1.1466 3.80 0.109 Material*CoolTemp 1 0.0010 0.0010 0.00 0.956 InjPress*InjTemp 1 0.2046 0.2046 0.68 0.448 InjPress*CoolTemp 1 0.2642 0.2642 0.88 0.392 InjTemp*CoolTemp 1 0.0014 0.0014 0.00 0.948 Error 5 1.5072 0.3014 Total 15 67.0043 |
Interpretation |
For the insulation data, the analysis of variance table shows the following:
The p-value for the set of 2-way interactions (0.003) is less than 0.05. Therefore, there is significant evidence that at least one factor depends on the level of another factor. The individual interaction results indicate that the interaction between material and injection pressure interaction is significant (p-value = 0.000).
The p-value for the set of main effects (0.001) is less than 0.05. Therefore, there is significant evidence that at least one coefficient is not equal to zero. The individual results indicate that Material is the only significant main effect (p-value = 0.000).