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I got the following result of a ANOVA analysis of five independent variables with 5,8,2,2,6 levels and 5 replications for each combination:

                   Df Sum Sq Mean Sq F value Pr(>F)    
IV1                 4 129805   32451  243.35 <2e-16 ***
IV2                 7  67227    9604   72.02 <2e-16 ***
IV3                 1  64253   64253  481.83 <2e-16 ***
IV4                 1 396445  396445 2972.94 <2e-16 ***
IV5                 5  91672   18334  137.49 <2e-16 ***
Residuals        4781 637553     133                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I now want to quantifiy which parameter has how much effect on the variation of the results. In a statistics book and suggest by an answer to my previousmy previous question I found the suggestion to simply calculate the ratio of the Sum of Square of the total Sum of Square. In my case as the Total sum of squares = 1386955 this would lead to 9% influence of the IV1 and 45% influence due to errors.

Question: Why is the Sum of Squares used here and not the mean of the sum of squares? The mean is calculated by dividing by the number of degrees of freedom. This seems much more logical to me: With more degrees of freedom, the sum of squares automatically gets higher so this must be compenstated somehow.

The results would be totaly different: IV1 would have 6% influence and the error would go down to below 1%.

Side question: Why is the Df for the residuals such a strange number? I thought the df of the residuals is abcd...(n-1), in my case 582264=3840

I got the following result of a ANOVA analysis of five independent variables with 5,8,2,2,6 levels and 5 replications for each combination:

                   Df Sum Sq Mean Sq F value Pr(>F)    
IV1                 4 129805   32451  243.35 <2e-16 ***
IV2                 7  67227    9604   72.02 <2e-16 ***
IV3                 1  64253   64253  481.83 <2e-16 ***
IV4                 1 396445  396445 2972.94 <2e-16 ***
IV5                 5  91672   18334  137.49 <2e-16 ***
Residuals        4781 637553     133                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I now want to quantifiy which parameter has how much effect on the variation of the results. In a statistics book and suggest by an answer to my previous question I found the suggestion to simply calculate the ratio of the Sum of Square of the total Sum of Square. In my case as the Total sum of squares = 1386955 this would lead to 9% influence of the IV1 and 45% influence due to errors.

Question: Why is the Sum of Squares used here and not the mean of the sum of squares? The mean is calculated by dividing by the number of degrees of freedom. This seems much more logical to me: With more degrees of freedom, the sum of squares automatically gets higher so this must be compenstated somehow.

The results would be totaly different: IV1 would have 6% influence and the error would go down to below 1%.

Side question: Why is the Df for the residuals such a strange number? I thought the df of the residuals is abcd...(n-1), in my case 582264=3840

I got the following result of a ANOVA analysis of five independent variables with 5,8,2,2,6 levels and 5 replications for each combination:

                   Df Sum Sq Mean Sq F value Pr(>F)    
IV1                 4 129805   32451  243.35 <2e-16 ***
IV2                 7  67227    9604   72.02 <2e-16 ***
IV3                 1  64253   64253  481.83 <2e-16 ***
IV4                 1 396445  396445 2972.94 <2e-16 ***
IV5                 5  91672   18334  137.49 <2e-16 ***
Residuals        4781 637553     133                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I now want to quantifiy which parameter has how much effect on the variation of the results. In a statistics book and suggest by an answer to my previous question I found the suggestion to simply calculate the ratio of the Sum of Square of the total Sum of Square. In my case as the Total sum of squares = 1386955 this would lead to 9% influence of the IV1 and 45% influence due to errors.

Question: Why is the Sum of Squares used here and not the mean of the sum of squares? The mean is calculated by dividing by the number of degrees of freedom. This seems much more logical to me: With more degrees of freedom, the sum of squares automatically gets higher so this must be compenstated somehow.

The results would be totaly different: IV1 would have 6% influence and the error would go down to below 1%.

Side question: Why is the Df for the residuals such a strange number? I thought the df of the residuals is abcd...(n-1), in my case 582264=3840

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user88
user88

Why compare Sum Of Squaressum of squares with ANOVA (and not mean)?

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theomega
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Why compare Sum Of Squares with ANOVA (and not mean)?

I got the following result of a ANOVA analysis of five independent variables with 5,8,2,2,6 levels and 5 replications for each combination:

                   Df Sum Sq Mean Sq F value Pr(>F)    
IV1                 4 129805   32451  243.35 <2e-16 ***
IV2                 7  67227    9604   72.02 <2e-16 ***
IV3                 1  64253   64253  481.83 <2e-16 ***
IV4                 1 396445  396445 2972.94 <2e-16 ***
IV5                 5  91672   18334  137.49 <2e-16 ***
Residuals        4781 637553     133                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I now want to quantifiy which parameter has how much effect on the variation of the results. In a statistics book and suggest by an answer to my previous question I found the suggestion to simply calculate the ratio of the Sum of Square of the total Sum of Square. In my case as the Total sum of squares = 1386955 this would lead to 9% influence of the IV1 and 45% influence due to errors.

Question: Why is the Sum of Squares used here and not the mean of the sum of squares? The mean is calculated by dividing by the number of degrees of freedom. This seems much more logical to me: With more degrees of freedom, the sum of squares automatically gets higher so this must be compenstated somehow.

The results would be totaly different: IV1 would have 6% influence and the error would go down to below 1%.

Side question: Why is the Df for the residuals such a strange number? I thought the df of the residuals is abcd...(n-1), in my case 582264=3840