Posted by John on Wednesday, May 9, 2012 at 8:14am.
Let's use an example, and perhaps you can use this as a guideline for similar ANOVA problems.
Suppose we wish to establish if a difference exists between three different levels of something (it doesn't matter what it is...this is just for purposes of illustration).
Here are the values for each level:
Level 1: 0, 1, 1, 2, 2
Level 2: 1, 2, 1, 3, 3
Level 3: 3, 3, 2, 4, 3
Totals for each level are:
Level 1: 6 -->add each value in the level for a total.
Level 2: 10
Level 3: 15
Total of squared values for each level are:
Level 1: 10 -->square each value in the level, then add for total.
Level 2: 24
Level 3: 47
To calculate SS within for the ANOVA table, take (sum of total squared values) minus (sum of totals divided by number of values per level). It would look like this:
(10 + 24 + 47) - [(6^2 + 10^2 + 15^2)/5] = 8.8
SS between would be calculated this way:
(6^2 + 10^2 + 15^2)/5 - (6 + 10 + 15)^2/15 = 8.13
Note: 15 = total number of values in all levels.
SS total = SS between + SS within = 16.93
To calculate df between:
k - 1 = 3 - 1 = 2
Note: k = number of levels.
To calculate df within:
N - k = 15 - 3 = 12
Note: N = total number of values in all levels.
df total = df between + df within = 14
To calculate MS between:
SS between/df between = 8.13/2 = 4.065
To calculate MS within:
SS within/df within = 8.8/12 = 0.73
To calculate F-ratio:
MS between/MS within = 4.065/0.73 = 5.57
Therefore, the ANOVA table would look like this:
Source......SS......df....MS......F
Between....8.13......2....4.065...5.57
Within.....8.8......12....0.73
Totals....16.93.....14
I hope this example will help. It looks like you already have the values needed to do the calculations for each of your 4 groups. Remember to check the appropriate table for your critical value to compare to your F-ratio. Once you do the comparison, you will either fail to reject the null or you will reject the null. You can then form your conclusions.
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