Given the summary statistics, the calculations of the hypothesis test  are shown in tabular form. While two columns of SS are shown for their explanatory value, only one column is required to display results.  formula_22 is the estimate of variance corresponding to formula_23 of the  model. The core ANOVA analysis consists of a series of calculations. The  data is collected in tabular form. Then  If the experiment is balanced, all of the formula_5 terms are  equal so the SS equations simplify. In a more complex experiment, where the experimental units (or  environmental effects) are not homogeneous, row statistics are also  used in the analysis. The model includes terms dependent on  formula_15. Determining the extra terms reduces the number of degrees of freedom available. Consider an experiment to study the effect of three different levels of a factor on a response (e.g. three levels of a fertilizer on plant growth). If we had 6 observations for each level, we could write the outcome of the experiment in a table like this, where "a"1, "a"2, and "a"3 are the three levels of the factor being studied. The null hypothesis, denoted H0, for the overall "F"-test for this experiment would be that all three levels of the factor produce the same response, on average. To calculate the "F"-ratio:  Step 1: Calculate the mean within each group:  Step 2: Calculate the overall mean:  Step 3: Calculate the "between-group" sum of squared differences:  where "n" is the number of data values per group. The between-group degrees of freedom is one less than the number of groups  so the between-group mean square value is  Step 4: Calculate the "within-group" sum of squares. Begin by centering the data in each group  The within-group sum of squares is the sum of squares of all 18 values in this table  The within-group degrees of freedom is  Thus the within-group mean square value is  Step 5: The "F"-ratio is  The critical value is the number that the test statistic must exceed to reject the test. In this case, "F"crit(2,15) = 3.68 at "α" = 0.05. Since "F"=9.3 > 3.68, the results are significant at the 5% significance level. One would reject the null hypothesis, concluding that there is strong evidence that the expected values in the three groups differ.