Three groups of seven participants each gave sum X = 36, 17 and 13 with sum X^2 = 200, 59 and 35 (overall sum X = 66, sum X^2 = 294, N = 21). Perform a one-way ANOVA and report the effect size eta squared.

Step-by-step solution

1. 1

   Compute the correction term once and reuse it. Every sum of squares in a one-way ANOVA is built from the same grand-total term:

   $\frac{\left(\sum X_{\text{tot}}\right)^{2}}{N}=\frac{66^{2}}{21}=\frac{4356}{21}=207.4286$

   Calculating it once and carrying extra decimals prevents the rounding drift that makes $SS_{\text{within}}$ come out negative in student work.

2. 2

   Total variability.

   $SS_{\text{total}}=\sum X^{2}_{\text{tot}}-\frac{\left(\sum X_{\text{tot}}\right)^{2}}{N}=294-207.4286=86.5714$

3. 3

   Variability explained by the grouping. Replace each score by its group mean, which in summary form means squaring each group total and dividing by that group's $n$:

   $SS_{\text{between}}=\sum\frac{\left(\sum X_j\right)^{2}}{n_j}-\frac{\left(\sum X_{\text{tot}}\right)^{2}}{N}=\frac{36^{2}+17^{2}+13^{2}}{7}-207.4286$

   $=\frac{1296+289+169}{7}-207.4286=250.5714-207.4286=43.1429$

4. 4

   Effect size: the share of variance the grouping accounts for.

   $\eta^{2}=\frac{SS_{\text{between}}}{SS_{\text{total}}}=\frac{43.1429}{86.5714}=0.4983\approx0.50$

   As an exact fraction this is $\frac{302/7}{606/7}=\frac{151}{303}$. Against the usual benchmarks ($0.01$ small, $0.06$ medium, $0.14$ large) an $\eta^{2}$ of about $0.50$ is a very large effect: half the variability in the scores is associated with which group a participant was in.

5. 5

   Finish the ANOVA table to confirm the result is significant. The leftover variability is

   $SS_{\text{within}}=86.5714-43.1429=43.4286,\qquad df_{\text{between}}=2,\quad df_{\text{within}}=18$

   $MS_{\text{between}}=\frac{43.1429}{2}=21.571,\qquad MS_{\text{within}}=\frac{43.4286}{18}=2.413$

   $F=\frac{21.571}{2.413}=8.94$

6. 6

   Compare $F$ with its critical value. For $df=(2,18)$, $F_{.05}=3.55$ and $F_{.01}=6.01$. Since $8.94$ exceeds both (exact $p=0.002$), the null hypothesis that all three groups come from the same population is rejected at the $.01$ level — consistent with the very large $\eta^{2}$. The group means themselves, $5.14$, $2.43$ and $1.86$, show where the difference lies.