Question 2 of 5: Two-Way ANOVA — Fill in the Blanks

Question 2: Two-Way ANOVA — Fill in the Blanks (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A completely randomized two-factor design: Factor A (Location) has $a=3$ levels, Factor B (Season) has $b=4$ levels, with $n=2$ replications per treatment combination ($N=abn=24$ observations total).

Find. Every blank cell in the table above, plus the significance conclusion for Location, Season and their interaction at α=0.05.

Approach. Every degree of freedom follows directly from the design structure alone; every other blank then follows from the two identities $SS=MS\times df$ and $SS_{total}=\sum$ (all other SS), so the table is solved without any raw data or simultaneous equations.

1. Step 1 — degrees of freedom from the design. With $a=3$, $b=4$, $n=2$: $$df_A=a-1=2,\quad df_B=b-1=3,\quad df_{AB}=(a-1)(b-1)=6,\quad df_{err}=ab(n-1)=12,\quad df_{tot}=N-1=23$$ As a check, $df_A+df_B+df_{AB}+df_{err}=2+3+6+12=23=df_{tot}$. ✓
2. Step 2 — recover SS wherever its own MS is given. $SS=MS\times df$: $$SS_A = MS_A\times df_A = 60.20\times 2 = 120.40, \qquad SS_{err} = MS_{err}\times df_{err} = 3.91\times 12 = 46.92$$
3. Step 3 — recover MS wherever its own SS is given. $MS=SS/df$: $$MS_B = SS_B/df_B = 12.46/3 = 4.153$$
4. Step 4 — recover the interaction SS by subtraction from the total. Every other source's SS is now known, so $$SS_{AB} = SS_{tot} - SS_A - SS_B - SS_{err} = 245.32 - 120.40 - 12.46 - 46.92 = 65.54$$ $$MS_{AB} = SS_{AB}/df_{AB} = 65.54/6 = 10.923$$
5. Step 5 — F-ratios and significance test. Every effect is tested against the error mean square, $F=MS_{effect}/MS_{err}$: $$F_A = \frac{60.20}{3.91} = 15.40, \qquad F_B = \frac{4.153}{3.91} = 1.062, \qquad F_{AB} = \frac{10.923}{3.91} = 2.794$$ Comparing to the exam's own supplied F-table at $\alpha=0.05$: $F_{crit}(2,12)=3.885$, $F_{crit}(3,12)=3.490$, $F_{crit}(6,12)=2.996$. $$\boxed{F_A=15.40 > 3.885 \Rightarrow \text{Location significant}}\qquad F_B=1.062 < 3.490 \Rightarrow \text{Season not significant}$$ $$F_{AB}=2.794 < 2.996 \Rightarrow \text{Interaction not significant (close to the threshold)}$$

Conclusion. At the 5% significance level, the sampled variable differs significantly among Locations ($F=15.40$, well above $F_{crit}=3.885$), but not significantly among Seasons ($F=1.06 < 3.49$) and there is no significant Location×Season interaction ($F=2.79 < 3.00$, although this is the closest of the three to its threshold). Practically: which location a sample is taken at is the dominant source of variability in this study, while the season of sampling — and whether the location effect itself changes from season to season — are not statistically distinguishable from random error at this sample size.