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 field study with $a=3$ levels of Factor A (Location), $b=4$ levels of Factor B (Season) and $n=2$ replications in every treatment combination, giving $N=abn=24$ observations. Four cells of the ANOVA table are printed; the rest are blank.

Find. Every blank cell of the table, and the significance verdict for Location, for Season and for the Location×Season interaction at $\alpha=0.05$, with the study conclusions stated in plain terms.

Approach. Every degree of freedom follows from the design structure alone, after which the two identities $SS=MS\times df$ and $SS_{tot}=SS_A+SS_B+SS_{AB}+SS_{err}$ recover every remaining cell in a fixed order — no raw replicate data and no simultaneous equations are needed, because the table is over-determined by construction.

1. Step 1 — write down every degree of freedom from the design. The df of a crossed two-factor design with replication depend only on $a$, $b$ and $n$: $$df_A=a-1=2,\quad df_B=b-1=3,\quad df_{AB}=(a-1)(b-1)=6,\quad df_{err}=ab(n-1)=12,\quad df_{tot}=N-1=23$$ The partition must be exact, and it is: $2+3+6+12=23=df_{tot}$. This check is worth doing first, because every mean square below divides by one of these numbers.
2. Step 2 — recover the sums of squares wherever the mean square is printed. Rearranging $MS=SS/df$ gives $SS=MS\times df$, which unlocks the Location and Error rows: $$SS_A = MS_A \times df_A = 80.17\times 2 = 160.34,\qquad SS_{err} = MS_{err}\times df_{err} = 3.79\times 12 = 45.48$$
3. Step 3 — recover the mean square wherever the sum of squares is printed. The Season row is the mirror image of Step 2: $$MS_B = \frac{SS_B}{df_B} = \frac{12.46}{3} = 4.153$$
4. Step 4 — obtain the interaction by subtraction. The sums of squares are additive, so with $SS_A$, $SS_B$ and $SS_{err}$ all now known the interaction is whatever the total has left over: $$SS_{AB} = SS_{tot} - SS_A - SS_B - SS_{err} = 262.96 - 160.34 - 12.46 - 45.48 = 44.68$$ $$MS_{AB} = \frac{SS_{AB}}{df_{AB}} = \frac{44.68}{6} = 7.447$$ That $SS_{AB}$ comes out positive is itself a confirmation that the printed cells have been read from the correct columns — a column misread would drive it negative.
5. Step 5 — form the F-ratios and test each against the supplied table. In a fixed-effects design every effect is tested against the error mean square, $F=MS_{effect}/MS_{err}$: $$F_A = \frac{80.17}{3.79} = 21.153,\qquad F_B = \frac{4.153}{3.79} = 1.096,\qquad F_{AB} = \frac{7.447}{3.79} = 1.965$$ Reading the $\alpha=0.05$ critical values from the F-table supplied with the exam, using each effect's own numerator df against $df_{err}=12$: $F_{0.05}(2,12)=3.8853$, $F_{0.05}(3,12)=3.4903$ and $F_{0.05}(6,12)=2.9961$. Comparing each statistic with its own critical value: $$\boxed{F_A=21.153 > 3.8853 \Rightarrow \text{Location effect significant}}$$ $$F_B=1.096 < 3.4903 \Rightarrow \text{Season effect not significant};\qquad F_{AB}=1.965 < 2.9961 \Rightarrow \text{interaction not significant}$$

Conclusions from the study. At the 5% significance level the interaction is tested first, because a significant interaction would make the main effects uninterpretable on their own. It is not significant here ($F_{AB}=1.97$ against a critical 3.00), so the Location and Season effects may be read directly and independently. The Location effect is highly significant ($F_A=21.15$ against a critical 3.89 — more than five times the threshold): the measured variable differs genuinely among the three sampling locations, and location accounts for $160.34/262.96 = 61\%$ of the total variability in the study. The Season effect is not significant ($F_B=1.10$ against a critical 3.49); with $MS_B=4.15$ barely above the error mean square of 3.79, season-to-season variation is essentially indistinguishable from experimental error at this sample size. Practically, the study says that where a sample is taken drives the response and when it is taken does not, and — because the interaction is absent — the ranking among locations is stable from season to season. A follow-up multiple-comparison procedure (Tukey's HSD or Fisher's LSD on the three location means) would be the appropriate next step to identify which locations differ, since the F-test establishes only that at least one does.