8.1 Two-Way Contingency Tables, Relative Frequencies, and Venn Diagrams

School leaders and classroom teachers routinely cross-classify student cohorts across two or more categorical variables—such as Year Level vs. Attendance Tier, Campus vs. Transport Mode, or EAL/D Status vs. Benchmark Achievement. On the LANTITE Numeracy test, questions involving two-way contingency tables and Venn diagrams rarely require difficult arithmetic; instead, they test whether you can track missing marginal totals accurately and select the exact denominator demanded by the wording of the question.

Reading and Completing Two-Way Contingency Tables

A two-way (contingency) table organises counts across two categorical variables simultaneously. One variable defines the rows and the other defines the columns:

• Interior cells (joint frequencies): Each inner cell shows how many observations satisfy both that specific row category and that specific column category at the same time.
• Row marginal totals: The rightmost column records the total count for each row across all column categories.
• Column marginal totals: The bottom row records the total count for each column across all row categories.
• Grand total ($N$): The bottom-right corner cell gives the total number of observations in the entire dataset. The sum of all row totals must equal the sum of all column totals, which equals $N$.

When a LANTITE question presents a partially completed two-way table, never guess or solve cells out of order. Scan the table for any row or column that has only one missing entry, compute that missing value by subtraction, and repeat until every required cell is filled.

Worked Example 1: Completing a School Transport Table

Consider a survey of $400$ students in Years 7 to 9 at Greenvale Secondary College regarding their primary mode of transport to school:

Suppose the cells for Year 7 Car ($40$), Year 8 Row Total ($140$), and Year 9 Bus ($60$) were initially blank:

1. Year 7 Car / Drop-off: The Year 7 row total is $150$, so $150 - (72 + 38) = 150 - 110 = 40$.
2. Year 8 Row Total: All three interior cells in Year 8 are known, so $68 + 42 + 30 = 140$.
3. Year 9 Bus: The Bus column total is $200$, so $200 - (72 + 68) = 200 - 140 = 60$. (You can cross-check this using the Year 9 row total: $110 - (20 + 30) = 60$.)

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The Three Denominators in a Two-Way Table

The single most common trap in two-way table questions is dividing the correct numerator by the wrong total. A single interior cell—such as the $42$ Year 8 students who walk to school in the table above—can produce three completely different relative frequencies depending on which reference group is specified in the prompt:

Before touching the calculator, always underline the phrase immediately following "of", "among", or "given that":

• "What percentage of all students..." $\rightarrow$ Divide by the Grand Total ($400$).
• "What percentage of Year 8 students..." $\rightarrow$ Restrict your universe to the Year 8 Row Total ($140$).
• "Of the students who walk to school, what percentage..." $\rightarrow$ Restrict your universe to the Walk Column Total ($100$).

Notice also that converse conditional percentages are not equal: $30.0\%$ of Year 8 students walk to school, whereas $42.0\%$ of walking students are in Year 8.

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Translating Between Two-Way Tables and Venn Diagrams

When two categorical variables are binary (Yes/No, Member/Non-Member), a $2 \times 2$ two-way table and a 2-set Venn diagram represent the exact same information:

The Inclusion-Exclusion Principle and Working Outward

If you add the total count in Set $A$ and the total count in Set $B$, every student in the overlap $A \cap B$ is counted twice. To find the number of students in at least one of the two sets ($A \cup B$), subtract the intersection once:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

Including the exterior region of students in neither set gives the universal total $N$:

$$N = n(A \cup B) + n(\text{Neither}) = n(A) + n(B) - n(A \cap B) + n(\text{Neither})$$

When constructing or reading any Venn diagram, follow one ironclad rule: always work from the innermost intersection outward.

Worked Example 2: 2-Set Co-Curricular Survey

In a cohort of $120$ Year 6 students, $68$ students sing in the Choir ($C$), $54$ play in the Concert Band ($B$), and $18$ participate in neither ensemble. How many students participate in Choir only?

1. Subtract "Neither" from the universal total $N$ to find the union $n(C \cup B)$:

$$n(C \cup B) = 120 - 18 = 102\text{ students in at least one ensemble}$$

1. Find the intersection $n(C \cap B)$ using inclusion-exclusion:

$$n(C \cap B) = n(C) + n(B) - n(C \cup B) = 68 + 54 - 102 = 122 - 102 = 20\text{ students in both}$$

1. Work outward to find the exclusive crescent regions:
   • Choir only: $n(C) - n(C \cap B) = 68 - 20 = 48$ students.
   • Band only: $n(B) - n(C \cap B) = 54 - 20 = 34$ students.
2. Check all four mutually exclusive regions sum to $N$: $48 + 20 + 34 + 18 = 120$.

Worked Example 3: 3-Set Venn Diagram (Working Outward from the Triple Overlap)

Among $100$ Year 11 students selecting science electives, $45$ take Biology ($B$), $40$ take Chemistry ($C$), and $35$ take Physics ($P$). Suppose $18$ take both Biology and Chemistry, $13$ take both Biology and Physics, $15$ take both Chemistry and Physics, and $8$ take all three sciences. To find how many take none of the three sciences:

1. Innermost triple intersection: $n(B \cap C \cap P) = 8$.
2. Subtract $8$ from each pairwise overlap to find the "exact two sciences" regions:
   • Biology and Chemistry only: $18 - 8 = 10$.
   • Biology and Physics only: $13 - 8 = 5$.
   • Chemistry and Physics only: $15 - 8 = 7$.
3. Subtract all three overlap pieces from each circle total to find the "single science only" regions:
   • Biology only: $45 - (10 + 8 + 5) = 22$.
   • Chemistry only: $40 - (10 + 8 + 7) = 15$.
   • Physics only: $35 - (5 + 8 + 7) = 15$.
4. Sum all seven interior regions and subtract from $100$:

$$n(B \cup C \cup P) = 22 + 15 + 15 + 10 + 5 + 7 + 8 = 82 \implies n(\text{None}) = 100 - 82 = 18\text{ students}$$