Q.The following are the paired scores $(x, y)$ of 12 students in two short tests, where $x$ is the score in Test A and $y$ the score in Test B, each taking the values $1, 2, 3$: $(1,1),\ (1,2),\ (2,1),\ (2,2),\ (2,3),\ (3,2),\ (3,3),\ (1,1),\ (2,2),\ (3,3),\ (2,1),\ (1,2)$ Construct a bivariate frequency table with $x$ along the rows and $y$ along the columns.

This is bivariate data, so we place each pair in the cell where its $x$-row meets its $y$-column, then count.

Step 1 — Set up an empty $3\times3$ grid. Rows are $x = 1, 2, 3$; columns are $y = 1, 2, 3$.

Step 2 — Tally each pair into its cell. Reading the 12 pairs in order and marking the cell where the $x$-value (row) meets the $y$-value (column):

• $x=1$: the pairs are $(1,1), (1,2), (1,1), (1,2)$ — so $y=1$ occurs 2 times, $y=2$ occurs 2 times, $y=3$ occurs 0 times.
• $x=2$: the pairs are $(2,1), (2,2), (2,3), (2,2), (2,1)$ — so $y=1$: 2, $y=2$: 2, $y=3$: 1.
• $x=3$: the pairs are $(3,2), (3,3), (3,3)$ — so $y=1$: 0, $y=2$: 1, $y=3$: 2.

Step 3 — Enter the frequencies and add the totals.

Independent check. The row totals add to $4+5+3 = 12$ and the column totals add to $4+5+3 = 12$, and both equal the count of pairs given ($N = 12$) — the three independent counts agreeing confirms no pair was misplaced or double-counted.

A quick way to guard against a tallying slip is to count how many pairs have each $x$-value first (here $x=1$: 4 pairs, $x=2$: 5, $x=3$: 3), which fixes the row totals in advance; the individual cells of each row must then add up to that pre-counted row total.

Placing a pair $(a, b)$ in the cell for row $b$, column $a$ (swapping which value is the row and which the column) is the most common error — always keep $x$ on the rows and $y$ on the columns throughout. Miscounting a repeated pair (e.g. $(1,1)$ and $(2,2)$ each appear twice here) is another slip that the row/column total cross-check catches.