Maths › Further Statistics 1 › Contingency tables
Contingency tables
Cross-tabulate two categorical variables and one chi-squared test asks whether they are independent: expected counts come from the margins, and the degrees of freedom from the shape of the table.
Builds on Goodness-of-fit tests and Conditional probability.
IN THIS TOPIC
- Compute expected frequencies as row total × column total ÷ grand total.
- Use (rows − 1)(columns − 1) degrees of freedom.
- State the hypotheses as independence and association, and conclude in context.
WHAT YOU PROBABLY THINK
Expected frequencies in a contingency table come from assuming every cell is equally likely.
Expectations from the margins
The null hypothesis is that the two variables are independent, not that the cells are equal. Under independence, the probability of a cell is the product of its row and column probabilities, so multiplying by the grand total gives:
Equal cells would be a far stronger claim, and one nobody tests here: a row holding twice as many observations should expect twice as much in every column, which is exactly what the formula delivers. The opening claim confuses independence with uniformity.
WORKED EXAMPLE
Testing an association
A survey cross-tabulates two variables as rows (20, 30, 10) and (30, 20, 40). Test at the 5% level whether the variables are independent.
Row totals 60 and 90; column totals 50, 50, 50; grand total 150. Expected: 60 × 50/150 = 20 across the first row, and 30 across the second.
χ² = 0 + 5 + 5 + 0 + 3.33 + 3.33 = 16.67.
Degrees of freedom (2 − 1)(3 − 1) = 2, critical value 5.991.
16.67 > 5.991, so reject H₀: there is evidence of an association between the two variables.
Counting the freedom in a grid
Once the margins are fixed, filling in one cell of a 2 × 2 table forces every other. In general only (rows − 1)(columns − 1) cells are free, and that is the degrees of freedom. No extra subtraction is needed for estimated parameters: the margins have already absorbed them.
YOUR TURN
Degrees of freedom and pooling
A contingency table has 4 rows and 3 columns, and no expected frequency falls below 5. State the degrees of freedom and the 5% critical value's degrees of freedom if two rows had to be combined.
Show the working
As it stands: (4 − 1)(3 − 1) = 6 degrees of freedom.
Combining two rows leaves 3 rows: (3 − 1)(3 − 1) = 4.
Pooling always costs degrees of freedom, because it removes cells that were free to vary.
THE EXAM BIT
- Write the hypotheses as 'no association' against 'some association', naming both variables.
- Compute expected frequencies to one decimal place and show at least one calculation in full.
- Degrees of freedom are (r − 1)(c − 1); do not subtract again for estimated parameters.
- If any expected frequency is below 5, combine categories before computing the statistic.
CHECK YOURSELF
A contingency table has 3 rows and 4 columns. State the degrees of freedom, and find the expected frequency for a cell whose row total is 40 and column total is 30, with grand total 200.
Show a hint
(r − 1)(c − 1), and row × column ÷ total.
Show the answer
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Expected = row total × column total ÷ grand total, which is independence written as arithmetic.
Degrees of freedom are (rows − 1)(columns − 1): the margins have already used up the rest.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Compute expected frequencies as row total × column total ÷ grand total.
- Use (rows − 1)(columns − 1) degrees of freedom.
- State the hypotheses as independence and association, and conclude in context.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.