Why is "Forgetting the extra $-1$ in the df when a parameter is estimated from the data" a common mistake in χ² Tests for Categorical Data?

Why it happens: Memorising ' = k-1' as a blanket rule and not noticing that or p was found from the sample (e.g. = x). How to avoid it: After computing E, ask: 'did I estimate any parameter from the data?' If yes, subtract one more for each. Write 'parameter estimated' next to your df line. Source: 9231 Paper 4 Examiner Reports — most common df error

Why is "Counting degrees of freedom from the original number of classes instead of after combining cells" a common mistake in χ² Tests for Categorical Data?

Why it happens: Finding df immediately from the frequency table before checking whether any E<5 forces a merge. How to avoid it: Always combine any cell with E<5 FIRST, recount the classes, THEN compute = (classes after combining) - 1 - (params). Source: 9231 Paper 4 mark schemes

Why is "Leaving an expected frequency below 5 without pooling" a common mistake in χ² Tests for Categorical Data?

Why it happens: Not checking the small tail classes; assuming all E are 'big enough'. How to avoid it: Scan every E before computing ^2. Merge any E<5 with an adjacent class until all combined expected frequencies are 5. Source: 9231 Paper 4 Examiner Reports

Why is "Using $\nu = rc - 1$ (or $rc$) instead of $(r-1)(c-1)$ for a contingency table" a common mistake in χ² Tests for Categorical Data?

Why it happens: Carrying over the goodness-of-fit 'k-1' habit to a two-way table. How to avoid it: For an r× c table the rule is = (r-1)(c-1). For a 2×2 table this is 1, for 3×2 it is 2. Source: 9231 Paper 4 mark schemes

Why is "Stating 'reject $H_0$' without a conclusion in context" a common mistake in χ² Tests for Categorical Data?

Why it happens: Treating the test as pure arithmetic and stopping at the comparison; forgetting the final descriptive sentence. How to avoid it: Always finish with a sentence naming the variables: 'there is (insufficient) evidence at the … level that …'. The final A1 mark is for context. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Rounding expected frequencies too early, shifting $\chi^2$ across the critical value" a common mistake in χ² Tests for Categorical Data?

Why it happens: Rounding each E to the nearest integer before forming (O-E)^2/E, accumulating error. How to avoid it: Keep E to at least 2-3 significant figures throughout, and only round the final ^2 to 3 s.f. for the comparison. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

χ² Tests for Categorical Data

Q: Why is "Leaving an expected frequency below 5 without pooling" a common mistake in χ² Tests for Categorical Data?

Why it happens: Not checking the small tail classes; assuming all E are 'big enough'. How to avoid it: Scan every E before computing ^2. Merge any E<5 with an adjacent class until all combined expected frequencies are 5. Source: 9231 Paper 4 Examiner Reports

Q: Why is "Using $\nu = rc - 1$ (or $rc$) instead of $(r-1)(c-1)$ for a contingency table" a common mistake in χ² Tests for Categorical Data?

Why it happens: Carrying over the goodness-of-fit 'k-1' habit to a two-way table. How to avoid it: For an r× c table the rule is = (r-1)(c-1). For a 2×2 table this is 1, for 3×2 it is 2. Source: 9231 Paper 4 mark schemes

Q: Why is "Stating 'reject $H_0$' without a conclusion in context" a common mistake in χ² Tests for Categorical Data?

Why it happens: Treating the test as pure arithmetic and stopping at the comparison; forgetting the final descriptive sentence. How to avoid it: Always finish with a sentence naming the variables: 'there is (insufficient) evidence at the … level that …'. The final A1 mark is for context. Source: 9231 Paper 4 Examiner Reports — recurring comment

Q: Why is "Rounding expected frequencies too early, shifting $\chi^2$ across the critical value" a common mistake in χ² Tests for Categorical Data?

Why it happens: Rounding each E to the nearest integer before forming (O-E)^2/E, accumulating error. How to avoid it: Keep E to at least 2-3 significant figures throughout, and only round the final ^2 to 3 s.f. for the comparison. Source: 9231 Paper 4 Examiner Reports

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Detailed Study Notes

Detailed notes on χ² - Tests for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Chi-squared (χ²) Tests for Categorical Data — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 (2026-2027 syllabus)

The $\chi^2$ test compares the frequencies you actually observe with the frequencies a model predicts. Master goodness-of-fit (does the data follow a stated distribution?) and contingency tables (are two variables independent?) — getting the degrees of freedom and the cell-combining right is where the marks live.

At a glance

Test statistic: $\chi^2 = \sum \dfrac{(O - E)^2}{E}$ , where $O$ = observed and $E$ = expected frequency. A big value means a poor fit.
Goodness-of-fit: tests whether data follow a stated distribution (uniform, binomial, Poisson, geometric, a given ratio). Expected frequencies are $E = n \times P(\text{category})$ .
Degrees of freedom (goodness-of-fit): $\nu = (\text{categories}) - 1 - (\text{parameters estimated from the data})$ .
Combine cells with $E < 5$ FIRST (so every $E \ge 5$ ), THEN count categories for $\nu$ .
Contingency table (independence): $E = \dfrac{(\text{row total})(\text{column total})}{\text{grand total}}$ , and $\nu = (r-1)(c-1)$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

4.3a — Fit a theoretical distribution, as prescribed by a given hypothesis, to given data, and calculate the expected frequencies $E = n \times P$ for each class.
4.3b — Use a $\chi^2$ goodness-of-fit test, including pooling classes so that each expected frequency is at least 5, and determine the correct number of degrees of freedom.
4.3c — Use a $\chi^2$ test for independence in a contingency table, calculating expected frequencies from row and column totals and concluding in context.

The χ² statistic — comparing observed with expected

Add up (O − E)²/E across all categories; a large total signals the model fits badly.

The big idea. Every $\chi^2$ test asks the same question: do the frequencies I observed match the frequencies my model predicts? You measure the gap between observed ( $O$ ) and expected ( $E$ ) frequencies and add up a standardised version of it.

The test statistic: $\chi^2 = \sum \frac{(O - E)^2}{E}$

The numerator $(O-E)^2$ measures how far each observed count is from its expected count — squared so positives and negatives do not cancel.
Dividing by $E$ standardises the gap: being 5 out from an expected 100 barely matters, but 5 out from an expected 6 is huge.
A small $\chi^2$ means $O \approx E$ everywhere (good fit). A large $\chi^2$ means at least one category is badly off (poor fit).

The hypothesis test structure (always state these):

$H_0$ : the data fit the stated model (the distribution holds / the two variables are independent).
$H_1$ : the data do not fit the stated model.
Compare your calculated $\chi^2$ with the critical value from the MF19 $\chi^2$ table at the stated significance level and the correct degrees of freedom $\nu$ .
Reject $H_0$ if $\chi^2 >$ critical value. Otherwise there is insufficient evidence to reject $H_0$ .

Each category contributes $(O-E)^2/E$. Categories where the blue and red bars differ most drive the $\chi^2$ total up.

Cambridge tip. Always lay the calculation out as a table with columns $O$ , $E$ , $O-E$ , $(O-E)^2$ and $(O-E)^2/E$ . The final column is summed to give $\chi^2$ . Examiners award method marks for a clearly structured table even if one arithmetic entry slips.

$\chi^2 = \sum \frac{(O-E)^2}{E}$ — divide by $E$ to standardise each gap.
Small $\chi^2$ = good fit; large $\chi^2$ = reject $H_0$ .
Reject $H_0$ when $\chi^2 >$ critical value from MF19.

See the full worked example for χ² tests for categorical data →

Goodness-of-fit — does the data follow a distribution?

Compute E = n × P for each class, then count df as categories − 1 − parameters estimated.

A goodness-of-fit test checks whether a set of data follows a particular theoretical distribution — a uniform, a binomial, a Poisson, a geometric, or simply a stated ratio.

Step 1 — find the expected frequencies. For each category, multiply the total number of observations $n$ by the probability the model gives that category: $E = n \times P(\text{category})$ The probabilities $P$ come from the distribution in $H_0$ (e.g. from the Poisson formula $P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}$ , or from a stated ratio like $9:3:3:1$ ).

Step 2 — degrees of freedom. This is the part candidates get wrong most often: $\nu = (\text{number of categories}) - 1 - (\text{number of parameters estimated from the data})$

You always subtract 1 (because the expected frequencies must sum to $n$ , which is one constraint).
Subtract one extra for each parameter you had to estimate from the data.
- Fitting a Poisson where $\lambda$ is given: subtract only 1.
- Fitting a Poisson where you estimate $\lambda = \bar{x}$ from the data: subtract 1 (sum) + 1 (mean) = 2.
- Fitting a binomial $\text{B}(n,p)$ where $p$ is estimated from the data: subtract an extra 1 for $p$ .

Step 3 — combine small cells. Whenever an expected frequency $E < 5$ , pool it with an adjacent class so the combined expected frequency is at least 5. Reduce the number of categories accordingly. Do this before counting df — the df is based on the number of classes after combining.

Pool the observed and the expected frequencies of the small classes together. Two classes become one, so the category count drops by 1 before you compute $\nu$.

Cambridge tip. When you estimate a parameter (e.g. $\lambda = \bar{x}$ ), write a one-line note " $\lambda$ estimated from data, so $\nu$ loses an extra degree of freedom." This signals the examiner you have applied the rule correctly and protects the df mark.

$E = n \times P(\text{category})$ from the distribution in $H_0$ .
$\nu = k - 1 - (\text{parameters estimated})$ ; subtract an extra 1 per estimated parameter.
Combine cells with $E < 5$ BEFORE counting categories for $\nu$ .

See the full worked example for χ² tests for categorical data →

Contingency tables — testing for independence

Expected = (row total × column total) / grand total; df = (r−1)(c−1).

A contingency table cross-classifies data by two categorical variables (e.g. gender × preferred subject). A $\chi^2$ test of independence asks whether the two variables are associated.

$H_0$ : the two variables are independent (no association).
$H_1$ : the two variables are not independent (there is an association).

Expected frequencies. Under independence, the expected count in any cell is: $E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$ Compute one $E$ for every cell, then form $\chi^2 = \sum \frac{(O-E)^2}{E}$ over all cells.

Degrees of freedom. For an $r \times c$ table: $\nu = (r - 1)(c - 1)$ where $r$ = number of rows and $c$ = number of columns. (The row and column totals are fixed, which removes the other degrees of freedom.)

Each cell's expected count is (its row total × its column total) ÷ grand total. With 2 rows and 3 columns, $\nu = (2-1)(3-1) = 2$.

Cell-combining still applies. If any expected cell is below 5, combine rows or columns (and adjust $r$ or $c$ ) before computing df — though Cambridge usually arranges the data so this is not needed in contingency questions.

Cambridge tip. Yates' correction is not required in 9231 — do not apply it. Just use the plain $\frac{(O-E)^2}{E}$ for every cell, and always finish with a conclusion in context ("there is/is not sufficient evidence of an association between … and …").

$E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ for every cell.
$\nu = (r-1)(c-1)$ for an $r \times c$ table.
No Yates' correction; conclude in context every time.

See the full worked example for χ² tests for categorical data →

Quick recap

$\chi^2 = \sum \frac{(O-E)^2}{E}$ measures the gap between observed and expected frequencies.
Goodness-of-fit: $E = n \times P$ ; $\nu = k - 1 - (\text{parameters estimated})$ .
Subtract one extra degree of freedom for each parameter estimated from the data (e.g. $\lambda = \bar{x}$ ).
Combine any class with $E < 5$ into an adjacent class FIRST, then count $\nu$ .
Contingency table: $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ , $\nu = (r-1)(c-1)$ .
Compare with the MF19 critical value; reject $H_0$ if $\chi^2 >$ critical; conclude in context. No Yates' correction.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\chi^2 = \sum \frac{(O-E)^2}{E}$ (in MF19, but know it cold).
Goodness-of-fit df: $\nu = (\text{categories}) - 1 - (\text{parameters estimated})$ .
Contingency df: $\nu = (r-1)(c-1)$ , with $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ .
Combine cells with $E<5$ before counting df.
Reject $H_0$ when $\chi^2 >$ MF19 critical value; always conclude in context.

How it’s examined

Topic 4.3 is a staple of Paper 4 (Further Probability & Statistics), typically worth 8-12 marks. A goodness-of-fit question gives a frequency table and asks you to fit a stated distribution (often Poisson or binomial), compute expected frequencies, combine small cells, find $\chi^2$ and test at a stated significance level. A contingency-table question asks you to test for independence. Marks are lost almost entirely through (i) the wrong degrees of freedom — forgetting the extra $-1$ for an estimated parameter, or counting df before combining cells — and (ii) failing to conclude in context. Showing the full $O,E,(O-E)^2/E$ table and stating $H_0$ / $H_1$ secures the structure marks.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 4 — Further Probability & Statistics); 9231/41, 9231/42, 9231/43 Further Probability & Statistics — past papers, mark schemes and MF19 list of formulae & tables. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — χ² Tests for Categorical Data

Step-by-step solutions to past-paper-style questions on χ² tests for categorical data, written exactly the way a tutor would explain them at the board.

Question type:

1A single expected frequency and one χ² contribution
Getting startedDirect calculation• expected-frequency, goodness-of-fit
Question
A fair six-sided die is rolled $120$ times. Under $H_0$ (the die is fair), find the expected frequency $E$ for each face. The face '6' came up $14$ times. Find that face's contribution $\frac{(O-E)^2}{E}$ to the $\chi^2$ statistic.
Step-by-step solution
1. Step 1
  Expected frequency. Under a fair die each face has probability $\frac{1}{6}$ , so $E = n \times P = 120 \times \frac{1}{6}$ .
  $E = 120 \times \frac{1}{6} = 20$
2. Step 2
  The gap for face '6': $O = 14$ , $E = 20$ .
  $O - E = 14 - 20 = -6$
3. Step 3
  Standardised contribution.
  $\frac{(O-E)^2}{E} = \frac{(-6)^2}{20} = \frac{36}{20} = 1.8$
Answer
Each $E = 20$ ; the contribution of face '6' is $\frac{(O-E)^2}{E} = 1.8$ .
Examiner tip
The full $\chi^2$ would sum all six such contributions. Note that the sign of $O-E$ is irrelevant once squared — only its magnitude matters.
2Goodness-of-fit to a stated ratio
Getting startedMulti-step problem• ratio, goodness-of-fit
Question
Genetic theory predicts offspring in the ratio $9:3:3:1$ across four phenotypes. In a sample of $160$ offspring the observed counts are $86, 32, 28, 14$ . Calculate the expected frequencies and the $\chi^2$ statistic. State the number of degrees of freedom.
Step-by-step solution
1. Step 1
  Expected frequencies. The ratio parts total $9+3+3+1 = 16$ , so $E = 160 \times \frac{\text{part}}{16}$ .
  $E_1 = 160\times\tfrac{9}{16}=90,\; E_2 = 160\times\tfrac{3}{16}=30,\; E_3 = 30,\; E_4 = 160\times\tfrac{1}{16}=10$
2. Step 2
  Contributions $\frac{(O-E)^2}{E}$ . ( $O$ : $86,32,28,14$ ; $E$ : $90,30,30,10$ .)
  $\frac{(-4)^2}{90}+\frac{2^2}{30}+\frac{(-2)^2}{30}+\frac{4^2}{10}=0.1778+0.1333+0.1333+1.6$
3. Step 3
  Sum to get the statistic (3 s.f.).
  $\chi^2 = 0.1778+0.1333+0.1333+1.6 = 2.04$
4. Step 4
  Degrees of freedom. $4$ categories, no parameters estimated (the ratio is fully given): $\nu = 4 - 1 = 3$ .
Answer
$E = 90, 30, 30, 10$ ; $\chi^2 = 2.04$ (3 s.f.); $\nu = 3$ .
Examiner tip
All $E \ge 5$ , so no combining is needed. Because the ratio is stated (no parameter estimated), $\nu = k-1 = 3$ . Keep at least 4 d.p. in the contributions before rounding the final total.
3Poisson fit with λ estimated and cells combined
Building confidenceMulti-step problem• poisson, parameter-estimated, cell-combining
Question
The number of flaws per metre of cable was recorded for $100$ metres:

Flaws 0 1 2 3 ≥4
Frequency 38 36 18 6 2

A Poisson model is proposed. Estimate $\lambda$ , find the expected frequencies, and state the degrees of freedom for a $\chi^2$ test after any necessary combining.
Step-by-step solution
1. Step 1
  Estimate $\lambda$ from the data as the sample mean $\bar{x}$ . Total flaws $= 0(38)+1(36)+2(18)+3(6)+4(2)=98$ over $100$ metres.
  $\lambda = \bar{x} = \frac{98}{100} = 0.98$
2. Step 2
  Poisson probabilities $P(X=r)=\frac{e^{-0.98}0.98^r}{r!}$ , then $E = 100\times P$ . ( $P(X\ge4)=1-P(X\le3)$ .)
  $E_0=37.5,\; E_1=36.8,\; E_2=18.0,\; E_3=5.89,\; E_{\ge4}=1.81$
3. Step 3
  Combine the small cell. $E_{\ge4}=1.81 < 5$ , so pool ' $\ge4$ ' with '3': combined $E = 5.89+1.81 = 7.70 \ge 5$ . Classes drop from $5$ to $4$ .
4. Step 4
  Degrees of freedom. After combining there are $4$ classes. Subtract $1$ (sum constraint) and a further $1$ because $\lambda$ was estimated from the data.
  $\nu = 4 - 1 - 1 = 2$
Answer
$\lambda = 0.98$ ; $E \approx 37.5, 36.8, 18.0$ and (combined $3$ & $\ge4$ ) $7.70$ ; $\nu = 2$ .
Examiner tip
Two separate reductions to df: $-1$ for the sum constraint and $-1$ for estimating $\lambda$ . Combining the last cell happens BEFORE the df is counted — that is the standard exam trap.
4Goodness-of-fit to a binomial with p given
Building confidenceMulti-step problem• binomial, goodness-of-fit
Question
In $80$ families of $3$ children, the number of girls per family is recorded as $9, 27, 32, 12$ (for $0,1,2,3$ girls). Test whether the data fit a $\text{B}(3, 0.5)$ distribution by computing the expected frequencies, $\chi^2$ and the degrees of freedom. ( $p = 0.5$ is given, not estimated.)
Step-by-step solution
1. Step 1
  Binomial probabilities $\text{B}(3,0.5)$ : $P(0)=\frac{1}{8}, P(1)=\frac{3}{8}, P(2)=\frac{3}{8}, P(3)=\frac{1}{8}$ . Then $E=80\times P$ .
  $E = 10,\; 30,\; 30,\; 10$
2. Step 2
  Check cells. All $E \ge 5$ , so no combining is needed.
3. Step 3
  Contributions ( $O$ : $9,27,32,12$ ; $E$ : $10,30,30,10$ ).
  $\frac{(-1)^2}{10}+\frac{(-3)^2}{30}+\frac{2^2}{30}+\frac{2^2}{10}=0.1+0.3+0.1333+0.4$
4. Step 4
  Sum, and find df. $p$ is given, so no parameter estimated: $\nu = 4-1=3$ .
  $\chi^2 = 0.933,\qquad \nu = 3$
Answer
$E = 10,30,30,10$ ; $\chi^2 = 0.933$ (3 s.f.); $\nu = 3$ .
Examiner tip
Because $p = 0.5$ is stated rather than estimated, you only subtract the single sum constraint: $\nu = k-1 = 3$ . Had $p$ been estimated from the data, $\nu$ would be $4-1-1=2$ .
5Contingency table — full test for independence
StretchMulti-step problem• contingency, independence, full-test
Question
A survey records preferred drink (Tea / Coffee) by age group (Young / Old):

Tea Coffee Total
Young 30 50 80
Old 40 30 70
Total 70 80 150

Test at the $5\%$ level whether drink preference is independent of age. (MF19: $\chi^2_{0.95}(1) = 3.841$ .)
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : drink preference and age are independent. $H_1$ : they are not independent.
2. Step 2
  Expected frequencies $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ .
  $E_{YT}=\tfrac{80\times70}{150}=37.33,\; E_{YC}=\tfrac{80\times80}{150}=42.67,\; E_{OT}=\tfrac{70\times70}{150}=32.67,\; E_{OC}=\tfrac{70\times80}{150}=37.33$
3. Step 3
  Contributions $\frac{(O-E)^2}{E}$ over the four cells ( $O$ : $30,50,40,30$ ).
  $\frac{(-7.33)^2}{37.33}+\frac{7.33^2}{42.67}+\frac{7.33^2}{32.67}+\frac{(-7.33)^2}{37.33}=1.440+1.260+1.645+1.440$
4. Step 4
  Sum to the statistic and find df $\nu=(r-1)(c-1)=(2-1)(2-1)=1$ .
  $\chi^2 = 5.79,\qquad \nu = 1$
5. Step 5
  Compare and conclude. $5.79 > 3.841$ , so reject $H_0$ .
Answer
$\chi^2 = 5.79 > 3.841$ , so reject $H_0$ : there is evidence at the $5\%$ level that drink preference is associated with age.
Examiner tip
Carry the $E$ values to at least 2 d.p. — premature rounding to $37,43,33,37$ can shift $\chi^2$ across the critical value. No Yates' correction is applied. Always conclude in context.
6Goodness-of-fit needing both estimation and combining
StretchMulti-step problem• geometric, parameter-estimated, cell-combining
Question
The number of rolls of a die until the first six was recorded for $200$ trials. A geometric distribution $P(X=r)=(1-p)^{r-1}p$ is proposed. The data are grouped as $X=1,2,3,4,\ge5$ with frequencies $33, 30, 25, 18, 94$ , and $p$ is estimated from the data as $\hat p = 0.167$ . After computing expected frequencies, state how the degrees of freedom are found.
Step-by-step solution
1. Step 1
  Geometric probabilities with $\hat p = 0.167$ : $P(X=r)=(0.833)^{r-1}(0.167)$ , and $P(X\ge5)=(0.833)^4$ . Then $E = 200\times P$ .
  $E_1=33.4,\; E_2=27.8,\; E_3=23.2,\; E_4=19.3,\; E_{\ge5}=96.3$
2. Step 2
  Check cells. Every $E \ge 5$ here, so no combining is required in this grouping (the open class ' $\ge5$ ' already has a large $E$ ). Classes remain at $5$ .
3. Step 3
  Degrees of freedom. Start from $5$ classes; subtract $1$ (sum constraint) and a further $1$ because $p$ was estimated from the data.
  $\nu = 5 - 1 - 1 = 3$
4. Step 4
  Why the extra $-1$ matters. Had a candidate forgotten that $p$ was estimated, they would wrongly use $\nu=4$ and read the wrong MF19 critical value — a guaranteed lost mark even with correct arithmetic.
Answer
All $E \ge 5$ (no combining); $\nu = 5 - 1 - 1 = 3$ (one for the sum, one for the estimated $p$ ).
Examiner tip
Whenever a class is below 5 it must be pooled FIRST; here the open tail keeps every $E\ge5$ , so the only subtleties are the two reductions to df. State 'p estimated from data' explicitly to protect the df mark.

Flaws	0	1	2	3	≥4
Frequency	38	36	18	6	2

	Tea	Coffee	Total
Young	30	50	80
Old	40	30	70
Total	70	80	150

Model Answers — χ² Tests for Categorical Data

High-scoring sample answers for χ² tests for categorical data on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
6 marks
A die is rolled $300$ times giving the face frequencies $42, 55, 38, 57, 64, 44$ . Test at the $5\%$ significance level whether the die is fair. (MF19: $\chi^2_{0.95}(5) = 11.07$ .)
Model answer
Hypotheses. $H_0$ : the die is fair (each face equally likely). $H_1$ : the die is not fair.

Expected frequencies. Under $H_0$ , each $E = 300 \times \frac{1}{6} = 50$ for all six faces.

Contributions $\frac{(O-E)^2}{E}$ with $O = 42,55,38,57,64,44$ and $E = 50$ : $\frac{(-8)^2}{50}+\frac{5^2}{50}+\frac{(-12)^2}{50}+\frac{7^2}{50}+\frac{14^2}{50}+\frac{(-6)^2}{50}$ $= 1.28 + 0.50 + 2.88 + 0.98 + 3.92 + 0.72 = 10.28$

Degrees of freedom. $6$ categories, no parameters estimated: $\nu = 6 - 1 = 5$ .

Conclusion. $\chi^2 = 10.3 < 11.07$ , so do not reject $H_0$ . There is insufficient evidence at the $5\%$ level that the die is unfair.
Why this scores
B1 hypotheses; M1 all $E=50$ ; M1 the contributions; A1 $\chi^2 = 10.3$ ; B1 $\nu = 5$ ; A1 correct comparison and in-context conclusion. The value sits just below the critical value — keep full accuracy.
Question 2
7 marks
A theory predicts four outcomes in the ratio $5:3:1:1$ . Of $200$ trials the observed counts are $108, 68, 14, 10$ . Test at the $10\%$ level whether the data fit the theory. (MF19: $\chi^2_{0.90}(3) = 6.251$ .)
Model answer
Hypotheses. $H_0$ : the data follow the ratio $5:3:1:1$ . $H_1$ : they do not.

Expected frequencies. Ratio total $= 5+3+1+1 = 10$ , so $E = 200 \times \frac{\text{part}}{10}$ : $E = 100,\; 60,\; 20,\; 20$ All $E \ge 5$ — no combining needed.

Contributions with $O = 108, 68, 14, 10$ : $\frac{8^2}{100}+\frac{8^2}{60}+\frac{(-6)^2}{20}+\frac{(-10)^2}{20} = 0.64 + 1.0667 + 1.8 + 5.0 = 8.51$

Degrees of freedom. $4$ categories, ratio fully given (no parameter estimated): $\nu = 4 - 1 = 3$ .

Conclusion. $\chi^2 = 8.51 > 6.251$ , so reject $H_0$ . There is evidence at the $10\%$ level that the data do not fit the $5:3:1:1$ ratio.
Why this scores
M1 expected frequencies from the ratio; A1 all four $E$ ; M1 contributions; A1 $\chi^2 = 8.51$ ; B1 $\nu = 3$ ; A1 reject; A1 in-context conclusion. The last category drives the result — show the $5.0$ contribution clearly.
Question 3
9 marks
The number of calls per minute at a switchboard over $100$ minutes is: $0\!:\!13,\;1\!:\!27,\;2\!:\!28,\;3\!:\!18,\;4\!:\!9,\;\ge5\!:\!5$ . Fit a Poisson distribution (estimating the mean from the data) and test at the $5\%$ level. (MF19: $\chi^2_{0.95}(4) = 9.488$ .)
Model answer
Hypotheses. $H_0$ : the number of calls per minute follows a Poisson distribution. $H_1$ : it does not.

Estimate the mean. Total calls $= 0(13)+1(27)+2(28)+3(18)+4(9)+5(5) = 198$ (taking $\ge5$ as $5$ ), so $\hat\lambda = \bar{x} = \frac{198}{100} = 1.98 \approx 2.0$ .

Expected frequencies $E = 100 \times P(X=r)$ with $\lambda = 1.98$ : $E_0 = 13.8,\; E_1 = 27.3,\; E_2 = 27.1,\; E_3 = 17.9,\; E_4 = 8.85,\; E_{\ge5} = 5.07$ All $E \ge 5$ , so no combining is required.

Contributions with $O = 13,27,28,18,9,5$ : $\frac{(-0.8)^2}{13.8}+\frac{(-0.3)^2}{27.3}+\frac{0.9^2}{27.1}+\frac{0.1^2}{17.9}+\frac{0.15^2}{8.85}+\frac{(-0.07)^2}{5.07}$ $= 0.0464 + 0.0033 + 0.0299 + 0.0006 + 0.0025 + 0.0010 = 0.0837$

Degrees of freedom. $6$ classes, less $1$ (sum) and less $1$ ( $\lambda$ estimated): $\nu = 6 - 1 - 1 = 4$ .

Conclusion. $\chi^2 = 0.0837 < 9.488$ , so do not reject $H_0$ . The Poisson model fits the data very well at the $5\%$ level.
Why this scores
B1 hypotheses; M1 $\hat\lambda = 1.98$ from the data; M1 Poisson $E$ values; A1 all $E$ ; M1 contributions; A1 $\chi^2$ ; B1 $\nu = 4$ (the $-1$ for estimating $\lambda$ is the discriminating mark); A1 conclusion. Forgetting the extra $-1$ would give $\nu=5$ and the wrong critical value.
Question 4
8 marks
Accidents per week at a factory over $50$ weeks: $0\!:\!28,\;1\!:\!14,\;2\!:\!6,\;\ge3\!:\!2$ . A Poisson distribution with mean $\lambda = 0.7$ (given) is proposed. Test at the $5\%$ level. (MF19: $\chi^2_{0.95}(2) = 5.991$ .)
Model answer
Hypotheses. $H_0$ : accidents per week follow $\text{Po}(0.7)$ . $H_1$ : they do not.

Expected frequencies $E = 50 \times P(X=r)$ , $\lambda = 0.7$ (given, not estimated): $E_0 = 50e^{-0.7} = 24.83,\quad E_1 = 17.38,\quad E_2 = 6.08,\quad E_{\ge3} = 1.71$

Combine cells. $E_{\ge3} = 1.71 < 5$ , so pool ' $\ge3$ ' with '2': combined $E = 6.08 + 1.71 = 7.79 \ge 5$ , combined $O = 6 + 2 = 8$ . Classes drop from $4$ to $3$ .

Contributions ( $O = 28, 14, 8$ ; $E = 24.83, 17.38, 7.79$ ): $\frac{3.17^2}{24.83}+\frac{(-3.38)^2}{17.38}+\frac{0.21^2}{7.79} = 0.405 + 0.657 + 0.006 = 1.07$

Degrees of freedom. $3$ classes after combining; $\lambda$ is given (not estimated): $\nu = 3 - 1 = 2$ .

Conclusion. $\chi^2 = 1.07 < 5.991$ , so do not reject $H_0$ . The data are consistent with $\text{Po}(0.7)$ at the $5\%$ level.
Why this scores
M1 $E$ values; M1 combining the last cell so combined $E\ge5$ ; A1 combined frequencies; M1 contributions; A1 $\chi^2 = 1.07$ ; B1 $\nu = 2$ (counted AFTER combining, with no extra $-1$ since $\lambda$ given); A1 conclusion. Combining first then $\nu = (\text{3 classes}) - 1$ is the key discipline.
Question 5
9 marks
A $3 \times 2$ table records exam grade (Pass/Fail) by teaching method (A, B, C):

Pass Fail Total
A 45 15 60
B 35 25 60
C 40 20 60
Total 120 60 180

Test at the $5\%$ level whether grade is independent of teaching method. (MF19: $\chi^2_{0.95}(2) = 5.991$ .)
Model answer
Hypotheses. $H_0$ : grade is independent of teaching method. $H_1$ : grade is not independent of teaching method.

Expected frequencies $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ . Each row total is $60$ : $E_{\text{Pass}} = \frac{60 \times 120}{180} = 40,\qquad E_{\text{Fail}} = \frac{60 \times 60}{180} = 20$ so every "Pass" cell has $E = 40$ and every "Fail" cell has $E = 20$ . All $E \ge 5$ .

Contributions $\frac{(O-E)^2}{E}$ :

Row A: $\frac{(45-40)^2}{40} + \frac{(15-20)^2}{20} = \frac{25}{40} + \frac{25}{20} = 0.625 + 1.25 = 1.875$

Row B: $\frac{(35-40)^2}{40} + \frac{(25-20)^2}{20} = 0.625 + 1.25 = 1.875$

Row C: $\frac{(40-40)^2}{40} + \frac{(20-20)^2}{20} = 0 + 0 = 0$

$\chi^2 = 1.875 + 1.875 + 0 = 3.75$

Degrees of freedom. $\nu = (r-1)(c-1) = (3-1)(2-1) = 2$ .

Conclusion. $\chi^2 = 3.75 < 5.991$ , so do not reject $H_0$ . There is insufficient evidence at the $5\%$ level of an association between grade and teaching method.
Why this scores
B1 hypotheses; M1 at least one correct $E$ from row×column/grand; A1 all $E$ ; M1 contributions; A1 $\chi^2 = 3.75$ ; B1 $\nu = (3-1)(2-1) = 2$ ; A1 in-context conclusion. A common slip is $\nu = rc - 1 = 5$ ; the correct rule is $(r-1)(c-1)$ .
Question 6
10 marks
In $100$ groups of $4$ seeds, the number germinating per group is: $0\!:\!2,\;1\!:\!10,\;2\!:\!30,\;3\!:\!38,\;4\!:\!20$ . Fit a binomial $\text{B}(4,p)$ with $p$ estimated from the data and test the fit at the $5\%$ level. (MF19: $\chi^2_{0.95}(2) = 5.991$ .)
Model answer
Hypotheses. $H_0$ : the number germinating follows a $\text{B}(4, p)$ distribution. $H_1$ : it does not.

Estimate $p$ . Total germinated $= 0(2)+1(10)+2(30)+3(38)+4(20) = 264$ out of $4 \times 100 = 400$ seeds, so $\hat p = \frac{264}{400} = 0.66$ .

Expected frequencies $E = 100 \times \binom{4}{r}(0.66)^r(0.34)^{4-r}$ : $E_0 = 1.34,\; E_1 = 10.4,\; E_2 = 30.2,\; E_3 = 39.1,\; E_4 = 18.97$

Combine cells. $E_0 = 1.34 < 5$ , so pool category '0' with '1': combined $E = 1.34 + 10.4 = 11.7 \ge 5$ , combined $O = 2 + 10 = 12$ . Classes drop from $5$ to $4$ .

Contributions ( $O = 12, 30, 38, 20$ ; $E = 11.7, 30.2, 39.1, 18.97$ ): $\frac{0.3^2}{11.7}+\frac{(-0.2)^2}{30.2}+\frac{(-1.1)^2}{39.1}+\frac{1.03^2}{18.97} = 0.0077 + 0.0013 + 0.0310 + 0.0559 = 0.0959$

Degrees of freedom. $4$ classes after combining; less $1$ (sum) and less $1$ ( $p$ estimated): $\nu = 4 - 1 - 1 = 2$ .

Conclusion. $\chi^2 = 0.0959 < 5.991$ , so do not reject $H_0$ . The binomial model fits the germination data well at the $5\%$ level.
Why this scores
B1 hypotheses; M1 $\hat p = 0.66$ ; M1 binomial $E$ values; A1 all $E$ ; M1 combine $E_0$ so combined $\ge5$ ; A1 combined frequencies; M1 contributions; A1 $\chi^2$ ; B1 $\nu = 2$ (the $-1$ for estimating $p$ AND counting after combining); A1 conclusion. This question traps BOTH high-frequency errors at once — combine first, then $\nu = 4 - 1 - 1$ .

	Pass	Fail	Total
A	45	15	60
B	35	25	60
C	40	20	60
Total	120	60	180

Key Formulae — χ² Tests for Categorical Data

The formulae you need to memorise for χ² tests for categorical data on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Chi-squared test statistic
$\chi^2 = \sum \frac{(O - E)^2}{E}$
$O$
observed frequency in a category
$E$
expected frequency in that category under $H_0$
When to use
Every $\chi^2$ test — goodness-of-fit and contingency. Sum over all categories (or all cells). In MF19, but know it cold.
Expected frequency (goodness-of-fit)
$E = n \times P(\text{category})$
$n$
total number of observations
$P(\text{category})$
probability of the category under the model in $H_0$
When to use
Fitting a uniform, binomial, Poisson, geometric or stated-ratio distribution. Take $P$ from the distribution in $H_0$ .
Degrees of freedom (goodness-of-fit)
$\nu = (\text{number of categories}) - 1 - (\text{parameters estimated})$
$\nu$
degrees of freedom
$parameters estimated$
number of model parameters found from the data (e.g. $\lambda=\bar{x}$ )
When to use
After combining any cells with $E<5$ . Subtract 1 for the sum constraint, then one more for each estimated parameter.
Expected frequency (contingency table)
$E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$
$row total$
sum of the cell's row
$column total$
sum of the cell's column
$grand total$
total of all observations
When to use
Independence test in an $r \times c$ table. Compute one $E$ per cell under the assumption of independence.
Degrees of freedom (contingency table)
$\nu = (r - 1)(c - 1)$
$r$
number of rows
$c$
number of columns
When to use
Independence test in an $r \times c$ contingency table. NOT $rc - 1$ .

Key Definitions and Keywords — χ² Tests for Categorical Data

Definitions to memorise and the exact keywords mark schemes credit for χ² tests for categorical data answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Observed frequency ( $O$ )
Examiner keyword
The actual count recorded in a category (or cell) from the collected data.
Expected frequency ( $E$ )
Examiner keyword
The count a category (or cell) would have if $H_0$ were exactly true: $E = n\times P$ for goodness-of-fit, or $E = \frac{\text{row}\times\text{column}}{\text{grand}}$ for a contingency table.
Goodness-of-fit test
Examiner keyword
A $\chi^2$ test of whether observed frequencies are consistent with a stated theoretical distribution (uniform, binomial, Poisson, geometric, or a given ratio).
Contingency table / test for independence
Examiner keyword
A two-way table cross-classifying data by two categorical variables; the $\chi^2$ test checks whether the two variables are independent (unassociated).
Degrees of freedom ( $\nu$ )
Examiner keyword
The parameter selecting the correct MF19 $\chi^2$ critical value. Goodness-of-fit: $\nu = k - 1 - (\text{parameters estimated})$ . Contingency: $\nu = (r-1)(c-1)$ .
Pooling (combining) cells
Examiner keyword
Merging adjacent classes so every expected frequency is at least 5. Done before counting degrees of freedom, because the $\chi^2$ approximation is unreliable when any $E<5$ .

Common Mistakes and Misconceptions — χ² Tests for Categorical Data

The traps other students keep falling into on χ² tests for categorical data questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Forgetting the extra $-1$ in the df when a parameter is estimated from the data
9231 Paper 4 Examiner Reports — most common df error
Why it happens
Memorising ' $\nu = k-1$ ' as a blanket rule and not noticing that $\lambda$ or $p$ was found from the sample (e.g. $\lambda = \bar{x}$ ).
How to avoid it
After computing $E$ , ask: 'did I estimate any parameter from the data?' If yes, subtract one more for each. Write 'parameter estimated' next to your df line.
✕Counting degrees of freedom from the original number of classes instead of after combining cells
9231 Paper 4 mark schemes
Why it happens
Finding df immediately from the frequency table before checking whether any $E<5$ forces a merge.
How to avoid it
Always combine any cell with $E<5$ FIRST, recount the classes, THEN compute $\nu = (\text{classes after combining}) - 1 - (\text{params})$ .
✕Leaving an expected frequency below 5 without pooling
9231 Paper 4 Examiner Reports
Why it happens
Not checking the small tail classes; assuming all $E$ are 'big enough'.
How to avoid it
Scan every $E$ before computing $\chi^2$ . Merge any $E<5$ with an adjacent class until all combined expected frequencies are $\ge 5$ .
✕Using $\nu = rc - 1$ (or $rc$ ) instead of $(r-1)(c-1)$ for a contingency table
9231 Paper 4 mark schemes
Why it happens
Carrying over the goodness-of-fit ' $k-1$ ' habit to a two-way table.
How to avoid it
For an $r\times c$ table the rule is $\nu = (r-1)(c-1)$ . For a $2\times2$ table this is $1$ , for $3\times2$ it is $2$ .
✕Stating 'reject $H_0$ ' without a conclusion in context
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Treating the test as pure arithmetic and stopping at the comparison; forgetting the final descriptive sentence.
How to avoid it
Always finish with a sentence naming the variables: 'there is (insufficient) evidence at the … level that …'. The final A1 mark is for context.
✕Rounding expected frequencies too early, shifting $\chi^2$ across the critical value
9231 Paper 4 Examiner Reports
Why it happens
Rounding each $E$ to the nearest integer before forming $\frac{(O-E)^2}{E}$ , accumulating error.
How to avoid it
Keep $E$ to at least 2-3 significant figures throughout, and only round the final $\chi^2$ to 3 s.f. for the comparison.

χ² Tests for Categorical Data

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1A single expected frequency and one χ² contribution

2Goodness-of-fit to a stated ratio

3Poisson fit with λ estimated and cells combined

4Goodness-of-fit to a binomial with p given

5Contingency table — full test for independence

6Goodness-of-fit needing both estimation and combining

Chi-squared test statistic

Expected frequency (goodness-of-fit)

Degrees of freedom (goodness-of-fit)

Expected frequency (contingency table)

Degrees of freedom (contingency table)

Observed frequency (OOO)

Expected frequency (EEE)

Goodness-of-fit test

Contingency table / test for independence

Degrees of freedom (ν\nuν)

Pooling (combining) cells

✕Forgetting the extra −1-1−1 in the df when a parameter is estimated from the data

✕Counting degrees of freedom from the original number of classes instead of after combining cells

✕Leaving an expected frequency below 5 without pooling

✕Using ν=rc−1\nu = rc - 1ν=rc−1 (or rcrcrc) instead of (r−1)(c−1)(r-1)(c-1)(r−1)(c−1) for a contingency table

✕Stating 'reject H0H_0H0​' without a conclusion in context

✕Rounding expected frequencies too early, shifting χ2\chi^2χ2 across the critical value

Observed frequency ( $O$ )

Expected frequency ( $E$ )

Degrees of freedom ( $\nu$ )

✕Forgetting the extra $-1$ in the df when a parameter is estimated from the data

✕Using $\nu = rc - 1$ (or $rc$ ) instead of $(r-1)(c-1)$ for a contingency table

✕Stating 'reject $H_0$ ' without a conclusion in context

✕Rounding expected frequencies too early, shifting $\chi^2$ across the critical value