FS1-6 Chi squared tests — revision question pack

1 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FS1-6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FS1-6.1 · Goodness of fit tests and contingency tables. The null and alternative hypotheses. Degrees of freedom.

Explanation

For goodness of fit, $H_0$ states in context that the proposed distribution fits; for a contingency table, both hypotheses must name the two classifications, for example $H_0$ : travel method is independent of year group and $H_1$ : travel method is associated with year group. Use $\chi^2=\sum\dfrac{(O-E)^2}{E}$ .
Goodness-of-fit degrees of freedom are $k-1-m$ after combining, where $m$ parameters were estimated from the data. For an $r\times c$ contingency table, $df=(r-1)(c-1)$ and each expected frequency is row total times column total divided by grand total.
Cells should be combined sensibly when $E<5$ .
In a critical-value test, state the degrees of freedom and the critical value explicitly before comparing it with $\chi^2$ ; the critical value can carry its own mark.
Finish with a contextual conclusion.

Worked example

A $2\times3$ contingency table has row totals $60,40$ , column totals $20,30,50$ and grand total $100$ . Find the expected frequencies and degrees of freedom for a test of independence.

1.First row expectations: $60(20,30,50)/100=(12,18,30)$ .
2.Second row expectations: $40(20,30,50)/100=(8,12,20)$ .
3. $df=(2-1)(3-1)=2$ .

Answer: Expected table $\begin{pmatrix}12&18&30\\8&12&20\end{pmatrix}$ ; $2$ degrees of freedom.

Common mistakes

Don't use observed rather than expected frequency in the denominator of each contribution.
Don't calculate contingency-table degrees of freedom as $rc-1$ .
Don't combine a cell with $E<5$ but forget to recalculate the number of groups and degrees of freedom.

Exam tip

Show one expected-frequency calculation and state the degrees-of-freedom formula before using the calculator.

Tier 1 · Easy

1.
A goodness-of-fit test has four categories with model probabilities $0.1,0.2,0.3,0.4$ and a sample size of $120$ . No parameter is estimated. Find the expected frequencies and the degrees of freedom.
(3)
(Total for Question 1 is 3 marks)
2.
A study records whether each of $80$ people completes a task. The observed counts are $\begin{array}{c|rr}&\text{completed}&\text{not completed}\\ \hline \text{method A}&24&16\\ \text{method B}&18&22\end{array}$ . Use a $5\%$ significance level to investigate whether completion is independent of method. The critical value for $1$ degree of freedom is $3.841$ .
(6)
(Total for Question 2 is 6 marks)

Tier 2 · Standard

1.
After fitting a one-parameter model from the same data, five ordered categories have observed frequencies $41,33,17,7,2$ and expected frequencies $40,32,18,7,3$ . Carry out a goodness-of-fit test at the $5\%$ level of significance. The critical value for $2$ degrees of freedom is $5.991$ .
(8)
(Total for Question 1 is 8 marks)
2.
The observed frequencies of $X$ are $14$ at $x=0$ , $26$ at $x=1$ , $28$ at $x=2$ , $18$ at $x=3$ , $9$ at $x=4$ , $3$ at $x=5$ , $1$ at $x=6$ and $1$ at $x=7$ ; no larger values occur. It is thought that a Poisson distribution is a suitable model for $X$ . Find an estimate for $\lambda$ and carry out a goodness-of-fit test at the $5\%$ level of significance, combining classes where necessary. The $5\%$ critical values with $4$ and $6$ degrees of freedom are $9.488$ and $12.592$ .
(7)
(Total for Question 2 is 7 marks)
3.
A machine repeats independent operations until its first success. For $120$ runs, the observed frequencies for a first success on operation $1$ , $2$ , $3$ , $4$ and $5$ or later are $35,28,20,14,23$ . Carry out a goodness-of-fit test for a geometric distribution with parameter $0.35$ at the $5\%$ level of significance. State your hypotheses clearly. The critical value for $4$ degrees of freedom is $9.488$ .
(7)
(Total for Question 3 is 7 marks)

Tier 3 · Hard

1.
The observed counts for travel method by year group are $\begin{array}{c|rrrr}&\text{walk}&\text{cycle}&\text{bus}&\text{car}\\ \hline \text{Year 10}&24&18&20&18\\ \text{Year 11}&16&22&20&22\\ \text{Year 12}&10&20&30&20\end{array}$ . Test independence at the $5\%$ level of significance. The critical value for $6$ degrees of freedom is $12.592$ .
(10)
(Total for Question 1 is 10 marks)
2.
The observed counts for response category by training group are $\begin{array}{c|rrrr}&A&B&C&D\\ \hline \text{group 1}&24&4&1&1\\ \text{group 2}&10&25&12&3\\ \text{group 3}&16&6&11&7\end{array}$ . Test independence at the $5\%$ level of significance, combining adjacent response categories where required. The $5\%$ critical values with $4$ and $6$ degrees of freedom are $9.488$ and $12.592$ .
(8)
(Total for Question 2 is 8 marks)
3.
In $200$ sets of six trials, the observed frequencies for $0,1,2,3,4,5,6$ successes are $13,72,72,34,3,6,0$ . A binomial model $B(6,p)$ is proposed, with $p$ estimated from these data. Carry out a goodness-of-fit test at the $1\%$ level of significance, combining classes where necessary. The $1\%$ critical values for $3$ and $5$ degrees of freedom are $11.345$ and $15.086$ respectively.
(10)
(Total for Question 3 is 10 marks)
4.
On each search, a wildlife researcher inspects independently selected sites from a large region until finding an occupied nest. Across $200$ searches, the numbers ending at sites $1$ , $2$ , $3$ , $4$ , $5$ and $6$ or later are $76,51,31,16,10,16$ respectively. The complete records show that $500$ sites were inspected altogether. Use $9.488$ as the $5\%$ critical value for $4$ degrees of freedom. Estimate the parameter of a geometric distribution from these data and carry out a goodness-of-fit test at the $5\%$ level of significance. State your hypotheses clearly.
(9)
(Total for Question 4 is 9 marks)
5.
A horticulturist classifies seedlings by stem-height band and leaf-colour category. The observed counts are $\begin{array}{c|rrr}&\text{green}&\text{pale}&\text{mottled}\\ \hline 0\text{--}4\text{ cm}&24&18&8\\ 5\text{--}9\text{ cm}&18&22&20\\ 10\text{--}14\text{ cm}&10&14&16\\ 15\text{--}19\text{ cm}&2&3&5\end{array}$ . At the $1\%$ level of significance, test whether leaf-colour category is independent of stem-height band, combining adjacent stem-height bands where required. State your hypotheses clearly. The $1\%$ critical values for $4$ and $6$ degrees of freedom are $13.277$ and $16.812$ respectively.
(10)
(Total for Question 5 is 10 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FS1-6.1 · Goodness of fit tests and contingency tables. The null and alternative hypotheses. Degrees of freedom.

Tier 1 · Easy

Mark scheme for FS1-6.1 Tier 1 · Easy
Question	Scheme	Marks
1	Expected frequencies $12,24,36,48$ $df=3$	3
(3 marks)		3
Notes
Multiply $120$ by each model probability to obtain $12,24,36,48$ . With $k=4$ groups and no estimated parameter, $df=k-1=3$ .
2	$H_0$ : completion is independent of method; $H_1$ : completion is associated with method Row totals are $40,40$ and column totals are $42,38$ Expected rows are both $(21,19)$ $\chi^2=1.80$ to $2$ decimal places $df=1$ and $1.80<3.841$ Do not reject $H_0$ ; there is insufficient evidence of an association between completion and method	6
(6 marks)		6
Notes
Its row totals are $40,40$ , its column totals are $42,38$ , and its grand total is $80$ . Under $H_0$ , each expected row is $(40/80)(42,38)=(21,19)$ , so every expected frequency exceeds $5$ . The statistic is $2(3^2/21+3^2/19)=1.804511278\ldots$ , which is $1.80$ to $2$ d.p. With $df=(2-1)(2-1)=1$ , $1.80<3.841$ , so do not reject independence.

Tier 2 · Standard

Mark scheme for FS1-6.1 Tier 2 · Standard
Question	Scheme	Marks
1	Combine the final two categories, giving observed $41,33,17,9$ and expected $40,32,18,10$ $H_0$ : the fitted model describes the category frequencies; $H_1$ : the fitted model does not describe the category frequencies $\chi^2=0.2118$ $df=2$ Critical value $=5.991$ Do not reject $H_0$ ; there is insufficient evidence that the model does not fit	8
(8 marks)		8
Notes
The final expected frequency is $3<5$ , so combine it with the adjacent fourth category. This gives $O=(41,33,17,9)$ and $E=(40,32,18,10)$ . Then $\chi^2=\dfrac{1^2}{40}+\dfrac{1^2}{32}+\dfrac{(-1)^2}{18}+\dfrac{(-1)^2}{10}=0.211806\ldots$ . After combining there are $k=4$ groups, and one parameter was estimated, so $df=k-1-1=2$ . Since $0.2118<5.991$ , do not reject $H_0$ .
2	$H_0$ : a Poisson distribution fits; $H_1$ : a Poisson distribution does not fit $\widehat\lambda=\overline x=200/100=2$ For $0,1,2,3,4,5+$ , $E=(13.5335\ldots,27.0671\ldots,27.0671\ldots,18.0447\ldots,9.02235\ldots,5.26530\ldots)$ Pool $5,6,7,\ldots$ , giving $O=(14,26,28,18,9,5)$ $\chi^2=0.103834929\ldots$ $df=k-1-m=6-1-1=4$ and $0.1038<9.488$ Do not reject $H_0$ ; there is insufficient evidence that a Poisson model is unsuitable	7
(7 marks)		7
Notes
The frequencies total $100$ and their weighted sum is $200$ , so the fitted parameter is $\widehat\lambda=2$ . For $X\sim\operatorname{Po}(2)$ , the expected frequencies for $0,1,2,3,4$ are $13.5335283237\ldots,27.0670566473\ldots,27.0670566473\ldots,18.0447044315\ldots,9.0223522158\ldots$ . Pooling the upper tail gives observed frequency $5$ and expected frequency $100$ minus the first five expectations, namely $5.2653017344\ldots$ . All six expected frequencies now exceed $5$ . Summing $(O-E)^2/E$ gives $\chi^2=0.1038349292\ldots$ . Since one parameter was estimated, $df=k-1-m=6-1-1=4$ . As $0.1038<9.488$ , do not reject $H_0$ .
3	$H_0$ : a geometric distribution with parameter $0.35$ fits the waiting times; $H_1$ : it does not fit the waiting times $E_x=120P(X=x)$ ; for example, $E_2=120(0.35)(0.65)=27.3$ The expected frequencies are $42,27.3,17.745,11.53425,21.42075$ $\chi^2=2.114726\ldots$ $df=5-1=4$ $2.114726\ldots<9.488$ Do not reject $H_0$ ; there is insufficient evidence that the geometric model fails to describe the machine's waiting times	7
(7 marks)		7
Notes
For a geometric variable with $p=0.35$ , the probabilities for $1,2,3,4,5+$ are $p,pq,pq^2,pq^3,q^4$ , where $q=0.65$ . Multiplication by $120$ gives $E=(42,27.3,17.745,11.53425,21.42075)$ , all above $5$ . The cell contributions are $1.1666667,0.0179487,0.2865610,0.5271191,0.1164306$ , whose sum is $\chi^2=2.1147260417\ldots$ . No parameter was estimated, so $df=5-1=4$ . Since $2.114726\ldots<9.488$ , do not reject the geometric model; the data do not provide sufficient evidence that it fails to describe these waiting times.

Tier 3 · Hard

Mark scheme for FS1-6.1 Tier 3 · Hard
Question	Scheme	Marks
1	$H_0$ : travel method is independent of year group; $H_1$ : travel method is associated with year group Expected row in each case: $(16.667,20,23.333,20)$ $\chi^2=9.577$ $df=6$ Critical value $=12.592$ Do not reject $H_0$ ; there is insufficient evidence of an association between travel method and year group	10
(10 marks)		10
Notes
Each row total is $80$ , the column totals are $50,60,70,60$ , and the grand total is $240$ . Thus every expected row is $(80/240)(50,60,70,60)=(16.6667,20,23.3333,20)$ . Summing $(O-E)^2/E$ over all $12$ cells gives $\chi^2=9.5771429\ldots$ . The degrees of freedom are $(3-1)(4-1)=6$ and the stated critical value is $12.592$ . Since $9.577<12.592$ , do not reject independence; there is insufficient evidence of an association between travel method and year group.
2	$H_0$ : response category is independent of training group; $H_1$ : response category is associated with training group Column $D$ has expected frequencies below $5$ , so combine adjacent response categories $C$ and $D$ The combined observed rows are $(24,4,2),(10,25,15),(16,6,18)$ Expected rows are $(12.5,8.75,8.75),(20.8333\ldots,14.5833\ldots,14.5833\ldots),(16.6667\ldots,11.6667\ldots,11.6667\ldots)$ $\chi^2=37.6686$ $df=(3-1)(3-1)=4$ $37.6686>9.488$ Reject $H_0$ ; there is evidence of an association between response category and training group	8
(8 marks)		8
Notes
Its row totals are $(30,50,40)$ , its column totals are $(50,35,24,11)$ and its grand total is $120$ . Column $D$ would have expectations $(11/4,55/12,11/3)$ , all below $5$ , so combine adjacent response categories $C$ and $D$ . The resulting observed rows are $(24,4,2),(10,25,15),(16,6,18)$ and column totals are $(50,35,35)$ . Recomputed expected rows are $(12.5,8.75,8.75)$ , $(125/6,175/12,175/12)$ and $(50/3,35/3,35/3)$ ; all exceed $5$ . Summing all nine contributions gives $\chi^2=6592/175=37.668571\ldots$ . The combined table has $df=(3-1)(3-1)=4$ . Since $37.6686>9.488$ , reject independence; there is evidence of an association between response category and training group.
3	$H_0$ : a binomial distribution $B(6,p)$ fits the numbers of successes; $H_1$ : a binomial distribution does not fit the numbers of successes There are $360$ successes in $1200$ trials, so $\widehat p=0.3$ $P(X=x)=\binom6x(0.3)^x(0.7)^{6-x}$ The expected frequencies for $x=0,1,2$ are $23.5298,60.5052,64.8270$ The expected frequencies for $x=3,4,5,6$ are $37.0440,11.9070,2.0412,0.1458$ Combine $x=4,5,6$ , giving pooled $O=(13,72,72,34,9)$ and $E=(23.5298,60.5052,64.8270,37.0440,14.0940)$ The five cell contributions are $4.712181,2.183786,0.793681,0.250133,1.841126$ $\chi^2=9.780908\ldots$ $df=5-1-1=3$ , so $9.780908\ldots<11.345$ Do not reject $H_0$ ; there is insufficient evidence at the $1\%$ level that a binomial model fails to fit the numbers of successes in the sets of trials	10
(10 marks)		10
Notes
The observations contain $72+2(72)+3(34)+4(3)+5(6)=360$ successes, so $\widehat p=360/(200\times6)=0.3$ . Under $B(6,0.3)$ the seven expected frequencies are $23.5298,60.5052,64.8270,37.0440,11.9070,2.0412,0.1458$ . The last two expectations are below $5$ , and combining only $x=5,6$ still gives $2.187<5$ , so combine the adjacent classes $x=4,5,6$ . This reduces the number of classes from $7$ to $5$ and hence changes the degrees of freedom from $7-1-1=5$ to $5-1-1=3$ . For the pooled arrays $O=(13,72,72,34,9)$ and $E=(23.5298,60.5052,64.8270,37.0440,14.0940)$ , the contributions are $4.7121814907,2.1837863033,0.7936805498,0.2501332469,1.8411264368$ . Their sum is $\chi^2=9.7809080275\ldots$ . Since this is below the $1\%$ pooled critical value $11.345$ , do not reject $H_0$ ; the data provide insufficient evidence at the $1\%$ level against the binomial model.
4	$H_0$ : a geometric distribution fits the search lengths; $H_1$ : a geometric distribution does not fit the search lengths $\widehat p=200/500=0.4$ $P(X=x)=0.4(0.6)^{x-1}$ for $x=1,2,3,\ldots$ The expected frequencies for $x=1,2,3,4,5,6+$ are $80,48,28.8,17.28,10.368,15.552$ For the open class $6+$ , the expected frequency is $200\times0.6^5=15.552$ The six cell contributions are $0.2,0.1875,0.1680556,0.0948148,0.0130617,0.0129053$ $\chi^2=0.676337\ldots$ $df=6-1-1=4$ and $0.676337\ldots<9.488$ Do not reject $H_0$ ; there is insufficient evidence at the $5\%$ level that the geometric model fails to describe the search lengths	9
(9 marks)		9
Notes
The fitted geometric mean is $500/200=2.5$ , so $\widehat p=1/2.5=0.4$ , equivalently the number of searches divided by the total number of sites inspected. The probabilities for $1,2,3,4,5,6+$ are $p,pq,pq^2,pq^3,pq^4,q^5$ , where $q=0.6$ . Multiplying by $200$ gives $E=(80,48,28.8,17.28,10.368,15.552)$ , all above $5$ . The six contributions sum to $\chi^2=3287/4860=0.6763374486\ldots$ . One parameter was estimated, so $df=6-1-1=4$ . Since $0.676337\ldots<9.488$ , do not reject $H_0$ ; at the $5\%$ level the data do not provide sufficient evidence against the fitted geometric model.
5	$H_0$ : leaf-colour category is independent of stem-height band; $H_1$ : leaf-colour category is associated with stem-height band The row totals are $50,60,40,10$ , the column totals are $54,57,49$ , and the grand total is $160$ The last row has expected frequencies $3.375,3.5625,3.0625$ , all below $5$ Combine the adjacent $10$ -- $14$ cm and $15$ -- $19$ cm groups The combined observed rows are $(24,18,8),(18,22,20),(12,17,21)$ The expected rows are $(16.875,17.8125,15.3125)$ , $(20.25,21.375,18.375)$ and $(16.875,17.8125,15.3125)$ $\chi^2=10.472276\ldots$ $df=(3-1)(3-1)=4$ $10.472276\ldots<13.277$ Do not reject $H_0$ ; there is insufficient evidence at the $1\%$ level of an association between leaf-colour category and stem-height band	10
(10 marks)		10
Notes
Before pooling, the final row has total $10$ , so its expected frequencies are $10(54,57,49)/160=(3.375,3.5625,3.0625)$ and do not meet the expected-frequency condition. Combining the two adjacent tallest stem-height bands gives observed rows $(24,18,8),(18,22,20),(12,17,21)$ with row totals $50,60,50$ . Using row total times column total divided by $160$ gives expected rows $(16.875,17.8125,15.3125)$ , $(20.25,21.375,18.375)$ and $(16.875,17.8125,15.3125)$ . Their nine contributions sum to $\chi^2=10.472275928\ldots$ . The pooled table is $3\times3$ , so $df=4$ . Since $10.472276\ldots<13.277$ , do not reject independence; at the $1\%$ level there is insufficient evidence of an association between leaf colour and stem height.