| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Cell combining required |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test application with standard procedures: part (a) tests understanding of degrees of freedom formula (11-1)(11-1)=100, reduced by 10 for combining cells; part (b) is routine hypothesis testing comparing 124.8 to critical value; part (c) simply requires identifying the largest contribution. All parts are direct recall/application of standard chi-squared procedures with no problem-solving or novel insight required. |
| Spec | 5.06a Chi-squared: contingency tables |
| Group | \(\frac { ( O - E ) ^ { 2 } } { E }\) |
| Attends school 3 and chose activity 1 | 0.01 |
| Attends school 8 and chose activity 3 | 18.5 |
| Attends school 8 and chose activity 7 | 24.2 |
| Attends school 11 and chose activity 7 | 49.0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Two rows or two columns need to be merged | E1 | Deduces test requires two rows or two columns to be merged |
| So \(\text{dof} = (11-1)(10-1) = 90\) | E1 | Completes reasoned argument showing merging results in dof \(= (11-1)(10-1) = 90\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between school and activity chosen; \(H_1\): There is an association between school and activity chosen | B1 | States both hypotheses using correct language; variables needed in at least null hypothesis |
| \(\chi^2\) cv for 90 dof \(= 124.116\); or corresponding probability AWRT 0.009 | B1 | AWRT 124.1 |
| \(124.8 > 124.116\) | R1 | Evaluates \(\chi^2\) test statistic by correctly comparing critical value with test statistic |
| Reject \(H_0\) | E1F | Infers \(H_0\) rejected; FT their comparison using \(\chi^2\) model |
| Sufficient evidence to suggest there is an association between school and activity chosen | E1F | Conclusion in context; must not be definite; FT consistent with their conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Students attending school 11 choosing activity 7 as they have the highest value of \(\frac{(O-E)^2}{E}\) | B1 | Correctly identifies strongest source of association as group with highest \(\frac{(O-E)^2}{E}\); condone "the highest value" |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Two rows or two columns need to be merged | E1 | Deduces test requires two rows or two columns to be merged |
| So $\text{dof} = (11-1)(10-1) = 90$ | E1 | Completes reasoned argument showing merging results in dof $= (11-1)(10-1) = 90$ |
**Subtotal: 2 marks**
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between school and activity chosen; $H_1$: There is an association between school and activity chosen | B1 | States both hypotheses using correct language; variables needed in at least null hypothesis |
| $\chi^2$ cv for 90 dof $= 124.116$; or corresponding probability AWRT 0.009 | B1 | AWRT 124.1 |
| $124.8 > 124.116$ | R1 | Evaluates $\chi^2$ test statistic by correctly comparing critical value with test statistic |
| Reject $H_0$ | E1F | Infers $H_0$ rejected; FT their comparison using $\chi^2$ model |
| Sufficient evidence to suggest there is an association between school and activity chosen | E1F | Conclusion in context; must not be definite; FT consistent with their conclusion |
**Subtotal: 5 marks**
---
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Students attending school 11 choosing activity 7 as they have the highest value of $\frac{(O-E)^2}{E}$ | B1 | Correctly identifies strongest source of association as group with highest $\frac{(O-E)^2}{E}$; condone "the highest value" |
**Subtotal: 1 mark**
**Question 5 total: 8 marks**
---5 A school management team oversees 11 different schools.\\
The school management team allows each student in the schools to choose one enrichment activity from 11 possible activities.
The school management team count the number of students in each school choosing each of the possible activities. They then conduct a $\chi ^ { 2 }$-test for association with the data they have gathered.
5
\begin{enumerate}[label=(\alph*)]
\item Exactly one of the calculated expected frequencies for the $\chi ^ { 2 }$-test is less than 5\\
Explain why the number of degrees of freedom for the test is 90\\
5
\item The school management team claims that there is an association between the school a student attends and the activity they choose.
The test statistic is 124.8
Test the claim using the $1 \%$ level of significance.\\
5
\item During the hypothesis test, the value of $\frac { ( O - E ) ^ { 2 } } { E }$, where $O$ is the observed frequency and $E$ is the expected frequency, was calculated for each group of students.
The values for four groups of students are shown in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Group & $\frac { ( O - E ) ^ { 2 } } { E }$ \\
\hline
Attends school 3 and chose activity 1 & 0.01 \\
\hline
Attends school 8 and chose activity 3 & 18.5 \\
\hline
Attends school 8 and chose activity 7 & 24.2 \\
\hline
Attends school 11 and chose activity 7 & 49.0 \\
\hline
\end{tabular}
\end{center}
State, with a reason, which of the four groups of students represents the strongest source of association.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2023 Q5 [8]}}