| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Interpret association after test |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly presented data in a 2×4 contingency table. Students must calculate expected frequencies, compute the test statistic, compare to critical value, and interpret results. While it requires multiple steps and contextual interpretation in part (b), it follows a routine procedure taught explicitly in Further Statistics with no novel problem-solving required. Slightly easier than average due to straightforward setup and provided critical value tables. |
| Spec | 5.06a Chi-squared: contingency tables |
| Number of errors made per day | |||||
| 0 | 1 | 2 | 3 or more | Total | |
| \(A\) | 8 | 10 | 20 | 2 | 40 |
| B | 18 | 4 | 15 | 3 | 40 |
| Total | 26 | 14 | 35 | 5 | 80 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between employee and number of errors per day; \(H_1\): There is an association between employee and number of errors per day | B1 | States both hypotheses using correct language. Variables need to be stated in at least the null hypothesis |
| Expected contingency table: \(E=0\): \(A=13\), \(B=13\); \(E=1\): \(A=7\), \(B=7\); \(E=2\): \(A=17.5\), \(B=17.5\); \(E=3+\): \(A=2.5\), \(B=2.5\) | M1 | Translate situation into expected contingency table for \(\chi^2\) model |
| Combines columns for "2" and "3 or more" for both observed and expected values: \(2+\): \(O_A=22\), \(E_A=20\); \(O_B=18\), \(E_B=20\) | A1 | |
| \(\sum\dfrac{(O-E)^2}{E} = \dfrac{(8-13)^2}{13}+\dfrac{(18-13)^2}{13}+\dfrac{(10-7)^2}{7}+\dfrac{(4-7)^2}{7}+\dfrac{(22-20)^2}{20}+\dfrac{(18-20)^2}{20}\) | M1 | Uses \(\chi^2\) model to calculate test statistic |
| \(= 6.8\) | A1 | Obtains correct value of \(\sum\dfrac{(O-E)^2}{E}\), AWRT 6.8 |
| \(\chi^2\) cv for 2 df \(= 5.991\) | B1 | Obtains correct critical value for the test, AWRT 6.0; or corresponding probability of test statistic AWRT 0.03 |
| \(5.991 < 6.8\), Reject \(H_0\) | R1 | Evaluates \(\chi^2\)-test statistic by comparing the critical value with the test statistic or the probability with 0.05 |
| \(H_0\) rejected | E1F | Infers \(H_0\) rejected, FT 'their' comparison using the \(\chi^2\) model |
| Some evidence to suggest/support that there is an association between employee and number of errors per day | E1F | Concludes in context. The conclusion must not be definite. FT their incorrect acceptance of \(H_0\) if stated or 'their' comparison if not |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Largest sources of association: employee A/0 errors and employee B/0 errors; \(\dfrac{(O-E)^2}{E} = 1.9\ldots\) | E1 | Explains reasoning by considering \((O-E)\) or \(\dfrac{(O-E)^2}{E}\) to identify largest sources of association |
| Employee A makes no errors per day less often than expected | E1 | Interprets main source of association in context |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between employee and number of errors per day; $H_1$: There is an association between employee and number of errors per day | B1 | States both hypotheses using correct language. Variables need to be stated in at least the null hypothesis |
| Expected contingency table: $E=0$: $A=13$, $B=13$; $E=1$: $A=7$, $B=7$; $E=2$: $A=17.5$, $B=17.5$; $E=3+$: $A=2.5$, $B=2.5$ | M1 | Translate situation into expected contingency table for $\chi^2$ model |
| Combines columns for "2" and "3 or more" for both observed and expected values: $2+$: $O_A=22$, $E_A=20$; $O_B=18$, $E_B=20$ | A1 | |
| $\sum\dfrac{(O-E)^2}{E} = \dfrac{(8-13)^2}{13}+\dfrac{(18-13)^2}{13}+\dfrac{(10-7)^2}{7}+\dfrac{(4-7)^2}{7}+\dfrac{(22-20)^2}{20}+\dfrac{(18-20)^2}{20}$ | M1 | Uses $\chi^2$ model to calculate test statistic |
| $= 6.8$ | A1 | Obtains correct value of $\sum\dfrac{(O-E)^2}{E}$, AWRT 6.8 |
| $\chi^2$ cv for 2 df $= 5.991$ | B1 | Obtains correct critical value for the test, AWRT 6.0; or corresponding probability of test statistic AWRT 0.03 |
| $5.991 < 6.8$, Reject $H_0$ | R1 | Evaluates $\chi^2$-test statistic by comparing the critical value with the test statistic or the probability with 0.05 |
| $H_0$ rejected | E1F | Infers $H_0$ rejected, FT 'their' comparison using the $\chi^2$ model |
| Some evidence to suggest/support that there is an association between employee and number of errors per day | E1F | Concludes in context. The conclusion must not be definite. FT their incorrect acceptance of $H_0$ if stated or 'their' comparison if not |
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# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Largest sources of association: employee A/0 errors and employee B/0 errors; $\dfrac{(O-E)^2}{E} = 1.9\ldots$ | E1 | Explains reasoning by considering $(O-E)$ or $\dfrac{(O-E)^2}{E}$ to identify largest sources of association |
| Employee A makes no errors per day less often than expected | E1 | Interprets main source of association in context |7 Two employees, $A$ and $B$, both produce the same toy for a company.
The company records the total number of errors made per day by each employee during a 40-day period. The results are summarised in the following table.
Employee
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{Number of errors made per day} \\
\hline
& 0 & 1 & 2 & 3 or more & Total \\
\hline
$A$ & 8 & 10 & 20 & 2 & 40 \\
\hline
B & 18 & 4 & 15 & 3 & 40 \\
\hline
Total & 26 & 14 & 35 & 5 & 80 \\
\hline
\end{tabular}
\end{center}
The company claims that there is an association between employee and number of errors made per day.
7
\begin{enumerate}[label=(\alph*)]
\item Test the company's claim, using the $5 \%$ level of significance.\\
7
\item By considering observed and expected frequencies, interpret in context the association between employee and number of errors made per day.\\
\includegraphics[max width=\textwidth, alt={}, center]{9be40ed6-6df8-426a-8afd-fefc17287de6-12_2492_1723_217_150}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9be40ed6-6df8-426a-8afd-fefc17287de6-16_2496_1721_214_148}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2021 Q7 [11]}}