| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test of independence with a 2×2 contingency table. Students need to calculate expected frequencies, compute the test statistic using the standard formula, and compare to critical values. While it requires careful calculation, it's a routine application of a standard technique with no conceptual complications or novel insights required. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | Direction of journey | |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | To school | From school |
| Delayed | 64 | 56 |
| Not delayed | 74 | 106 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): no association between delay and direction of journey. \(H_1\): there is association | B1 | Or: delay and direction of journey are independent, etc. Context needed. Ignore any statements involving parameters (e.g. \(\rho\)) |
| \(E_f\): \(55.2,\ 64.8,\ 82.8,\ 97.2\) | B1 | At least 2 correct to 2 SF. Can be implied by correct \(X^2\) |
| \(X^2 = 1.248 + 1.063 + 0.832 + 0.709\) | M1 | M0 if formula stated incorrectly. Allow from no Yates (4.33). E.g. \(5.045\) from no modulus: M0. \(1.403 + 1.195 + 0.935 + 0.797 = 4.33\): M1A0 |
| \(= 3.85\) | A1 | No Yates: loses this mark only |
| \(> 2.706\) | A1 | Explicit comparison with \(2.706\) |
| Reject \(H_0\). Significant evidence of association between delays and direction of journey. | M1ft, A1ft [7] | Consistent, ft on their TS. Context needed, not over-assertive. M1 and A1 need comparison with \(2.7(06)\) or \(3.84(1)\) |
# Question 4:
$H_0$: no association between delay and direction of journey. $H_1$: there is association | **B1** | Or: delay and direction of journey are independent, etc. Context needed. Ignore any statements involving parameters (e.g. $\rho$)
$E_f$: $55.2,\ 64.8,\ 82.8,\ 97.2$ | **B1** | At least 2 correct to 2 SF. Can be implied by correct $X^2$
$X^2 = 1.248 + 1.063 + 0.832 + 0.709$ | **M1** | M0 if formula stated incorrectly. Allow from no Yates (4.33). E.g. $5.045$ from no modulus: M0. $1.403 + 1.195 + 0.935 + 0.797 = 4.33$: M1A0
$= 3.85$ | **A1** | No Yates: loses this mark only
$> 2.706$ | **A1** | Explicit comparison with $2.706$
Reject $H_0$. Significant evidence of association between delays and direction of journey. | **M1ft, A1ft [7]** | Consistent, ft on their TS. Context needed, not over-assertive. M1 and A1 need comparison with $2.7(06)$ or $3.84(1)$
---4 A school pupil keeps a note of whether her journeys to school and from school are delayed. The results for a random sample of journeys are shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{Direction of journey} \\
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & To school & From school \\
\hline
Delayed & 64 & 56 \\
\hline
Not delayed & 74 & 106 \\
\hline
\end{tabular}
\end{center}
Test at the 10\% significance level whether there is association between delays and the direction of the journey.
\hfill \mbox{\textit{OCR Further Statistics AS 2022 Q4 [7]}}