| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly presented data in a 2×3 contingency table. Students must calculate expected frequencies using row/column totals, compute the test statistic, and compare to critical values—all routine S3 procedures requiring no novel insight, though slightly easier than average due to straightforward setup and 10% significance level. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 5 } \multicolumn{2}{c|}{} | Inspirational message | \multirow{2}{*}{Total} | |||||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | \(\boldsymbol { A }\) | \(\boldsymbol { B }\) | \(\boldsymbol { C }\) | ||||
\multirow{2}{*}{
| Male | 25 | 37 | 45 | 107 | ||
| \cline { 2 - 6 } | Female | 32 | 50 | 36 | 118 | ||
| Total | 57 | 87 | 81 | 225 | |||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between gender and (inspirational) message (independent). \(H_1\): There is an association between gender and (inspirational) message (dependent). | B1 | Must mention "gender" and "message" at least once. Use of "relationship"/"correlation"/"connection"/"link" is B0. |
| Expected values: Male: \(27.106..., 41.373..., 38.52\); Female: \(29.893..., 45.626..., 42.48\) using \(\dfrac{\text{(Row Total)(Column Total)}}{\text{(Grand Total)}}\) | M1, A1 | M1: Some attempt at formula. A1: At least 5 expected frequencies correct awrt/trunc. 2 d.p. No fractions. |
| At least 2 correct terms for \(\dfrac{(O-E)^2}{E}\) or \(\dfrac{O^2}{E}\) with their \(E_i\) | dM1 | Dependent on 1st M1. Accept 2 sf accuracy. |
| At least 5 correct \(\dfrac{(O-E)^2}{E}\) or \(\dfrac{O^2}{E}\) terms to 1 d.p. or better | A1 | Allow truncation |
| \(X^2 = \sum\dfrac{(O-E)^2}{E}\) or \(\sum\dfrac{O^2}{E} - 225\); awrt \(\mathbf{3.27}\) | dM1, A1 | Dependent on 2nd M1 |
| \(v = (2-1)(3-1) = 2\) | B1 | Can be implied by correct critical value of \(4.605\) |
| \(\chi^2_2(0.10) = 4.605 \Rightarrow\) CR: \(X^2 \geqslant 4.605\) | B1ft | ft their \(v\) |
| Does not lie in CR / not significant / do not reject \(H_0\) | — | |
| There is insufficient evidence to support the headteacher's belief / there is no association between gender and inspirational message | A1 | Dependent on 3rd M1 and \(4.605\). Must mention "headteacher's belief" or "gender" and "message". Contradictory statements score A0. |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between gender and (inspirational) message (independent). $H_1$: There is an association between gender and (inspirational) message (dependent). | B1 | Must mention "gender" **and** "message" at least once. Use of "relationship"/"correlation"/"connection"/"link" is B0. |
| Expected values: Male: $27.106..., 41.373..., 38.52$; Female: $29.893..., 45.626..., 42.48$ using $\dfrac{\text{(Row Total)(Column Total)}}{\text{(Grand Total)}}$ | M1, A1 | M1: Some attempt at formula. A1: At least 5 expected frequencies correct awrt/trunc. 2 d.p. No fractions. |
| At least 2 correct terms for $\dfrac{(O-E)^2}{E}$ or $\dfrac{O^2}{E}$ with their $E_i$ | dM1 | Dependent on 1st M1. Accept 2 sf accuracy. |
| At least 5 correct $\dfrac{(O-E)^2}{E}$ or $\dfrac{O^2}{E}$ terms to 1 d.p. or better | A1 | Allow truncation |
| $X^2 = \sum\dfrac{(O-E)^2}{E}$ or $\sum\dfrac{O^2}{E} - 225$; awrt $\mathbf{3.27}$ | dM1, A1 | Dependent on 2nd M1 |
| $v = (2-1)(3-1) = 2$ | B1 | Can be implied by correct critical value of $4.605$ |
| $\chi^2_2(0.10) = 4.605 \Rightarrow$ CR: $X^2 \geqslant 4.605$ | B1ft | ft their $v$ |
| Does not lie in CR / not significant / do not reject $H_0$ | — | |
| There is **insufficient evidence** to support the headteacher's belief / there is **no association** between gender and inspirational message | A1 | Dependent on 3rd M1 and $4.605$. Must mention "headteacher's belief" or "gender" **and** "message". Contradictory statements score A0. |
**[10]**2. A school uses online report cards to promote both hard work and good behaviour of its pupils. Each card details a pupil's recent achievement and contains exactly one of three inspirational messages $A , B$ or $C$, chosen by the pupil's teacher.
The headteacher believes that there is an association between the pupil's gender and the inspirational message chosen. He takes a random sample of 225 pupils and examines the card for each pupil. His results are shown in Table 1.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | l | c | c | c | c | }
\cline { 2 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{c|}{Inspirational message} & \multirow{2}{*}{Total} \\
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & $\boldsymbol { A }$ & $\boldsymbol { B }$ & $\boldsymbol { C }$ & \\
\hline
\multirow{2}{*}{\begin{tabular}{ l }
Pupil's \\
gender \\
\end{tabular}} & Male & 25 & 37 & 45 & 107 \\
\cline { 2 - 6 }
& Female & 32 & 50 & 36 & 118 \\
\hline
\multicolumn{2}{|c|}{Total} & 57 & 87 & 81 & 225 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
Stating your hypotheses clearly, test, at the $10 \%$ level of significance, whether or not there is evidence to support the headteacher's belief. Show your working clearly. You should state your expected frequencies correct to 2 decimal places.
\hfill \mbox{\textit{Edexcel S3 2017 Q2 [10]}}