| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| 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 a 2×3 contingency table. Students need to calculate expected frequencies, compute the test statistic, find critical value from tables, and state a conclusion. While it requires careful arithmetic and proper hypothesis testing procedure, it follows a completely routine template with no conceptual challenges or novel elements beyond textbook application. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Standard achieved | ||
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Above average | Average | Below average |
| Scheme \(A\) | 31 | 35 | 22 |
| Scheme \(B\) | 19 | 50 | 43 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected values: \(31\ E=22\), \(35\ E=37.4\), \(22\ E=28.6\), \(19\ E=28\), \(50\ E=47.6\), \(43\ E=36.4\) | M1A1 | |
| \(\frac{(31-22)^2}{22}+\frac{(35-37.4)^2}{37.4}+\frac{(22-28.6)^2}{28.6}+\frac{(19-28)^2}{28}+\frac{(50-47.6)^2}{47.6}+\frac{(43-36.4)^2}{36.4}\) | M1 | |
| \(9.569\ (9.57)\) | A1 | |
| Tabular value \(= 5.991\) and comparing \(9.569 > 5.991\) | M1 | |
| So standard achieved is dependent on reading scheme used | A1 | |
| Total | 6 |
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected values: $31\ E=22$, $35\ E=37.4$, $22\ E=28.6$, $19\ E=28$, $50\ E=47.6$, $43\ E=36.4$ | M1A1 | |
| $\frac{(31-22)^2}{22}+\frac{(35-37.4)^2}{37.4}+\frac{(22-28.6)^2}{28.6}+\frac{(19-28)^2}{28}+\frac{(50-47.6)^2}{47.6}+\frac{(43-36.4)^2}{36.4}$ | M1 | |
| $9.569\ (9.57)$ | A1 | |
| Tabular value $= 5.991$ and comparing $9.569 > 5.991$ | M1 | |
| So standard achieved is dependent on reading scheme used | A1 | |
| **Total** | **6** | |
---1 Young children are learning to read using two different reading schemes, $A$ and $B$. The standards achieved are measured against the national average standard achieved and classified as above average, average or below average. For two randomly chosen groups of young children, the numbers in each category are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & \multicolumn{3}{|c|}{Standard achieved} \\
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Above average & Average & Below average \\
\hline
Scheme $A$ & 31 & 35 & 22 \\
\hline
Scheme $B$ & 19 & 50 & 43 \\
\hline
\end{tabular}
\end{center}
Test at the $5 \%$ significance level whether standard achieved is independent of the reading scheme used.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q1 [6]}}