| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Larger contingency table (4+ categories) |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 2×4 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 with multiple cells, it follows a completely routine procedure taught in Further Statistics with no conceptual challenges or novel elements. |
| Spec | 5.06a Chi-squared: contingency tables |
| Instructor \(A\) | Instructor \(B\) | Instructor \(C\) | Instructor \(D\) | Total | |
| Pass | 72 | 42 | 52 | 68 | 234 |
| Fail | 33 | 34 | 41 | 58 | 166 |
| Total | 105 | 76 | 93 | 126 | 400 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected values: 72(61.425), 42(44.46), 52(54.405), 68(73.71), 33(43.575), 34(31.54), 41(38.595), 58(52.29) | M1 | At least 3 correct to 4sf |
| Calculates E values (in brackets above) | A1 | All correct to 4sf (if 6.037 not found) |
| Test value: \(\sum\frac{(O-E)^2}{E} = 1.8206 + 0.1361 + 0.1063 + 0.4423 + 2.5664 + 0.1919 + 0.1499 + 0.6235\) | M1 | Correct formula used |
| \(= 6.037\) | A1 | Correct to 3 sf |
| \(H_0\): (driving) test success is independent of instructor | B1 | |
| Tabular value: 3 df \(10\% = 6.251\); \(6.04 < 6.251\); Accept \(H_0\) | M1 | Compare with correct tabular value and conclusion |
| Sufficient evidence to suggest that (driving) test success is independent of instructor | A1 | Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow \(\pm 1\) in the third significant figure of test value |
| Total | 7 |
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected values: 72(61.425), 42(44.46), 52(54.405), 68(73.71), 33(43.575), 34(31.54), 41(38.595), 58(52.29) | M1 | At least 3 correct to 4sf |
| Calculates E values (in brackets above) | A1 | All correct to 4sf (if 6.037 not found) |
| Test value: $\sum\frac{(O-E)^2}{E} = 1.8206 + 0.1361 + 0.1063 + 0.4423 + 2.5664 + 0.1919 + 0.1499 + 0.6235$ | M1 | Correct formula used |
| $= 6.037$ | A1 | Correct to 3 sf |
| $H_0$: (driving) test success is independent of instructor | B1 | |
| Tabular value: 3 df $10\% = 6.251$; $6.04 < 6.251$; Accept $H_0$ | M1 | Compare with correct tabular value and conclusion |
| Sufficient evidence to suggest that (driving) test success is independent of instructor | A1 | Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow $\pm 1$ in the third significant figure of test value |
| **Total** | **7** | |
---2 A driving school employs four instructors to prepare people for their driving test. The allocation of people to instructors is random. For each of the instructors, the following table gives the number of people who passed and the number who failed their driving test last year.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
& Instructor $A$ & Instructor $B$ & Instructor $C$ & Instructor $D$ & Total \\
\hline
Pass & 72 & 42 & 52 & 68 & 234 \\
\hline
Fail & 33 & 34 & 41 & 58 & 166 \\
\hline
Total & 105 & 76 & 93 & 126 & 400 \\
\hline
\end{tabular}
\end{center}
Test at the 10\% significance level whether success in the driving test is independent of the instructor.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q2 [7]}}