| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a standard paired t-test with clearly defined hypotheses (one-tailed test for mean difference > 0.4). Students must calculate differences, find mean and standard deviation, compute the test statistic, and compare to critical value. While it requires careful arithmetic and understanding of paired tests, it follows a routine procedure with no conceptual surprises. The 10% significance level and small sample size are standard features. Part (b) is straightforward recall (normality assumption). Slightly above average difficulty due to the non-zero hypothesized difference and being Further Maths content, but still a textbook exercise. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.07d Paired vs two-sample: selection |
| Employee | \(A\) | \(B\) | \(C\) | D | \(E\) | \(F\) | G | \(H\) | I | J |
| Time before new technology | 10.2 | 9.8 | 12.4 | 11.6 | 10.8 | 11.2 | 14.6 | 10.6 | 12.3 | 11.0 |
| Time after new technology | 9.6 | 8.5 | 12.4 | 10.9 | 10.2 | 10.6 | 12.8 | 10.8 | 12.5 | 10.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differences: \(0.6 \quad 1.3 \quad 0.0 \quad 0.7 \quad 0.6 \quad 0.6 \quad 1.8 \quad -0.2 \quad -0.2 \quad 0.4\) | M1 | Attempt at differences, allow one error |
| \(\sum d = 5.6 \quad \sum d^2 = 6.74\) <br> \(s^2 = \frac{1}{9}\left(6.74 - \frac{5.6^2}{10}\right) = 0.400(4)\) | M1 | |
| \(H_0: \mu_B - \mu_A = 0.4\) and \(H_1: \mu_B - \mu_A > 0.4\) | B1 | Allow use of \(\mu_d\) |
| \(t = \dfrac{0.56 - 0.4}{\sqrt{\dfrac{0.4004}{10}}} \quad (0.800)\) | M1 | |
| \(t = 0.7996 \quad (0.800)\) | A1 | |
| Critical value \(= t_{0.90}(9) = 1.383\) <br> Compare: \(0.7996 < 1.383\) <br> Accept \(H_0\) | M1 | Compare calculated value with \(1.383\) and correct FT conclusion |
| Insufficient evidence to support manager's claim. | A1 | Correct conclusion in context. Level of uncertainty in language is used. |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distribution of population differences is normal | B1 | |
| 1 |
## Question 1:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differences: $0.6 \quad 1.3 \quad 0.0 \quad 0.7 \quad 0.6 \quad 0.6 \quad 1.8 \quad -0.2 \quad -0.2 \quad 0.4$ | M1 | Attempt at differences, allow one error |
| $\sum d = 5.6 \quad \sum d^2 = 6.74$ <br> $s^2 = \frac{1}{9}\left(6.74 - \frac{5.6^2}{10}\right) = 0.400(4)$ | M1 | |
| $H_0: \mu_B - \mu_A = 0.4$ and $H_1: \mu_B - \mu_A > 0.4$ | B1 | Allow use of $\mu_d$ |
| $t = \dfrac{0.56 - 0.4}{\sqrt{\dfrac{0.4004}{10}}} \quad (0.800)$ | M1 | |
| $t = 0.7996 \quad (0.800)$ | A1 | |
| Critical value $= t_{0.90}(9) = 1.383$ <br> Compare: $0.7996 < 1.383$ <br> Accept $H_0$ | M1 | Compare calculated value with $1.383$ and correct FT conclusion |
| Insufficient evidence to support manager's claim. | A1 | Correct conclusion in context. Level of uncertainty in language is used. |
| | **7** | |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distribution of **population** differences is normal | B1 | |
| | **1** | |1 A manager is investigating the times taken by employees to complete a particular task as a result of the introduction of new technology. He claims that the mean time taken to complete the task is reduced by more than 0.4 minutes. He chooses a random sample of 10 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Employee & $A$ & $B$ & $C$ & D & $E$ & $F$ & G & $H$ & I & J \\
\hline
Time before new technology & 10.2 & 9.8 & 12.4 & 11.6 & 10.8 & 11.2 & 14.6 & 10.6 & 12.3 & 11.0 \\
\hline
Time after new technology & 9.6 & 8.5 & 12.4 & 10.9 & 10.2 & 10.6 & 12.8 & 10.8 & 12.5 & 10.6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Test at the 10\% significance level whether the manager's claim is justified.
\item State an assumption that is necessary for this test to be valid.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q1 [8]}}