| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question requiring standard procedures: identifying test type (two-tailed because testing 'different from'), calculating sample mean (597.1/50), computing z-statistic using known formula, and comparing to critical value at 1% level. All steps are routine applications of memorized procedures with no conceptual challenges or novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Two-tailed because looking for difference | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 12.7 \quad H_1: \mu \neq 12.7\) | B1 | No ft from part (a) |
| \(\dfrac{\frac{597.1}{50} - 12.7}{\frac{2.3}{\sqrt{50}}}\) | M1 | |
| \(= -2.330\); Accept \(-2.336\) or \(-2.337\) | A1 | or 0.00989 or 0.0099, or 0.0097 if area comparison used |
| \(\text{'-2.330'} > -2.576\) or \(\text{'2.330'} < 2.576\) | M1 | Accept 2.574 to 2.579. Or use of CV: \(12.7 - 2.576 \times (2.3/\sqrt{50}) = 11.862\) M1A1. \(11.942 > 11.862\) M1A1 |
| [Not reject \(H_0\)] There is insufficient evidence to suggest that \(\mu\) is not 12.7 | A1 FT | OE ft their \(z_\text{calc}\). In context, not definite. No contradictions. SC use of 1 tailed test can score B0M1A1M1 for comparison with 0.01 A0 max 3/5 |
| Total: 5 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Two-tailed because looking for difference | B1 | |
| **Total: 1** | | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 12.7 \quad H_1: \mu \neq 12.7$ | B1 | No ft from part (a) |
| $\dfrac{\frac{597.1}{50} - 12.7}{\frac{2.3}{\sqrt{50}}}$ | M1 | |
| $= -2.330$; Accept $-2.336$ or $-2.337$ | A1 | or 0.00989 or 0.0099, or 0.0097 if area comparison used |
| $\text{'-2.330'} > -2.576$ or $\text{'2.330'} < 2.576$ | M1 | Accept 2.574 to 2.579. Or use of CV: $12.7 - 2.576 \times (2.3/\sqrt{50}) = 11.862$ **M1A1**. $11.942 > 11.862$ **M1A1** |
| [Not reject $H_0$] There is insufficient evidence to suggest that $\mu$ is not 12.7 | A1 FT | OE ft their $z_\text{calc}$. In context, not definite. No contradictions. SC use of 1 tailed test can score **B0M1A1M1** for comparison with 0.01 **A0** max 3/5 |
| **Total: 5** | | |5 The lengths, in centimetres, of worms of a certain kind are normally distributed with mean $\mu$ and standard deviation 2.3. An article in a magazine states that the value of $\mu$ is 12.7 . A scientist wishes to test whether this value is correct. He measures the lengths, $x \mathrm {~cm}$, of a random sample of 50 worms of this kind and finds that $\sum x = 597.1$. He plans to carry out a test, at the $1 \%$ significance level, of whether the true value of $\mu$ is different from 12.7 .
\begin{enumerate}[label=(\alph*)]
\item State, with a reason, whether he should use a one-tailed or a two-tailed test.
\item Carry out the test.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [6]}}