| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Find minimum sample size |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: calculating sample size from a given z-statistic (reverse application of the z-formula), performing a one-tailed test with given significance level, and explaining CLT application. All steps are routine A-level statistics techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume standard deviation for the region is \(7.1\) | B1 | Or standard deviation is same as for whole population OE |
| \(\frac{63.2 - 65.2}{\frac{7.1}{\sqrt{n}}} = -2.182\) | M1 | Attempt to find correct equation (accept \(+2.182\)) |
| \(n = \{-2.182 \times 7.1 \div (-2)\}^2\) | A1 | Any correct expression for \(n\) or \(\sqrt{n}\). SOI |
| \(n = 60\) | A1 | CWO. Must be an integer |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): population mean (or \(\mu\)) \(= 65.2\) ; \(H_1\): population mean (or \(\mu\)) \(< 65.2\) | B1 | Not just 'mean' |
| \(2.182 > 1.751\) | M1 | Or valid area comparison |
| There is evidence that animals are shorter in this region | A1 | CWO. No contradictions |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Population unknown or population not given as normal | B1 | Allow population not normal. Accept distribution of \(X\) unknown |
| Total: 1 |
## Question 2:
**Part (i)(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume standard deviation for the region is $7.1$ | B1 | Or standard deviation is same as for whole population OE |
| $\frac{63.2 - 65.2}{\frac{7.1}{\sqrt{n}}} = -2.182$ | M1 | Attempt to find correct equation (accept $+2.182$) |
| $n = \{-2.182 \times 7.1 \div (-2)\}^2$ | A1 | Any correct expression for $n$ or $\sqrt{n}$. SOI |
| $n = 60$ | A1 | CWO. Must be an integer |
| | **Total: 4** | |
**Part (i)(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: population mean (or $\mu$) $= 65.2$ ; $H_1$: population mean (or $\mu$) $< 65.2$ | B1 | Not just 'mean' |
| $2.182 > 1.751$ | M1 | Or valid area comparison |
| There is evidence that animals are shorter in this region | A1 | CWO. No contradictions |
| | **Total: 3** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Population unknown or population not given as normal | B1 | Allow population not normal. Accept distribution of $X$ unknown |
| | **Total: 1** | |2 The heights of a certain species of animal have been found to have mean 65.2 cm and standard deviation 7.1 cm . A researcher suspects that animals of this species in a certain region are shorter on average than elsewhere. She takes a large random sample of $n$ animals of this species from this region and finds that their mean height is 63.2 cm . She then carries out an appropriate hypothesis test.\\
(i) She finds that the value of the test statistic $z$ is - 2.182 , correct to 3 decimal places.
\begin{enumerate}[label=(\alph*)]
\item Stating a necessary assumption, calculate the value of $n$.
\item Carry out the hypothesis test at the $4 \%$ significance level.\\
(ii) Explain why it was necessary to use the Central Limit theorem in carrying out the test.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2019 Q2 [8]}}