| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Find critical region for test |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: stating hypotheses, finding a critical region using normal distribution tables, and understanding when CLT applies. The calculations are routine (finding critical value from z-table and converting to sample mean scale), and part (c) tests basic conceptual understanding that CLT isn't needed when the population is already normal. Slightly easier than average due to being a standard textbook-style question with no novel problem-solving required. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): population mean [of \(H\)] \(= 4.23\); \(H_1\): population mean [of \(H\)] \(> 4.23\) | B1 | Allow \(\mu = 4.23\) or population mean of \(h = 4.23\) but NOT \(h = 4.23\) or \(H = 4.23\) or \(\bar{h} = 4.23\) or \(\bar{H} = 4.23\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{\bar{h} - 4.23}{\dfrac{0.67}{\sqrt{200}}} = 1.645\) | M1 | For standardising and forming an equation. Must have \(\sqrt{200}\). Allow \(\pm 1.645\) or \(\pm 1.96\). Accept \(>\) and \(<\). Allow \(\bar{H}\) or any letter instead of \(\bar{h}\). |
| \(\bar{h} = 4.31\) (3 sf) | A1 | May be implied by \(\bar{h} > 4.31\). Allow \(\bar{h} < 4.31\) for this A1 only, condone 4.15 also seen. |
| \(\bar{h} > 4.31\) or \(\bar{h} \geqslant 4.31\) (3 sf) | A1 | Condone any letter instead of \(\bar{h}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Incorrect, because the population of \(H\) is given as normally distributed [with known variance]. | B1 | Allow \(h\) instead of \(H\) or just 'The population is normal.' Must use 'population' or 'underlying distribution'. |
## Question 4(a):
$H_0$: population mean [of $H$] $= 4.23$; $H_1$: population mean [of $H$] $> 4.23$ | B1 | Allow $\mu = 4.23$ or population mean of $h = 4.23$ but NOT $h = 4.23$ or $H = 4.23$ or $\bar{h} = 4.23$ or $\bar{H} = 4.23$.
---
## Question 4(b):
$\dfrac{\bar{h} - 4.23}{\dfrac{0.67}{\sqrt{200}}} = 1.645$ | M1 | For standardising and forming an equation. Must have $\sqrt{200}$. Allow $\pm 1.645$ or $\pm 1.96$. Accept $>$ and $<$. Allow $\bar{H}$ or any letter instead of $\bar{h}$.
$\bar{h} = 4.31$ (3 sf) | A1 | May be implied by $\bar{h} > 4.31$. Allow $\bar{h} < 4.31$ for this A1 only, condone 4.15 also seen.
$\bar{h} > 4.31$ or $\bar{h} \geqslant 4.31$ (3 sf) | A1 | Condone any letter instead of $\bar{h}$.
---
## Question 4(c):
Incorrect, because the population of $H$ is given as normally distributed [with known variance]. | B1 | Allow $h$ instead of $H$ or just 'The population is normal.' Must use 'population' or 'underlying distribution'.
---4 The height $H$, in metres, of mature trees of a certain variety is normally distributed with standard deviation 0.67. In order to test whether the population mean of $H$ is greater than 4.23, the heights of a random sample of 200 trees are measured.
\begin{enumerate}[label=(\alph*)]
\item Write down suitable null and alternative hypotheses for the test.\\
The sample mean height, $\bar { h }$ metres, of the 200 trees is found and the test is carried out. The result of the test is to reject the null hypothesis at the 5\% significance level.
\item Find the set of possible values of $\bar { h }$.
\item Ajit said, 'In (b) we had to assume that $\bar { H }$ is normally distributed, so it was necessary to use the Central Limit Theorem.'
Explain whether you agree with Ajit.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q4 [5]}}