| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State probability of Type I error |
| Difficulty | Moderate -0.3 This is a straightforward hypothesis testing question covering standard definitions and procedures. Part (a) requires recalling that Type I error probability equals the significance level. Part (b) is a routine one-sample z-test with given summary statistics. Part (c) tests understanding that Type II errors only occur when Hâ‚€ is not rejected. All components are textbook applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.02\) or \(2\%\) | B1 | \(<0.02\) scores B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 2.3\), \(H_1: \mu > 2.3\) | B1 | Accept 'population mean' for \(\mu\) (not just mean). If not seen here, can be awarded if correctly seen in part (a) |
| \(s^2 = \frac{100}{99}\left(\frac{580}{100} - (2.38)^2\right)\) or \(\frac{1}{99}(580 - \frac{238^2}{100})\) | M1 | Correct substitution in \(s^2\) or \(\sqrt{s^2}\) formula |
| \(= 0.137 = 113/825\) or \(s = 0.370\) (3 sf) and \(\bar{x} = 238/100\ [=2.38]\) | A1 | \(\bar{x}\) and \(s^2\) (or \(s\)) correct. (SC biased estimate \(0.1356\) and \(\bar{x} = 2.38\) scores B1) |
| \(\frac{2.38 - 2.3}{\sqrt{\frac{0.137}{100}}}\ [= 2.161\ \text{or}\ 2.162]\) | M1 | |
| \(= 2.16\) (3 sf) OR \(0.0153/0.0154\) if area comparison used | A1 | |
| \(`2.16' > 2.054\) (or \(2.055\)) OR \(`0.0153\) or \(0.0154' < 0.02\) | M1 | Valid comparison |
| There is evidence to reject \(H_0\). There is sufficient evidence to suggest that the [mean] height in scientist's region is greater than \(2.3\) [m] OR there is sufficient evidence to suggest that the scientist's claim is justified. | A1FT | No contradictions. In context, non-definite. Accept CV method \(x = 2.376 < 2.38\) or \(x = 2.304 > 2.3\) M1 A1 for \(x\) and M1 A1ft for comparison and conclusion. Two tail test can score B0 M1 A1 M1 A1 M1 (comparison with \(0.01\)oe) A0ft max 5/7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not possible since \(H_0\) was rejected. | B1FT | Need both. Accept "No" as \(H_0\) was rejected. Follow through their conclusion in (b). Condone a definite statement. |
## Question 7:
### Part 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.02$ or $2\%$ | **B1** | $<0.02$ scores **B0** |
---
### Part 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 2.3$, $H_1: \mu > 2.3$ | **B1** | Accept 'population mean' for $\mu$ (not just mean). If not seen here, can be awarded if correctly seen in part (a) |
| $s^2 = \frac{100}{99}\left(\frac{580}{100} - (2.38)^2\right)$ or $\frac{1}{99}(580 - \frac{238^2}{100})$ | **M1** | Correct substitution in $s^2$ or $\sqrt{s^2}$ formula |
| $= 0.137 = 113/825$ or $s = 0.370$ (3 sf) and $\bar{x} = 238/100\ [=2.38]$ | **A1** | $\bar{x}$ and $s^2$ (or $s$) correct. (SC biased estimate $0.1356$ and $\bar{x} = 2.38$ scores **B1**) |
| $\frac{2.38 - 2.3}{\sqrt{\frac{0.137}{100}}}\ [= 2.161\ \text{or}\ 2.162]$ | **M1** | |
| $= 2.16$ (3 sf) OR $0.0153/0.0154$ if area comparison used | **A1** | |
| $`2.16' > 2.054$ (or $2.055$) OR $`0.0153$ or $0.0154' < 0.02$ | **M1** | Valid comparison |
| There is evidence to reject $H_0$. There is sufficient evidence to suggest that the [mean] height in scientist's region is greater than $2.3$ [m] OR there is sufficient evidence to suggest that the scientist's claim is justified. | **A1FT** | No contradictions. In context, non-definite. Accept CV method $x = 2.376 < 2.38$ or $x = 2.304 > 2.3$ **M1 A1** for $x$ and M1 A1ft for comparison and conclusion. Two tail test can score **B0 M1 A1 M1 A1 M1** (comparison with $0.01$oe) A0ft max 5/7 |
---
### Part 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not possible since $H_0$ was rejected. | **B1FT** | Need both. Accept "No" as $H_0$ was rejected. Follow through their conclusion in (b). Condone a definite statement. |7 The heights of one-year-old trees of a certain variety are known to have mean 2.3 m . A scientist believes that, on average, trees of this age and variety in her region are slightly taller than in other places. She plans to carry out a hypothesis test, at the $2 \%$ significance level, in order to test her belief.
\begin{enumerate}[label=(\alph*)]
\item State the probability that she will make a Type I error.\\
She takes a random sample of 100 such trees in her region and measures their heights, $h \mathrm {~m}$. Her results are summarised below.
$$n = 100 \quad \sum h = 238 \quad \sum h ^ { 2 } = 580$$
\item Carry out the test at the $2 \%$ significance level.\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac74d4c-f5e0-4c5d-ab25-5692dfb06f0b-10_2717_35_109_2012}
\item The scientist carries out the test correctly, but another scientist claims that she has made a Type II error.
Comment on this claim.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q7 [9]}}