| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State probability of Type I error |
| Difficulty | Standard +0.8 This question requires understanding of Type I/II errors in hypothesis testing with normal distributions, including calculating Type II error probability which involves finding the critical region, standardizing under the alternative hypothesis, and using the normal distribution. While conceptually demanding for S2 level, the calculations are relatively straightforward with a large sample size making the CLT application clear. |
| 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 | Mark | Guidance |
| \(0.01\) or \(1\%\) | B1 | Note: \(x \leqslant 0.01\) scores B0. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2.326 = \dfrac{\bar{h}-163.21}{6.95 \div \sqrt{300}}\) | M1 | Accept any \(z\) (\(\pm\)). |
| \(\bar{h} = 164.14\). Rejection region is \(\bar{h} > 164.14\). \([\text{P(Type II)} = \text{P}(\bar{h} < 164.14 \mid \mu = 164.91)]\) | A1 | Accept 3 sf accuracy here. |
| \(\dfrac{\textit{their } '164.14' - 164.91}{6.95 \div \sqrt{300}}\ [= -1.919]\) | M1 | For standardising \(164.91\) with their \(164.14\) (could be \(163.21\)). |
| \(\Phi(\textit{their}\ '-1.919') = 1 - \Phi(\textit{their}\ '1.919')\) | M1 | For area attempt consistent with their values. |
| \(= 0.0275\) or \(0.0276\) or \(0.028[0]\) (3 s.f.) | A1 | Accept anything in range \(0.0275\) to \(0.028[0]\). |
| Total: 5 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.01$ or $1\%$ | B1 | Note: $x \leqslant 0.01$ scores B0. |
| **Total: 1** | | |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2.326 = \dfrac{\bar{h}-163.21}{6.95 \div \sqrt{300}}$ | M1 | Accept any $z$ ($\pm$). |
| $\bar{h} = 164.14$. Rejection region is $\bar{h} > 164.14$. $[\text{P(Type II)} = \text{P}(\bar{h} < 164.14 \mid \mu = 164.91)]$ | A1 | Accept 3 sf accuracy here. |
| $\dfrac{\textit{their } '164.14' - 164.91}{6.95 \div \sqrt{300}}\ [= -1.919]$ | M1 | For standardising $164.91$ with their $164.14$ (could be $163.21$). |
| $\Phi(\textit{their}\ '-1.919') = 1 - \Phi(\textit{their}\ '1.919')$ | M1 | For area attempt consistent with their values. |
| $= 0.0275$ or $0.0276$ or $0.028[0]$ (3 s.f.) | A1 | Accept anything in range $0.0275$ to $0.028[0]$. |
| **Total: 5** | | |7 The heights, in centimetres, of adult females in Litania have mean $\mu$ and standard deviation $\sigma$. It is known that in 2004 the values of $\mu$ and $\sigma$ were 163.21 and 6.95 respectively. The government claims that the value of $\mu$ this year is greater than it was in 2004. In order to test this claim a researcher plans to carry out a hypothesis test at the $1 \%$ significance level. He records the heights of a random sample of 300 adult females in Litania this year and finds the value of the sample mean.
\begin{enumerate}[label=(\alph*)]
\item State the probability of a Type I error.\\
\includegraphics[max width=\textwidth, alt={}]{ff3433b0-baab-45e3-845e-56a794739bba-12_74_1577_557_322} ........................................................................................................................................
You should assume that the value of $\sigma$ after 2004 remains at 6.95 .
\item Given that the value of $\mu$ this year is actually 164.91 , find the probability of a Type II error.\\
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 [6]}}