| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Hypothesis test then Type II error probability |
| Difficulty | Standard +0.3 This is a standard two-part hypothesis testing question requiring (a) a routine one-tailed z-test with known variance and (b) calculation of Type II error probability given a specific alternative mean. Both parts follow textbook procedures with straightforward calculations, making it slightly easier than average but requiring understanding of error types. |
| 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 mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 64.6\); \(H_1: \mu < 64.6\) | B1 | Allow population mean, not just 'mean'. |
| \([\pm]\frac{63.5 - 64.6}{5.2 \div \sqrt{100}}\) | M1 | Standardising. Must have \(\sqrt{100}\). |
| \([\pm]-2.115\) | A1 | Accept \(-2.12\) (3sf) |
| \(\text{'}-2.115\text{'} > 1.96\) or \(\text{'}-2.115\text{'} < -1.96\) [do not accept \(H_0\)] | M1 | Valid comparison (\(0.0172 < 0.025\) for area comparison). |
| There is evidence that \(\mu < 64.6\) | A1FT | Not definite, e.g. not '\(\mu < 64.6\)', in context. No contradictions. Accept critical value method leading to \(63.5 < 63.58\) or \(64.6 > 64.52\). |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{m - 64.6}{5.2 \div \sqrt{100}} = -1.96\) | M1 | Finding the critical value using \(N\!\left(64.6,\, \frac{5.2}{\sqrt{100}}\right)\) and a \(z\) value. |
| \(m = 63.5808\) | A1 | |
| \(\frac{63.5808 - 62.7}{5.2 \div \sqrt{100}} [= 1.694]\) | M1 | Standardising using \(N\!\left(62.7,\, \frac{5.2}{\sqrt{100}}\right)\) and a critical value. |
| \(1 - \Phi(\text{'1.694'})\) | M1 | For area consistent with *their* values. |
| \(0.0451\) | A1 | Accept answers that round to \(0.045\). |
| 5 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 64.6$; $H_1: \mu < 64.6$ | B1 | Allow population mean, not just 'mean'. |
| $[\pm]\frac{63.5 - 64.6}{5.2 \div \sqrt{100}}$ | M1 | Standardising. Must have $\sqrt{100}$. |
| $[\pm]-2.115$ | A1 | Accept $-2.12$ (3sf) |
| $\text{'}-2.115\text{'} > 1.96$ or $\text{'}-2.115\text{'} < -1.96$ [do not accept $H_0$] | M1 | Valid comparison ($0.0172 < 0.025$ for area comparison). |
| There is evidence that $\mu < 64.6$ | A1FT | Not definite, e.g. not '$\mu < 64.6$', in context. No contradictions. Accept critical value method leading to $63.5 < 63.58$ or $64.6 > 64.52$. |
| | **5** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{m - 64.6}{5.2 \div \sqrt{100}} = -1.96$ | M1 | Finding the critical value using $N\!\left(64.6,\, \frac{5.2}{\sqrt{100}}\right)$ and a $z$ value. |
| $m = 63.5808$ | A1 | |
| $\frac{63.5808 - 62.7}{5.2 \div \sqrt{100}} [= 1.694]$ | M1 | Standardising using $N\!\left(62.7,\, \frac{5.2}{\sqrt{100}}\right)$ and a critical value. |
| $1 - \Phi(\text{'1.694'})$ | M1 | For area consistent with *their* values. |
| $0.0451$ | A1 | Accept answers that round to $0.045$. |
| | **5** | |7 The masses, in grams, of apples from a certain farm have mean $\mu$ and standard deviation 5.2. The farmer says that the value of $\mu$ is 64.6. A quality control inspector claims that the value of $\mu$ is actually less than 64.6. In order to test his claim he chooses a random sample of 100 apples from the farm.
\begin{enumerate}[label=(\alph*)]
\item The mean mass of the 100 apples is found to be 63.5 g .
Carry out the test at the $2.5 \%$ significance level.
\item Later another test of the same hypotheses at the $2.5 \%$ significance level, with another random sample of 100 apples from the same farm, is carried out.
Given that the value of $\mu$ is in fact 62.7 , calculate the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q7 [10]}}