| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Find minimum sample size for Type II error constraint |
| Difficulty | Challenging +1.2 This is a Further Maths S4 question on hypothesis testing requiring understanding of Type I/II errors and power calculations. Part (a) is straightforward recall (Type I error = significance level). Part (b) requires setting up inequalities involving the critical region and alternative distribution, then solving for n using normal tables—a multi-step calculation but following a standard procedure taught in S4. While more advanced than core A-level, it's a routine application of taught techniques rather than requiring novel insight. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.05\) or \(5\%\) | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let CR be \(\bar{X} > k\); \(P\left(\bar{X} > k \,\Big | \, \bar{X} \text{ is } N\left(150, \frac{16}{n}\right)\right) = 0.05\) | |
| \(\frac{\bar{k}-150}{\frac{4}{\sqrt{n}}} = 1.6449\) | M1B1A1 | \(\frac{\bar{k}-150}{\frac{4}{\sqrt{n}}} = z\)-value, \( |
| \(\bar{k} = 150 + 1.6449 \times \frac{4}{\sqrt{n}}\) | ||
| \(\frac{\bar{k}-152}{\frac{4}{\sqrt{n}}} = -1.2816\) | M1B1A1 | \(\frac{\bar{k}-152}{\frac{4}{\sqrt{n}}} = z\)-value, \(1< |
| \(\bar{k} = 152 - 1.2816 \times \frac{4}{\sqrt{n}}\) | ||
| \(150 + 1.6449\times\frac{4}{\sqrt{n}} < 152 - 1.2816\times\frac{4}{\sqrt{n}}\) or \(\frac{150 + \frac{6.5796}{\sqrt{n}} - 152}{\frac{4}{\sqrt{n}}} = -1.2816\) | M1dd | Dependent on both previous M marks; forming equation and solving leading to \(n=...\) or \(\sqrt{n}=...\) |
| \(\left[\sqrt{n}\right] > 5.853\) | A1 | awrt 5.85 |
| \(\left[n >\right] 34.25\) | M1 | For squaring |
| \(n = 35\) | A1cso (10) | 35 only |
## Question 5:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.05$ or $5\%$ | B1 (1) | |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let CR be $\bar{X} > k$; $P\left(\bar{X} > k \,\Big|\, \bar{X} \text{ is } N\left(150, \frac{16}{n}\right)\right) = 0.05$ | | |
| $\frac{\bar{k}-150}{\frac{4}{\sqrt{n}}} = 1.6449$ | M1B1A1 | $\frac{\bar{k}-150}{\frac{4}{\sqrt{n}}} = z$-value, $|z|>1.5$; B1 awrt $\pm1.6449$; A1 correct equation = awrt 1.65/1.64 |
| $\bar{k} = 150 + 1.6449 \times \frac{4}{\sqrt{n}}$ | | |
| $\frac{\bar{k}-152}{\frac{4}{\sqrt{n}}} = -1.2816$ | M1B1A1 | $\frac{\bar{k}-152}{\frac{4}{\sqrt{n}}} = z$-value, $1<|z|<1.5$; B1 $\pm$ awrt $1.2816$; A1 correct equation = awrt $-1.28$ |
| $\bar{k} = 152 - 1.2816 \times \frac{4}{\sqrt{n}}$ | | |
| $150 + 1.6449\times\frac{4}{\sqrt{n}} < 152 - 1.2816\times\frac{4}{\sqrt{n}}$ or $\frac{150 + \frac{6.5796}{\sqrt{n}} - 152}{\frac{4}{\sqrt{n}}} = -1.2816$ | M1dd | Dependent on both previous M marks; forming equation and solving leading to $n=...$ or $\sqrt{n}=...$ |
| $\left[\sqrt{n}\right] > 5.853$ | A1 | awrt 5.85 |
| $\left[n >\right] 34.25$ | M1 | For squaring |
| $n = 35$ | A1cso (10) | 35 only |
**Total: 11**
---\begin{enumerate}
\item A machine makes posts. The length of a post is normally distributed with unknown mean $\mu$ and standard deviation 4 cm .
\end{enumerate}
A random sample of size $n$ is taken to test, at the $5 \%$ significance level, the hypotheses
$$\mathrm { H } _ { 0 } : \mu = 150 \quad \mathrm { H } _ { 1 } : \mu > 150$$
(a) State the probability of a Type I error for this test.
The manufacturer requires the probability of a Type II error to be less than 0.1 when the actual value of $\mu$ is 152\\
(b) Calculate the minimum value of $n$.\\
\hfill \mbox{\textit{Edexcel S4 2018 Q5 [11]}}