| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| 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 +1.8 This Further Statistics question requires understanding of hypothesis testing framework (Type I/II errors, significance levels), normal distribution calculations, and iterative problem-solving to find minimum n. Part (a) is straightforward recall, but part (b)(i) demands setting up inequalities involving both null and alternative distributions, requiring z-score manipulation and trial-and-error. The conceptual understanding needed for (b)(ii) about the trade-off between error types elevates this beyond routine exercises, though the calculations themselves are standard for Further Maths students. |
| 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 |
| Size of the test \(= 0.01\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{k-15}{\frac{0.2}{\sqrt{n}}} = -2.3263\) | M1 | Finding CR using Normal distribution, must have \(1.5 < |
| \(k = 15 - \frac{0.46526}{\sqrt{n}}\) | A1 | Correct equation in form \(k = \ldots\), using awrt 2.326 |
| \(\frac{\text{"}15 - \frac{0.46526}{\sqrt{n}}\text{"} - 14.9}{\frac{0.2}{\sqrt{n}}} > 1.6449\) | M1d, A1ft | Dependent on previous M; standardising using their \(k\) and equating to \(z\) value \(1.5 < |
| \(\frac{0.79424}{\sqrt{n}} < 0.1 \quad \Rightarrow \sqrt{n} > 7.9424\) oe | M1d | Dependent on previous M; isolating \(\sqrt{n}\) or squaring; condone \(n = 7.9424\) |
| \(n = 64\) | A1cso | 64 with correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The probability of a Type II error would decrease | B1 | Suitable comment |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{k-14.9}{\frac{0.2}{\sqrt{n}}} = 1.6449\) | M1 | |
| \(k = 14.9 + \frac{0.32898}{\sqrt{n}}\) | A1 | |
| \(\frac{\text{"}14.9 + \frac{0.32898}{\sqrt{n}}\text{"} - 15}{\frac{0.2}{\sqrt{n}}} > -2.3263\) | M1d, A1ft | |
| \(\frac{0.79424}{\sqrt{n}} < 0.1 \quad \Rightarrow \sqrt{n} > 7.9424\) oe | M1d | |
| \(n = 64\) | A1cso |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Size of the test $= 0.01$ | B1 | |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{k-15}{\frac{0.2}{\sqrt{n}}} = -2.3263$ | M1 | Finding CR using Normal distribution, must have $1.5 < |z| < 3.5$ |
| $k = 15 - \frac{0.46526}{\sqrt{n}}$ | A1 | Correct equation in form $k = \ldots$, using awrt 2.326 |
| $\frac{\text{"}15 - \frac{0.46526}{\sqrt{n}}\text{"} - 14.9}{\frac{0.2}{\sqrt{n}}} > 1.6449$ | M1d, A1ft | Dependent on previous M; standardising using their $k$ and equating to $z$ value $1.5 < |z| < 3$; ft their $k$ for correct equation with awrt 1.645 |
| $\frac{0.79424}{\sqrt{n}} < 0.1 \quad \Rightarrow \sqrt{n} > 7.9424$ oe | M1d | Dependent on previous M; isolating $\sqrt{n}$ or squaring; condone $n = 7.9424$ |
| $n = 64$ | A1cso | 64 with correct working |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| The probability of a Type II error would decrease | B1 | Suitable comment |
---
# ALT Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{k-14.9}{\frac{0.2}{\sqrt{n}}} = 1.6449$ | M1 | |
| $k = 14.9 + \frac{0.32898}{\sqrt{n}}$ | A1 | |
| $\frac{\text{"}14.9 + \frac{0.32898}{\sqrt{n}}\text{"} - 15}{\frac{0.2}{\sqrt{n}}} > -2.3263$ | M1d, A1ft | |
| $\frac{0.79424}{\sqrt{n}} < 0.1 \quad \Rightarrow \sqrt{n} > 7.9424$ oe | M1d | |
| $n = 64$ | A1cso | |\begin{enumerate}
\item A manufacturer has a machine that produces lollipop sticks.
\end{enumerate}
The length of a lollipop stick produced by the machine is normally distributed with unknown mean $\mu$ and standard deviation 0.2
Farhan believes that the machine is not working properly and the mean length of the lollipop sticks has decreased.\\
He takes a random sample of size $n$ to test, at the 1\% level of significance, the hypotheses
$$\mathrm { H } _ { 0 } : \mu = 15 \quad \mathrm { H } _ { 1 } : \mu < 15$$
(a) Write down the size of this test.
Given that the actual value of $\mu$ is 14.9\\
(b) (i) calculate the minimum value of $n$ such that the probability of a Type II error is less than 0.05\\
Show your working clearly.\\
(ii) Farhan uses the same sample size, $n$, but now carries out the test at a $5 \%$ level of significance. Without doing any further calculations, state how this would affect the probability of a Type II error.
\hfill \mbox{\textit{Edexcel FS1 2021 Q7 [8]}}