| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Simultaneous critical region and Type II error |
| Difficulty | Standard +0.8 This S4 question requires understanding of hypothesis testing mechanics including critical regions and Type II error calculations with normal distributions. While the individual components (finding critical values, probability calculations) are standard, combining them—especially calculating P(Type II error) by finding the probability of falling in the acceptance region under the alternative hypothesis—requires solid conceptual understanding and careful execution across multiple steps, making it moderately challenging but within reach for well-prepared Further Maths students. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks |
|---|---|
| \(\frac{\bar{X} - 250}{4/\sqrt{15}} > 2.3263\) or \(\frac{\bar{X} - 250}{4/\sqrt{15}} < -2.3263\) | B1 M1 |
| \(\bar{X} > 252.40...\) or \(\bar{X} < 247.6...\) | A1 |
| 252 and 248 | awrt |
Total: 3 marks
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\bar{X} < 252.4 / \mu = 254) - P(\bar{X} < 247.6 / \mu = 254)\) | M1 | using their '252.4' and '247.6' |
| \(= P\left(Z < \frac{252.4 - 254}{4/\sqrt{15}}\right) - P\left(Z < \frac{247.6 - 254}{4/\sqrt{15}}\right)\) | M1 | stand using \(4/\sqrt{15}\), 254 their '252.4' or '247.6' |
| \(= P(Z < -1.5492) - P(Z < -6.20)\) | A1 | -1.5492 and -6.20 o.e. |
| \(= (1 - 0.9394) - (1 - 1)\) | M1 | |
| \(= 0.0606\) | A1 |
Total: 5 marks
**Part a:**
$\frac{\bar{X} - 250}{4/\sqrt{15}} > 2.3263$ or $\frac{\bar{X} - 250}{4/\sqrt{15}} < -2.3263$ | B1 M1 |
$\bar{X} > 252.40...$ or $\bar{X} < 247.6...$ | A1 |
252 and 248 | awrt |
Total: 3 marks
**Part b:**
$P(\bar{X} < 252.4 / \mu = 254) - P(\bar{X} < 247.6 / \mu = 254)$ | M1 | using their '252.4' and '247.6' | |
$= P\left(Z < \frac{252.4 - 254}{4/\sqrt{15}}\right) - P\left(Z < \frac{247.6 - 254}{4/\sqrt{15}}\right)$ | M1 | stand using $4/\sqrt{15}$, 254 their '252.4' or '247.6' |
$= P(Z < -1.5492) - P(Z < -6.20)$ | A1 | -1.5492 and -6.20 o.e. |
$= (1 - 0.9394) - (1 - 1)$ | M1 | |
$= 0.0606$ | A1 |
Notes:
- (a) only needs to try and find one side for M1
- (b) only need to see one of the standardisation for second M1. if consider only 252.4 and get 0.0606 they get M0 M1 A0 M1 A1 ie they can get 3/5
Total: 5 marks
---6. A butter packing machine cuts butter into blocks. The weight of a block of butter is normally distributed with a mean weight of 250 g and a standard deviation of 4 g . A random sample of 15 blocks is taken to monitor any change in the mean weight of the blocks of butter.
\begin{enumerate}[label=(\alph*)]
\item Find the critical region of a suitable test using a $2 \%$ level of significance.\\
(3)
\item Assuming the mean weight of a block of butter has increased to 254 g , find the probability of a Type II error.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2007 Q6 [8]}}