| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | CI with known population variance |
| Difficulty | Standard +0.8 This question requires understanding of hypothesis testing with two independent normal samples, working with the distribution of the difference of sample means, and calculating Type II error probability. While the individual components are S3 standard material, part (iii) requiring explicit Type II error calculation with a specific alternative hypothesis value is more demanding than typical textbook exercises and requires careful manipulation of the sampling distribution under both null and alternative hypotheses. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_2 = \mu_1\), \(H_1: \mu_2 > \mu_1\), where \(\mu_1\) and \(\mu_2\) are the mean concentrations in the lake before and after the spillage respectively | B1, B1 2 | For both hypotheses. Allow in words if population mean used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{X}_2 - \bar{X}_1 \geq zs\) | M1 | |
| \(z = 1.645\) | A1 | Accept \(>, =, <, \leq, ts\) |
| \(s = 0.24\sqrt{1/5 + 1/6}\) | B1 | |
| \(\geq 0.2391\) | A1 4 | Or \(>\); 0.239 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\bar{X}_2 - \bar{X}_1 < 0.2391)\) | M1 | May be implied |
| \(z = [0.2391 - 0.3]/s\) | M1 | |
| \(p = 0.3376\) | A1 | ART 0.337 or 0.338 |
| This is a large probability for this error | B1 4 | Relevant comment |
# Question 5:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_2 = \mu_1$, $H_1: \mu_2 > \mu_1$, where $\mu_1$ and $\mu_2$ are the mean concentrations in the lake before and after the spillage respectively | B1, B1 **2** | For both hypotheses. Allow in words if population mean used |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{X}_2 - \bar{X}_1 \geq zs$ | M1 | |
| $z = 1.645$ | A1 | Accept $>, =, <, \leq, ts$ |
| $s = 0.24\sqrt{1/5 + 1/6}$ | B1 | |
| $\geq 0.2391$ | A1 **4** | Or $>$; 0.239 |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\bar{X}_2 - \bar{X}_1 < 0.2391)$ | M1 | May be implied |
| $z = [0.2391 - 0.3]/s$ | M1 | |
| $p = 0.3376$ | A1 | ART 0.337 or 0.338 |
| This is a large probability for this error | B1 **4** | Relevant comment |
---5 The concentration level of mercury in a large lake is known to have a normal distribution with standard deviation 0.24 in suitable units. At the beginning of June 2008, the mercury level was measured at five randomly chosen places on the lake, and the sample mean is denoted by $\bar { x } _ { 1 }$. Towards the end of June 2008 there was a spillage in the lake which may have caused the mercury level to rise. Because of this the level was then measured at six randomly chosen points of the lake, and the mean of this sample is denoted by $\bar { x } _ { 2 }$.\\
(i) State hypotheses for a test based on the two samples for whether, on average, the level of mercury had increased. Define any parameters that you use.\\
(ii) Find the set of values of $\bar { x } _ { 2 } - \bar { x } _ { 1 }$ for which there would be evidence at the 5\% significance level that, on average, the level of mercury had increased.\\
(iii) Given that the average level had actually increased by 0.3 units, find the probability of making a Type II error in your test, and comment on its value.
\hfill \mbox{\textit{OCR S3 2009 Q5 [10]}}