OCR H240/02 2019 June — Question 10 7 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyStandard +0.3 This is a straightforward one-sample z-test with variance known, requiring students to set up hypotheses, calculate a test statistic using the given standard deviation, and compare to critical values. While it involves careful handling of small numbers, it follows a standard procedure with no conceptual challenges beyond applying the normal distribution hypothesis test formula—slightly above average difficulty due to being a complete hypothesis test rather than just recall.
Spec2.05e Hypothesis test for normal mean: known variance

10 The level, in grams per millilitre, of a pollutant at different locations in a certain river is denoted by the random variable \(X\), where \(X\) has the distribution \(\mathrm { N } ( \mu , 0.0000409 )\). In the past the value of \(\mu\) has been 0.0340 . This year the mean level of the pollutant at 50 randomly chosen locations was found to be 0.0325 grams per millilitre. Test, at the 5\% significance level, whether the mean level of pollutant has changed.

Question 10: Exemplars
Hypotheses Section
Exemplar A:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 0.034\)B1 Correct null hypothesis with parameter notation
\(H_1: \mu \neq 0.034\) where \(\mu\) = (pop) mean pollutant levelB1 Correct two-tailed alternative with definition of \(\mu\)
Exemplar B:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 0.034\)B1 Correct null hypothesis
\(H_1: \mu \neq 0.034\)B0 Missing definition of \(\mu\)
Exemplar C:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): The (pop) mean pollutant level is \(0.034\)B1 Correct in words
\(H_1\): The (pop) mean pollutant level is not \(0.034\)B0 Must be in terms of parameter values — see Specimen paper mark scheme
Exemplar D:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0 = 0.034\), \(H_0 \neq 0.034\)B0B0 No parameter used
Exemplar E:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 0.034\)B1 Correct null hypothesis
\(H_1: \mu = 0.0325\) where \(\mu\) = (pop) mean pollutant levelB0 \(H_1\) must be an inequality
Probability and Conclusion Section
Exemplar F:
AnswerMarks Guidance
AnswerMarks Guidance
No statement of distribution
\(P(\bar{X} = 0.0325) = 0.0486\)M1A1 Method and correct probability
\(0.0486 > 0.025\)A1 Correct comparison
Don't reject \(H_0\)M1 Correct decision
Likely that mean level of pollutant hasn't changedA1 Correct conclusion in context
Exemplar G:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\bar{X} = 0.0325) = 0.0486\)M1A1
\(0.0486 > 0.025\)A1
Accept \(H_0\)M1
There is evidence that mean level of pollutant hasn't changedA0 Conclusion not stated correctly
Exemplar H:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\bar{X} < 0.0325) = 0.951\)M1A0 Wrong probability used
\(0.951 > 0.025\)A0
Insufficient evidence that poll't level has changedM0A0
Exemplar I:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\bar{X} > 0.0325) = 0.951\)M1A1
\(0.951 > 0.025\)A0
Sufficient evidence that mean poll't level has changedM0A0 Incorrect conclusion
Exemplar J:
AnswerMarks Guidance
AnswerMarks Guidance
\(\bar{X} \sim N(0.034, 0.000000818)\)
\(P(\bar{X} < 0.0325) = 0.013\)M1A0 Incorrect variance used
\(0.013 < 0.025\)A0
Sufficient evidence that level has changedM1A1 Follow through conclusion
Exemplar K:
AnswerMarks Guidance
AnswerMarks Guidance
\(\mu \pm 1.96\sigma = 0.0322\) to \(0.0358\)M1A1 BOD
\(0.0325\) lies within this rangeA1
Reject \(H_1\)M1
Insufficient evidence that level of poll't has decreasedA0
Exemplar L:
AnswerMarks Guidance
AnswerMarks Guidance
\(CV = 0.0322\)M1A1
\(0.0325 > 0.0322\)A1
Reject \(H_0\). Evidence that level of poll't has changed.M0A0 Incorrect decision leads to wrong conclusion
Exemplar M:
AnswerMarks Guidance
AnswerMarks Guidance
\((0.0322 - 0.034) \div \sqrt{0.0000409/50} = -1.66\)M1A1
\(1.66 < 1.96\)A1
Don't reject \(H_0\). Level of poll't hasn't changed.M1A0 Conclusion lacks context detail
1-tail Hypotheses
Exemplar N:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 0.034\)B1
\(H_1: \mu < 0.034\) where \(\mu\) = (pop) mean pollutant levelB0 Wrong tail
Exemplar O:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 0.034\)B0
\(H_1: \mu < 0.034\)B0 Wrong tail and no definition
Exemplar P:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): The (pop) mean pollutant level is \(0.034\)B0
\(H_1\): The (pop) mean pollutant level is less than \(0.034\)B0 Wrong tail
# Question 10: Exemplars

## Hypotheses Section

### Exemplar A:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 0.034$ | B1 | Correct null hypothesis with parameter notation |
| $H_1: \mu \neq 0.034$ where $\mu$ = (pop) mean pollutant level | B1 | Correct two-tailed alternative with definition of $\mu$ |

### Exemplar B:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 0.034$ | B1 | Correct null hypothesis |
| $H_1: \mu \neq 0.034$ | B0 | Missing definition of $\mu$ |

### Exemplar C:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: The (pop) mean pollutant level is $0.034$ | B1 | Correct in words |
| $H_1$: The (pop) mean pollutant level is not $0.034$ | B0 | Must be in terms of parameter values — see Specimen paper mark scheme |

### Exemplar D:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0 = 0.034$, $H_0 \neq 0.034$ | B0B0 | No parameter used |

### Exemplar E:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 0.034$ | B1 | Correct null hypothesis |
| $H_1: \mu = 0.0325$ where $\mu$ = (pop) mean pollutant level | B0 | $H_1$ must be an inequality |

---

## Probability and Conclusion Section

### Exemplar F:
| Answer | Marks | Guidance |
|--------|-------|----------|
| No statement of distribution | — | |
| $P(\bar{X} = 0.0325) = 0.0486$ | M1A1 | Method and correct probability |
| $0.0486 > 0.025$ | A1 | Correct comparison |
| Don't reject $H_0$ | M1 | Correct decision |
| Likely that mean level of pollutant hasn't changed | A1 | Correct conclusion in context |

### Exemplar G:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\bar{X} = 0.0325) = 0.0486$ | M1A1 | |
| $0.0486 > 0.025$ | A1 | |
| Accept $H_0$ | M1 | |
| There is evidence that mean level of pollutant hasn't changed | A0 | Conclusion not stated correctly |

### Exemplar H:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\bar{X} < 0.0325) = 0.951$ | M1A0 | Wrong probability used |
| $0.951 > 0.025$ | A0 | |
| Insufficient evidence that poll't level has changed | M0A0 | |

### Exemplar I:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\bar{X} > 0.0325) = 0.951$ | M1A1 | |
| $0.951 > 0.025$ | A0 | |
| Sufficient evidence that mean poll't level has changed | M0A0 | Incorrect conclusion |

### Exemplar J:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{X} \sim N(0.034, 0.000000818)$ | — | |
| $P(\bar{X} < 0.0325) = 0.013$ | M1A0 | Incorrect variance used |
| $0.013 < 0.025$ | A0 | |
| Sufficient evidence that level has changed | M1A1 | Follow through conclusion |

### Exemplar K:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu \pm 1.96\sigma = 0.0322$ to $0.0358$ | M1A1 | BOD |
| $0.0325$ lies within this range | A1 | |
| Reject $H_1$ | M1 | |
| Insufficient evidence that level of poll't has decreased | A0 | |

### Exemplar L:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $CV = 0.0322$ | M1A1 | |
| $0.0325 > 0.0322$ | A1 | |
| Reject $H_0$. Evidence that level of poll't has changed. | M0A0 | Incorrect decision leads to wrong conclusion |

### Exemplar M:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.0322 - 0.034) \div \sqrt{0.0000409/50} = -1.66$ | M1A1 | |
| $1.66 < 1.96$ | A1 | |
| Don't reject $H_0$. Level of poll't hasn't changed. | M1A0 | Conclusion lacks context detail |

---

## 1-tail Hypotheses

### Exemplar N:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 0.034$ | B1 | |
| $H_1: \mu < 0.034$ where $\mu$ = (pop) mean pollutant level | B0 | Wrong tail |

### Exemplar O:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 0.034$ | B0 | |
| $H_1: \mu < 0.034$ | B0 | Wrong tail and no definition |

### Exemplar P:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: The (pop) mean pollutant level is $0.034$ | B0 | |
| $H_1$: The (pop) mean pollutant level is less than $0.034$ | B0 | Wrong tail |
10 The level, in grams per millilitre, of a pollutant at different locations in a certain river is denoted by the random variable $X$, where $X$ has the distribution $\mathrm { N } ( \mu , 0.0000409 )$.

In the past the value of $\mu$ has been 0.0340 .

This year the mean level of the pollutant at 50 randomly chosen locations was found to be 0.0325 grams per millilitre.

Test, at the 5\% significance level, whether the mean level of pollutant has changed.

\hfill \mbox{\textit{OCR H240/02 2019 Q10 [7]}}