OCR S2 2011 June — Question 6 12 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks12
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Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyStandard +0.3 This is a straightforward two-tail z-test with given summary statistics. Students must calculate the sample mean (26.28), sample standard deviation (4.67), perform a z-test against μ=24.3, and compare to critical values at 1% level. The CLT question is standard bookwork. While it requires multiple steps, each is routine for S2 students with no novel problem-solving required.
Spec5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean
6 Records show that before the year 1990 the maximum daily temperature \(T ^ { \circ } \mathrm { C }\) at a seaside resort in August can be modelled by a distribution with mean 24.3. The maximum temperatures of a random sample of 50 August days since 1990 can be summarised by $$n = 50 , \quad \Sigma t = 1314.0 , \quad \Sigma t ^ { 2 } = 36602.17 .$$
  1. Test, at the \(1 \%\) significance level, whether there is evidence of a change in the mean maximum daily temperature in August since 1990.
  2. Give a reason why it is possible to use the Central Limit Theorem in your test.
Question 6(i): Specific Examples
Case α (Wrong symbol, implied \(\bar{t}\))
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \bar{t} = 24.3\); \(H_1: \bar{t} \neq 24.3\)B0B0 *[wrong symbol]*
\(\bar{t}\) not seen explicitlyB1 *[implied by ...]*
\(\hat{\sigma}^2 = \left[\frac{36602.17}{50} - 26.28^2\right] = 41.405\)M1, M0 *[biased est]*
\(z = \frac{26.28 - 24.3}{\sqrt{41.405/50}} = 2.1758\)A0, M1
\(< 2.576\)A0
Accept \(H_0\), maximum temp unchangedA1, M1A0 *[over-assertive, otherwise A1]* 5 marks total
Case β (Wrong hypotheses μ = 26.28)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 26.28\); \(H_1: \mu \neq 26.28\)B0B0 *[WRONG]*
\(\bar{t} = 24.3\)B0 *[explicitly]*
\(\hat{\sigma}^2 = \ldots = 42.25\)M1M1
\(z = \frac{26.28 - 24.3}{\sqrt{42.25/50}} = 2.154\)A1, M1 *[allow this – BOD]*
\(< 2.576\)A1
Accept \(H_0\). Insufficient evidence of a change in maximum daily temperatureA1, M1, A1 8 marks total
Case γ (Wrong hypotheses, \(\bar{t}\) not seen separately)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 26.28\); \(H_1: \mu \neq 26.28\)B0B0 *[WRONG]*
\(\bar{t}\) not seen separatelyB1 *[implied]*
\(\hat{\sigma}^2 = \ldots = 42.25\)M1M1
\(z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154\)A1, M1 *[DON'T allow this]*
\(> -2.576\)A0
Accept \(H_0\). Insufficient evidence of a change in maximum daily temperatureA0, M0, A0 5 marks total
Case δ (Missing symbol)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0 = 24.3\); \(H_1 \neq 24.3\)B1 only *[missing symbol]*
\(\bar{t} = 26.28\)B1
\(\hat{\sigma}^2 = \ldots = 42.25\)M1M1
\(z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154\)A1, M1 *[loses 1]*
\(> -2.576\)A0, A1
Insufficient evidence to reject \(H_0\). No change in maximum daily temperatureM1, A1 *[OK]* 9 marks total
Case ε (One-tail, B(10,0.4))
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 24.3\); \(H_1: \mu > 24.3\)B1B0 *[one-tail]*
\(\bar{t} = 26.28\)B1
\(\hat{\sigma}^2 = \ldots = 42.25\)M1M1
\(z = \frac{26.28 - 24.3}{\sqrt{42.25/50}} = 2.154\)A1, M1
\(< 2.326\)A0, A1
Accept \(H_0\). Insufficient evidence of a change in maximum daily temperatureM1, A1 9 marks total
Case ζ (p-value approach)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154\)M1, A1
So \(p = 0.0156 > 0.005\)A1, M1 *[OK here]*
Accept \(H_0\). Insufficient evidence of a change in maximum daily temperatureA1 (11 marks total)
Case η (No \(\sqrt{50}\))
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(z = \frac{26.28 - 24.3}{\sqrt{42.25}} = 0.3046\)M0 *[no \(\sqrt{50}\)]*
\(< 2.576\)A0
Accept \(H_0\). Insufficient evidence of a change in maximum daily temperatureA0, M0, A0 (6 marks total) 
# Question 6(i): Specific Examples

## Case α (Wrong symbol, implied $\bar{t}$)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \bar{t} = 24.3$; $H_1: \bar{t} \neq 24.3$ | B0B0 | *[wrong symbol]* |
| $\bar{t}$ not seen explicitly | B1 | *[implied by ...]* |
| $\hat{\sigma}^2 = \left[\frac{36602.17}{50} - 26.28^2\right] = 41.405$ | M1, M0 | *[biased est]* |
| $z = \frac{26.28 - 24.3}{\sqrt{41.405/50}} = 2.1758$ | A0, M1 | |
| $< 2.576$ | A0 | |
| Accept $H_0$, maximum temp unchanged | A1, M1A0 | *[over-assertive, otherwise A1]* **5 marks total** |

## Case β (Wrong hypotheses μ = 26.28)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 26.28$; $H_1: \mu \neq 26.28$ | B0B0 | *[WRONG]* |
| $\bar{t} = 24.3$ | B0 | *[explicitly]* |
| $\hat{\sigma}^2 = \ldots = 42.25$ | M1M1 | |
| $z = \frac{26.28 - 24.3}{\sqrt{42.25/50}} = 2.154$ | A1, M1 | *[allow this – BOD]* |
| $< 2.576$ | A1 | |
| Accept $H_0$. Insufficient evidence of a change in maximum daily temperature | A1, M1, A1 | **8 marks total** |

## Case γ (Wrong hypotheses, $\bar{t}$ not seen separately)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 26.28$; $H_1: \mu \neq 26.28$ | B0B0 | *[WRONG]* |
| $\bar{t}$ not seen separately | B1 | *[implied]* |
| $\hat{\sigma}^2 = \ldots = 42.25$ | M1M1 | |
| $z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154$ | A1, M1 | *[DON'T allow this]* |
| $> -2.576$ | A0 | |
| Accept $H_0$. Insufficient evidence of a change in maximum daily temperature | A0, M0, A0 | **5 marks total** |

## Case δ (Missing symbol)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0 = 24.3$; $H_1 \neq 24.3$ | B1 only | *[missing symbol]* |
| $\bar{t} = 26.28$ | B1 | |
| $\hat{\sigma}^2 = \ldots = 42.25$ | M1M1 | |
| $z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154$ | A1, M1 | *[loses 1]* |
| $> -2.576$ | A0, A1 | |
| Insufficient evidence to reject $H_0$. No change in maximum daily temperature | M1, A1 | *[OK]* **9 marks total** |

## Case ε (One-tail, B(10,0.4))
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 24.3$; $H_1: \mu > 24.3$ | B1B0 | *[one-tail]* |
| $\bar{t} = 26.28$ | B1 | |
| $\hat{\sigma}^2 = \ldots = 42.25$ | M1M1 | |
| $z = \frac{26.28 - 24.3}{\sqrt{42.25/50}} = 2.154$ | A1, M1 | |
| $< 2.326$ | A0, A1 | |
| Accept $H_0$. Insufficient evidence of a change in maximum daily temperature | M1, A1 | **9 marks total** |

## Case ζ (p-value approach)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{24.3 - 26.28}{\sqrt{42.25/50}} = -2.154$ | M1, A1 | |
| So $p = 0.0156 > 0.005$ | A1, M1 | *[OK here]* |
| Accept $H_0$. Insufficient evidence of a change in maximum daily temperature | A1 | **(11 marks total)** |

## Case η (No $\sqrt{50}$)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{26.28 - 24.3}{\sqrt{42.25}} = 0.3046$ | M0 | *[no $\sqrt{50}$]* |
| $< 2.576$ | A0 | |
| Accept $H_0$. Insufficient evidence of a change in maximum daily temperature | A0, M0, A0 | **(6 marks total)** |
6 Records show that before the year 1990 the maximum daily temperature $T ^ { \circ } \mathrm { C }$ at a seaside resort in August can be modelled by a distribution with mean 24.3. The maximum temperatures of a random sample of 50 August days since 1990 can be summarised by

$$n = 50 , \quad \Sigma t = 1314.0 , \quad \Sigma t ^ { 2 } = 36602.17 .$$

(i) Test, at the $1 \%$ significance level, whether there is evidence of a change in the mean maximum daily temperature in August since 1990.\\
(ii) Give a reason why it is possible to use the Central Limit Theorem in your test.

\hfill \mbox{\textit{OCR S2 2011 Q6 [12]}}
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