| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for hypothesis testing |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test for a mean with known population standard deviation, requiring students to state the CLT assumption (normality of sample mean for n=30) and explain why CLT is needed (original distribution unknown). The calculation is routine and the conceptual understanding required is standard S2 material, making it slightly easier than average A-level questions. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Scenario | Working | Marks |
| α | \(H_0: \bar{x} = 28.0\); \(H_1: \bar{x} > 28.0\) | B0B0 |
| \(z = \dfrac{28.98 - 28.0}{\sqrt{12/30}} = 1.550\) | M1 | *wrong \(\sqrt{}\)* |
| \(< 1.645\) | A0, A1 | |
| Accept \(H_0\), no increase in average score | M1A0 | 3 |
| β | \(H_0: \mu = 28.98\); \(H_1: \mu < 28.98\) | B0B0 |
| \(z = \dfrac{28.98 - 28.0}{12/\sqrt{30}} = 0.447\) | M1, A1 | *allow this – BOD* |
| \(< 1.645\) | A1 | |
| Accept \(H_0\). Insufficient evidence of a change in maximum daily temperature. | M1, A1 | 5 |
| γ | \(H_0: \mu = 28.98\); \(H_1: \mu < 28.98\) | B0B0 |
| \(z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447\) | M1, A0 | *DON'T allow this* |
| \(> -1.645\) | A1 | |
| Accept \(H_0\). Insufficient evidence of a change in average score. | M0, A0 | 2 |
| δ | \(H_0 = 28.0\); \(H_1 > 28.0\) | B1 only |
| \(z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447\) | M1, A0 | *loses 1* |
| \(> -1.645\) | A1 | |
| Insufficient evidence to reject \(H_0\). No change in average score. | M1, A1 | 5 |
| ε | \(H_0: \mu = 28.0\); \(H_1: \mu \neq 28.0\) | B1B0 |
| \(z = \dfrac{28.98 - 28.0}{12/\sqrt{30}} = 0.447\) | M1, A1 | |
| \(< 1.96\) [also if \(< 1.645\)] | A0 | |
| Accept \(H_0\). Insufficient evidence of a change in average score. | M1, A1 | 5 |
| ζ | \(H_0: \mu = 28.0\); \(H_1: \mu > 28.0\) | B2 |
| \(z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447\) *but then...* | M1, A1 | |
| So \(p = 0.327 > 0.05\) | A1 | *OK here* |
| Accept \(H_0\). Insufficient evidence of a change in average score. | M1, A1 | 7 |
| η | \(H_0: \mu = 28.0\); \(H_1: \mu > 28.0\) | B2 |
| \(z = \dfrac{28.98 - 28.0}{12} = 0.0817\) | M0 | *no \(\sqrt{30}\)* |
| \(< 1.645\) | A0, A0 | |
| Accept \(H_0\). Insufficient evidence of a change in average score. | M0, A0 | 2 |
## Question 6(i): [marks out of 7, condition not included]
| Scenario | Working | Marks | Notes |
|----------|---------|-------|-------|
| α | $H_0: \bar{x} = 28.0$; $H_1: \bar{x} > 28.0$ | B0B0 | *wrong symbol* |
| | $z = \dfrac{28.98 - 28.0}{\sqrt{12/30}} = 1.550$ | M1 | *wrong $\sqrt{}$* |
| | $< 1.645$ | A0, A1 | |
| | Accept $H_0$, no increase in average score | M1A0 | **3** | *over-assertive, otherwise A1* |
| β | $H_0: \mu = 28.98$; $H_1: \mu < 28.98$ | B0B0 | *WRONG* |
| | $z = \dfrac{28.98 - 28.0}{12/\sqrt{30}} = 0.447$ | M1, A1 | *allow this – BOD* |
| | $< 1.645$ | A1 | |
| | Accept $H_0$. Insufficient evidence of a change in maximum daily temperature. | M1, A1 | **5** | |
| γ | $H_0: \mu = 28.98$; $H_1: \mu < 28.98$ | B0B0 | *WRONG* |
| | $z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447$ | M1, A0 | *DON'T allow this* |
| | $> -1.645$ | A1 | |
| | Accept $H_0$. Insufficient evidence of a change in average score. | M0, A0 | **2** | |
| δ | $H_0 = 28.0$; $H_1 > 28.0$ | B1 only | *missing symbol* |
| | $z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447$ | M1, A0 | *loses 1* |
| | $> -1.645$ | A1 | |
| | Insufficient evidence to reject $H_0$. No change in average score. | M1, A1 | **5** | *OK* |
| ε | $H_0: \mu = 28.0$; $H_1: \mu \neq 28.0$ | B1B0 | *two-tail* |
| | $z = \dfrac{28.98 - 28.0}{12/\sqrt{30}} = 0.447$ | M1, A1 | |
| | $< 1.96$ [also if $< 1.645$] | A0 | |
| | Accept $H_0$. Insufficient evidence of a change in average score. | M1, A1 | **5** | |
| ζ | $H_0: \mu = 28.0$; $H_1: \mu > 28.0$ | B2 | |
| | $z = \dfrac{28.0 - 28.98}{12/\sqrt{30}} = -0.447$ *but then...* | M1, A1 | |
| | So $p = 0.327 > 0.05$ | A1 | *OK here* |
| | Accept $H_0$. Insufficient evidence of a change in average score. | M1, A1 | **7** | |
| η | $H_0: \mu = 28.0$; $H_1: \mu > 28.0$ | B2 | |
| | $z = \dfrac{28.98 - 28.0}{12} = 0.0817$ | M0 | *no $\sqrt{30}$* |
| | $< 1.645$ | A0, A0 | |
| | Accept $H_0$. Insufficient evidence of a change in average score. | M0, A0 | **2** | |
---6 Gordon is a cricketer. Over a long period he knows that his population mean score, in number of runs per innings, is 28 , and the population standard deviation is 12 . In a new season he adopts a different batting style and he finds that in 30 innings using this style his mean score is 28.98 .\\
(i) Stating a necessary assumption, test at the $5 \%$ significance level whether his population mean score has increased.\\
(ii) Explain whether it was necessary to use the Central Limit Theorem in part (i).
\hfill \mbox{\textit{OCR S2 2013 Q6 [10]}}