| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Hypothesis test for mean |
| Difficulty | Standard +0.3 This is a straightforward one-sample z-test with summary statistics provided. Part (i) requires calculating sample mean and standard deviation, then performing a standard hypothesis test—routine A-level procedure. Part (ii) tests conceptual understanding of CLT (large sample size n=64 justifies normality assumption). Slightly above average difficulty due to the two-part structure and need to justify the statistical method, but all steps are standard S2 content with no novel problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\hat{\mu} = \bar{t} = 13.7\) | B1 | 13.7 stated |
| \(\frac{12657.28}{64} - 13.7^2\ [=10.08];\ \times\frac{64}{63}\) | M1, M1 | Correct formula for biased estimate; \(\times\frac{64}{63}\) used, or equivalent |
| \(= 10.24\) | A1 | Variance or SD 10.24 or 10.2 |
| \(H_0: \mu = 13.1,\ H_1: \mu > 13.1\) | B2 | Both correct. [SR: One error, B1, but \(x\) or \(t\) or \(\bar{x}\) or \(\bar{t}\), 0] |
| \(\frac{13.7 - 13.1}{\sqrt{10.24/64}} = 1.5\) or \(p = 0.0668\) | M1, A1 | Standardise, or find CV, with \(\sqrt{64}\) or 64; \(z =\) a.r.t. 1.50, or \(p = 0.0668\), or CV 13.758 [\(\sqrt{}\) on \(z\)] |
| \(1.5 < 1.645\) or \(0.0668 > 0.05\) | B1 | Compare \(z\) & 1.645, or \(p\) & 0.05 (must be correct tail), or \(z = 1.645\) & 13 with CV |
| Do not reject \(H_0\). Insufficient evidence that time taken on average is greater than 13.1 min | M1, A1 11 | Correct comparison & conclusion, needs 64, *not* \(\mu = 13.7\); Contextualised, some acknowledgement of uncertainty |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Yes, not told that dist is normal | B1 1 | Equivalent statement, *not* "\(n\) is large", don't need "yes" |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\hat{\mu} = \bar{t} = 13.7$ | B1 | 13.7 stated |
| $\frac{12657.28}{64} - 13.7^2\ [=10.08];\ \times\frac{64}{63}$ | M1, M1 | Correct formula for biased estimate; $\times\frac{64}{63}$ used, or equivalent |
| $= 10.24$ | A1 | Variance or SD 10.24 or 10.2 |
| $H_0: \mu = 13.1,\ H_1: \mu > 13.1$ | B2 | Both correct. [SR: One error, B1, but $x$ or $t$ or $\bar{x}$ or $\bar{t}$, 0] |
| $\frac{13.7 - 13.1}{\sqrt{10.24/64}} = 1.5$ or $p = 0.0668$ | M1, A1 | Standardise, or find CV, with $\sqrt{64}$ or 64; $z =$ a.r.t. 1.50, or $p = 0.0668$, or CV 13.758 [$\sqrt{}$ on $z$] |
| $1.5 < 1.645$ or $0.0668 > 0.05$ | B1 | Compare $z$ & 1.645, or $p$ & 0.05 (must be correct tail), or $z = 1.645$ & 13 with CV |
| Do not reject $H_0$. Insufficient evidence that time taken on average is greater than 13.1 min | M1, A1 **11** | Correct comparison & conclusion, needs 64, *not* $\mu = 13.7$; Contextualised, some acknowledgement of uncertainty |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Yes, not told that dist is normal | B1 **1** | Equivalent statement, *not* "$n$ is large", don't need "yes" |
---7 A motorist records the time taken, $T$ minutes, to drive a particular stretch of road on each of 64 occasions. Her results are summarised by
$$\Sigma t = 876.8 , \quad \Sigma t ^ { 2 } = 12657.28$$
(i) Test, at the $5 \%$ significance level, whether the mean time for the motorist to drive the stretch of road is greater than 13.1 minutes.\\
(ii) Explain whether it is necessary to use the Central Limit Theorem in your test.
\hfill \mbox{\textit{OCR S2 2009 Q7 [12]}}