| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Hypothesis test for mean |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test using the normal distribution with known standard deviation. The calculation is routine (compute z-statistic, compare to critical value), requiring only standard formulas. Part (ii) tests conceptual understanding of CLT but is a simple recall question. Slightly above average difficulty due to the two-tailed test and CLT justification, but still a standard textbook exercise 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 |
|---|---|---|
| \(Either: z = \frac{213.4 - 230}{45/\sqrt{50}} = \frac{-2.608}{-2.576 \text{ or } 0.0047} < 0.005\) | M1 | Standardise \(z\) with \(\sqrt{50}\), ignore sign or \(\sqrt{\text{}}\) or squaring errors |
| A1 | \(z\)-value, a.r.t. \(-2.61\), or \(p\) in range [0.0044, 0.005) | |
| B1 | Correctly compare \((-2)2.576\), signs consistent, or \(p\) explicitly with 0.005 | |
| *Or* \(CV \text{ is } 230 - 2.576 \times \frac{45}{\sqrt{50}} = 213.6\) | M1 | \(230 - z\sigma\sqrt{50}\), allow \(\sqrt{\text{}}\) or squaring errors, allow \(\pm\) but not just +; \(z = 2.576\) |
| \(213.4 < 213.6\) | A1 | Explicitly compare 213.4 with 213.6 |
| Reject \(H_0\). Significant evidence that population mean is less than 230 | M1 FT | "Reject", FT, needs correct method and form of comparison; acknowledge uncertainty |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Yes, population distribution is not known to be normal | B2 | Not, "yes, sample size is large" but ignore "can use it as..." |
| 2 | SR: Both right and wrong answers: B1; α "Yes as it must be assumed normal": B1 |
**(i)**
$Either: z = \frac{213.4 - 230}{45/\sqrt{50}} = \frac{-2.608}{-2.576 \text{ or } 0.0047} < 0.005$ | M1 | Standardise $z$ with $\sqrt{50}$, ignore sign or $\sqrt{\text{}}$ or squaring errors
| A1 | $z$-value, a.r.t. $-2.61$, or $p$ in range [0.0044, 0.005)
| B1 | Correctly compare $(-2)2.576$, signs consistent, or $p$ explicitly with 0.005
*Or* $CV \text{ is } 230 - 2.576 \times \frac{45}{\sqrt{50}} = 213.6$ | M1 | $230 - z\sigma\sqrt{50}$, allow $\sqrt{\text{}}$ or squaring errors, allow $\pm$ but not just +; $z = 2.576$
$213.4 < 213.6$ | A1 | Explicitly compare 213.4 with 213.6
**Reject $H_0$. Significant evidence that population mean is less than 230** | M1 FT | "Reject", FT, needs correct method and form of comparison; acknowledge uncertainty
| | 5
**(ii)**
Yes, population distribution is not known to be normal | B2 | Not, "yes, sample size is large" but ignore "can use it as..."
| | 2 | SR: Both right and wrong answers: B1; α "Yes as it must be assumed normal": B14 The continuous random variable $X$ has mean $\mu$ and standard deviation 45. A significance test is to be carried out of the null hypothesis $\mathrm { H } _ { 0 } : \mu = 230$ against the alternative hypothesis $\mathrm { H } _ { 1 } : \mu \neq 230$, at the $1 \%$ significance level. A random sample of size 50 is obtained, and the sample mean is found to be 213.4.\\
(i) Carry out the test.\\
(ii) Explain whether it is necessary to use the Central Limit Theorem in your test.
\hfill \mbox{\textit{OCR S2 2011 Q4 [7]}}