| 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.\\
\begin{enumerate}[label=(\roman*)]
\item Carry out the test.
\item Explain whether it is necessary to use the Central Limit Theorem in your test.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2011 Q4 [7]}}