| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a straightforward two-tail z-test with summary statistics provided. Students must calculate the sample mean, perform a hypothesis test at 10% significance, and explain the Central Limit Theorem justification for not requiring normality (n=120 is large). All steps are standard S2 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{t} = 11.76\) | B1 | 11.76 seen or implied |
| \(\hat{\sigma}^2 = \frac{120}{119}\!\left(\frac{18737.712}{120} - 11.76^2\right) = 18\) | M1, M1 | Biased estimate (\(= 17.85\)). \(\times 120/119\), or single formula with 119 divisor |
| A1 | Answer \(18 \pm 0.05\) | |
| \(H_0: \mu = 11.0\), \(H_1: \mu \neq 11.0\) | B2 | One error, B1 |
| \(\alpha\): \(z = \frac{11.76 - 11.0}{\sqrt{18/120}} = \mathbf{1.9623}\) | M1 | Standardise with 120, ignore cc or \(\sqrt{}\) errors |
| \(> 1.645\) | A1, A1 | A.r.t. \((\pm)1.96\) or \(p \in [0.0245, 0.025]\) www. Compare explicitly with \((\pm)1.645\) or 0.05, consistent with their \(z\) or \(p\) |
| \(\beta\): CV \(11.0 \pm 1.645 \times \sqrt{18/120} = 11.637\) (or 10.363). \(11.76 > 11.64\) | M1, A1, A1 | \(11.0 + z\sigma/\sqrt{120}\), needs 120 and \(+\) or \(\pm\). Ignore 10.363. Explicit comparison, consistent tail |
| Reject \(H_0\). Significant evidence that the average time has changed. | M1, A1ft | Correct first conclusion. Contextualised, acknowledge uncertainty. FT on wrong CR/\(z\)/\(p\) |
| Total: 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No, the Central Limit Theorem applies | B1 | or "No, large sample". Withhold if extra wrong or irrelevant reason(s) given |
| Total: 1 | Needs both "no" and reason |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{t} = 11.76$ | B1 | 11.76 seen or implied |
| $\hat{\sigma}^2 = \frac{120}{119}\!\left(\frac{18737.712}{120} - 11.76^2\right) = 18$ | M1, M1 | Biased estimate ($= 17.85$). $\times 120/119$, or single formula with 119 divisor |
| | A1 | Answer $18 \pm 0.05$ |
| $H_0: \mu = 11.0$, $H_1: \mu \neq 11.0$ | B2 | One error, B1 |
| $\alpha$: $z = \frac{11.76 - 11.0}{\sqrt{18/120}} = \mathbf{1.9623}$ | M1 | Standardise with 120, ignore cc or $\sqrt{}$ errors |
| $> 1.645$ | A1, A1 | A.r.t. $(\pm)1.96$ or $p \in [0.0245, 0.025]$ www. Compare explicitly with $(\pm)1.645$ or 0.05, consistent with their $z$ or $p$ |
| $\beta$: CV $11.0 \pm 1.645 \times \sqrt{18/120} = 11.637$ (or 10.363). $11.76 > 11.64$ | M1, A1, A1 | $11.0 + z\sigma/\sqrt{120}$, needs 120 and $+$ or $\pm$. Ignore 10.363. Explicit comparison, consistent tail |
| Reject $H_0$. Significant evidence that the average time has changed. | M1, A1ft | Correct first conclusion. Contextualised, acknowledge uncertainty. FT on wrong CR/$z$/$p$ |
| **Total: 11** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No, the Central Limit Theorem applies | B1 | or "No, large sample". Withhold if extra wrong or irrelevant reason(s) given |
| **Total: 1** | | Needs both "no" and reason |
---6 Records for a doctors' surgery over a long period suggest that the time taken for a consultation, $T$ minutes, has a mean of 11.0. Following the introduction of new regulations, a doctor believes that the average time has changed. She finds that, with new regulations, the consultation times for a random sample of 120 patients can be summarised as
$$n = 120 , \Sigma t = 1411.20 , \Sigma t ^ { 2 } = 18737.712 .$$
(i) Test, at the $10 \%$ significance level, whether the doctor's belief is correct.\\
(ii) Explain whether, in answering part (i), it was necessary to assume that the consultation times were normally distributed.
\hfill \mbox{\textit{OCR S2 2015 Q6 [12]}}