| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (lower tail) |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard S2 content. Part (i) is a routine one-sample z-test with given population variance. Part (ii) requires recognizing when to use a t-test instead, which is a common textbook scenario. Part (iii) tests basic understanding of CLT applicability. The question requires no novel insight—just systematic application of learned procedures with clear signposting of what to do at each stage. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 13.3, \quad H_1: \mu < 13.3\) | B2 | Both correct: B2. One error [e.g. \(p, z\), no symbol] B1, but \(\bar{x}\) etc B0 |
| \(\alpha: \quad z = \frac{12.48 - 13.3}{\sqrt{1225/50}} = -1.6566 \quad [p = 0.0488]\) | M1 | Standardise with \(\sqrt{50}\), allow \(\sqrt{}\) errors, allow cc, allow 13.3 \(-\) 12.48 |
| A1 | \(z\) in range \([-1.66, -1.65]\), or \(p\) in range \([0.04875, 0.0489]\), allow 0.9512 only if consistent | |
| B1 | Compare with \(-1.645\), allow \(+1.6566\) with \(+1.645\), or \(p\) with \(0.05/0.95\) as consistent | |
| \(\beta: \quad \text{CV } 13.3 - 1.645\sqrt{\frac{12.25}{50}} = 12.4857...\) | M1 | \(13.3 - z\sigma/\sqrt{50}\), any recognisable \(z\), allow \(\sqrt{}\) errors etc, ignore 13.3 \(+\) ... |
| B1 | \(z = -1.645\) | |
| A1 | Compare 12.49 (or better) with 12.48, ignore 13.3 \(+\) ... | |
| SC: 2-tailed, 12.33 gets B1B0 M1B0A1ft M1A1 (6 marks) | ||
| Consistent, needs \(\sqrt{50}\), like-with-like comparison, hypotheses \(\textit{not}\) 12.48 | ||
| Contextualised, acknowledge uncertainty, their \(z\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sigma^2 = \frac{50}{49} \times 12.25 = [=12.5]\) | M1 | Multiply 12.25 by 50/49, allow \(\sqrt{}\) etc, allow if done in part (i) but then 0 |
| \(z = \frac{12.48 - 13.3}{\sqrt{12.5/50}} = -1.64\) | M1 | Standardise with \(\sqrt{50}\) |
| A1 | Obtain a.r.t. \(-1.64\), allow \(+1.64\) if consistent with (i). | |
| \(> -1.645\) | B1 | Compare with same CV as in (i) |
| \(\text{Opposite conclusion}\) | A1ft | State opposite conclusion (ft), any form, allow \(\bar{x}/\mu\) here , needs M1M1 (5 marks) |
| Identical mark scheme for method \(\beta\). CV 12.4775 | ||
| SC1: 50 omitted consistently in both: M1M0A0B1A1 max 3/5 | ||
| SC2: no \(\sqrt{50}\) in (i), \(\sqrt{50}\) but not 49/49 in (ii): M0M1A0B1A1 max 3/5 | ||
| SC3: \(\bar{x}\) and \(\mu\) confused consistently: can get B0 M1A1 B1 M0 (5 marks) | ||
| SC4: 50/49 used in (i): can get B2 M1A0B1 M1 in (ii), M1 in (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Yes as population not known to be normal | B1 | Not "\(n\) large" unless "Yes, not known normal, but \(n\) large so can use" (1 mark) |
| No wrong extras, e.g. "depends on whether it's sample or population" |
## (i)
$H_0: \mu = 13.3, \quad H_1: \mu < 13.3$ | B2 | Both correct: B2. One error [e.g. $p, z$, no symbol] B1, but $\bar{x}$ etc B0
$\alpha: \quad z = \frac{12.48 - 13.3}{\sqrt{1225/50}} = -1.6566 \quad [p = 0.0488]$ | M1 | Standardise with $\sqrt{50}$, allow $\sqrt{}$ errors, allow cc, allow 13.3 $-$ 12.48
| A1 | $z$ in range $[-1.66, -1.65]$, or $p$ in range $[0.04875, 0.0489]$, allow 0.9512 only if consistent
| | | B1 | Compare with $-1.645$, allow $+1.6566$ with $+1.645$, or $p$ with $0.05/0.95$ as consistent
$\beta: \quad \text{CV } 13.3 - 1.645\sqrt{\frac{12.25}{50}} = 12.4857...$ | M1 | $13.3 - z\sigma/\sqrt{50}$, any recognisable $z$, allow $\sqrt{}$ errors etc, ignore 13.3 $+$ ...
| B1 | $z = -1.645$
| A1 | Compare 12.49 (or better) with 12.48, ignore 13.3 $+$ ...
| | SC: 2-tailed, 12.33 gets B1B0 M1B0A1ft M1A1 (6 marks)
| | Consistent, needs $\sqrt{50}$, like-with-like comparison, hypotheses $\textit{not}$ 12.48
| | Contextualised, acknowledge uncertainty, their $z$
## (ii)
$\sigma^2 = \frac{50}{49} \times 12.25 = [=12.5]$ | M1 | Multiply 12.25 by 50/49, allow $\sqrt{}$ etc, allow if done in part (i) but then 0
$z = \frac{12.48 - 13.3}{\sqrt{12.5/50}} = -1.64$ | M1 | Standardise with $\sqrt{50}$
| A1 | Obtain a.r.t. $-1.64$, allow $+1.64$ if consistent with (i).
$> -1.645$ | B1 | Compare with same CV as in (i)
$\text{Opposite conclusion}$ | A1ft | State opposite conclusion (ft), any form, allow $\bar{x}/\mu$ here , needs M1M1 (5 marks)
| | | Identical mark scheme for method $\beta$. CV 12.4775
| | SC1: 50 omitted consistently in both: M1M0A0B1A1 max 3/5
| | SC2: no $\sqrt{50}$ in (i), $\sqrt{50}$ but not 49/49 in (ii): M0M1A0B1A1 max 3/5
| | SC3: $\bar{x}$ and $\mu$ confused consistently: can get B0 M1A1 B1 M0 (5 marks)
| | SC4: 50/49 used in (i): can get B2 M1A0B1 M1 in (ii), M1 in (ii)
## (iii)
Yes as population not known to be normal | B1 | Not "$n$ large" unless "Yes, not known normal, but $n$ large so can use" (1 mark)
| | No wrong extras, e.g. "depends on whether it's sample or population"It is known that the lifetime of a certain species of animal in the wild has mean 13.3 years. A zoologist reads a study of 50 randomly chosen animals of this species that have been kept in zoos. According to the study, for these 50 animals the sample mean lifetime is 12.48 years and the population variance is 12.25 years$^2$.
\begin{enumerate}[label=(\roman*)]
\item Test at the 5% significance level whether these results provide evidence that animals of this species that have been kept in zoos have a shorter expected lifetime than those in the wild. [7]
\item Subsequently the zoologist discovered that there had been a mistake in the study. The quoted variance of 12.25 years$^2$ was in fact the sample variance. Determine whether this makes a difference to the conclusion of the test. [5]
\item Explain whether the Central Limit Theorem is needed in these tests. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2016 Q8 [13]}}