| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Unbiased estimates calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of standard hypothesis testing procedures: calculating an unbiased variance estimate (multiplying by n/(n-1)) and performing a routine one-sample t-test with clearly stated hypotheses. The question requires only direct application of formulas with no conceptual challenges or problem-solving insight, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sigma^2 = \frac{50}{49} \times 0.0967 = 0.0987\) | M1 | Use \(\frac{n}{n-1} \times s^2\) or \(s^2\), allow \(\sqrt{}\) |
| A1 2 | Answer, a.r.t. 0.0987 | |
| (ii) \(H_0: \mu = 1.8; H_1: \mu \neq 1.8\) | B1B1 | Hypotheses correctly stated in terms of μ [SR: μ wrong/omitted: B1 both, but \(\bar{X}\): B0] |
| where μ is the population mean | ||
| \((1.72 - 1.8) = -1.8(006)\) | M1 | Standardise with √n, allow +, biased σ, √ errors |
| \(\alpha, \beta: z = \frac{\sqrt{50}}{6/\sqrt{50}}\) | ||
| \(\alpha: \quad -1.8 < -1.645\) | B1 | Compare \(\pm z\) with ±1.645, align consistent |
| \(\beta: \quad \Phi(-1.8) = 1 - 0.9641 < 0.05\) | B1 | Explicitly compare Φ(z) with 0.05, correct tail |
| \(\gamma: \quad CV 1.8 - k\sigma/\sqrt{50}\) | M1 | Correct expression for CV, \(– or \pm\), k from \(\Phi^{-1}\) |
| \(k = 1.645, CV = 1.727\) | A1 | CV = 1.727, √ on their k, ignore upper limit |
| \(1.72 < 1.727\) | B1V | \(k = 1.645\) and compare CV with 1.72 |
| M1 | Reject \(H_0√\), correct method, needs √50, μ = 1.8; | |
| Reject \(H_0\) | allow cc, or k error or biased σ estimate | |
| Significant evidence that mean height is not 1.8 | A1V 7 | Conclusion stated in context [SR: 1.8, 1.72 interchanged: B0B0M1A0B1M0] |
**(i)** $\sigma^2 = \frac{50}{49} \times 0.0967 = 0.0987$ | M1 | Use $\frac{n}{n-1} \times s^2$ or $s^2$, allow $\sqrt{}$
| A1 2 | Answer, a.r.t. 0.0987
**(ii)** $H_0: \mu = 1.8; H_1: \mu \neq 1.8$ | B1B1 | Hypotheses correctly stated in terms of μ [SR: μ wrong/omitted: B1 both, but $\bar{X}$: B0]
where μ is the population mean | |
$(1.72 - 1.8) = -1.8(006)$ | M1 | Standardise with √n, allow +, biased σ, √ errors
$\alpha, \beta: z = \frac{\sqrt{50}}{6/\sqrt{50}}$ | |
$\alpha: \quad -1.8 < -1.645$ | B1 | Compare $\pm z$ with ±1.645, align consistent
$\beta: \quad \Phi(-1.8) = 1 - 0.9641 < 0.05$ | B1 | Explicitly compare Φ(z) with 0.05, correct tail
$\gamma: \quad CV 1.8 - k\sigma/\sqrt{50}$ | M1 | Correct expression for CV, $– or \pm$, k from $\Phi^{-1}$
$k = 1.645, CV = 1.727$ | A1 | CV = 1.727, √ on their k, ignore upper limit
$1.72 < 1.727$ | B1V | $k = 1.645$ and compare CV with 1.72
| M1 | Reject $H_0√$, correct method, needs √50, μ = 1.8;
Reject $H_0$ | | allow cc, or k error or biased σ estimate
Significant evidence that mean height is not 1.8 | A1V 7 | Conclusion stated in context [SR: 1.8, 1.72 interchanged: B0B0M1A0B1M0]4 The height of sweet pea plants grown in a nursery is a random variable. A random sample of 50 plants is measured and is found to have a mean height 1.72 m and variance $0.0967 \mathrm {~m} ^ { 2 }$.\\
(i) Calculate an unbiased estimate for the population variance of the heights of sweet pea plants.\\
(ii) Hence test, at the $10 \%$ significance level, whether the mean height of sweet pea plants grown by the nursery is 1.8 m , stating your hypotheses clearly.
\hfill \mbox{\textit{OCR S2 2005 Q4 [9]}}