| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a standard two-tail z-test with large sample (n=120) allowing use of normal approximation. Part (i) is routine hypothesis testing procedure. Part (ii) requires understanding of Type I error and sample size calculation, which is slightly beyond basic application but still follows a standard formula. The question is slightly above average difficulty due to the Type I error concept and sample size estimation, but remains a textbook-style exercise with no novel insight 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 78.0\) | B1 | Both correct, B2 |
| \(H_1: \mu \neq 78.0\) | B1 | One error, B1, but \(x\) or \(\bar{x}\): B0 |
| \(z = \dfrac{76.4 - 78.0}{\sqrt{68.9/120}} = -2.1115\) | M1, A1 | Needs \(\pm(76.4-78)/\sqrt{(\sigma \div 120)}\), allow \(\sqrt{}\) errors. art \(-2.11\), or \(p = 0.0173 \pm 0.0002\) |
| \(> -2.576\) or \(0.0173 > 0.005\) | B1 | Compare \(z\) with \((-)2.576\), or \(p\) with 0.005 |
| \(78 \pm z\sqrt{68.9/120} = 76.048\) | M1 | Needs 78 and 120, can be \(-\) only |
| \(76.4 > 76.048\) | A1\(\sqrt{}\), B1 | Correct CV to 3 sf, \(\sqrt{}\) on \(z\). \(z = 2.576\) and compare 76.4, allow from \(78 \leftrightarrow 76.4\) |
| Do not reject \(H_0\). Insufficient evidence that the mean time has changed | M1, A1\(\sqrt{}\) 7 | Correct comparison & conclusion, needs 120. "like with like", correct tail, \(\bar{x}\) and \(\mu\) right way round. Contextualised, some indication of uncertainty |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1}{\sqrt{68.9/n}} > 2.576\) | M1 | IGNORE INEQUALITIES THROUGHOUT. Standardise 1 with \(n\) and 2.576, allow \(\sqrt{}\) errors, cc etc but *not* 2.326 |
| \(\sqrt{n} > 21.38\) | M1 | Correct method to solve for \(\sqrt{n}\) (*not* from \(n\)) |
| \(n_{\min} = \mathbf{458}\) | A1 | 458 only (*not* 457), *or* 373 from 2.326, signs correct |
| Variance is estimated | B1 4 | Equivalent statement, allow "should use \(t\)". In principle nothing superfluous, but "variance stays same" B1 bod |
# Question 8:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 78.0$ | B1 | Both correct, B2 |
| $H_1: \mu \neq 78.0$ | B1 | One error, B1, but $x$ or $\bar{x}$: B0 |
| $z = \dfrac{76.4 - 78.0}{\sqrt{68.9/120}} = -2.1115$ | M1, A1 | Needs $\pm(76.4-78)/\sqrt{(\sigma \div 120)}$, allow $\sqrt{}$ errors. art $-2.11$, or $p = 0.0173 \pm 0.0002$ |
| $> -2.576$ or $0.0173 > 0.005$ | B1 | Compare $z$ with $(-)2.576$, or $p$ with 0.005 |
| $78 \pm z\sqrt{68.9/120} = 76.048$ | M1 | Needs 78 and 120, can be $-$ only |
| $76.4 > 76.048$ | A1$\sqrt{}$, B1 | Correct CV to 3 sf, $\sqrt{}$ on $z$. $z = 2.576$ and compare 76.4, allow from $78 \leftrightarrow 76.4$ |
| Do not reject $H_0$. Insufficient evidence that the mean time has changed | M1, A1$\sqrt{}$ **7** | Correct comparison & conclusion, needs 120. "like with like", correct tail, $\bar{x}$ and $\mu$ right way round. Contextualised, some indication of uncertainty |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1}{\sqrt{68.9/n}} > 2.576$ | M1 | IGNORE INEQUALITIES THROUGHOUT. Standardise 1 with $n$ and 2.576, allow $\sqrt{}$ errors, cc etc but *not* 2.326 |
| $\sqrt{n} > 21.38$ | M1 | Correct method to solve for $\sqrt{n}$ (*not* from $n$) |
| $n_{\min} = \mathbf{458}$ | A1 | 458 only (*not* 457), *or* 373 from 2.326, signs correct |
| Variance is estimated | B1 **4** | Equivalent statement, allow "should use $t$". In principle nothing superfluous, but "variance stays same" B1 bod |8 In a large company the time taken for an employee to carry out a certain task is a normally distributed random variable with mean 78.0 s and unknown variance. A new training scheme is introduced and after its introduction the times taken by a random sample of 120 employees are recorded. The mean time for the sample is 76.4 s and an unbiased estimate of the population variance is $68.9 \mathrm {~s} ^ { 2 }$.\\
(i) Test, at the $1 \%$ significance level, whether the mean time taken for the task has changed.\\
(ii) It is required to redesign the test so that the probability of making a Type I error is less than 0.01 when the sample mean is 77.0 s . Calculate an estimate of the smallest sample size needed, and explain why your answer is only an estimate.
\hfill \mbox{\textit{OCR S2 2009 Q8 [11]}}