| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single normal population sample mean |
| Difficulty | Standard +0.3 This is a straightforward application of sampling distribution of the mean and hypothesis testing. Part (i) requires standardizing a sample mean using the Central Limit Theorem (routine calculation), while part (ii) involves finding critical values for a two-tailed test at 10% significance level. Both parts use standard procedures with no novel insight required, making it slightly easier than average for an S2 question. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a 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 |
|---|---|---|
| (i) \(z = \frac{2.55 - 2.62}{0.3/\sqrt{45}} = -1.565\) | M1 | Standardising no cc |
| \(P(z > -1.565) = 0.941\) | M1, A1 [3] | Dividing 0.3 by \(\sqrt{45}\) as denominator; Correct answer (Accept equivalent method using totals) |
| (ii) rejection region is \(m < a_1\) and \(m > a_2\) where \(\frac{a_1 - 2.62}{0.3/\sqrt{30}} = -1.645\) and \(\frac{a_2 - 2.62}{0.3/\sqrt{30}} = 1.645\) | B1, M1, M1 | ±1.645 seen; one correct unsimplified equation of correct form; second unsimplified equation of correct form (or clear use of 1-tail test and ±1.282 used) |
| \(m < 2.53\) and \(m > 2.71\) | A1 [4] | correct answer |
**(i)** $z = \frac{2.55 - 2.62}{0.3/\sqrt{45}} = -1.565$ | M1 | Standardising no cc
$P(z > -1.565) = 0.941$ | M1, A1 [3] | Dividing 0.3 by $\sqrt{45}$ as denominator; Correct answer (Accept equivalent method using totals)
**(ii)** rejection region is $m < a_1$ and $m > a_2$ where $\frac{a_1 - 2.62}{0.3/\sqrt{30}} = -1.645$ and $\frac{a_2 - 2.62}{0.3/\sqrt{30}} = 1.645$ | B1, M1, M1 | ±1.645 seen; one correct unsimplified equation of correct form; second unsimplified equation of correct form (or clear use of 1-tail test and ±1.282 used)
$m < 2.53$ and $m > 2.71$ | A1 [4] | correct answer3 Metal bolts are produced in large numbers and have lengths which are normally distributed with mean 2.62 cm and standard deviation 0.30 cm .\\
(i) Find the probability that a random sample of 45 bolts will have a mean length of more than 2.55 cm .\\
(ii) The machine making these bolts is given an annual service. This may change the mean length of bolts produced but does not change the standard deviation. To test whether the mean has changed, a random sample of 30 bolts is taken and their lengths noted. The sample mean length is $m \mathrm {~cm}$. Find the set of values of $m$ which result in rejection at the $10 \%$ significance level of the hypothesis that no change in the mean length has occurred.
\hfill \mbox{\textit{CAIE S2 2010 Q3 [7]}}