| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard calculations (sample mean/variance, CI formula) and basic interpretation of whether a claimed value lies within the interval. The multi-part structure and interpretation component add slight complexity beyond pure recall, but all techniques are routine for S2 level with no novel problem-solving required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Est}(\mu) = 331(.125)\) | B1 | |
| \(\text{Est}(\sigma^2) = \dfrac{8}{7}\left(\dfrac{\text{"877179"}}{8} - \text{"331.125"}^2\right)\) | M1 | Allow their \(\Sigma x^2\) |
| \(= 4.125\) or \(4.13\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 2.326\) | B1 | |
| \(331 \pm z \times \sqrt{\dfrac{4.2}{50}}\) | M1 | Allow incorrect \(z\) \((\neq 1, 0)\), not a prob |
| \(= 330\) to \(332\) (3 sf) | A1 [3] | Ignore brackets, if given. CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No, because \(333\) is not within CI | B1ft [1] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Est}(\mu) = 331(.125)$ | B1 | |
| $\text{Est}(\sigma^2) = \dfrac{8}{7}\left(\dfrac{\text{"877179"}}{8} - \text{"331.125"}^2\right)$ | M1 | Allow their $\Sigma x^2$ |
| $= 4.125$ or $4.13$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 2.326$ | B1 | |
| $331 \pm z \times \sqrt{\dfrac{4.2}{50}}$ | M1 | Allow incorrect $z$ $(\neq 1, 0)$, not a prob |
| $= 330$ to $332$ (3 sf) | A1 [3] | Ignore brackets, if given. CWO |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No, because $333$ is not within CI | B1ft [1] | |
---4 The volumes of juice in bottles of Apricola are normally distributed. In a random sample of 8 bottles, the volumes of juice, in millilitres, were found to be as follows.
$$\begin{array} { l l l l l l l l }
332 & 334 & 330 & 328 & 331 & 332 & 329 & 333
\end{array}$$
(i) Find unbiased estimates of the population mean and variance.
A random sample of 50 bottles of Apricola gave unbiased estimates of 331 millilitres and 4.20 millilitres ${ } ^ { 2 }$ for the population mean and variance respectively.\\
(ii) Use this sample of size 50 to calculate a $98 \%$ confidence interval for the population mean.\\
(iii) The manufacturer claims that the mean volume of juice in all bottles is 333 millilitres. State, with a reason, whether your answer to part (ii) supports this claim.
\hfill \mbox{\textit{CAIE S2 2011 Q4 [7]}}