| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Moderate -0.8 This is a straightforward confidence interval question requiring standard formulas for mean and variance, then a routine interpretation. The calculations are mechanical (finding sample mean/variance, applying z-critical value), and part (ii) simply asks whether 100 lies in the interval. Part (iii) is basic statistical understanding. No complex reasoning or novel problem-solving required, making it easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Est}(\mu) = \frac{14910}{150}\) \((= 99.4)\) | B1 | |
| \(\text{Est}(\sigma^2) = \frac{150}{149}\left(\frac{1525000}{150} - \text{"99.4"}^2\right) = 288.228\) | M1 A1 | Allow M1 if \(\frac{150}{149}\) omitted |
| \(z = 2.576\) | B1 | Accept 2.574–2.579 |
| \(\text{"99.4"} \pm z \times \sqrt{288.228 \div 150}\) | M1 | Any \(z\) |
| \(\text{CI} = 95.8\) to \(103\) (3 sf) | A1 [6] | NB use of biased Var can score 5/6 max |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 100 lies within this CI, hence yes | B1\(\checkmark\) [1] | Both needed, ft their CI |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| To avoid bias OR necessary to enable statistical inference | B1 [1] | Or any equivalent |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Est}(\mu) = \frac{14910}{150}$ $(= 99.4)$ | **B1** | |
| $\text{Est}(\sigma^2) = \frac{150}{149}\left(\frac{1525000}{150} - \text{"99.4"}^2\right) = 288.228$ | **M1 A1** | Allow **M1** if $\frac{150}{149}$ omitted |
| $z = 2.576$ | **B1** | Accept 2.574–2.579 |
| $\text{"99.4"} \pm z \times \sqrt{288.228 \div 150}$ | **M1** | Any $z$ |
| $\text{CI} = 95.8$ to $103$ (3 sf) | **A1** [6] | NB use of biased Var can score 5/6 max |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 100 lies within this CI, hence yes | **B1$\checkmark$** [1] | Both needed, ft their CI |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| To avoid bias OR necessary to enable statistical inference | **B1** [1] | Or any equivalent |
---5 The 150 oranges in a random sample from a certain supplier were weighed and the masses, $X$ grams, were recorded. The results are summarised below.
$$n = 150 \quad \Sigma x = 14910 \quad \Sigma x ^ { 2 } = 1525000$$
(i) Calculate a $99 \%$ confidence interval for the population mean of $X$.\\
(ii) The supplier claims that the mean mass of his oranges is 100 grams. Use your answer to part (i) to explain whether this claim should be accepted.\\
(iii) State briefly why the sample should be random.
\hfill \mbox{\textit{CAIE S2 2016 Q5 [8]}}