| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Moderate -0.3 Part (i) is a standard confidence interval calculation requiring sample mean/variance computation and z-value lookup—routine A-level statistics. Part (ii) tests conceptual understanding that higher confidence requires wider intervals, which is basic theory. Both parts are textbook exercises with no novel problem-solving required, making this slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{est}(\mu) = \frac{25110}{50} = 502.2\) | B1 | |
| \(\text{est}(\sigma^2) = \frac{50}{49}\left(\frac{12610300}{50} - \left(\frac{25110}{50}\right)^2\right) = \frac{50}{49} \times \frac{58}{50} = 1.1836\) | M1 | OE |
| \(1.18\) (3 sf) or \(\frac{58}{49}\) | A1 | Accept SD \(= 1.0879\) |
| \(z = 2.054\) or \(2.055\) | B1 | |
| \(502.2 \pm z \times \frac{\sqrt{1.1836}}{\sqrt{50}}\) | M1 | Must be of correct form |
| \(501.9\) to \(502.5\) (1dp) | A1 | CWO. Must be in interval. SC accept use of biased variance (1.16) for M1 A1 |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| More confident or \(z\) would be greater, hence wider. | B1 | OE. Reason needed |
| 1 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{est}(\mu) = \frac{25110}{50} = 502.2$ | B1 | |
| $\text{est}(\sigma^2) = \frac{50}{49}\left(\frac{12610300}{50} - \left(\frac{25110}{50}\right)^2\right) = \frac{50}{49} \times \frac{58}{50} = 1.1836$ | M1 | OE |
| $1.18$ (3 sf) or $\frac{58}{49}$ | A1 | Accept SD $= 1.0879$ |
| $z = 2.054$ or $2.055$ | B1 | |
| $502.2 \pm z \times \frac{\sqrt{1.1836}}{\sqrt{50}}$ | M1 | Must be of correct form |
| $501.9$ to $502.5$ (1dp) | A1 | CWO. Must be in interval. SC accept use of biased variance (1.16) for M1 A1 |
| | **6** | |
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| More confident **or** $z$ would be greater, hence wider. | B1 | OE. Reason needed |
| | **1** | |3 The masses, in grams, of bags of flour are normally distributed with mean $\mu$. The masses, $m$ grams, of a random sample of 50 bags are summarised by $\Sigma m = 25110$ and $\Sigma m ^ { 2 } = 12610300$.\\
(i) Calculate a $96 \%$ confidence interval for $\mu$, giving the end-points correct to 1 decimal place.\\
Another random sample of 50 bags of flour is taken and a $99 \%$ confidence interval for $\mu$ is calculated.\\
(ii) Without calculation, state whether this confidence interval will be wider or narrower than the confidence interval found in part (i). Give a reason for your answer.\\
\hfill \mbox{\textit{CAIE S2 2019 Q3 [7]}}