| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring standard formulas: calculating sample mean/variance, constructing a confidence interval using the t-distribution, and applying the definition of confidence level to find P(both intervals miss μ) = 0.02². All steps are routine applications of learned procedures with no conceptual challenges beyond understanding what a confidence interval means. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{est}(\mu) = 9750/150 = (65)\) | B1 | |
| \(\text{est}(\sigma^2) = \frac{1}{149}\left(647500 - \frac{9750^2}{150}\right)\) | M1 | Correct subst. in correct formula |
| \(= 92.3\) (3 s.f.) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = 2.326\) | B1 | |
| \(`65' \pm z \times \frac{\sqrt{92.28188'}}{\sqrt{150}}\) | M1 | Any \(z\) |
| \(= 63.2\) to \(66.8\) (3 s.f.) | A1 [3] | Use of 'biased' can still score here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.02^2\) | M1 | Allow M1 for \(0.02\) seen |
| \(= 0.0004\) o.e. | A1 [2] |
## Question 4:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{est}(\mu) = 9750/150 = (65)$ | B1 | |
| $\text{est}(\sigma^2) = \frac{1}{149}\left(647500 - \frac{9750^2}{150}\right)$ | M1 | Correct subst. in correct formula |
| $= 92.3$ (3 s.f.) | A1 [3] | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = 2.326$ | B1 | |
| $`65' \pm z \times \frac{\sqrt{92.28188'}}{\sqrt{150}}$ | M1 | Any $z$ |
| $= 63.2$ to $66.8$ (3 s.f.) | A1 [3] | Use of 'biased' can still score here |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.02^2$ | M1 | Allow M1 for $0.02$ seen |
| $= 0.0004$ o.e. | A1 [2] | |
---4 The masses, in grams, of a certain type of plum are normally distributed with mean $\mu$ and variance $\sigma ^ { 2 }$. The masses, $m$ grams, of a random sample of 150 plums of this type were found and the results are summarised by $\Sigma m = 9750$ and $\Sigma m ^ { 2 } = 647500$.\\
(i) Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.\\
(ii) Calculate a 98\% confidence interval for $\mu$.
Two more random samples of plums of this type are taken and a $98 \%$ confidence interval for $\mu$ is calculated from each sample.\\
(iii) Find the probability that neither of these two intervals contains $\mu$.
\hfill \mbox{\textit{CAIE S2 2013 Q4 [8]}}