| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Moderate -0.8 This question tests routine confidence interval calculation for a proportion (part a), standard unbiased estimator formulas (part b), and basic understanding of random sampling (part c). All parts are direct application of memorized formulas and definitions with no problem-solving or novel insight required. The calculations are straightforward and this represents a typical, below-average difficulty S2 question. |
| Spec | 2.01a Population and sample: terminology5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\frac{102}{250} \times \frac{250-102}{250}}{250}\) (= 0.000966144) | M1 | Any \(z\) but must be a \(z\) value. One side of the interval scores M1 |
| \(\frac{102}{250} \pm z\sqrt{0.00096614}\) | ||
| \(z = 1.645\) | B1 | |
| Confidence Interval is 0.357 to 0.459 (3 sf) | A1 | Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Estimate of mean \(\left(\frac{50460}{250}\right) = \\)201.84\( | B1 | Allow without units. Allow 3 s.f. \)\\(202\) |
| \(\frac{250}{249}\left(\frac{19854200}{250} - \left(\frac{50460}{250}\right)^2\right)\) or \(\frac{1}{249}\left(19854200 - \frac{50460^2}{250}\right)\) | M1 | |
| Estimate of variance = 38 832.75 dollars\(^2\) or 38 800 (3 sf) | A1 | Allow with missing units. Calculation of biased gives 38 700 scores M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. Every house doesn't have an equal chance of being selected or most houses have no chance of being selected | B1 | Or other similar e.g. Houses in streets with few houses are more likely to be selected. Not just 'biased', OE, without explanation |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\frac{102}{250} \times \frac{250-102}{250}}{250}$ (= 0.000966144) | M1 | Any $z$ but must be a $z$ value. One side of the interval scores M1 |
| $\frac{102}{250} \pm z\sqrt{0.00096614}$ | | |
| $z = 1.645$ | B1 | |
| Confidence Interval is 0.357 to 0.459 (3 sf) | A1 | Must be an interval |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Estimate of mean $\left(\frac{50460}{250}\right) = \$201.84$ | B1 | Allow without units. Allow 3 s.f. $\$202$ |
| $\frac{250}{249}\left(\frac{19854200}{250} - \left(\frac{50460}{250}\right)^2\right)$ or $\frac{1}{249}\left(19854200 - \frac{50460^2}{250}\right)$ | M1 | |
| Estimate of variance = 38 832.75 dollars$^2$ or 38 800 (3 sf) | A1 | Allow with missing units. Calculation of biased gives 38 700 scores M0A0 |
## Question 2(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. Every house doesn't have an equal chance of being selected or most houses have no chance of being selected | B1 | Or other similar e.g. Houses in streets with few houses are more likely to be selected. Not just 'biased', OE, without explanation |
---2 In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel.
\begin{enumerate}[label=(\alph*)]
\item It was found that 102 adults in the sample travel by bus. Find an approximate $90 \%$ confidence interval for the proportion of all the adults in Fromleigh who travel by bus.
\item The survey included a question about the amount, $x$ dollars, spent on travel per year. The results are summarised as follows.
$$n = 250 \quad \Sigma x = 50460 \quad \Sigma x ^ { 2 } = 19854200$$
Find unbiased estimates of the population mean and variance of the amount spent per year on travel.\\
A councillor wanted to select a random sample of houses in Fromleigh. He planned to select the first house on each of the 143 streets in Fromleigh.
\item Explain why this would not provide a random sample.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q2 [7]}}