| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (lower tail, H₁: p < p₀) |
| Difficulty | Standard +0.3 Part (i) is a standard one-tailed binomial hypothesis test with clearly stated hypotheses and significance level - routine S2 material. Part (ii) requires working backwards from a confidence interval limit to find the confidence level, which adds a modest problem-solving element but uses standard formulas. Overall slightly easier than average due to straightforward setup and small sample size. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: p = 0.15\), \(H_1: p < 0.15\) | B1 | Accept \(H_0: \mu = 9\), \(H_1: \mu < 9\). Use of Normal approximation: \(N(0.15, \frac{0.15 \times 0.85}{60})\) |
| \((N(60 \times 0.15,\ 60 \times 0.15 \times 0.85)) = N(9, 7.65)\) | \(= N(0.15, 0.002125)\) | |
| \(\frac{6.5 - 9}{\sqrt{7.65}}\) | M1 | For standardising (or \(\frac{\frac{6}{60}+0.5-0.15}{\sqrt{0.002125}} = -0.904\)). Allow wrong or no cc |
| \(= -0.904\) | A1 | Accept \(\pm\) |
| \(0.904 < 1.282\) | M1 | Valid comparison of z values or \(\phi(-0.904) = 0.183 > 0.1\); ft their 0.904 |
| No evidence train late less often | A1ft | Use of \(\text{Bin}(60, 0.15)\) to give \(P(\leq 6) = 0.1848\) M1A1; valid comparison with 0.1 M1 conclusion A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.1 + z \times \sqrt{\frac{0.1 \times 0.9}{60}} = 0.150\) | M1 | For \(\sqrt{0.1 \times 0.9 / 60}\) seen |
| M1 | for \(0.1 + z \times \ldots = 0.150\) or \(2z\ldots = 0.1\) | |
| \(z = 1.291\) | A1 | |
| \(\phi(1.291) = 0.90(16)\) | M1 | for correct method to find \(\alpha\) |
| \(\alpha = 80\) | A1ft | ft their \(z\). Must be a +ve non-zero integer \(< 100\) |
## Question 6(i):
| $H_0: p = 0.15$, $H_1: p < 0.15$ | B1 | Accept $H_0: \mu = 9$, $H_1: \mu < 9$. Use of Normal approximation: $N(0.15, \frac{0.15 \times 0.85}{60})$ |
|---|---|---|
| $(N(60 \times 0.15,\ 60 \times 0.15 \times 0.85)) = N(9, 7.65)$ | | $= N(0.15, 0.002125)$ |
| $\frac{6.5 - 9}{\sqrt{7.65}}$ | M1 | For standardising (or $\frac{\frac{6}{60}+0.5-0.15}{\sqrt{0.002125}} = -0.904$). Allow wrong or no cc |
| $= -0.904$ | A1 | Accept $\pm$ |
| $0.904 < 1.282$ | M1 | Valid comparison of z values or $\phi(-0.904) = 0.183 > 0.1$; ft their 0.904 |
| No evidence train late less often | A1ft | Use of $\text{Bin}(60, 0.15)$ to give $P(\leq 6) = 0.1848$ M1A1; valid comparison with 0.1 M1 conclusion A1ft |
---
## Question 6(ii):
| $0.1 + z \times \sqrt{\frac{0.1 \times 0.9}{60}} = 0.150$ | M1 | For $\sqrt{0.1 \times 0.9 / 60}$ seen |
|---|---|---|
| | M1 | for $0.1 + z \times \ldots = 0.150$ or $2z\ldots = 0.1$ |
| $z = 1.291$ | A1 | |
| $\phi(1.291) = 0.90(16)$ | M1 | for correct method to find $\alpha$ |
| $\alpha = 80$ | A1ft | ft their $z$. Must be a +ve non-zero integer $< 100$ |
---6 In the past, Angus found that his train was late on $15 \%$ of his daily journeys to work. Following a timetable change, Angus found that out of 60 randomly chosen days, his train was late on 6 days.\\
(i) Test at the $10 \%$ significance level whether Angus' train is late less often than it was before the timetable change.\\
Angus used his random sample to find an $\alpha \%$ confidence interval for the proportion of days on which his train is late. The upper limit of his interval was 0.150 , correct to 3 significant figures.\\
(ii) Calculate the value of $\alpha$ correct to the nearest integer.\\
\hfill \mbox{\textit{CAIE S2 2018 Q6 [10]}}