| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (upper tail) |
| Difficulty | Moderate -0.3 This is a straightforward one-tail z-test with clearly stated parameters and a large sample size. The main steps (stating H₀ and H₁, calculating z-statistic, comparing to critical value) are routine for S2 level. Part (ii) tests conceptual understanding but requires only brief reasoning about normality already being given. Slightly easier than average due to all information being explicitly provided and no complex interpretation needed. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume sd still 4.8 or is unchanged | B1 | or: the 150 times can be treated as a random sample / are independent |
| \(H_0\): Pop mean \(= 26.5\); \(H_1\): Pop mean \(> 26.5\) | B1 | Allow '\(\mu\)' but not just 'mean' |
| \(\dfrac{27.5 - 26.5}{\frac{4.8}{\sqrt{150}}}\) | M1 | Standardise, with \(\sqrt{}\); accept CV method |
| \(= 2.552\) | A1 | |
| Compare with \(z\)-value: \(2.552 > 2.326\) | M1 | or comp \(1 - \Phi(2.552)\) with \(0.01\); \(1 - 0.9946 = 0.0054 < 0.01\) |
| There is evidence time has increased | A1ft | No contradictions; (2 tail test scores max. B1 B0 M1 A1 M1 for comparison with 2.576, A0 no ft) |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No because pop is normal so distr of \(\bar{X}\) is normal | B1 | Condone just 'No because pop is normal' |
| Total: 1 |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume sd still 4.8 or is unchanged | **B1** | or: the 150 times can be treated as a random sample / are independent |
| $H_0$: Pop mean $= 26.5$; $H_1$: Pop mean $> 26.5$ | **B1** | Allow '$\mu$' but not just 'mean' |
| $\dfrac{27.5 - 26.5}{\frac{4.8}{\sqrt{150}}}$ | **M1** | Standardise, with $\sqrt{}$; accept CV method |
| $= 2.552$ | **A1** | |
| Compare with $z$-value: $2.552 > 2.326$ | **M1** | or comp $1 - \Phi(2.552)$ with $0.01$; $1 - 0.9946 = 0.0054 < 0.01$ |
| There is evidence time has increased | **A1ft** | No contradictions; (2 tail test scores max. B1 B0 M1 A1 M1 for comparison with 2.576, A0 no ft) |
| | **6** | |
## Question 2(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| No because pop is normal so distr of $\bar{X}$ is normal | B1 | Condone just 'No because pop is normal' |
| **Total: 1** | | |
---2 The time, in minutes, that John takes to travel to work has a normal distribution. Last year the mean and standard deviation were 26.5 and 4.8 respectively. This year John uses a different route and he finds that the mean time for his first 150 journeys is 27.5 minutes.\\
(i) Stating a necessary assumption, test at the $1 \%$ significance level whether the mean time for his journey to work has increased.\\
(ii) State, with a reason, whether it was necessary to use the Central Limit theorem in your answer to part (i).\\
\hfill \mbox{\textit{CAIE S2 2019 Q2 [7]}}