| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Hypothesis test for mean |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem with standard normal distribution calculations and a routine two-tailed hypothesis test. All steps are procedural with clear guidance given in the question structure, requiring only direct application of learned techniques rather than problem-solving insight. |
| Spec | 2.05c Significance levels: one-tail and two-tail5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{40 - 38.4}{\frac{6.9}{\sqrt{30}}} = 1.270 \quad \frac{38 - 38.4}{\frac{6.9}{\sqrt{30}}} = -0.3175\) | M1 | M1 for either correct expression must have \(\sqrt{30}\) (condone continuity correction) |
| A1 | A1 for \(\pm 1.270\) or for 1.27 or AWRT | |
| A1 | A1 for \(\pm(-0.3175)\) must be opposite sign or for 0.317 or 0.318 or AWRT | |
| \(\Phi(\text{'1.270'}) - (1 - \phi(\text{'0.3175'}))\) | M1 | For correct method consistent with *their* values |
| \(= 0.523\) (3 sf) or 0.522 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 2-tail because looking for 'change', not decrease or increase | B1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Population mean journey time (or \(\mu\)) \(= 38.4\) ; \(H_1\): Population mean journey time (or \(\mu\)) \(\neq 38.4\) | B1 | Not just 'mean journey time' |
| \(\dfrac{40.2 - 38.4}{\dfrac{6.9}{\sqrt{30}}}\) | M1 | For standardising (must have \(\sqrt{30}\)) |
| \(= 1.429\) | A1 | |
| \('1.429' < 1.645\) | M1 | For valid comparison (area comparison \(0.0765 > 0.05\)) |
| There is no evidence that mean journey time has changed. | A1 FT | In context. Not definite (e.g. not 'mean journey time has not changed'). No contradictions. FT *their* \('1.429'\) (Note use of 1-tail test scores B0 M1A1M1(comparison with 1.282) A0 max) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Population mean journey time (or \(\mu\)) \(= 38.4\) ; \(H_1\): Population mean journey time (or \(\mu\)) \(\neq 38.4\) | B1 | Not just 'mean journey time' |
| \(38 + 1.645\left(\dfrac{6.9}{\sqrt{30}}\right)\) | M1 | |
| \(= 40.47\) | A1 | |
| \(40.2 < 40.47\) | M1 | For valid comparison |
| There is no evidence that mean journey time has changed. | A1 FT | In context. Not definite (e.g. not 'mean journey time has not changed'). No contradictions. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Yes, because population distribution unknown. | B1 | Allow: Yes, because population distribution not normal. |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{40 - 38.4}{\frac{6.9}{\sqrt{30}}} = 1.270 \quad \frac{38 - 38.4}{\frac{6.9}{\sqrt{30}}} = -0.3175$ | M1 | M1 for either correct expression must have $\sqrt{30}$ (condone continuity correction) |
| | A1 | A1 for $\pm 1.270$ or for 1.27 or AWRT |
| | A1 | A1 for $\pm(-0.3175)$ must be opposite sign or for 0.317 or 0.318 or AWRT |
| $\Phi(\text{'1.270'}) - (1 - \phi(\text{'0.3175'}))$ | M1 | For correct method consistent with *their* values |
| $= 0.523$ (3 sf) or 0.522 | A1 | |
## Question 6(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| 2-tail because looking for 'change', not decrease or increase | B1 | OE |
## Question 6(b)(ii):
**Standard Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Population mean journey time (or $\mu$) $= 38.4$ ; $H_1$: Population mean journey time (or $\mu$) $\neq 38.4$ | B1 | Not just 'mean journey time' |
| $\dfrac{40.2 - 38.4}{\dfrac{6.9}{\sqrt{30}}}$ | M1 | For standardising (must have $\sqrt{30}$) |
| $= 1.429$ | A1 | |
| $'1.429' < 1.645$ | M1 | For valid comparison (area comparison $0.0765 > 0.05$) |
| There is no evidence that mean journey time has changed. | A1 FT | In context. Not definite (e.g. not 'mean journey time has not changed'). No contradictions. FT *their* $'1.429'$ (Note use of 1-tail test scores B0 M1A1M1(comparison with 1.282) A0 max) |
**Alternative Method – Critical Values Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Population mean journey time (or $\mu$) $= 38.4$ ; $H_1$: Population mean journey time (or $\mu$) $\neq 38.4$ | B1 | Not just 'mean journey time' |
| $38 + 1.645\left(\dfrac{6.9}{\sqrt{30}}\right)$ | M1 | |
| $= 40.47$ | A1 | |
| $40.2 < 40.47$ | M1 | For valid comparison |
| There is no evidence that mean journey time has changed. | A1 FT | In context. Not definite (e.g. not 'mean journey time has not changed'). No contradictions. |
**Total: 5 marks**
---
## Question 6(b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Yes, because population distribution unknown. | B1 | Allow: Yes, because population distribution not normal. |
**Total: 1 mark**6 The time, in minutes, for Anjan's journey to work on Mondays has mean 38.4 and standard deviation 6.9.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Anjan's mean journey time for a random sample of 30 Mondays is between 38 and 40 minutes.\\
Anjan wishes to test whether his mean journey time is different on Tuesdays. He chooses a random sample of 30 Tuesdays and finds that his mean journey time for these 30 Tuesdays is 40.2 minutes. Assume that the standard deviation for his journey time on Tuesdays is 6.9 minutes.
\item \begin{enumerate}[label=(\roman*)]
\item State, with a reason, whether Anjan should use a one-tail or a two-tail test.
\item Carry out the test at the $10 \%$ significance level.
\item Explain whether it was necessary to use the Central Limit theorem in part (b)(ii).\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q6 [12]}}