| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for hypothesis testing |
| Difficulty | Standard +0.8 This question requires understanding of CLT justification, hypothesis testing mechanics, and Type I/II error calculations. Part (ii) demands conceptual understanding of when CLT is needed (non-normal population, large n). Part (iv) involves calculating P(Type II error) which requires finding the probability of failing to reject H₀ when μ has shifted—a multi-step calculation involving standardization under the alternative hypothesis. This goes beyond routine hypothesis testing to test deeper statistical understanding. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Assume sd unchanged or 4500; \(H_0\): Pop mean \(= 34600\); \(H_1\): Pop mean \(> 34600\); \(\frac{35400 - 34600}{\frac{4500}{\sqrt{90}}}\); \(= 1.687/1.686\ (1.69)\); cf \(1.645 < 1.686\); Evidence that mean wkly profit has increased | B1, B1, M1, A1, M1, A1f [6] |
| (ii) | Distr'n of \(X\) unknown; Yes | B1*, B1* dep [2] |
| (iii) | \(0.05\) or \(5\%\) | B1 [1] |
| (iv) | \(\frac{a - 34600}{\frac{4500}{\sqrt{90}}} = 1.645\); \(a = 35380\); \(\frac{35380 - 36500}{\frac{4500}{\sqrt{90}}}\ (= -2.361)\); \(1 - \Phi(2.361)\); \(= 0.0091\) | M1, A1, M1, M1, A1 [6] |
## Question 7:
**(i)** | Assume sd unchanged or 4500; $H_0$: Pop mean $= 34600$; $H_1$: Pop mean $> 34600$; $\frac{35400 - 34600}{\frac{4500}{\sqrt{90}}}$; $= 1.687/1.686\ (1.69)$; cf $1.645 < 1.686$; Evidence that mean wkly profit has increased | B1, B1, M1, A1, M1, A1f [6] | Both, allow just $\mu$ but not just "mean"; allow without $\sqrt{90}$; valid comparison (or $0.0458/0.0459 < 0.05$ or $35380 < 35400$ or $34600 < 34620$); if $H_1: \neq$ and 1.96 used, max B1B0M1A1M1A1f, no contradictions |
**(ii)** | Distr'n of $X$ unknown; Yes | B1*, B1* dep [2] | Allow not Normal |
**(iii)** | $0.05$ or $5\%$ | B1 [1] |
**(iv)** | $\frac{a - 34600}{\frac{4500}{\sqrt{90}}} = 1.645$; $a = 35380$; $\frac{35380 - 36500}{\frac{4500}{\sqrt{90}}}\ (= -2.361)$; $1 - \Phi(2.361)$; $= 0.0091$ | M1, A1, M1, M1, A1 [6] | Attempt to find cv, must see $(+)\ 1.645$, allow without $\sqrt{90}$; if found in (i) award when used; standardising with their "CV", must use $\sqrt{90}$; correct tail |7 In the past the weekly profit at a store had mean $\$ 34600$ and standard deviation $\$ 4500$. Following a change of ownership, the mean weekly profit for 90 randomly chosen weeks was $\$ 35400$.\\
(i) Stating a necessary assumption, test at the $5 \%$ significance level whether the mean weekly profit has increased.\\
(ii) State, with a reason, whether it was necessary to use the Central Limit theorem in part (i).
The mean weekly profit for another random sample of 90 weeks is found and the same test is carried out at the 5\% significance level.\\
(iii) State the probability of a Type I error.\\
(iv) Given that the population mean weekly profit is now $\$ 36500$, calculate the probability of a Type II error.
\hfill \mbox{\textit{CAIE S2 2013 Q7 [14]}}