| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Find CI width or confidence level |
| Difficulty | Moderate -0.5 This question tests basic understanding of hypothesis testing concepts (one-tail vs two-tail, interpreting z-values, finding significance levels) with straightforward calculations. Part (i) is pure recall, part (ii) requires comparing z = -1.750 to critical values (±1.645 for 10% two-tailed), and part (iii) involves finding P(|Z| > 1.750) = 2×0.0401 ≈ 8%. While it requires understanding of the normal distribution and hypothesis testing framework, all steps are routine applications of standard procedures with no problem-solving insight needed. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) 2-tail; \(H_1: \mu \neq 35\) | B1 [1] | |
| (ii) comp –1.75 with –1.645 (or 1.75 with 1.645) | M1 | Allow "Accept \(\mu = 35\)". No contradictions |
| Evidence that \(\mu\) is not 35 or reject \(\mu = 35\) | A1 | [2] |
| (iii) 8 | B2 [2] | SR B1 for 4, 8.02, or 92% |
**(i)** 2-tail; $H_1: \mu \neq 35$ | B1 [1] |
**(ii)** comp –1.75 with –1.645 (or 1.75 with 1.645) | M1 | Allow "Accept $\mu = 35$". No contradictions
Evidence that $\mu$ is not 35 or reject $\mu = 35$ | A1 | [2]
**(iii)** 8 | B2 [2] | SR B1 for 4, 8.02, or 92%2 Dipak carries out a test, at the $10 \%$ significance level, using a normal distribution. The null hypothesis is $\mu = 35$ and the alternative hypothesis is $\mu \neq 35$.\\
(i) Is this a one-tail or a two-tail test? State briefly how you can tell.
Dipak finds that the value of the test statistic is $z = - 1.750$.\\
(ii) Explain what conclusion he should draw.\\
(iii) This result is significant at the $\alpha \%$ level. Find the smallest possible value of $\alpha$, correct to the nearest whole number.
\hfill \mbox{\textit{CAIE S2 2010 Q2 [5]}}