| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | One-tailed hypothesis test |
| Difficulty | Moderate -0.3 This is a straightforward application of a one-tailed hypothesis test using normal approximation to binomial. Students must identify H₀: p=0.25, H₁: p>0.25, calculate np=15, apply continuity correction (21.5), find z-score, and compare to critical value. While it requires multiple steps, each is routine for S2 level with no conceptual challenges or novel problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0: \mu = 15\) or \(p = 0.25\); \(H_1: \mu > 15\) or \(p > 0.25\) | B1 | For \(H_0\) and \(H_1\) correct |
| (ii) Test statistic: \(z = \pm\frac{21.5 - 15}{\sqrt{60 \times 0.25 \times 0.75}} = 1.938\) OR \(z = \pm\frac{22 + \frac{60}{2} - 15}{\sqrt{\frac{0.25 \times 0.75}{60}}} = 1.938\) | M1 A1 | For attempt at standardising with or without cc, must have \(\sqrt{\text{something}}\) with 60 in denominator; For 1.94 (1.938) |
| Critical value: \(z = 1.645\) | M1 | For comparing with 1.645 or 1.96 if 2-tailed, signs consistent, or comparing areas to 5% |
| In CR Claim justified | A1ft | For correct answer (ft only for correct one-tail test) |
| 4 |
**(i)** $H_0: \mu = 15$ or $p = 0.25$; $H_1: \mu > 15$ or $p > 0.25$ | B1 | For $H_0$ and $H_1$ correct
**(ii)** Test statistic: $z = \pm\frac{21.5 - 15}{\sqrt{60 \times 0.25 \times 0.75}} = 1.938$ OR $z = \pm\frac{22 + \frac{60}{2} - 15}{\sqrt{\frac{0.25 \times 0.75}{60}}} = 1.938$ | M1 A1 | For attempt at standardising with or without cc, must have $\sqrt{\text{something}}$ with 60 in denominator; For 1.94 (1.938)
Critical value: $z = 1.645$ | M1 | For comparing with 1.645 or 1.96 if 2-tailed, signs consistent, or comparing areas to 5%
In CR Claim justified | A1ft | For correct answer (ft only for correct one-tail test)
| | **4** | |1 Each multiple choice question in a test has 4 suggested answers, exactly one of which is correct. Rehka knows nothing about the subject of the test, but claims that she has a special method for answering the questions that is better than just guessing. There are 60 questions in the test, and Rehka gets 22 correct.\\
(i) State null and alternative hypotheses for a test of Rehka's claim.\\
(ii) Using a normal approximation, test at the $5 \%$ significance level whether Rehka's claim is justified.
\hfill \mbox{\textit{CAIE S2 2004 Q1 [5]}}