| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with clearly defined parameters (n=50, p=1/5 under H₀, x=13). Students need to state H₀: p=0.2 vs H₁: p>0.2, then calculate P(X≥13) using binomial tables or normal approximation and compare to 10% significance level. While it requires understanding of hypothesis testing framework, it's a standard textbook application with no conceptual tricks or complex reasoning—slightly easier than average due to the generous 10% significance level and clear context. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.2\) or \(\mu = 10\); \(H_1: p > 0.2\) or \(\mu > 10\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N(10, 8)\) seen or implied | B1 | or \(N\!\left(0.2,\, \frac{0.2 \times 0.8}{50}\right)\) |
| \(\frac{12.5 - 10}{\sqrt{8}}\) or \(\frac{\frac{125}{50} - 0.2}{\sqrt{\frac{0.2\times0.8}{50}}}\) | M1 | For standardising, allow with no or wrong cc |
| \(= 0.884\) | A1 | |
| comp 1.282 | M1f | Allow area comparison with 0.188 or comp 1.645 if \(H_1: p \neq 0.2\) |
| Claim not justified / No evidence to support claim | A1f [5] | Allow accept \(H_0\) provided correctly defined. Follow through their test statistic; dep 1-tail test. No contradictions. SR: Use of \(B(50, 0.2)\) scores B1 provided at least two probabilities calculated. M1 for finding \(P(X \geq 13)\), allow one end error. A1 for 0.186 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2$ or $\mu = 10$; $H_1: p > 0.2$ or $\mu > 10$ | B1 [1] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N(10, 8)$ seen or implied | B1 | or $N\!\left(0.2,\, \frac{0.2 \times 0.8}{50}\right)$ |
| $\frac{12.5 - 10}{\sqrt{8}}$ or $\frac{\frac{125}{50} - 0.2}{\sqrt{\frac{0.2\times0.8}{50}}}$ | M1 | For standardising, allow with no or wrong cc |
| $= 0.884$ | A1 | |
| comp 1.282 | M1f | Allow area comparison with 0.188 or comp 1.645 if $H_1: p \neq 0.2$ |
| Claim not justified / No evidence to support claim | A1f [5] | Allow accept $H_0$ provided correctly defined. Follow through their test statistic; dep 1-tail test. No contradictions. SR: Use of $B(50, 0.2)$ scores B1 provided at least two probabilities calculated. M1 for finding $P(X \geq 13)$, allow one end error. A1 for 0.186 |
---2 Sami claims that he can read minds. He asks each of 50 people to choose one of the 5 letters A, B, C, D or E. He then tells each person which letter he believes they have chosen. He gets 13 correct. Sami says "This shows that I can read minds, because 13 is more than I would have got right if I were just guessing."\\
(i) State null and alternative hypotheses for a test of Sami's claim.\\
(ii) Test at the $10 \%$ significance level whether Sami's claim is justified.
\hfill \mbox{\textit{CAIE S2 2015 Q2 [6]}}