| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Explain Type I or II error |
| Difficulty | Standard +0.3 This is a straightforward one-tailed binomial hypothesis test with standard structure: (i) requires setting up H₀: p=0.2 vs H₁: p<0.2, calculating P(X≤1) for B(25,0.2), and comparing to 2.5%; (ii) tests knowledge that large n allows normal approximation; (iii) asks for standard Type II error definition. All parts are routine applications of S2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.05a 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 |
| \(H_0: p = 0.2\); \(H_1: p < 0.2\) | B1 | (Allow \(\pi\)) |
| \(P(0\ \text{or}\ 1\ \text{5s in}\ 25 \mid H_0)\) | M1 | \(0.8^{25} + 25 \times 0.8^{24} \times 0.2\); use of B(25,1/5) and P(0) or P(1) or both – may be implied by "0.0274" |
| \(= 0.0274\) (3 s.f.) | A1 | |
| Compare with 0.025 | M1 | Valid comparison |
| No evidence (at 2.5% level) to support claim | A1\(\checkmark\) [5] | No contradictions; SR: Use of Normal N(5,4) leading to \(z=1.75\) or 0.0401 B1\* \(H_0\ \mu=5\), \(H_1\ \mu<5\) B1. Comparison \(1.75 < 1.96\) or \(0.0401 > 0.025\) B1\* dep |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Normal | B1 | |
| \(\mu = 200,\ \sigma^2 = 160\) or \(\sigma = \sqrt{160}\) | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Concluding that the machine produces the right proportion of 5s, although it doesn't. | B1 [1] | Not concluding that the machine produces too few 5s although it does. Must be in context. o.e. No contradictions |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2$; $H_1: p < 0.2$ | B1 | (Allow $\pi$) |
| $P(0\ \text{or}\ 1\ \text{5s in}\ 25 \mid H_0)$ | M1 | $0.8^{25} + 25 \times 0.8^{24} \times 0.2$; use of B(25,1/5) and P(0) or P(1) or both – may be implied by "0.0274" |
| $= 0.0274$ (3 s.f.) | A1 | |
| Compare with 0.025 | M1 | Valid comparison |
| No evidence (at 2.5% level) to support claim | A1$\checkmark$ [5] | No contradictions; SR: Use of Normal N(5,4) leading to $z=1.75$ or 0.0401 **B1**\* $H_0\ \mu=5$, $H_1\ \mu<5$ B1. Comparison $1.75 < 1.96$ or $0.0401 > 0.025$ **B1**\* dep |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal | B1 | |
| $\mu = 200,\ \sigma^2 = 160$ or $\sigma = \sqrt{160}$ | B1 [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Concluding that the machine produces the right proportion of 5s, although it doesn't. | B1 [1] | Not concluding that the machine produces too few 5s although it does. Must be in context. o.e. No contradictions |
---6 A machine is designed to generate random digits between 1 and 5 inclusive. Each digit is supposed to appear with the same probability as the others, but Max claims that the digit 5 is appearing less often than it should. In order to test this claim the manufacturer uses the machine to generate 25 digits and finds that exactly 1 of these digits is a 5 .\\
(i) Carry out a test of Max's claim at the $2.5 \%$ significance level.\\
(ii) Max carried out a similar hypothesis test by generating 1000 digits between 1 and 5 inclusive. The digit 5 appeared 180 times. Without carrying out the test, state the distribution that Max should use, including the values of any parameters.\\
(iii) State what is meant by a Type II error in this context.
\hfill \mbox{\textit{CAIE S2 2014 Q6 [8]}}