| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: defining Type I/II errors, calculating their probabilities using binomial distribution with small n=5, and constructing a confidence interval using normal approximation. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Conclude die is biased when it isn't oe | B1 | In context |
| \({}^5C_3\!\left(\dfrac{1}{6}\right)^3\!\left(\dfrac{5}{6}\right)^2 + 5\!\left(\dfrac{1}{6}\right)^4\!\left(\dfrac{5}{6}\right) + \left(\dfrac{1}{6}\right)^5 + 5\) | M1 | or \(1 - \left({}^5C_2\!\left(\dfrac{1}{6}\right)^2\!\left(\dfrac{5}{6}\right)\!3 + 5\!\left(\dfrac{1}{6}\right)\!\left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^5\right)\) |
| \(= \dfrac{23}{648}\) or \(0.0355\) (3 sf) | A1 [3] | allow 1 end error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or attempt \(P(0,1,2)\) with \(p = \dfrac{2}{3}\) | M1 | Or \(1 - P(3,4,5)\) |
| \({}^5C_2\!\left(\dfrac{2}{3}\right)^2\!\left(\dfrac{1}{3}\right)^3 + 5\!\left(\dfrac{2}{3}\right)\!\left(\dfrac{1}{3}\right)^4 + \left(\dfrac{1}{3}\right)^5\) | M1 | Attempt at correct expression |
| \(= \dfrac{17}{81}\) or \(0.210\) (3 sf) | A1 [3] | Allow 0.21 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Est Var}(P_s) = \dfrac{0.625 \times (1 - 0.625)}{80} = \dfrac{3}{1024}\) | M1 | |
| \(z = 2.054\) (or \(2.055\)) | B1 | |
| \(0.625 \pm z \times \sqrt{\dfrac{3}{1024}}\) | M1 | Any \(z\) |
| \(= 0.514\) to \(0.736\) (3 sf) | A1 [4] |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Conclude die is biased when it isn't oe | B1 | In context |
| ${}^5C_3\!\left(\dfrac{1}{6}\right)^3\!\left(\dfrac{5}{6}\right)^2 + 5\!\left(\dfrac{1}{6}\right)^4\!\left(\dfrac{5}{6}\right) + \left(\dfrac{1}{6}\right)^5 + 5$ | M1 | or $1 - \left({}^5C_2\!\left(\dfrac{1}{6}\right)^2\!\left(\dfrac{5}{6}\right)\!3 + 5\!\left(\dfrac{1}{6}\right)\!\left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^5\right)$ |
| $= \dfrac{23}{648}$ or $0.0355$ (3 sf) | A1 [3] | allow 1 end error |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or attempt $P(0,1,2)$ with $p = \dfrac{2}{3}$ | M1 | Or $1 - P(3,4,5)$ |
| ${}^5C_2\!\left(\dfrac{2}{3}\right)^2\!\left(\dfrac{1}{3}\right)^3 + 5\!\left(\dfrac{2}{3}\right)\!\left(\dfrac{1}{3}\right)^4 + \left(\dfrac{1}{3}\right)^5$ | M1 | Attempt at correct expression |
| $= \dfrac{17}{81}$ or $0.210$ (3 sf) | A1 [3] | Allow 0.21 |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Est Var}(P_s) = \dfrac{0.625 \times (1 - 0.625)}{80} = \dfrac{3}{1024}$ | M1 | |
| $z = 2.054$ (or $2.055$) | B1 | |
| $0.625 \pm z \times \sqrt{\dfrac{3}{1024}}$ | M1 | Any $z$ |
| $= 0.514$ to $0.736$ (3 sf) | A1 [4] | |7 Leila suspects that a particular six-sided die is biased so that the probability, $p$, that it will show a six is greater than $\frac { 1 } { 6 }$. She tests the die by throwing it 5 times. If it shows a six on 3 or more throws she will conclude that it is biased.\\
(i) State what is meant by a Type I error in this situation and calculate the probability of a Type I error.\\
(ii) Assuming that the value of $p$ is actually $\frac { 2 } { 3 }$, calculate the probability of a Type II error.
Leila now throws the die 80 times and it shows a six on 50 throws.\\
(iii) Calculate an approximate $96 \%$ confidence interval for $p$.
\hfill \mbox{\textit{CAIE S2 2013 Q7 [10]}}