| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type II error probability |
| Difficulty | Standard +0.3 This is a straightforward application of binomial hypothesis testing with standard procedures: setting up H₀ and H₁, finding critical region at 1% level, comparing test statistic, and calculating Type I and Type II error probabilities using binomial tables. All steps are routine for S2 level with no novel problem-solving required, making it slightly easier than average. |
| 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 | Mark | Guidance |
| \(H_0: p = \frac{1}{4}\) and \(H_1: p > \frac{1}{4}\) | B1 | |
| \(^{10}C_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^4 + ^{10}C_7\left(\frac{1}{4}\right)^7\left(\frac{3}{4}\right)^3 + ^{10}C_8\left(\frac{1}{4}\right)^8\left(\frac{3}{4}\right)^2 + 10\left(\frac{1}{4}\right)^9\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^{10}\) | M1 | Correct terms; allow one term incorrect or omitted or extra. Or summing all correct terms from 0 to 5, allow one term incorrect or omitted or extra |
| \(= 0.0197\) | A1 | or 0.9803 |
| Compare \(0.0197\) with \(0.01\) | M1 | Valid comparison with 0.01, or valid comparison with 0.99 |
| No evidence to conclude \(p > \frac{1}{4}\) | A1 | FT No contradictions. Use of two-tail test can score B0M1A1M1 (comparison with 0.005) A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^{10}C_7\left(\frac{1}{4}\right)^7\left(\frac{3}{4}\right)^3 + ^{10}C_8\left(\frac{1}{4}\right)^8\left(\frac{3}{4}\right)^2 + 10\left(\frac{1}{4}\right)^9\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^{10}\) | M1 | Their \(P(X \geqslant 6) - ^{10}C_6(0.25)^6(0.75)^4\) |
| \(P(\text{Type I}) = 0.00351\) (3 sf) | A1 | Accept 0.00348 to 0.00351 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| C.R. is \(X \geqslant 7\); \(P(\text{Type II}) = 1 - P\!\left(X \geqslant 7 \mid p = \frac{3}{5}\right)\) | M1 | May be implied |
| \(1 - \left(^{10}C_7\left(\frac{3}{5}\right)^7\left(\frac{2}{5}\right)^3 + ^{10}C_8\left(\frac{3}{5}\right)^8\left(\frac{2}{5}\right)^2 + 10\left(\frac{3}{5}\right)^9\left(\frac{2}{5}\right) + \left(\frac{3}{5}\right)^{10}\right)\) | M1 | Accept \(1 - P(X \geqslant 8 \mid p = \frac{3}{5})\) or \(1 - P(X \geqslant 6 \mid p = \frac{3}{5})\) |
| \(= 0.618\) | A1 |
## Question 8:
### Part 8(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = \frac{1}{4}$ and $H_1: p > \frac{1}{4}$ | **B1** | |
| $^{10}C_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^4 + ^{10}C_7\left(\frac{1}{4}\right)^7\left(\frac{3}{4}\right)^3 + ^{10}C_8\left(\frac{1}{4}\right)^8\left(\frac{3}{4}\right)^2 + 10\left(\frac{1}{4}\right)^9\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^{10}$ | **M1** | Correct terms; allow one term incorrect or omitted or extra. Or summing all correct terms from 0 to 5, allow one term incorrect or omitted or extra |
| $= 0.0197$ | **A1** | or 0.9803 |
| Compare $0.0197$ with $0.01$ | **M1** | Valid comparison with 0.01, or valid comparison with 0.99 |
| No evidence to conclude $p > \frac{1}{4}$ | **A1** | **FT** No contradictions. Use of two-tail test can score B0M1A1M1 (comparison with 0.005) A0 |
**Total: 5 marks**
---
### Part 8(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{10}C_7\left(\frac{1}{4}\right)^7\left(\frac{3}{4}\right)^3 + ^{10}C_8\left(\frac{1}{4}\right)^8\left(\frac{3}{4}\right)^2 + 10\left(\frac{1}{4}\right)^9\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^{10}$ | **M1** | Their $P(X \geqslant 6) - ^{10}C_6(0.25)^6(0.75)^4$ |
| $P(\text{Type I}) = 0.00351$ (3 sf) | **A1** | Accept 0.00348 to 0.00351 |
**Total: 2 marks**
---
### Part 8(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| C.R. is $X \geqslant 7$; $P(\text{Type II}) = 1 - P\!\left(X \geqslant 7 \mid p = \frac{3}{5}\right)$ | **M1** | May be implied |
| $1 - \left(^{10}C_7\left(\frac{3}{5}\right)^7\left(\frac{2}{5}\right)^3 + ^{10}C_8\left(\frac{3}{5}\right)^8\left(\frac{2}{5}\right)^2 + 10\left(\frac{3}{5}\right)^9\left(\frac{2}{5}\right) + \left(\frac{3}{5}\right)^{10}\right)$ | **M1** | Accept $1 - P(X \geqslant 8 \mid p = \frac{3}{5})$ or $1 - P(X \geqslant 6 \mid p = \frac{3}{5})$ |
| $= 0.618$ | **A1** | |
**Total: 3 marks**8 The four sides of a spinner are $A , B , C , D$. The spinner is supposed to be fair, but Sonam suspects that the spinner is biased so that the probability, $p$, that it will land on side $A$ is greater than $\frac { 1 } { 4 }$. He spins the spinner 10 times and finds that it lands on side $A 6$ times.\\
(i) Test Sonam's suspicion using a $1 \%$ significance level.\\
Later Sonam carries out a similar test at the $1 \%$ significance level, using another 10 spins of the spinner.\\
(ii) Calculate the probability of a Type I error.\\
(iii) Assuming that the value of $p$ is actually $\frac { 3 } { 5 }$, calculate the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2019 Q8 [10]}}