| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with standard parameters (n=30, p=1/6 under H₀). Part (i) requires routine calculation of P(X≤2) and comparison to 5%, while part (ii) simply asks for the definition of Type I error probability (which equals the significance level). Both parts are textbook applications with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: P(6) = \frac{1}{6}\); \(H_1: P(6) < \frac{1}{6}\) | B1 | |
| \(\left(\frac{5}{6}\right)^{30} + 30\left(\frac{1}{6}\right)\times\left(\frac{5}{6}\right)^{29} + {}^{30}C_2\left(\frac{1}{6}\right)^2\times\left(\frac{5}{6}\right)^{28}\) | M1 | Allow one term incorrect, omitted or extra |
| \(= 0.103\) | A1 | |
| \('0.103' > 0.05\) | M1 | |
| No evidence (at 5% level) that die biased | A1ft | oe No contradictions |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{5}{6}\right)^{30} + 30\left(\frac{1}{6}\right)\times\left(\frac{5}{6}\right)^{29}\) | M1 | |
| \(P(\text{Type I}) = 0.0295\) | A1 | |
| Total: 2 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: P(6) = \frac{1}{6}$; $H_1: P(6) < \frac{1}{6}$ | B1 | |
| $\left(\frac{5}{6}\right)^{30} + 30\left(\frac{1}{6}\right)\times\left(\frac{5}{6}\right)^{29} + {}^{30}C_2\left(\frac{1}{6}\right)^2\times\left(\frac{5}{6}\right)^{28}$ | M1 | Allow one term incorrect, omitted or extra |
| $= 0.103$ | A1 | |
| $'0.103' > 0.05$ | M1 | |
| No evidence (at 5% level) that die biased | A1ft | oe No contradictions |
| **Total: 5** | | |
---
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{5}{6}\right)^{30} + 30\left(\frac{1}{6}\right)\times\left(\frac{5}{6}\right)^{29}$ | M1 | |
| $P(\text{Type I}) = 0.0295$ | A1 | |
| **Total: 2** | | |
---3 Sumitra has a six-sided die. She suspects that it is biased so that it shows a six less often than it would if it were fair. She decides to test the die by throwing it 30 times and noting the number of throws on which it shows a six.\\
(i) It shows a six on exactly 2 throws. Use a binomial distribution to carry out the test at the $5 \%$ significance level.\\
(ii) Later, Sumitra repeats the test at the $5 \%$ significance level by throwing the die 30 times again. Find the probability of a Type I error in this second test.\\
\hfill \mbox{\textit{CAIE S2 2019 Q3 [7]}}