| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| 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 application of Type I error calculation and hypothesis testing with binomial distributions. Part (i) requires understanding that Type I error means rejecting H₀ when it's true, so calculating P(X < 3) when p = 0.25. Part (ii) is a standard one-tailed binomial test at 1% level. Both parts use direct binomial probability calculations with no conceptual tricks, making this slightly easier than average for S2 level. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.75^{20} + 20 \times 0.75^{19} \times 0.25 + {}^{20}C_2 \times 0.75^{18} \times 0.25^2\) | M1 | No end errors |
| \(= 0.0913\) | A1 | As final answer |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop proportion \(= 0.25\); \(H_1\): Pop proportion \(< 0.25\) | B1 | Allow \(p\) or \(\pi\), not "proportion" (Accept anywhere in the question) |
| \(0.75^{25} + 25 \times 0.75^{24} \times 0.25\) | M1 | Must be \(B(25, 0.25)\). No end errors |
| \(= 0.00702\) | A1 | |
| comp \(0.01\) | M1 | Valid comparison |
| There is evidence that the claim is not justified | A1 FT | OE. No contradictions |
| Total: | 5 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.75^{20} + 20 \times 0.75^{19} \times 0.25 + {}^{20}C_2 \times 0.75^{18} \times 0.25^2$ | M1 | No end errors |
| $= 0.0913$ | A1 | As final answer |
| **Total:** | **2** | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop proportion $= 0.25$; $H_1$: Pop proportion $< 0.25$ | B1 | Allow $p$ or $\pi$, not "proportion" (Accept anywhere in the question) |
| $0.75^{25} + 25 \times 0.75^{24} \times 0.25$ | M1 | Must be $B(25, 0.25)$. No end errors |
| $= 0.00702$ | A1 | |
| comp $0.01$ | M1 | Valid comparison |
| There is evidence that the claim is not justified | A1 FT | OE. No contradictions |
| **Total:** | **5** | |4 It is claimed that 1 in every 4 packets of certain biscuits contains a free gift. Marisa and André both suspect that the true proportion is less than 1 in 4.\\
(i) Marisa chooses 20 packets at random. She decides that if fewer than 3 contain free gifts, she will conclude that the claim is not justified. Use a binomial distribution to find the probability of a Type I error.\\
(ii) André chooses 25 packets at random. He decides to carry out a significance test at the $1 \%$ level, using a binomial distribution. Given that only 1 of the 25 packets contains a free gift, carry out the test.\\
\hfill \mbox{\textit{CAIE S2 2017 Q4 [7]}}