| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| 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 a one-tailed binomial hypothesis test with standard steps (state hypotheses, find critical region at 5% level, compare test statistic). The calculations are routine for S2 level, requiring only binomial probability tables or calculator work. While it tests understanding of Type I error, this is a standard textbook concept at this level, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): \(P(\text{free gift}) = 0.3\) or \(p = 0.3\) | B1 | [1] |
| \(H_1\): \(P(\text{free gift}) < 0.3\) or \(p < 0.3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X \leqslant 2) = 0.7^{20} + 20 \times 0.7^{19} \times 0.3 + {}^{20}C_2 \times 0.7^{18} \times 0.3^2\) | M1* | \(P(X \leqslant 2)\) attempted |
| \(= 0.03548\) or \(0.0355\) | A1 | |
| \(P(X \leqslant 3) = \text{'0.03548'} + {}^{20}C_3 \times 0.7^{17} \times 0.3^3\ (= 0.107)\) | M1* | \(P(X \leqslant 3)\) attempted |
| One comparison with 0.05 seen | M1* | or implied by fully correct methods for \(P(X\leqslant 2)\) and \(P(X\leqslant 3)\) |
| \(P(\text{Type I error}) = 0.0355\) (3 sf) | DA1\(\checkmark\) | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X \leqslant 3) = \text{'0.107'}\); \(\text{'0.107'} > 0.05\) or cv \(= 2\) and compare \(3 > 2\) | M1 | Compare their \(P(X \leqslant 3)\) with 0.05 |
| No evidence to reject claim oe | A1\(\checkmark\) | [2] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: $P(\text{free gift}) = 0.3$ or $p = 0.3$ | B1 | [1] |
| $H_1$: $P(\text{free gift}) < 0.3$ or $p < 0.3$ | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \leqslant 2) = 0.7^{20} + 20 \times 0.7^{19} \times 0.3 + {}^{20}C_2 \times 0.7^{18} \times 0.3^2$ | M1* | $P(X \leqslant 2)$ attempted |
| $= 0.03548$ or $0.0355$ | A1 | |
| $P(X \leqslant 3) = \text{'0.03548'} + {}^{20}C_3 \times 0.7^{17} \times 0.3^3\ (= 0.107)$ | M1* | $P(X \leqslant 3)$ attempted |
| One comparison with 0.05 seen | M1* | or implied by fully correct methods for $P(X\leqslant 2)$ and $P(X\leqslant 3)$ |
| $P(\text{Type I error}) = 0.0355$ (3 sf) | DA1$\checkmark$ | [5] | dep on all 3 Ms |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \leqslant 3) = \text{'0.107'}$; $\text{'0.107'} > 0.05$ or cv $= 2$ and compare $3 > 2$ | M1 | Compare their $P(X \leqslant 3)$ with 0.05 |
| No evidence to reject claim oe | A1$\checkmark$ | [2] | No evidence that 30% is not correct oe ft their 0.107 |
---5 It is claimed that $30 \%$ of packets of Froogum contain a free gift. Andre thinks that the actual proportion is less than $30 \%$ and he decides to carry out a hypothesis test at the $5 \%$ significance level. He buys 20 packets of Froogum and notes the number of free gifts he obtains.\\
(i) State null and alternative hypotheses for the test.\\
(ii) Use a binomial distribution to find the probability of a Type I error.
Andre finds that 3 of the 20 packets contain free gifts.\\
(iii) Carry out the test.
\hfill \mbox{\textit{CAIE S2 2016 Q5 [8]}}