| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 9 |
| 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 one-tailed binomial hypothesis test with standard structure: identify test type, find critical region at 5% significance, and make a conclusion. The question follows a routine template requiring recall of hypothesis testing procedure with minimal problem-solving, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Looking for decrease (or improvement) | B1 | OE |
| \(H_0: P(\text{not arrive}) = 0.2\); \(H_1: P(\text{not arrive}) < 0.2\) | B1 | Allow '\(p = 0.2\)' |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=0)\) and \(P(X=1)\) attempted | M1 | B(30, 0.2) Not necessarily added; may be implied by calculation \(P(X \leq 2)\) or \(P(X \leq 3)\) |
| \(P(X \leq 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2\ (= 0.0442)\) | M1 | Attempt \(P(X \leq 2)\) |
| \(P(X \leq 3) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123\) | B1 | Or '\(0.0442 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123\)' |
| Critical region (CR) is \(X \leq 2\) | A1 | |
| \(P(\text{Type I}) = 0.0442\) (3 sf) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 3 is outside CR | M1 | Comparison of 3 with their CR or \(P(X \leq 3) = 0.123\) which is \(> 0.05\) |
| No evidence that \(p\) has decreased (or that publicity has worked) | A1 | Correct conclusion. No contradictions |
| Total: 2 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Looking for decrease (or improvement) | B1 | OE |
| $H_0: P(\text{not arrive}) = 0.2$; $H_1: P(\text{not arrive}) < 0.2$ | B1 | Allow '$p = 0.2$' |
| **Total: 2** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=0)$ and $P(X=1)$ attempted | M1 | B(30, 0.2) Not necessarily added; may be implied by calculation $P(X \leq 2)$ or $P(X \leq 3)$ |
| $P(X \leq 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2\ (= 0.0442)$ | M1 | Attempt $P(X \leq 2)$ |
| $P(X \leq 3) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123$ | B1 | Or '$0.0442 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123$' |
| Critical region (CR) is $X \leq 2$ | A1 | |
| $P(\text{Type I}) = 0.0442$ (3 sf) | A1 | |
| **Total: 5** | | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 3 is outside CR | M1 | Comparison of 3 with their CR or $P(X \leq 3) = 0.123$ which is $> 0.05$ |
| No evidence that $p$ has decreased (or that publicity has worked) | A1 | Correct conclusion. No contradictions |
| **Total: 2** | | |
---6 At a certain hospital it was found that the probability that a patient did not arrive for an appointment was 0.2 . The hospital carries out some publicity in the hope that this probability will be reduced. They wish to test whether the publicity has worked.
A random sample of 30 appointments is selected and the number of patients that do not arrive is noted. This figure is used to carry out a test at the $5 \%$ significance level.\\
(a) Explain why the test is one-tailed and state suitable null and alternative hypotheses.\\
(b) Use a binomial distribution to find the critical region, and find the probability of a Type I error.\\
(c) In fact 3 patients out of the 30 do not arrive.
State the conclusion of the test, explaining your answer.\\
\hfill \mbox{\textit{CAIE S2 2020 Q6 [9]}}