| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Critique inappropriate sampling methods |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question requiring standard recall of sampling principles, test setup, and binomial critical regions. Part (i) tests basic understanding of sampling bias (time-based clustering), parts (ii)-(iv) are routine textbook procedures, and part (v) is simple comparison to critical region. No novel problem-solving or complex multi-step reasoning required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a 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 |
| Prob could be different later in day or on a different day | B1 | Or any explanation why not random; or "Not random" or "Not representative" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | 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 | Mark | Guidance |
| Concluding that prob has decreased (or publicity has worked) when it hasn't | B1 | In context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=0)\) and \(P(X=1)\) attempted | M1 | \(B(30, 0.2)\) not necessarily added; may be implied by calc \(P(X \leq 2)\) or \(P(X \leq 3)\) |
| \(P(X \leq 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 \quad (= 0.0442)\) | 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 \(\text{'0.0442'} + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123\) |
| CR is \(X \leq 2\) | A1 | |
| \(P(\text{Type I}) = 0.0442\) (3 sf) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | 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\(\checkmark\) | Correct conclusion. No contradictions |
| Total: 2 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Prob could be different later in day or on a different day | B1 | Or any explanation why not random; or "Not random" or "Not representative" |
---
## Question 7(ii):
| Answer | Mark | 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 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Concluding that prob has decreased (or publicity has worked) when it hasn't | B1 | In context |
---
## Question 7(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=0)$ and $P(X=1)$ attempted | M1 | $B(30, 0.2)$ not necessarily added; may be implied by calc $P(X \leq 2)$ or $P(X \leq 3)$ |
| $P(X \leq 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 \quad (= 0.0442)$ | 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 $\text{'0.0442'} + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123$ |
| CR is $X \leq 2$ | A1 | |
| $P(\text{Type I}) = 0.0442$ (3 sf) | A1 | |
| **Total: 5** | | |
---
## Question 7(v):
| Answer | Mark | 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$\checkmark$ | Correct conclusion. No contradictions |
| **Total: 2** | | |7 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.\\
(i) It is suggested that the first 30 appointments on a Monday should be used for the test. Give a reason why this is not an appropriate sample.\\
A suitable 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.\\
(ii) Explain why the test is one-tail and state suitable null and alternative hypotheses.\\
(iii) State what is meant by a Type I error in this context.\\
(iv) Use the binomial distribution to find the critical region, and find the probability of a Type I error.\\
(v) 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 Q7 [11]}}