| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Standard +0.3 This is a straightforward one-tailed binomial hypothesis test with standard parts: explaining why P(X=8) alone is insufficient (need cumulative probability), performing the test correctly using P(X≥8), explaining Type I errors conceptually, and calculating the significance level. All components are routine S2 material requiring recall and application of standard procedures rather than problem-solving insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Need to find \(P(X \geqslant 8)\) | B1 | oe (e.g. invalid because it should be a tail probability compared with 0.05). |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): \(P(\text{green}) = 0.5\), \(H_1\): \(P(\text{green}) > 0.5\) | B1 | Allow \(p = 0.5\). Allow \(p > 0.5\). |
| \(P(X \geqslant 8) = 0.0439 + {}^{10}C_9 \times (0.5) \times (0.5)^9 + 0.5^{10}\) | M1 | Attempt \(0.0439 + P(X=9) + P(X=10)\). Must see Binomial expressions \(B(10, 0.5)\). |
| \(= 0.0547\) or \(0.0546\) (3 sf) | A1 | SC B1 0.0547 or 0.0546 with no working. |
| \(0.0547 > 0.05\) | M1 | Valid comparison of tail probability with 0.05. |
| Do not reject \(H_0\). There is insufficient evidence [at the 5% level] to accept the hypothesis that boys prefer green. Or: There is sufficient evidence to support the researcher's claim. | A1FT | In context, not definite. No contradictions. Allow 'There is insufficient evidence to reject the hypothesis that boys like green and orange equally'. Not definite, e.g. not 'They don't prefer green' or 'Researchers claim true'. Any mention of 'claim' must be clear which claim it is. |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\) was not rejected | B1 | Mark independently. |
| Answer | Marks | Guidance |
|---|---|---|
| \({}^{10}C_9 \times (0.5) \times (0.5)^9 + 0.5^{10}\) or "\(0.0547\)" \(- 0.0439\) | M1 | Finding \(P(9,10)\) using \(B(10, 0.5)\). Could be seen in (a)(ii). |
| \(P(\text{Type I error}) = \frac{11}{1024}\) or \(0.0107\) (3 sf) or \(0.0108\) | A1 |
## Question 3(a)(i):
Need to find $P(X \geqslant 8)$ | B1 | oe (e.g. invalid because it should be a tail probability compared with 0.05).
---
## Question 3(a)(ii):
$H_0$: $P(\text{green}) = 0.5$, $H_1$: $P(\text{green}) > 0.5$ | B1 | Allow $p = 0.5$. Allow $p > 0.5$.
$P(X \geqslant 8) = 0.0439 + {}^{10}C_9 \times (0.5) \times (0.5)^9 + 0.5^{10}$ | M1 | Attempt $0.0439 + P(X=9) + P(X=10)$. Must see Binomial expressions $B(10, 0.5)$.
$= 0.0547$ or $0.0546$ (3 sf) | A1 | **SC B1** 0.0547 or 0.0546 with no working.
$0.0547 > 0.05$ | M1 | Valid comparison of tail probability with 0.05.
Do not reject $H_0$. There is insufficient evidence [at the 5% level] to accept the hypothesis that boys prefer green. Or: There is sufficient evidence to support the **researcher's** claim. | A1FT | In context, not definite. No contradictions. Allow 'There is insufficient evidence to reject the hypothesis that boys like green and orange equally'. Not definite, e.g. not 'They don't prefer green' or 'Researchers claim true'. Any mention of 'claim' must be clear which claim it is.
---
## Question 3(b):
$H_0$ was not rejected | B1 | Mark independently.
---
## Question 3(c):
${}^{10}C_9 \times (0.5) \times (0.5)^9 + 0.5^{10}$ or "$0.0547$" $- 0.0439$ | M1 | Finding $P(9,10)$ using $B(10, 0.5)$. Could be seen in (a)(ii).
$P(\text{Type I error}) = \frac{11}{1024}$ or $0.0107$ (3 sf) or $0.0108$ | A1 |
---3 A researcher read a magazine article which stated that boys aged 1 to 3 prefer green to orange. It claimed that, when offered a green cube and an orange cube to play with, a boy is more likely to choose the green one.
The researcher disagrees with this claim. She believes that boys of this age are equally likely to choose either colour. In order to test her belief, the researcher carried out a hypothesis test at the 5\% significance level. She offered a green cube and an orange cube to each of 10 randomly chosen boys aged 1 to 3 , and recorded the number, $X$, of boys who chose the green cube.
Out of the 10 boys, 8 boys chose the green cube.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Assuming that the researcher's belief that either colour cube is equally likely to be chosen is valid, a student correctly calculates that $\mathrm { P } ( X = 8 ) = 0.0439$, correct to 3 significant figures. He says that, because this value is less than 0.05 , the null hypothesis should be rejected.
Explain why this statement is incorrect.
\item Carry out the test on the researcher's claim that either colour cube is equally likely to be chosen.
\end{enumerate}\item Another researcher claims that a Type I error was made in carrying out the test.
Explain why this cannot be true.\\
A similar test, at the $5 \%$ significance level, was carried out later using 10 other randomly chosen boys aged 1 to 3 .
\item Find the probability of a Type I error.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q3 [9]}}