| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Two-tailed hypothesis test |
| Difficulty | Moderate -0.3 This is a straightforward two-tailed hypothesis test using normal approximation to binomial. Part (a) requires stating H₀: p = 0.25 and H₁: p ≠ 0.25 (routine). Part (b) involves standard procedure: checking np > 5, applying continuity correction, calculating z-score, and comparing to critical value at 2% level. All steps are textbook-standard with no conceptual challenges or novel problem-solving required. |
| Spec | 2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: p = \frac{1}{4}\) , \(H_1: p \neq \frac{1}{4}\) | B1 | or \(H_0: \mu = 25\) or \(H_1: \mu \neq 25\) |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N\left(25, \frac{75}{4}\right)\) | B1 | SOI. Allow B1 for \(N\left(25, \frac{75}{4}\right)\) or \(N(0.25, 0.001875)\) SOI |
| \(\pm \frac{15.5 - 25}{\sqrt{\frac{75}{4}}}\) or \(\frac{\frac{15.5}{100} - 0.25}{\sqrt{\frac{0.25 \times 0.75}{100}}}\) | M1 | Standardise with *their* \(N(25,\ldots)\). Allow with no or wrong continuity correction |
| \(\pm -2.194\) \((2.19)\) | A1 | |
| \(-2.326 < -2.194\) or \(0.0141 > 0.01\) or \(0.9859 < 0.99\) | M1 | For valid comparison (accept 2.326 to 2.329) |
| No evidence to reject that the probability is \(\frac{1}{4}\) | A1 FT | OE must be in context and not definite, e.g. not 'Claim untrue'. No contradictions. FT *their* \(z\); dependent on two-tailed test (one-tailed test can score B1 M1 A1 M1 A0). SC for use of Binomial \(B(100, 0.25)\) \(P = 0.0111\) for B1 and then comparison with 0.01 and correct conclusion for B1, maximum 2 out of 5 marks |
| Total: 5 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p = \frac{1}{4}$ , $H_1: p \neq \frac{1}{4}$ | B1 | or $H_0: \mu = 25$ or $H_1: \mu \neq 25$ |
| | **Total: 1** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N\left(25, \frac{75}{4}\right)$ | B1 | SOI. Allow B1 for $N\left(25, \frac{75}{4}\right)$ or $N(0.25, 0.001875)$ SOI |
| $\pm \frac{15.5 - 25}{\sqrt{\frac{75}{4}}}$ or $\frac{\frac{15.5}{100} - 0.25}{\sqrt{\frac{0.25 \times 0.75}{100}}}$ | M1 | Standardise with *their* $N(25,\ldots)$. Allow with no or wrong continuity correction |
| $\pm -2.194$ $(2.19)$ | A1 | |
| $-2.326 < -2.194$ or $0.0141 > 0.01$ or $0.9859 < 0.99$ | M1 | For valid comparison (accept 2.326 to 2.329) |
| No evidence to reject that the probability is $\frac{1}{4}$ | A1 FT | OE must be in context and not definite, e.g. not 'Claim untrue'. No contradictions. FT *their* $z$; dependent on two-tailed test (one-tailed test can score B1 M1 A1 M1 A0). **SC** for use of Binomial $B(100, 0.25)$ $P = 0.0111$ for B1 and then comparison with 0.01 and correct conclusion for B1, maximum 2 out of 5 marks |
| | **Total: 5** | |1 In a game, a ball is thrown and lands in one of 4 slots, labelled $A , B , C$ and $D$. Raju wishes to test whether the probability that the ball will land in slot $A$ is $\frac { 1 } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item State suitable null and alternative hypotheses for Raju's test.\\
The ball is thrown 100 times and it lands in slot $A 15$ times.
\item Use a suitable approximating distribution to carry out the test at the $2 \%$ significance level.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q1 [6]}}