| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type II error probability |
| Difficulty | Standard +0.3 This is a straightforward application of binomial hypothesis testing with standard procedures: part (a) requires a routine one-tailed test with given significance level, part (b) involves calculating P(Type II error) using a different probability (standard but requires careful setup), and part (c) asks for a definition in context. All parts follow textbook methods with no novel insight required, 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 | Mark | Guidance |
| \(H_0\): P(contains offer) \(= \frac{1}{3}\), \(H_1\): P(contains offer) \(< \frac{1}{3}\) | B1 | Allow \(p\) for P(contains offer) but not just proportion |
| \(P(0, 1 \text{ or } 2 \text{ offers in } 20 \mid H_0) = \left(\frac{2}{3}\right)^{20} + 20\left(\frac{2}{3}\right)^{19}\left(\frac{1}{3}\right) + {}^{20}C_2\left(\frac{2}{3}\right)^{18}\left(\frac{1}{3}\right)^2\) | M1 | |
| \(= 0.0176\) (3sf) | A1 | |
| \(0.0176 < 0.1\) | M1 | For valid comparison. SC comparison of \(0.982(4) > 0.9\) scores M1 and recovers the previous M1 A1 |
| (Reject \(H_0\)) No evidence (at 10% level) to support manufacturers claim | A1 FT | In context. Not definite. No contradictions. (Note 2 tail test scores max B0M1A1M1A0, max 3 out of 5). Accept critical region method: M1 A1 for correctly finding critical region of \(< 4\); 2 in critical region M1; A1 conclusion. SC Use of Normal approximation \(N\!\left(\frac{20}{3}, \frac{40}{9}\right)\) scores B1 M1 A0. M1 A1 max; the first M1 for \(\dfrac{2.5 - \frac{20}{3}}{\sqrt{\frac{40}{9}}}\) requires use of correct continuity correction and the comparison \(0.024 < 0.1\) OE must be a valid comparison |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - P(X \leqslant 3)\) | M1 | M1 for \(1-\) (one term omitted or extra or incorrect) or omit '\(1-\)' |
| \(= 1 - \left[\left(\frac{6}{7}\right)^{20} + 20\left(\frac{6}{7}\right)^{19}\left(\frac{1}{7}\right) + {}^{20}C_2\left(\frac{6}{7}\right)^{18}\left(\frac{1}{7}\right)^2 + {}^{20}C_3\left(\frac{6}{7}\right)^{17}\left(\frac{1}{7}\right)^3\right]\) | A1 | For all correct expression |
| \(= 0.318\) (3sf) | A1 | As final answer |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Concluding that prop is 1 in 3 when it is actually less (1 in 7) | B1 | OE, in context |
| 1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: P(contains offer) $= \frac{1}{3}$, $H_1$: P(contains offer) $< \frac{1}{3}$ | B1 | Allow $p$ for P(contains offer) but not just proportion |
| $P(0, 1 \text{ or } 2 \text{ offers in } 20 \mid H_0) = \left(\frac{2}{3}\right)^{20} + 20\left(\frac{2}{3}\right)^{19}\left(\frac{1}{3}\right) + {}^{20}C_2\left(\frac{2}{3}\right)^{18}\left(\frac{1}{3}\right)^2$ | M1 | |
| $= 0.0176$ (3sf) | A1 | |
| $0.0176 < 0.1$ | M1 | For valid comparison. **SC** comparison of $0.982(4) > 0.9$ scores M1 and recovers the previous M1 A1 |
| (Reject $H_0$) No evidence (at 10% level) to support manufacturers claim | A1 FT | In context. Not definite. No contradictions. (Note 2 tail test scores max B0M1A1M1A0, max 3 out of 5). Accept critical region method: M1 A1 for correctly finding critical region of $< 4$; 2 in critical region M1; A1 conclusion. **SC** Use of Normal approximation $N\!\left(\frac{20}{3}, \frac{40}{9}\right)$ scores B1 M1 A0. M1 A1 max; the first M1 for $\dfrac{2.5 - \frac{20}{3}}{\sqrt{\frac{40}{9}}}$ requires use of correct continuity correction and the comparison $0.024 < 0.1$ OE must be a valid comparison |
| Total: 5 | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - P(X \leqslant 3)$ | **M1** | M1 for $1-$ (one term omitted or extra or incorrect) or omit '$1-$' |
| $= 1 - \left[\left(\frac{6}{7}\right)^{20} + 20\left(\frac{6}{7}\right)^{19}\left(\frac{1}{7}\right) + {}^{20}C_2\left(\frac{6}{7}\right)^{18}\left(\frac{1}{7}\right)^2 + {}^{20}C_3\left(\frac{6}{7}\right)^{17}\left(\frac{1}{7}\right)^3\right]$ | **A1** | For all correct expression |
| $= 0.318$ (3sf) | **A1** | As final answer |
| | **3** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Concluding that prop is 1 in 3 when it is actually less (1 in 7) | **B1** | OE, in context |
| | **1** | |6 A biscuit manufacturer claims that, on average, 1 in 3 packets of biscuits contain a prize offer. Gerry suspects that the proportion of packets containing the prize offer is less than 1 in 3 . In order to test the manufacturer's claim, he buys 20 randomly selected packets. He finds that exactly 2 of these packets contain the prize offer.
\begin{enumerate}[label=(\alph*)]
\item Carry out the test at the $10 \%$ significance level.
\item Maria also suspects that the proportion of packets containing the prize offer is less than 1 in 3 . She also carries out a significance test at the $10 \%$ level using 20 randomly selected packets. She will reject the manufacturer's claim if she finds that there are 3 or fewer packets containing the prize offer.
Find the probability of a Type II error in Maria's test if the proportion of packets containing the prize offer is actually 1 in 7 .
\item Explain what is meant by a Type II error in this context.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q6 [9]}}