| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Interpret test conclusion or error |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard binomial test procedures (finding critical region, comparing observed value, stating conclusion) and understanding of Type I/II errors. Part (c) is particularly routine since P(Type I error) = significance level by definition. Slightly easier than average as it tests recall and basic application rather than problem-solving. |
| 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-tail5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): population proportion \(= 0.08\) OE; \(H_1\): population proportion \(> 0.08\) OE | B1 | Allow '\(p=0.08\)' etc. |
| \(P(X\geq4) = 1-P(X\leq3) = 1-(0.92^{25}+25\times0.92^{24}\times0.08+{}^{25}C_2\times0.92^{23}\times0.08^2+{}^{25}C_3\times0.92^{22}\times0.08^3)\) | M1 | Allow 1 – (one term omitted or extra or wrong) |
| \(0.135\) (3 sf) | A1 | |
| \(0.135 > 0.05\) | M1 | Valid comparison. Note: '\(0.865\)'\(<0.95\) can score M1 A1 and can recover previous M1 A1 for \(0.865\) |
| There is no evidence that proportion owning Chantor has increased | A1 FT | In context. Not definite. No contradictions |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\) was not rejected | *B1 FT | \(H_0\) was rejected (consistent with (a)) |
| Hence Type II might have been made | DB1 FT | Type I error |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X\geq5)=1-P(X\leq4) = 1-\!\left((1-0.1351)+{}^{25}C_4\times0.92^{21}\times0.08^4\right)[=0.0451]\) | *M1 | Attempted. Note: If critical region method used in (a) marks can be awarded here |
| \(0.0451 < 0.05\) | A1 | Comparison of \(0.045[1]\) with \(0.05\). Note: If critical region method used in (a) marks can be awarded here |
| \(P(\text{Type I error}) = 0.0451\) or \(0.0452\) | A1 | Dependent on M1* only. SC Unsupported answers score: B1 for \(0.0451<0.05\) and B1 for final answer \(0.0451\) only |
| 3 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: population proportion $= 0.08$ OE; $H_1$: population proportion $> 0.08$ OE | B1 | Allow '$p=0.08$' etc. |
| $P(X\geq4) = 1-P(X\leq3) = 1-(0.92^{25}+25\times0.92^{24}\times0.08+{}^{25}C_2\times0.92^{23}\times0.08^2+{}^{25}C_3\times0.92^{22}\times0.08^3)$ | M1 | Allow 1 – (one term omitted or extra or wrong) |
| $0.135$ (3 sf) | A1 | |
| $0.135 > 0.05$ | M1 | Valid comparison. Note: '$0.865$'$<0.95$ can score M1 A1 and can recover previous M1 A1 for $0.865$ |
| There is no evidence that proportion owning Chantor has increased | A1 FT | In context. Not definite. No contradictions |
| | **5** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$ was not rejected | *B1 FT | $H_0$ was rejected (consistent with **(a)**) |
| Hence Type II might have been made | DB1 FT | Type I error |
| | **2** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X\geq5)=1-P(X\leq4) = 1-\!\left((1-0.1351)+{}^{25}C_4\times0.92^{21}\times0.08^4\right)[=0.0451]$ | *M1 | Attempted. Note: If critical region method used in **(a)** marks can be awarded here |
| $0.0451 < 0.05$ | A1 | Comparison of $0.045[1]$ with $0.05$. Note: If critical region method used in **(a)** marks can be awarded here |
| $P(\text{Type I error}) = 0.0451$ or $0.0452$ | A1 | Dependent on M1* only. **SC** Unsupported answers score: **B1** for $0.0451<0.05$ and **B1** for final answer $0.0451$ only |
| | **3** | |6 It is known that $8 \%$ of adults in a certain town own a Chantor car. After an advertising campaign, a car dealer wishes to investigate whether this proportion has increased. He chooses a random sample of 25 adults from the town and notes how many of them own a Chantor car.
\begin{enumerate}[label=(\alph*)]
\item He finds that 4 of the 25 adults own a Chantor car.
Carry out a hypothesis test at the 5\% significance level.
\item Explain which of the errors, Type I or Type II, might have been made in carrying out the test in part (a).\\
Later, the car dealer takes another random sample of 25 adults from the town and carries out a similar hypothesis test at the 5\% significance level.
\item Find the probability of a Type I error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q6 [10]}}