| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with clearly stated hypotheses (H₀: p=0.05, H₁: p>0.05), small sample size making calculations manageable, and standard 5% significance level. While it requires understanding of hypothesis testing framework and binomial probability calculations, it follows a routine procedure with no conceptual complications or novel problem-solving required—slightly easier than average due to its textbook structure. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(p\) = probability that a randomly selected frame is faulty | B1 | Minimum needed: \(p\) = probability that frame/bike is faulty. Do not allow '\(p\) = probability that it is faulty'. Must include word probability/chance/proportion/percentage/likelihood but NOT possibility. Do NOT allow '\(p\) = the probability that faulty frames have increased' |
| \(H_0: p = 0.05\) | B1 | Allow \(p=5\%\), allow \(\theta\) or \(\pi\) and \(\rho\) but not \(x\). Allow \(H_0 = p=0.05\), Allow \(H_0: p = \frac{1}{20}\). Do not allow \(H_0: P(X=x)=0.05\), \(H_0: =0.05\), \(=5\%\), \(P(0.05)\), \(p(0052)\), \(p(x)=0.05\), \(x=0.05\) (unless \(x\) correctly defined as probability). Do not allow \(H_1: p \geq 0.05\). Allow NH and AH in place of \(H_0\) and \(H_1\) |
| \(H_1: p > 0.05\) | B1 | |
| \(P(X \geq 4)\) | B1 | For notation \(P(X \geq 4)\) or \(1 - P(X \leq 3)\). This mark may be implied by 0.0109 as long as no incorrect notation. No further marks if point probs used — \(P(X=4) = 0.0094\). DO NOT FT wrong \(H_1\) |
| \(= 1 - P(X \leq 3) = 1 - 0.9891 = 0.0109\) | B1* | For 0.0109, independent of previous mark. Or for \(1 - 0.9891\) |
| \(0.0109 < 0.05\), so reject \(H_0\) | M1* dep, A1* | For comparison with 5%. Or significant or 'accept \(H_1\)' |
| There is evidence to suggest that the proportion of faulty frames has increased. | E1* dep on A1 | Must include 'sufficient evidence' or something similar such as 'to suggest that'. 'Sufficient evidence' or similar can be seen in either the A mark or E mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(X \sim B(18, 0.05)\); \(P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.9419 = 0.0581 > 5\%\) | (B1) | For 0.0581. Do not insist on correct notation |
| \(P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9891 = 0.0109 < 5\%\) | (B1) | For 0.0109 |
| So critical region is \(\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\) | (M1), (A1) | For at least one correct comparison with 5%. CAO for critical region. Condone \(\{4,5\ldots\}\), \(X \geq 4\) oe |
| 4 lies in the critical region, so significant. There is evidence to suggest that the proportion of faulty frames has increased. | (E1) |
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $p$ = probability that a randomly selected frame is faulty | B1 | Minimum needed: $p$ = probability that frame/bike is faulty. Do not allow '$p$ = probability that it is faulty'. Must include word probability/chance/proportion/percentage/likelihood but NOT possibility. Do NOT allow '$p$ = the probability that faulty frames have increased' |
| $H_0: p = 0.05$ | B1 | Allow $p=5\%$, allow $\theta$ or $\pi$ and $\rho$ but not $x$. Allow $H_0 = p=0.05$, Allow $H_0: p = \frac{1}{20}$. Do not allow $H_0: P(X=x)=0.05$, $H_0: =0.05$, $=5\%$, $P(0.05)$, $p(0052)$, $p(x)=0.05$, $x=0.05$ (unless $x$ correctly defined as probability). Do not allow $H_1: p \geq 0.05$. Allow NH and AH in place of $H_0$ and $H_1$ |
| $H_1: p > 0.05$ | B1 | |
| $P(X \geq 4)$ | B1 | For notation $P(X \geq 4)$ or $1 - P(X \leq 3)$. This mark may be implied by 0.0109 as long as no incorrect notation. No further marks if point probs used — $P(X=4) = 0.0094$. DO NOT FT wrong $H_1$ |
| $= 1 - P(X \leq 3) = 1 - 0.9891 = 0.0109$ | B1* | For 0.0109, independent of previous mark. Or for $1 - 0.9891$ |
| $0.0109 < 0.05$, so reject $H_0$ | M1* dep, A1* | For comparison with 5%. Or significant or 'accept $H_1$' |
| There is evidence to suggest that the proportion of faulty frames has increased. | E1* dep on A1 | Must include 'sufficient evidence' or something similar such as 'to suggest that'. 'Sufficient evidence' or similar can be seen in either the A mark or E mark |
**[Total: 8 marks]**
**OR Critical Region Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $X \sim B(18, 0.05)$; $P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.9419 = 0.0581 > 5\%$ | (B1) | For 0.0581. Do not insist on correct notation |
| $P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9891 = 0.0109 < 5\%$ | (B1) | For 0.0109 |
| So critical region is $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$ | (M1), (A1) | For at least one correct comparison with 5%. CAO for critical region. Condone $\{4,5\ldots\}$, $X \geq 4$ oe |
| 4 lies in the critical region, so significant. There is evidence to suggest that the proportion of faulty frames has increased. | (E1) | |2 A manufacturer produces titanium bicycle frames. The bicycle frames are tested before use and on average $5 \%$ of them are found to be faulty. A cheaper manufacturing process is introduced and the manufacturer wishes to check whether the proportion of faulty bicycle frames has increased. A random sample of 18 bicycle frames is selected and it is found that 4 of them are faulty. Carry out a hypothesis test at the $5 \%$ significance level to investigate whether the proportion of faulty bicycle frames has increased.
\hfill \mbox{\textit{OCR MEI S1 Q2 [8]}}