| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| 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 | Standard +0.3 This is a straightforward hypothesis test using the binomial distribution with clear setup (H₀: p=0.55, H₁: p>0.55, n=8, x=6). Students need to calculate P(X≥6) and compare to 10% significance level—standard S2 procedure requiring only routine application of binomial probability formulas and understanding of one-tailed tests. Part (ii) tests basic understanding of independence assumption. Slightly easier than average due to small n making calculations manageable and clear question structure. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: p = 0.55, H_1: p > 0.55\) | B2 | All correct, B2. One error (e.g. \(\neq\), wrong or no letter) B1, but \(r, x\) etc: B0 |
| \(R \sim B(8, 0.55)\) where \(R\) is the number of girls | M1 | \(B(8, 0.55)\) stated or implied, e.g. \(N(4.4, 1.98)\) |
| \(\alpha: \quad P(R \geq 6) = 1 - 0.7799 = 0.2201\) | A1 | \(P(\geq 6) = 0.2201\), or \(P(< 6) = 0.7799\) |
| \(> 0.1\) | B1 | Compare \(P(\geq 6)\) with 0.1 or \(P(< 6)\) with 0.9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{CR is} \geq 7 \text{ and} < 7\) | B1 | Correct CR stated and explicit comparison with 6 |
| \(p = 0.0632\) | A1 | This probability seen, a.r.t. 0.0632. Award if 0.9368 seen and CR is correct. If CR not clearly stated, cannot get last M1A1 |
| Interpret in context, acknowledge uncertainty, double negative (7 marks) | ||
| M1 | Correct first conclusion, requires \(B(8, 0.55)\), not \(P(< 6) = [= 0.0632]\) | |
| A1 | or \(P(< 6) = 0.9368\) or \(P(< 6) = [= 0.1569]\). Allow 0.7799 if compared with 0.9 | |
| SC: Normal: max B2 M1 | ||
| SC: Two different attempts: max B2 M1 unless both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Assume that the last 8 years are a random sample of years when Head Student has been chosen | B1 | Refer to random sample, allow implied by any method described (1 mark) |
| Must be choosing \(\textit{years}\), not \(\textit{students}\) | ||
| Not quote conditions for random sample unless explicitly "years" | ||
| Extras: ignore unless clearly wrong, in which case B0 |
## (i)
$H_0: p = 0.55, H_1: p > 0.55$ | B2 | All correct, B2. One error (e.g. $\neq$, wrong or no letter) B1, but $r, x$ etc: B0
$R \sim B(8, 0.55)$ where $R$ is the number of girls | M1 | $B(8, 0.55)$ stated or implied, e.g. $N(4.4, 1.98)$
$\alpha: \quad P(R \geq 6) = 1 - 0.7799 = 0.2201$ | A1 | $P(\geq 6) = 0.2201$, or $P(< 6) = 0.7799$
$> 0.1$ | B1 | Compare $P(\geq 6)$ with 0.1 or $P(< 6)$ with 0.9
## $\beta:$
$\text{CR is} \geq 7 \text{ and} < 7$ | B1 | Correct CR stated and explicit comparison with 6
$p = 0.0632$ | A1 | This probability seen, a.r.t. 0.0632. Award if 0.9368 seen and CR is correct. If CR not clearly stated, cannot get last M1A1
| | Interpret in context, acknowledge uncertainty, double negative (7 marks)
| M1 | Correct first conclusion, requires $B(8, 0.55)$, not $P(< 6) = [= 0.0632]$
| A1 | or $P(< 6) = 0.9368$ or $P(< 6) = [= 0.1569]$. Allow 0.7799 if compared with 0.9
| | SC: Normal: max B2 M1
| | SC: Two different attempts: max B2 M1 unless both correct
## (ii)
Assume that the last 8 years are a random sample of years when Head Student has been chosen | B1 | Refer to random sample, allow implied by any method described (1 mark)
| | Must be choosing $\textit{years}$, not $\textit{students}$
| | Not quote conditions for random sample unless explicitly "years"
| | Extras: ignore unless clearly wrong, in which case B055% of the pupils in a large school are girls. A member of the student council claims that the probability that a girl rather than a boy becomes Head Student is greater than 0.55. As evidence for his claim he says that 6 of the last 8 Head Students have been girls.
\begin{enumerate}[label=(\roman*)]
\item Use an exact binomial distribution to test the claim at the 10% significance level. [7]
\item A statistics teacher says that considering only the last 8 Head Students may not be satisfactory. Explain what needs to be assumed about the data for the test to be valid. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2016 Q5 [8]}}