| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| 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 one-tailed binomial hypothesis test with clearly stated hypotheses and significance level. Parts (a) and (b) provide scaffolding for the main test in (c), which follows a standard procedure: state hypotheses, find critical region or p-value, and conclude. The context is accessible and the calculations are routine for AS-level statistics, making it slightly easier than average. |
| 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 |
|---|---|---|
| \(\left[\frac{19.8}{36} =\right] 0.55\) | B1 | or \(\frac{11}{20}\); must be decimal or simplified fraction; but condone 55% |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.0380\ (96943116\ldots)\) BC | B1 | Correct to at least 2 sf; allow truncated |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: p = 0.45\), \(H_1: p < 0.45\) | B1 | OR \(H_0: p = 0.55\), \(H_1: p > 0.55\); both hypotheses |
| \(p\) is the probability that a driver speeding/caught speeding/on speed awareness course is female | B1\( | Must not be inconsistent with \)H_1$ |
| Use or sight of \(B(25, 0.45)\) | M1* | OR use or sight of \(B(25, 0.55)\); can be implied by \(P(X=7)\) (OR \(P(X=18)\)) \(= 0.038\ldots\) |
| \(P(X \leq 7) = 0.06385\ (0773831)\) | A1 | \(P(X \geq 18) = 1 - P(X \leq 17) = 1 - 0.936149 = 0.06385\) |
| \(0.06385 > 0.05\) | *M1 | FT their probability |
| Result is NOT significant or "accept \(H_0\)" | B1 | Dep on \(0.06385\); may be incorporated in conclusion |
| There is insufficient evidence (at the 5% level) to suggest that drivers speeding/caught speeding/on speed awareness course are more likely to be male oe | E1 | Dep all previous marks except B1$; not assertive |
| SC: Use of \(p = 0.5\) can score B1B1M1 (can be implied by \(0.0216\) or \(0.0143\)) A0M1B0E0 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. Each driver speeding/caught speeding has the same (constant) probability of being selected for the speed awareness course | E1 | OR the probability of any particular driver caught speeding being selected for the speed awareness course is independent of any other driver caught speeding being selected |
# Question 8:
## Part (a):
$\left[\frac{19.8}{36} =\right] 0.55$ | B1 | or $\frac{11}{20}$; must be decimal or simplified fraction; but condone 55%
## Part (b):
$0.0380\ (96943116\ldots)$ BC | B1 | Correct to at least 2 sf; allow truncated
## Part (c):
$H_0: p = 0.45$, $H_1: p < 0.45$ | B1 | OR $H_0: p = 0.55$, $H_1: p > 0.55$; both hypotheses
$p$ is the probability that a driver speeding/caught speeding/on speed awareness course is female | B1$ | Must not be inconsistent with $H_1$
Use or sight of $B(25, 0.45)$ | M1* | OR use or sight of $B(25, 0.55)$; can be implied by $P(X=7)$ (OR $P(X=18)$) $= 0.038\ldots$
$P(X \leq 7) = 0.06385\ (0773831)$ | A1 | $P(X \geq 18) = 1 - P(X \leq 17) = 1 - 0.936149 = 0.06385$
$0.06385 > 0.05$ | *M1 | FT their probability
Result is NOT significant or "accept $H_0$" | B1 | Dep on $0.06385$; may be incorporated in conclusion
There is insufficient evidence (at the 5% level) to suggest that drivers speeding/caught speeding/on speed awareness course are more likely to be male oe | E1 | Dep all previous marks except B1$; not assertive
| | SC: Use of $p = 0.5$ can score B1B1M1 (can be implied by $0.0216$ or $0.0143$) A0M1B0E0
## Part (d):
e.g. Each driver speeding/caught speeding has the same (constant) probability of being selected for the speed awareness course | E1 | OR the probability of any particular driver caught speeding being selected for the speed awareness course is independent of any other driver caught speeding being selected
---8 According to the latest research there are 19.8 million male drivers and 16.2 million female drivers on the roads in the UK.
\begin{enumerate}[label=(\alph*)]
\item A driver in the UK is selected at random. Find the probability that the driver is male.
\item Calculate the probability that there are 7 female drivers in a random sample of 25 UK drivers.
When driving in a built-up area, Rebecca exceeded the speed limit and was obliged to attend a speed awareness course. Her husband said "It's nearly always male drivers who are speeding." When Rebecca attends the course, she finds that there are 25 drivers, 7 of whom are female. You should assume that the drivers on the speed awareness course constitute a random sample of drivers caught speeding.
\item In this question you must show detailed reasoning.
Conduct a hypothesis test to determine whether there is any evidence at the $5 \%$ level to suggest that male drivers are more likely to exceed the speed limit than female drivers.
\item State a modelling assumption that is necessary in order to conduct the hypothesis test in part (c).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2019 Q8 [10]}}