| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 15 |
| 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 (p > 0.2), a standard significance level (5%), and n=100, x=28. Students need to calculate P(X ≥ 28) under H₀: p=0.2 and compare to 0.05. While it requires understanding of hypothesis testing framework, the calculations are routine and the question provides all necessary information explicitly, making it slightly easier than average. |
| 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-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2p + q + 0.2 + 0.3 = 1\) soi oe | B1 (2.1) | |
| \(2 \times p \times q = 0.06\) soi | M1 (3.1a) | allow M1 if 2 omitted |
| eliminate \(p\) or \(q\) with a correct substitution from one of their equations | M1 (1.1) | |
| \(q^2 - 0.5q + 0.06 = 0\) or \(2p^2 - 0.5p + 0.03 = 0\) oe | A1 (1.1) | e.g. \(2 \times \frac{0.03}{q} + q = 0.5\) or \(2p + \frac{0.03}{p} = 0.5\); NB if 2 omitted, A0 for \(2p^2 - 0.5p + 0.06 = 0\) or \(2q^2 + q + 0.24 = 0\) which have no real roots |
| \(q = 0.2\) or \(0.3\) and \(p = 0.15\) or \(0.1\) | A1 (1.1) | may be implied by e.g. \(q = 0.2\) or \(0.3\) and \(2p = 0.3\) or \(0.2\) |
| (\(q < 2p\) so) \(q = 0.2\) and \(p = 0.15\) | A1 [6] (3.2a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(10 \times q \times (1-q)^9\) soi | M1 (1.1) | |
| \(0.27\) or \(0.268\) or awrt \(0.2684\) isw | A1 [2] (1.1) | FT their \(q\) where \(0 < q < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: p = 0.2\); \(H_1: p > 0.2\) | B1 (1.1) | both hypotheses; allow equivalent in words or e.g. \(P(1) = 0.2\); allow any parameter as long as clearly defined as probability |
| \(p\) is the probability that the spinner shows a 1 (on any given spin) oe | B1 (2.5) | |
| use of \(X \sim B(100, 0.2)\) where \(x\) is the number of 1s obtained in 100 spins to obtain \(P(X \geq k)\) or \(P(X \leq k)\) | M1 (3.3) | \(k = 27, 28\) or \(29\); M0 for \(P(X=k)\); NB \(P(X=28)=0.014\ldots\), \(P(X=27)=0.020168\ldots\) |
| \(P(X \leq 27) = \) awrt \(0.97\) or \(P(X \geq 28) = \) awrt \(0.034\) | A1 (1.1) | or critical region is \(X \geq 28\) |
| \(0.034 < 0.05\) or \(0.97 > 0.95\) | M1 (3.4) | or 28 is in critical region; FT their probability, dependent on award of first M1 |
| significant or reject \(H_0\) or accept \(H_1\); may be embedded in conclusion in context | A1 (1.1) | must have the correct probability or correct critical region for the last two A marks |
| there is sufficient evidence to suggest (at 5% level) that the probability of a score of 1 is greater than 0.2 | A1 [7] (2.2b) | do not allow e.g. conclude/prove/indicate or other assertive statement instead of suggest; A0 if answer spoiled |
## Question 12:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2p + q + 0.2 + 0.3 = 1$ soi oe | B1 (2.1) | |
| $2 \times p \times q = 0.06$ soi | M1 (3.1a) | allow M1 if 2 omitted |
| eliminate $p$ or $q$ with a correct substitution from one of their equations | M1 (1.1) | |
| $q^2 - 0.5q + 0.06 = 0$ or $2p^2 - 0.5p + 0.03 = 0$ oe | A1 (1.1) | e.g. $2 \times \frac{0.03}{q} + q = 0.5$ or $2p + \frac{0.03}{p} = 0.5$; NB if 2 omitted, A0 for $2p^2 - 0.5p + 0.06 = 0$ or $2q^2 + q + 0.24 = 0$ which have no real roots |
| $q = 0.2$ or $0.3$ and $p = 0.15$ or $0.1$ | A1 (1.1) | may be implied by e.g. $q = 0.2$ or $0.3$ and $2p = 0.3$ or $0.2$ |
| ($q < 2p$ so) $q = 0.2$ and $p = 0.15$ | A1 [6] (3.2a) | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $10 \times q \times (1-q)^9$ soi | M1 (1.1) | |
| $0.27$ or $0.268$ or awrt $0.2684$ isw | A1 [2] (1.1) | FT their $q$ where $0 < q < 1$ |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p = 0.2$; $H_1: p > 0.2$ | B1 (1.1) | both hypotheses; allow equivalent in words or e.g. $P(1) = 0.2$; allow any parameter as long as clearly defined as probability |
| $p$ is the probability that the spinner shows a 1 (on any given spin) oe | B1 (2.5) | |
| use of $X \sim B(100, 0.2)$ where $x$ is the number of 1s obtained in 100 spins to obtain $P(X \geq k)$ or $P(X \leq k)$ | M1 (3.3) | $k = 27, 28$ or $29$; M0 for $P(X=k)$; NB $P(X=28)=0.014\ldots$, $P(X=27)=0.020168\ldots$ |
| $P(X \leq 27) = $ awrt $0.97$ or $P(X \geq 28) = $ awrt $0.034$ | A1 (1.1) | or critical region is $X \geq 28$ |
| $0.034 < 0.05$ or $0.97 > 0.95$ | M1 (3.4) | or 28 is in critical region; FT their probability, dependent on award of first M1 |
| significant or reject $H_0$ or accept $H_1$; may be embedded in conclusion in context | A1 (1.1) | must have the correct probability or correct critical region for the last two A marks |
| there is sufficient evidence to **suggest** (at 5% level) that the probability of a score of 1 is greater than 0.2 | A1 [7] (2.2b) | do not allow e.g. conclude/prove/indicate or other assertive statement instead of suggest; A0 if answer spoiled |
---\begin{enumerate}[label=(\alph*)]
\item Given that $q < 2 p$, determine the values of $p$ and $q$.
\item The spinner is spun 10 times. Calculate the probability that exactly one 5 is obtained.
Elaine's teacher believes that the probability that the spinner shows a 1 is greater than 0.2 . The spinner is spun 100 times and gives a score of 1 on 28 occasions.
\item Conduct a hypothesis test at the $5 \%$ level to determine whether there is any evidence to suggest that the probability of obtaining a score of 1 is greater than 0.2 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2020 Q12 [15]}}