| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multiple binomial probability calculations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question covering basic binomial probability calculations, expectation/variance formulas, and a standard one-tailed hypothesis test. All parts use routine techniques with clear setup—calculating P(X=k) for small n, applying E(X)=np and Var(X)=np(1-p), and performing a textbook significance test. The multi-stage calculation in part (iv) requires careful tracking but no novel insight. Slightly above average difficulty due to length and the need for methodical work, but well within standard S1 scope. |
| Spec | 2.03a Mutually exclusive and independent events2.04a Discrete probability distributions2.04b 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 |
| (A) \(P(\text{both}) = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\) | B1 CAO | |
| (B) \(P(\text{one}) = 2 \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}\) | B1 CAO | |
| (C) \(P(\text{neither}) = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\) | B1 CAO | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Independence necessary because otherwise the probability of one seed germinating would change according to whether or not the other one germinates. | E1 | |
| May not be valid as the two seeds would have similar growing conditions e.g. temperature, moisture, etc. | E1 | NB Allow valid alternatives. Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected number \(= 2 \times \frac{2}{3} = \frac{4}{3}\) (= 1.33) | B1 FT | |
| \(E(X^2) = 0 \times \frac{1}{9} + 1 \times \frac{4}{9} + 4 \times \frac{4}{9} = \frac{20}{9}\) | M1 | for \(E(X^2)\) |
| \(\text{Var}(X) = \frac{20}{9} - \left(\frac{4}{3}\right)^2 = \frac{4}{9} = 0.444\) | A1 CAO | NB use of \(npq\) scores M1 for product, A1CAO. Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expect \(200 \times \frac{8}{9} = 177.8\) plants | M1 | for \(200 \times \frac{8}{9}\) |
| So expect \(0.85 \times 177.8 = 151\) onions | M1 dep | for \(\times 0.85\) |
| A1 CAO | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(X \sim B(18, p)\); Let \(p\) = probability of germination (for population) | B1 | for definition of \(p\) |
| \(H_0: p = 0.90\) | B1 | for \(H_0\) |
| \(H_1: p < 0.90\) | B1 | for \(H_1\) |
| \(P(X \leq 14) = 0.0982 > 5\%\) | M1 | for probability |
| So not enough evidence to reject \(H_0\) | M1 dep, A1 | for comparison |
| Conclude that there is not enough evidence to indicate that the germination rate is below 90%. | E1 | for conclusion in context |
| Note: critical region method — M1 for region \(\{0,1,2,\ldots,13\}\); M1 for 14 does not lie in critical region then A1 E1 as per scheme | Total: 7 marks |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (A) $P(\text{both}) = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$ | B1 CAO | |
| (B) $P(\text{one}) = 2 \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}$ | B1 CAO | |
| (C) $P(\text{neither}) = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$ | B1 CAO | Total: 3 marks |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Independence necessary because otherwise the probability of one seed germinating would change according to whether or not the other one germinates. | E1 | |
| May not be valid as the two seeds would have similar growing conditions e.g. temperature, moisture, etc. | E1 | NB Allow valid alternatives. Total: 2 marks |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected number $= 2 \times \frac{2}{3} = \frac{4}{3}$ (= 1.33) | B1 FT | |
| $E(X^2) = 0 \times \frac{1}{9} + 1 \times \frac{4}{9} + 4 \times \frac{4}{9} = \frac{20}{9}$ | M1 | for $E(X^2)$ |
| $\text{Var}(X) = \frac{20}{9} - \left(\frac{4}{3}\right)^2 = \frac{4}{9} = 0.444$ | A1 CAO | NB use of $npq$ scores M1 for product, A1CAO. Total: 3 marks |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expect $200 \times \frac{8}{9} = 177.8$ plants | M1 | for $200 \times \frac{8}{9}$ |
| So expect $0.85 \times 177.8 = 151$ onions | M1 dep | for $\times 0.85$ |
| | A1 CAO | Total: 3 marks |
### Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $X \sim B(18, p)$; Let $p$ = probability of germination (for population) | B1 | for definition of $p$ |
| $H_0: p = 0.90$ | B1 | for $H_0$ |
| $H_1: p < 0.90$ | B1 | for $H_1$ |
| $P(X \leq 14) = 0.0982 > 5\%$ | M1 | for probability |
| So not enough evidence to reject $H_0$ | M1 dep, A1 | for comparison |
| Conclude that there is not enough evidence to indicate that the germination rate is below 90%. | E1 | for conclusion in context |
| Note: critical region method — M1 for region $\{0,1,2,\ldots,13\}$; M1 for 14 does not lie in critical region then A1 E1 as per scheme | | Total: 7 marks |
---2 When onion seeds are sown outdoors, on average two-thirds of them germinate. A gardener sows seeds in pairs, in the hope that at least one will germinate.
\begin{enumerate}[label=(\roman*)]
\item Assuming that germination of one of the seeds in a pair is independent of germination of the other seed, find the probability that, if a pair of seeds is selected at random,\\
(A) both seeds germinate,\\
(B) just one seed germinates,\\
(C) neither seed germinates.
\item Explain why the assumption of independence is necessary in order to calculate the above probabilities. Comment on whether the assumption is likely to be valid.
\item A pair of seeds is sown. Find the expectation and variance of the number of seeds in the pair which germinate.
\item The gardener plants 200 pairs of seeds. If both seeds in a pair germinate, the gardener destroys one of the two plants so that only one is left to grow. Of the plants that remain after this, only $85 \%$ successfully grow to form an onion. Find the expected number of onions grown from the 200 pairs of seeds.
If the seeds are sown in a greenhouse, the germination rate is higher. The seed manufacturing company claims that the germination rate is $90 \%$. The gardener suspects that the rate will not be as high as this, and carries out a trial to investigate. 18 randomly selected seeds are sown in the greenhouse and it is found that 14 germinate.
\item Write down suitable hypotheses and carry out a test at the $5 \%$ level to determine whether there is any evidence to support the gardener's suspicions.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q2 [18]}}