| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Standard +0.3 This is a standard S1 binomial hypothesis testing question with straightforward calculations in parts (i)-(iv), though part (v) requires systematic trial of values to find the minimum n. All techniques are routine for this module, with no novel problem-solving required beyond careful application of binomial probability formulas and hypothesis test procedures. |
| 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 |
| \(X \sim B(15, 0.85)\) | ||
| \(P(\text{exactly 12 germinate}) = \binom{15}{12} \times 0.85^{12} \times 0.15^3\) | M1 | For \(0.85^{12} \times 0.15^3\) |
| M1 | For \(\binom{15}{12} \times p^{12} \times q^3\) | |
| \(= 0.2184\) | A1 | CAO |
| OR | OR | |
| from tables: \(0.3958 - 0.1773 = 0.2185\) | M2 | For \(0.3958 - 0.1773\) |
| A1 | CAO | |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 12) = P(X \leq 11) = 0.1773\) | M1 | For \(P(X \leq 11)\) or P(≤11) (With no extras) CAO (as final answer). May see alternative method: \(0.3958 - 0.2185 = 0.1773\). \(0.3958\) - their wrong answer to part (i) scores M1A0. |
| A1 | ||
| Total | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(p\) = probability of a seed germinating (for the population) | B1 | For definition of \(p\). See below for additional notes. |
| \(H_0: p = 0.85\) | B1 | For \(H_0\) |
| \(H_1: p < 0.85\) | B1 | For \(H_1\). Dep on \(< 0.85\) used in \(H_1\). Do not allow just 'Germination rate will be lower' or similar. E0 for simply stating \(H_1\) in words. |
| \(H_1\) has this form because the test is to investigate whether the proportion of seeds which germinate is lower. | E1 | For use of 0.15 as P(not germinating), contact team leader. E0 for simply stating \(H_1\) in words. |
| Total | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(X \sim B(20, 0.85)\) | ||
| \(P(X \leq 13) = 0.0219\) | M1* | For probability (provided not as part of finding \(P(X \leq 13)\)) Ignore notation. No further marks if point probs used - \(P(X = 13) = 0.0160\) DO NOT FT wrong \(H_1\), but see extra notes. Allow 'accept \(H_0\)' or 'reject \(H_1\)'. Must include 'sufficient evidence' or something similar such as 'to suggest that' ie an element of doubt either in the A or E mark. |
| \(0.0219 > 1\%\) | M1* dep | For comparison |
| So not enough evidence to reject \(H_0\). Not significant. | A1* | for not significant oe. |
| Conclude that there is not enough evidence to indicate that the proportion of seeds which have germinated has decreased. | E1* dep | For conclusion in context. Must mention decrease, not just change. |
| Total | [4] |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | For either probability |
| A1 | cao dep on at least one correct comparison with 1%. |
| A1* | |
| E1* dep | Could get M1A0A1E1 if poor notation for CR. Do not allow just '13 not in CR' - Must say 'not significant' or accept \(H_0\) or similar. Do not allow just '13 not in CR' - Must say 'not significant' or accept \(H_0\) or similar. If use a calculator to find \(P(X \leq 33) = 0.000661\) and compare with 1% then B2 for \(P(X \leq 33) = 0.000661 < 0.01\) so reject \(H_0\) then final E1 as per scheme. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(33 < 35\) | M1 | For comparison. Allow '33 lies in the CR'. Must include 'sufficient evidence' or something similar such as 'to suggest that' ie an element of doubt either in the A or E mark. Do not FT wrong \(H_1\).: In part (iv) ignore any interchanged \(H_0\) and \(H_1\) seen in part (ii). |
| So there is sufficient evidence to reject \(H_0\). | A1* | |
| Conclude that there is enough evidence to indicate that the proportion of seeds which have germinated has decreased. | E1* dep | Must mention decrease, not just change. |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For \(n = 3\), \(P(X \leq 0) = 0.0034 < 0.01\) | M1 | For \(P(X \leq 0) = 0.0034\) |
| For \(n = 2\), \(P(X \leq 0) = 0.0225 > 0.01\) | M1 | For \(P(X \leq 0) = 0.0225\) |
| So the least value of \(n\) for which the critical region is not empty and thus \(H_0\) could be rejected is 3. | A1 | CAO. Condone just 'n = 3' for final A mark dep on both M marks. If wrong \(H_1\) allow max M2A0 if correct probabilities seen. |
| ALTERNATIVE METHOD using logs | ||
| \(0.15^n < 0.01\) | M1 | |
| \(n > \log 0.01 / \log 0.15\) | M1 | |
| \(n > 2.427\) | ||
| Least \(n = 3\) | A1 | |
| Total | [3] |
## (i)(A)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $X \sim B(15, 0.85)$ | | |
| $P(\text{exactly 12 germinate}) = \binom{15}{12} \times 0.85^{12} \times 0.15^3$ | M1 | For $0.85^{12} \times 0.15^3$ |
| | M1 | For $\binom{15}{12} \times p^{12} \times q^3$ |
| $= 0.2184$ | A1 | CAO |
| **OR** | **OR** | |
| from tables: $0.3958 - 0.1773 = 0.2185$ | M2 | For $0.3958 - 0.1773$ |
| | A1 | CAO |
| **Total** | **[3]** | |
## (i)(B)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(X < 12) = P(X \leq 11) = 0.1773$ | M1 | For $P(X \leq 11)$ or P(≤11) (With no extras) CAO (as final answer). May see alternative method: $0.3958 - 0.2185 = 0.1773$. $0.3958$ - their wrong answer to part (i) scores M1A0. |
| | A1 | |
| **Total** | **[2]** | |
## (ii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Let $p$ = probability of a seed germinating (for the population) | B1 | For definition of $p$. See below for additional notes. |
| $H_0: p = 0.85$ | B1 | For $H_0$ |
| $H_1: p < 0.85$ | B1 | For $H_1$. Dep on $< 0.85$ used in $H_1$. Do not allow just 'Germination rate will be lower' or similar. E0 for simply stating $H_1$ in words. |
| $H_1$ has this form because the test is to investigate whether the proportion of seeds which germinate is lower. | E1 | For use of 0.15 as P(not germinating), contact team leader. E0 for simply stating $H_1$ in words. |
| **Total** | **[4]** | |
## (iii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Let $X \sim B(20, 0.85)$ | | |
| $P(X \leq 13) = 0.0219$ | M1* | For probability (provided not as part of finding $P(X \leq 13)$) Ignore notation. No further marks if point probs used - $P(X = 13) = 0.0160$ DO NOT FT wrong $H_1$, but see extra notes. Allow 'accept $H_0$' or 'reject $H_1$'. Must include 'sufficient evidence' or something similar such as 'to suggest that' ie an element of doubt either in the A or E mark. |
| $0.0219 > 1\%$ | M1* dep | For comparison |
| So not enough evidence to reject $H_0$. Not significant. | A1* | for not significant oe. |
| Conclude that there is not enough evidence to indicate that the proportion of seeds which have germinated has decreased. | E1* dep | For conclusion in context. Must mention decrease, not just change. |
| **Total** | **[4]** | |
---
## (iii) ALTERNATIVE METHOD – follow method above unless some mention of CR seen
### Critical region method
$P(X \leq 12) = 0.0059 < 1\%$
$P(X \leq 13) = 0.0219 > 1\%$
So critical region is $\{0, 1, 2, \ldots, 12\}$, $X \leq 12$, oe but not ≤ 12) oe
13 not in CR so not significant
There is insufficient evidence to indicate that the proportion of seeds which have germinated has decreased.
| **Marks** | **Guidance** |
|---|---|
| M1 | For either probability |
| A1 | cao dep on at least one correct comparison with 1%. |
| A1* | |
| E1* dep | Could get M1A0A1E1 if poor notation for CR. Do not allow just '13 not in CR' - Must say 'not significant' or accept $H_0$ or similar. Do not allow just '13 not in CR' - Must say 'not significant' or accept $H_0$ or similar. If use a calculator to find $P(X \leq 33) = 0.000661$ and compare with 1% then B2 for $P(X \leq 33) = 0.000661 < 0.01$ so reject $H_0$ then final E1 as per scheme. |
---
## (iv)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $33 < 35$ | M1 | For comparison. Allow '33 lies in the CR'. Must include 'sufficient evidence' or something similar such as 'to suggest that' ie an element of doubt either in the A or E mark. Do not FT wrong $H_1$.: In part (iv) ignore any interchanged $H_0$ and $H_1$ seen in part (ii). |
| So there is sufficient evidence to reject $H_0$. | A1* | |
| Conclude that there is enough evidence to indicate that the proportion of seeds which have germinated has decreased. | E1* dep | Must mention decrease, not just change. |
| **Total** | **[3]** | |
## (v)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| For $n = 3$, $P(X \leq 0) = 0.0034 < 0.01$ | M1 | For $P(X \leq 0) = 0.0034$ |
| For $n = 2$, $P(X \leq 0) = 0.0225 > 0.01$ | M1 | For $P(X \leq 0) = 0.0225$ |
| So the least value of $n$ for which the critical region is not empty and thus $H_0$ could be rejected is 3. | A1 | CAO. Condone just 'n = 3' for final A mark dep on both M marks. If wrong $H_1$ allow max M2A0 if correct probabilities seen. |
| **ALTERNATIVE METHOD using logs** | | |
| $0.15^n < 0.01$ | M1 | |
| $n > \log 0.01 / \log 0.15$ | M1 | |
| $n > 2.427$ | | |
| Least $n = 3$ | A1 | |
| **Total** | **[3]** | |It is known that on average 85% of seeds of a particular variety of tomato will germinate. Ramesh selects 15 of these seeds at random and sows them.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\Alph*)]
\item Find the probability that exactly 12 germinate. [3]
\item Find the probability that fewer than 12 germinate. [2]
\end{enumerate}
\end{enumerate}
The following year Ramesh finds that he still has many seeds left. Because the seeds are now one year old, he suspects that the germination rate will be lower. He conducts a trial by randomly selecting $n$ of these seeds and sowing them. He then carries out a hypothesis test at the 1% significance level to investigate whether he is correct.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Write down suitable null and alternative hypotheses for the test. Give a reason for your choice of alternative hypothesis. [4]
\item In a trial with $n = 20$, Ramesh finds that 13 seeds germinate. Carry out the test. [4]
\item Suppose instead that Ramesh conducts the trial with $n = 50$, and finds that 33 seeds germinate. Given that the critical value for the test in this case is 35, complete the test. [3]
\item If $n$ is small, there is no point in carrying out the test at the 1% significance level, as the null hypothesis cannot be rejected however many seeds germinate. Find the least value of $n$ for which the null hypothesis can be rejected, quoting appropriate probabilities to justify your answer. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2014 Q7 [19]}}